TITLE: Using the Scalar Electrostatic Potential to Calculate Transition Probabilities
QUESTION [3 upvotes]: transition probabilites of atomic systems prone to some time-varying electromagnetic field are very often calculated using perturbation theory leading to expressions including the vector potential $\mathbf{A}$. This approach seems to be very complicated and I am wondering:

Is using the electrostatic potential $U$ to calculate "electric transitions" (approximately) sufficient?

This was the short version of my question; below you will find some further notes that (hopefully) contain all information needed to clearify it.
In relativistic quantum mechanics, the action of an an electrodynamic field on an electron can be realized by minimal coupling using the replacement $\partial_\mu \rightarrow \partial_\mu - \mathrm{i}eA_\mu$. Thus, in a certain sense, (quantum) electrodynamics can be derived from the Dirac equation using the invariance of the free-electron action with respect to a phase change.
Electromagnetic Fields and the Schrödinger Equation
Nevertheless, often one must ignore the beauty of gauge theory to be able to practically calculate something under certain simplifications using the non-relativistic Schrödinger-equation.
There, one can incorporate an external electromagnetic field applying the replacements
$$H\left(\mathbf{p},\mathbf{x}\right) \rightarrow H\left(\mathbf{p}-e\mathbf{A}\left(t,\mathbf{x}\right),\mathbf{x}\right)+eU\left(t,\mathbf{x}\right)$$ to the Hamiltonian of the system where $\mathbf{A}$ is the known vector potential and $U$ is the potential corresponding to the electric field.
In the Coulomb gauge $\mathrm{div}\mathbf{A}=0$ one can always achieve $U\equiv 0$ if no sources are present and the overall Hamiltonian will look like $$H = H_0 -\frac{e}{m}\mathbf{A}\left(t,\mathbf{x}\right)\mathbf{p}$$
where $H_0 = \frac{\mathbf{p}^2}{2m} + V(\mathbf{x})$ and the term in $\mathbf{A}^2$, the ponderomotive potential has been neglected.
Time-Dependent Perturbation Theory
The last expression for $H$ has the form $$H = H_0(\mathbf{p},\mathbf{x}) + H_t(\mathbf{p},\mathbf{x},t)\ .$$
Using the Dyson series, one can formally write down the time-dependent solution in the interaction picture as
$$|\psi_I(t)> = T\exp\left[\frac{1}{\mathrm{i}\hbar}\int_{t_{0}}^{t}H_{t,I}\left(t^{\prime}\right)dt^{\prime}\right]|\psi_I(t_0)> .$$
Then, one takes only the first two terms into account to find that transition amplitudes are approximately given by $$A_{n\rightarrow m}\approx\frac{1}{\mathrm{i}\hbar}\int_{t_{0}}^{t}\left\langle m\right|H_{t,I}\left(t^{\prime}\right)\left|n\right\rangle dt^{\prime}\ \mathrm{with}$$
$$H_{t,I}\left(t\right) = e^{\mathrm{i}\left(t-t_{0}\right)H_{0}/\hbar}H_t\left(t\right)e^{-\mathrm{i}\left(t-t_{0}\right)H_{0}/\hbar}\ .$$
Relation to the Question
In the given case we have $$H_t = -\frac{e}{m}\mathbf{A}\left(t,\mathbf{x}\right)\mathbf{p}$$
and assuming that $$\mathbf{A}\left(\mathbf{x},t\right) = \mathbf{A}\left(\mathbf{x}\right) T(t)\ ,$$
and further using
$$\mathbf{p}=\frac{\mathrm{i}m}{\hbar}\left[H_{0},\mathbf{x}\right]\ ,$$
one will have to solve $$A_{n\rightarrow m}\propto <m|\mathbf{A}\left(\mathbf{x}\right)\cdot \mathbf{x}|n>\int_{t_0}^t e^{-\mathrm{i}\omega_{mn}\tau}T(\tau) d\tau\ +\ \mathrm{h.c.}$$
with $\hbar \omega_{mn} = E_m - E_n$, where the $E_i$ are energy levels of the unperturbed $H_0$.
Using here the vector potential $\mathbf{A}$ is the correct way to incorporate the field, of course. For me, it seems extremely complicated to do so since to calculate the transition probabilities for an arbitrary field distribution one will have to expand $\mathbf{A}$ into vector spherical harmonics.
So, the use of a scalar potential like $U$ would greatly simplify computations and my question can now be re-asked as

Why can't we just simply use $U(\mathbf{x})\cdot T(t)$ to calculate the transition amplitudes $A_{n\rightarrow m}$ for "electric excitations"?

Thank you very much for insights, comments and corrections.

REPLY [1 votes]: If the case is purely electrostatic, one does not need time dependent perturbation theory, since its static. So you can just do the usual time independent perturbation theory. If the field is not static then you can't have A = 0, and its not time independent, so you need all this technology. I don't think there is anything deeper going on.


text:
(*
 * Copyright Florian Haftmann
 *
 * SPDX-License-Identifier: BSD-2-Clause
 *)

(*<*)
theory Guide
  imports Word_Lib_Sumo Machine_Word_32 Machine_Word_64 Ancient_Numeral
begin

context semiring_bit_operations
begin

lemma bit_eq_iff:
  \<open>a = b \<longleftrightarrow> (\<forall>n. 2 ^ n \<noteq> 0 \<longrightarrow> bit a n \<longleftrightarrow> bit b n)\<close>
  using bit_eq_iff [of a b] by (simp add: possible_bit_def)

end

notation (output)  Generic_set_bit.set_bit (\<open>Generic'_set'_bit.set'_bit\<close>)

hide_const (open) Generic_set_bit.set_bit

no_notation bit  (infixl \<open>!!\<close> 100)

(*>*)
section \<open>A short overview over bit operations and word types\<close>

subsection \<open>Key principles\<close>

text \<open>
  When formalizing bit operations, it is tempting to represent
  bit values as explicit lists over a binary type. This however
  is a bad idea, mainly due to the inherent ambiguities in
  representation concerning repeating leading bits.

  Hence this approach avoids such explicit lists altogether
  following an algebraic path:

    \<^item> Bit values are represented by numeric types: idealized
      unbounded bit values can be represented by type \<^typ>\<open>int\<close>,
      bounded bit values by quotient types over \<^typ>\<open>int\<close>, aka \<^typ>\<open>'a word\<close>.

    \<^item> (A special case are idealized unbounded bit values ending
      in @{term [source] 0} which can be represented by type \<^typ>\<open>nat\<close> but
      only support a restricted set of operations).

  The fundamental principles are developed in theory \<^theory>\<open>HOL.Bit_Operations\<close>
  (which is part of \<^theory>\<open>Main\<close>):

    \<^item> Multiplication by \<^term>\<open>2 :: int\<close> is a bit shift to the left and

    \<^item> Division by \<^term>\<open>2 :: int\<close> is a bit shift to the right.

    \<^item> Concerning bounded bit values, iterated shifts to the left
    may result in eliminating all bits by shifting them all
    beyond the boundary.  The property \<^prop>\<open>(2 :: int) ^ n \<noteq> 0\<close>
    represents that \<^term>\<open>n\<close> is \<^emph>\<open>not\<close> beyond that boundary.

    \<^item> The projection on a single bit is then @{thm [mode=iff] bit_iff_odd [where ?'a = int, no_vars]}.

    \<^item> This leads to the most fundamental properties of bit values:

      \<^item> Equality rule: @{thm [display, mode=iff] bit_eq_iff [where ?'a = int, no_vars]}

      \<^item> Induction rule: @{thm [display, mode=iff] bits_induct [where ?'a = int, no_vars]}

    \<^item> Characteristic properties @{prop [source] \<open>bit (f x) n \<longleftrightarrow> P x n\<close>}
      are available in fact collection \<^text>\<open>bit_simps\<close>.

  On top of this, the following generic operations are provided:

    \<^item> Singleton \<^term>\<open>n\<close>th bit: \<^term>\<open>(2 :: int) ^ n\<close>

    \<^item> Bit mask upto bit \<^term>\<open>n\<close>: @{thm mask_eq_exp_minus_1 [where ?'a = int, no_vars]}

    \<^item> Left shift: @{thm push_bit_eq_mult [where ?'a = int, no_vars]}

    \<^item> Right shift: @{thm drop_bit_eq_div [where ?'a = int, no_vars]}

    \<^item> Truncation: @{thm take_bit_eq_mod [where ?'a = int, no_vars]}

    \<^item> Bitwise negation: @{thm [mode=iff] bit_not_iff_eq [where ?'a = int, no_vars]}

    \<^item> Bitwise conjunction: @{thm [mode=iff] bit_and_iff [where ?'a = int, no_vars]}

    \<^item> Bitwise disjunction: @{thm [mode=iff] bit_or_iff [where ?'a = int, no_vars]}

    \<^item> Bitwise exclusive disjunction: @{thm [mode=iff] bit_xor_iff [where ?'a = int, no_vars]}

    \<^item> Setting a single bit: @{thm set_bit_def [where ?'a = int, no_vars]}

    \<^item> Unsetting a single bit: @{thm unset_bit_def [where ?'a = int, no_vars]}

    \<^item> Flipping a single bit: @{thm flip_bit_def [where ?'a = int, no_vars]}

    \<^item> Signed truncation, or modulus centered around \<^term>\<open>0::int\<close>:
        @{thm [display] signed_take_bit_def [where ?'a = int, no_vars]}

    \<^item> (Bounded) conversion from and to a list of bits:
        @{thm [display] horner_sum_bit_eq_take_bit [where ?'a = int, no_vars]}

  Bit concatenation on \<^typ>\<open>int\<close> as given by
    @{thm [display] concat_bit_def [no_vars]}
  appears quite
  technical but is the logical foundation for the quite natural bit concatenation
  on \<^typ>\<open>'a word\<close> (see below).
\<close>

subsection \<open>Core word theory\<close>

text \<open>
  Proper word types are introduced in theory \<^theory>\<open>HOL-Library.Word\<close>, with
  the following specific operations:

    \<^item> Standard arithmetic:
        @{term \<open>(+) :: 'a::len word \<Rightarrow> 'a word \<Rightarrow> 'a word\<close>},
        @{term \<open>uminus :: 'a::len word \<Rightarrow> 'a word\<close>},
        @{term \<open>(-) :: 'a::len word \<Rightarrow> 'a word \<Rightarrow> 'a word\<close>},
        @{term \<open>(*) :: 'a::len word \<Rightarrow> 'a word \<Rightarrow> 'a word\<close>},
        @{term \<open>0 :: 'a::len word\<close>}, @{term \<open>1 :: 'a::len word\<close>}, numerals etc.

    \<^item> Standard bit operations: see above.

    \<^item> Conversion with unsigned interpretation of words:

        \<^item> @{term [source] \<open>unsigned :: 'a::len word \<Rightarrow> 'b::semiring_1\<close>}

        \<^item> Important special cases as abbreviations:

            \<^item> @{term [source] \<open>unat :: 'a::len word \<Rightarrow> nat\<close>}

            \<^item> @{term [source] \<open>uint :: 'a::len word \<Rightarrow> int\<close>}

            \<^item> @{term [source] \<open>ucast :: 'a::len word \<Rightarrow> 'b::len word\<close>}

    \<^item> Conversion with signed interpretation of words:

        \<^item> @{term [source] \<open>signed :: 'a::len word \<Rightarrow> 'b::ring_1\<close>}

        \<^item> Important special cases as abbreviations:

            \<^item> @{term [source] \<open>sint :: 'a::len word \<Rightarrow> int\<close>}

            \<^item> @{term [source] \<open>scast :: 'a::len word \<Rightarrow> 'b::len word\<close>}

    \<^item> Operations with unsigned interpretation of words:

        \<^item> @{thm [mode=iff] word_le_nat_alt [no_vars]}

        \<^item> @{thm [mode=iff] word_less_nat_alt [no_vars]}

        \<^item> @{thm unat_div_distrib [no_vars]}

        \<^item> @{thm unat_drop_bit_eq [no_vars]}

        \<^item> @{thm unat_mod_distrib [no_vars]}

        \<^item> @{thm [mode=iff] udvd_iff_dvd [no_vars]}

    \<^item> Operations with signed interpretation of words:

        \<^item> @{thm [mode=iff] word_sle_eq [no_vars]}

        \<^item> @{thm [mode=iff] word_sless_alt [no_vars]}

        \<^item> @{thm sint_signed_drop_bit_eq [no_vars]}

    \<^item> Rotation and reversal:

        \<^item> @{term [source] \<open>word_rotl :: nat \<Rightarrow> 'a::len word \<Rightarrow> 'a word\<close>}

        \<^item> @{term [source] \<open>word_rotr :: nat \<Rightarrow> 'a::len word \<Rightarrow> 'a word\<close>}

        \<^item> @{term [source] \<open>word_roti :: int \<Rightarrow> 'a::len word \<Rightarrow> 'a word\<close>}

        \<^item> @{term [source] \<open>word_reverse :: 'a::len word \<Rightarrow> 'a word\<close>}

    \<^item> Concatenation: @{term [source, display] \<open>word_cat :: 'a::len word \<Rightarrow> 'b::len word \<Rightarrow> 'c::len word\<close>}

  For proofs about words the following default strategies are applicable:

    \<^item> Using bit extensionality (facts \<^text>\<open>bit_eq_iff\<close>, \<^text>\<open>bit_word_eqI\<close>; fact
      collection \<^text>\<open>bit_simps\<close>).

    \<^item> Using the @{method transfer} method.
\<close>

subsection \<open>More library theories\<close>

text \<open>
  Note: currently, most theories listed here are hardly separate
  entities since they import each other in various ways.
  Always inspect them to understand what you pull in if you
  want to import one.

    \<^descr>[Syntax]

      \<^descr>[\<^theory>\<open>Word_Lib.Syntax_Bundles\<close>]
        Bundles to provide alternative syntax for various bit operations.

      \<^descr>[\<^theory>\<open>Word_Lib.Hex_Words\<close>]
        Printing word numerals as hexadecimal numerals.

      \<^descr>[\<^theory>\<open>Word_Lib.Type_Syntax\<close>]
        Pretty type-sensitive syntax for cast operations.

      \<^descr>[\<^theory>\<open>Word_Lib.Word_Syntax\<close>]
        Specific ASCII syntax for prominent bit operations on word.

    \<^descr>[Proof tools]

      \<^descr>[\<^theory>\<open>Word_Lib.Norm_Words\<close>]
        Rewriting word numerals to normal forms.

      \<^descr>[\<^theory>\<open>Word_Lib.Bitwise\<close>]
        Method @{method word_bitwise} decomposes word equalities and inequalities into bit propositions.

      \<^descr>[\<^theory>\<open>Word_Lib.Bitwise_Signed\<close>]
        Method @{method word_bitwise_signed} decomposes word equalities and inequalities into bit propositions.

      \<^descr>[\<^theory>\<open>Word_Lib.Word_EqI\<close>]
        Method @{method word_eqI_solve} decomposes word equalities and inequalities into bit propositions.

    \<^descr>[Operations]

      \<^descr>[\<^theory>\<open>Word_Lib.Signed_Division_Word\<close>]

        Signed division on word:

          \<^item> @{term [source] \<open>(sdiv) :: 'a::len word \<Rightarrow> 'a word \<Rightarrow> 'a word\<close>}

          \<^item> @{term [source] \<open>(smod) :: 'a::len word \<Rightarrow> 'a word \<Rightarrow> 'a word\<close>}

      \<^descr>[\<^theory>\<open>Word_Lib.Aligned\<close>] \

          \<^item> @{thm [mode=iff] is_aligned_iff_udvd [no_vars]}

      \<^descr>[\<^theory>\<open>Word_Lib.Least_significant_bit\<close>]

        The least significant bit as an alias:
        @{thm [mode=iff] lsb_odd [where ?'a = int, no_vars]}

      \<^descr>[\<^theory>\<open>Word_Lib.Most_significant_bit\<close>]

        The most significant bit:

          \<^item> @{thm [mode=iff] msb_int_def [of k]}

          \<^item> @{thm [mode=iff] word_msb_sint [no_vars]}

          \<^item> @{thm [mode=iff] msb_word_iff_sless_0 [no_vars]}

          \<^item> @{thm [mode=iff] msb_word_iff_bit [no_vars]}

      \<^descr>[\<^theory>\<open>Word_Lib.Bit_Shifts_Infix_Syntax\<close>]

        Bit shifts decorated with infix syntax:

          \<^item> @{thm Bit_Shifts_Infix_Syntax.shiftl_def [no_vars]}

          \<^item> @{thm Bit_Shifts_Infix_Syntax.shiftr_def [no_vars]}

          \<^item> @{thm Bit_Shifts_Infix_Syntax.sshiftr_def [no_vars]}

      \<^descr>[\<^theory>\<open>Word_Lib.Next_and_Prev\<close>] \

          \<^item> @{thm word_next_unfold [no_vars]}

          \<^item> @{thm word_prev_unfold [no_vars]}

      \<^descr>[\<^theory>\<open>Word_Lib.Enumeration_Word\<close>]

        More on explicit enumeration of word types.

      \<^descr>[\<^theory>\<open>Word_Lib.More_Word_Operations\<close>]

        Even more operations on word.

    \<^descr>[Types]

      \<^descr>[\<^theory>\<open>Word_Lib.Signed_Words\<close>]

          Formal tagging of word types with a \<^text>\<open>signed\<close> marker.

    \<^descr>[Lemmas]

      \<^descr>[\<^theory>\<open>Word_Lib.More_Word\<close>]

          More lemmas on words.

      \<^descr>[\<^theory>\<open>Word_Lib.Word_Lemmas\<close>]

          More lemmas on words, covering many other theories mentioned here.

    \<^descr>[Words of popular lengths].

      \<^descr>[\<^theory>\<open>Word_Lib.Word_8\<close>]

          for 8-bit words.

      \<^descr>[\<^theory>\<open>Word_Lib.Word_16\<close>]

          for 16-bit words.

      \<^descr>[\<^theory>\<open>Word_Lib.Word_32\<close>]

          for 32-bit words.

      \<^descr>[\<^theory>\<open>Word_Lib.Word_64\<close>]

          for 64-bit words. This theory is not part of  \<^text>\<open>Word_Lib_Sumo\<close>, because it shadows
          names from \<^theory>\<open>Word_Lib.Word_32\<close>. They can be used together, but then will have
          to use qualified names in applications.

      \<^descr>[\<^theory>\<open>Word_Lib.Machine_Word_32\<close> and \<^theory>\<open>Word_Lib.Machine_Word_64\<close>]

          provide lemmas for 32-bit words and 64-bit words under the same name,
          which can help to organize applications relying on some form
          of genericity.
\<close>

subsection \<open>More library sessions\<close>

text \<open>
  \<^descr>[\<^text>\<open>Native_Word\<close>] Makes machine words and machine arithmetic available for code generation.
  It provides a common abstraction that hides the differences between the different target languages.
  The code generator maps these operations to the APIs of the target languages.
\<close>

subsection \<open>Legacy theories\<close>

text \<open>
  The following theories contain material which has been
  factored out since it is not recommended to use it in
  new applications, mostly because matters can be expressed
  succinctly using already existing operations.

  This section gives some indication how to migrate away
  from those theories.  However theorem coverage may still
  be terse in some cases.

  \<^descr>[\<^theory>\<open>Word_Lib.Word_Lib_Sumo\<close>]

    An entry point importing any relevant theory in that session.  Intended
    for backward compatibility: start importing this theory when
    migrating applications to Isabelle2021, and later sort out
    what you really need. You may need to include
   \<^theory>\<open>Word_Lib.Word_64\<close> separately.

  \<^descr>[\<^theory>\<open>Word_Lib.Generic_set_bit\<close>]

    Kind of an alias: @{thm set_bit_eq [no_vars]}

  \<^descr>[\<^theory>\<open>Word_Lib.Typedef_Morphisms\<close>]

    A low-level extension to HOL typedef providing
    conversions along type morphisms.  The @{method transfer} method
    seems to be sufficient for most applications though.

  \<^descr>[\<^theory>\<open>Word_Lib.Bit_Comprehension\<close>]

    Comprehension syntax for bit values over predicates
    \<^typ>\<open>nat \<Rightarrow> bool\<close>.  For \<^typ>\<open>'a::len word\<close>, straightforward
    alternatives exist; difficult to handle for \<^typ>\<open>int\<close>.

  \<^descr>[\<^theory>\<open>Word_Lib.Reversed_Bit_Lists\<close>]

    Representation of bit values as explicit list in
    \<^emph>\<open>reversed\<close> order.

    This should rarely be necessary: the \<^const>\<open>bit\<close> projection
    should be sufficient in most cases.  In case explicit lists
    are needed, existing operations can be used:

    @{thm [display] horner_sum_bit_eq_take_bit [where ?'a = int, no_vars]}

  \<^descr>[\<^theory>\<open>Word_Lib.Many_More\<close>]

    Collection of operations and theorems which are kept for backward
    compatibility and not used in other theories in session \<^text>\<open>Word_Lib\<close>.
    They are used in applications of \<^text>\<open>Word_Lib\<close>, but should be migrated to there.
\<close>

section \<open>Changelog\<close>

text \<open>
  \<^descr>[Changes since AFP 2021] ~

    \<^item> Theory \<^theory>\<open>Word_Lib.Ancient_Numeral\<close> is not part of \<^theory>\<open>Word_Lib.Word_Lib_Sumo\<close>
      any longer.

    \<^item> Infix syntax for \<^term>\<open>(AND)\<close>, \<^term>\<open>(OR)\<close>, \<^term>\<open>(XOR)\<close> organized in
      syntax bundle \<^bundle>\<open>bit_operations_syntax\<close>.

    \<^item> Abbreviation \<^abbrev>\<open>max_word\<close> moved from distribution into theory
      \<^theory>\<open>Word_Lib.Legacy_Aliases\<close>.

    \<^item> Operation \<^const>\<open>test_bit\<close> replaced by input abbreviation \<^abbrev>\<open>test_bit\<close>.

    \<^item> Abbreviations \<^abbrev>\<open>bin_nth\<close>, \<^abbrev>\<open>bin_last\<close>, \<^abbrev>\<open>bin_rest\<close>,
      \<^abbrev>\<open>bintrunc\<close>, \<^abbrev>\<open>sbintrunc\<close>, \<^abbrev>\<open>norm_sint\<close>,
      \<^abbrev>\<open>bin_cat\<close> moved into theory \<^theory>\<open>Word_Lib.Legacy_Aliases\<close>.

    \<^item> Operations \<^abbrev>\<open>bshiftr1\<close>,
      \<^abbrev>\<open>setBit\<close>, \<^abbrev>\<open>clearBit\<close> moved from distribution into theory
      \<^theory>\<open>Word_Lib.Legacy_Aliases\<close> and replaced by input abbreviations.

    \<^item> Operations \<^const>\<open>shiftl1\<close>, \<^const>\<open>shiftr1\<close>, \<^const>\<open>sshiftr1\<close>
      moved here from distribution.

    \<^item> Operation \<^const>\<open>complement\<close> replaced by input abbreviation \<^abbrev>\<open>complement\<close>.
\<close>

(*<*)
end
(*>*)

subset_name: curated


TITLE: Is pressure exerted by different forces on a fluid particle same?
QUESTION [2 upvotes]: If the  closed container containing the fluid (not any air) is moving with some acceleration, can pressure due to a fluid particle's weight be equal to pressure due to a force exerted by another fluid particles so that so that the fluid particle moves with the same acceleration?

Here Pressure due to gravity is equal to pressure due to forces exerted by fluids in horizontal direction.Why?

REPLY [1 votes]: In the vertical direction, ff the particle is more dense than the fluid, it will tend to sink to the bottom of the container, and if it is less dense, it will tend to rise to the top of the container.  This just relates to the buoyancy effect of the particle's weight in combination with the hydrostatic pressure distribution within the fluid caused by gravity.
Now for the horizontal acceleration.  If the fluid and container are accelerating horizontally in the positive x direction, that is equivalent to imposing artificial gravity in the negative x direction.  If the particle is more dense than the liquid, the particle will move toward the back of the container (relative to the fluid), and if the particle is less dense than the liquid, it will move toward the front of the container (relative to the fluid).  In either case, if there is sufficient time for it to reach terminal velocity (relative to the fluid) before it hits the front or back of the container, it attain the same acceleration as the fluid itself relative to a laboratory frame of reference.  Before that, it's acceleration will be either higher or lower than the fluid acceleration, depending on its relative density.


TITLE: Proving that $X^6-15X^4-6X^3+75X^2-90X-116$ is irreducible over $\mathbb Q$
QUESTION [8 upvotes]: When asked to find the minimal polynomial of  $\sqrt[3]{3}+\sqrt[2]{5}$ over $ \mathbb Q$, I easily found out that $X^6-15X^4-6X^3+75X^2-90X-116$ has $\sqrt[3]{3}+\sqrt[2]{5}$ as a root. It's very likely that it is indeed the minimal polynomial.
All I need to prove now is that it is irreducible over $\mathbb Q$ and equivalently that it is irreducible over $\mathbb Z$. The rational root theorem at least proves (computations needed though) that the polynomial has no root in $\mathbb Q$.
What to do next ? Eisenstein criterion is of no use here.

REPLY [1 votes]: Known Result :

Let $I$ be a proper ideal in the integral domain $R$ and let $p(x)$ be a non constant polynomial in  $R[x]$. If the image of $p(x)$ in $(R/I)[x]$ is irreducible then $p(x)$ is irreducible in $R[x]$.

This is actually slightly general than Eisenstein. If you want some reference I would suggest you to look at "Dummit Foote -Abstract Algebra -Polynomial Rings -Irreducibility".
Now, $X^6-15X^4-6X^3+75X^2-90X-116$ after going modulo $2$ gives you $x^6-x^4+x^2$ which is seen to be reducible in the first sight.
Now, $X^6-15X^4-6X^3+75X^2-90X-116$ after going modulo $3$ gives you (??). This can be seen to be reducible (??) (Trust me i will not make  you do laborious work )
Now, $X^6-15X^4-6X^3+75X^2-90X-116$ after going modulo $5$ gives you (??).  This can be seen to be reducible (??) (Trust me i will not make you do laborious work )
Now, $X^6-15X^4-6X^3+75X^2-90X-116$ after going modulo $7$ gives you (??). As claimed by Mr.Martin, This is Irreducible.
I prefer going modulo some prime because this would give a finite field $\mathbb{Z}_p$ and it would not be so difficult to find irreducible polynomials of given degree.
Once you know all irreducible polynomials of degree $2$ and $3$ in $\mathbb{Z}_7$ then you are done...
Just make sure no irreducible polynomial you got  of degree $2$ and $3$ divides your $p(x)$ (on $\mathbb{Z}_7$).
Thus, you found some prime $p$ in this case it is $7$ such that $p(x)$ is irreducible in $\mathbb{Z}_7$ and by above stated result $p(x)$ is irreducible in $\mathbb{Z}[x]$ and so is in $\mathbb{Q}[x]$ (why ??)
P.S : This is not any better than other ways but I felt it would be a good way to conclude irreducibility.. choosing a prime $p$ in this case is as difficult as choosing prime $p$ in eisenstein :D so this is not the worst way :).


TITLE: Clarification on a proof that the rank of a locally free sheaf is the same everywhere if $X$ is connected.
QUESTION [0 upvotes]: I have seen the answer in this previous post.
My question is as follows.
Given a locally free sheaf $F$ over a connected scheme $X$. Why is it true that if $U$ and $V$ are two open sets in $X$ such that $F(U)$ and $F(V)$ are free $O_X(U), O_X(V)$ modules, respectfully, then their ranks are equal?
More simply, if the stalk of $F$ at a point $p$, $F_p$ is a free $O_X(p)$ module of rank $k$, why is there necessarily some open neighborhood $p\in U\subseteq X$, such that $F(U)$ is a free $O_X(U)$ module of rank $k$?
I am trying to think about this as follows. By taking an affine neighborhood of $p$ we can assume $X$ is an affine scheme. Taking generators $(m_1,U_1),...,(m_k,U_k)$ of $F_p$, and then defining $U=\cap_{i=1}^k U_i$, it becomes clear that $F(U)$ is not a free module of rank $n$ if there is a non-trivial combination of $f_i\in O_X(U)$, such that $\sum_{i=1}^k f_i\cdot m_i = 0$. Naively I would like to say that this implies the existence of a null non-trivial combination of the $m_i$ in $F_p$ by applying the restriction map from $U$ to $p$, but this argument is false, as the restriction map is not necessarily injective. I.e., $\rho_{p}^U(f_i)$ can be $0$ even if $f_i\neq 0$ in $O_X(U)$. Moreover, when I think about it, $O_X(U)$ can be comprised entirely of the $0$ section for all I know.
Thanks for reading.

REPLY [2 votes]: You are working with the wrong definition of a locally free sheaf.  The correct definition is that a sheaf $F$ is locally free if $X$ can be covered by open sets $U$ such that $F|_U$ is a free $O_X|_U$-module.  This is stronger than the stalks being free modules (though it is equivalent under appropriate finiteness conditions, e.g. if $F$ is coherent).
In particular, for any such open sets $U$, the stalks of $F$ are free of the same rank (namely, the rank of $F|_U$ over $O_X|_U$) at every point of $U$.  It follows that the rank of the stalks of $F$ is locally constant and thus constant since $X$ is connected.


TITLE: Fourier transform as change of basis
QUESTION [1 upvotes]: The Fourier transform $\hat{f}$ of some function $f$ is often presented as a change of basis to a basis of complex exponentials. This however begs the question: since $\hat{f}$ is expressed with respect to a basis, is $f$ too expressed with respect to some basis?
In a linear algebra setting, the initial and final bases are clearly defined, for example from the standard basis to some linear combination of it. Here, however, it seems that the initial basis is just left out of the discussion.

REPLY [0 votes]: Building on what mcd has written, the usual way to expand in the orthonormal basis $e_n$ is:
$$
              f = \sum_{n=-\infty}^{\infty}\langle f,e_n\rangle e_n.
$$
That's essentially what you're doing with the Fourier transform with regard to a basis with a continuous index 's' that ranges over the real numbers:
$$
            f = \int_{\mathbb{R}}\left\langle f,\frac{e^{isx'}}{\sqrt{2\pi}}\right\rangle \frac{e^{isx}}{\sqrt{2\pi}}ds
$$


TITLE: Has the additive identity to be same for all elements of a set V which is a vector space over a field F??
QUESTION [0 upvotes]: I am solving a problem in which it's given that V = $\mathbb R$ and F = $\mathbb R$ and the definition of vector addition and scalar multiplication is given. The problem is to check which axioms of vector space are satisfied  and which are not.
I am getting unique additive identity as $ \frac{-v}{2}$ $\forall$ $v\in V$.
My doubt is that has the additive identity to be same for $\forall$ $v\in V$, like the one in the usual addition (which is 0) or it can be different but unique for   different elements of V like the additive inverse in usual addition (which is $-v$ $\forall$ $v\in V$).

REPLY [1 votes]: Indeed the additive identity has to be the same for all elements of the vector space. By definition, the additive identity is an element $e\in V$ such that
$$e + v = v + e = v$$
for all $v\in V$.
This definition is what it means for an element to be an additive identity, and a natural question is: Can there only be one additive identity or can there be more?
To answer this, assume that $e_1$ and $e_2$ are two vectors that are both additive identities. Since $e_1$ is an additive identity, we have for all $v\in V$ that $e_1 + v = v$. Since this is true for all $v$, I can put $v=e_2$, so $e_1+e_2=e_2$.
On the other hand, since $e_2$ is an additive identity, we have for all $v$ that $v+e_2 = v$. In particular, we can put $v=e_1$, so $e_1+e_2=e_1$.
Ok, but now $e_1+e_2 = e_2$, and also $e_1+e_2=e_1$, so it turns out that $e_2=e_1$. In conclusion, there can only be one additive identity.
Thus for all vectors $v\in V$, the additive identity will be the same element (which is usually called "$0$" for vector spaces).


TITLE: Question regarding combinatorics formula
QUESTION [1 upvotes]: I was solving this math problem:
If we roll three standard dices what is the possibility that the sum of eyes will be equal to 7?
So I tried to approach this problem in two different ways:
1) The order in every sequence IS IMPORTANT (a,a,b) and (a,b,a) sequences are counted as different. In this case all the possible variants are counted using formula A*(6,3)=6^3=216. When we look for the favorable endings there are 4 different possible sums: (5,1,1) (3,2,2) (4,2,1) (3,3,1) . All that is left is to count the different variants in which these sums can be displayed and we will have all possible outcomes. This is done using formula P(3,2) = 3! / 2! = 3 and for sequence (4,2,1) formula 3! = 6 . We get 3+3+3+6 favorable sequence and the possibility is 15/216 = 5/72. 5/72 is also the answer according to the book.
2) However, It is way more logical to think that sequences (3,2,2) and (2,2,3) are same and should be counted as one because all that matters is the sum of eyes on the three dices. However, when I make the calculations here, the answer seems to be different. All the possible variants can be counted using formula C*(6,3)=C(6+3-1,C)=(6+3-1)!/(3!*(6-1)!) = 56. We know that the desired results can be achieved with 4 possible sequences (5,1,1) (3,2,2) (4,2,1) (3,3,1) and since they are all the same we do not need to count all the possible displays, therefore, we have 4 favorable endings. In this case the possiblity is 4/36 = 1/9. Obviously, the possibilities in both 1) and 2) cases should have matched, so, could You please tell me where have I made a mistake? Thank you in advance :)
P.S the suggested answer is in the first 1) variant, but the way it is being calculated does not seem the best approach since order in sequences should really not matter. I am particularly interested in this formula where you have to count all the possible endings where order does not matter but the same element can repeat. I fail to use it every single time, maybe there is something wrong with the way I use it?

REPLY [1 votes]: I'll just put in my own insight.
Let's say the die were rolled in order. Then we can look at the entire thing in terms of the first die.
By the way the total probability space is $6^3=216$. These are permutations that allow for repetition.
So the first die can never be $6$. Unless there's an option of $0$ on one of the other die you'd always add up to something greater than $7$.
If the first die has a $5$, then the number remaining to be added is $2$. You can't roll a $2$ on the die left because that would ruin our target sum of $7$. So the second die can either be $1$ or $1$ and the last can keep the remaining $1$. That gives us only one permutation.
If the first die has a $4$, then the second die can only possess values from $1$ to $2$, not $3$, because we need to leave something for the third die. That gives us $2$ permutations.
Repeating this for the other values of the first die:
$3 \implies 3 \text{ permutations}$
$2 \implies 4 \text{ permutations}$
$1 \implies 5 \text{ permutations}$
Combining this with:
$4 \implies 2 \text{ permutations}$
$5 \implies 1 \text{ permutations}$
$6 \implies 0 \text{ permutations}$
So the probability is:
$=\dfrac{0+1+2+3+4+5}{216}=\dfrac{5}{72}$
And the thing you were proposing. This system is full of permutations and not combinations. Think about it like this: would it be fair to serve the guy behind you in a line just because mathematically there is one combination of your arrangement on line? That's how the die feel. The sequences $1,5,1$ and $5,1,5$ exist on different die so they can't be equated as one. Changing the order changes which die have the number. And that's definitely a permutation.


text:
\begin{document}
\maketitle
\begin{center}
{\footnotesize Department of Mathematics, Faculty of Science, Chouaib Doukkali University, Morocco\\
Email: moucouf@hotmail.com}
\end{center}
\begin{abstract}
We derive a closed-form expression for the kth power of semicirculant matrices by using the determinant of certain matrices. As an application, a closed-form expression for the kth power of $r$-circulant matrices is also povided.
\end{abstract}

\section{Introduction}

The $n\times n$ $r$-circulant matrix $C_{n,r}$ over a unitary commutative ring $R$ is one having the
following form
\begin{equation}
C_{n,r}=\left(\begin{array}{cccccc}
c_{0}&c_{1}&c_{2}&\cdots&c_{n-2}&c_{n-1}\\
rc_{n-1}&c_{0}&c_{1}&\cdots&  c_{n-3}   &c_{n-2}\\
rc_{n-2}&rc_{n-1}&c_{0}&\cdots&c_{n-4}&c_{n-3}\\
\cdots&\cdots&\cdots&\cdots&\cdots  &\cdots\\
rc_{1}&rc_{2}&rc_{3}&\cdots&rc_{n-1}&c_{0}
\end{array}
\right),
\end{equation}
where $r\in R$ is a parameter. The $r$-circulant matrix $C_{n,r}$ is determined by $r$  and its first row elements $c_{0},\ldots,c_{n-1}$, thus we denote $C_{n,r}=\text{circ}_{n,r}(c_{0},\ldots,c_{n-1})$.
\\However, an infinite semicirculant matrix over $R$ is one having the following form (see e.g., Henrici~\cite{Hen} or Davis~\cite{Dav})
\begin{equation*}
A=[a_{0}, a_{1}, a_{2},\ldots]=
\begin{pmatrix}
a_{0}& a_{1}&a_{2}&a_{3}&\cdots\\
0&a_{0}&a_{1}&a_{2}&\cdots\\
0& 0&a_{0}&a_{1}& \cdots\\
0&0& 0&a_{0}& \cdots\\
\cdots&\cdots& \cdots&\cdots&\cdots \\
\end{pmatrix}.
\end{equation*}
In the work~\cite{Mou}, the general expression of the $k$th power $A^{k}=[a_{0}(k), a_{1}(k), a_{2}(k),\ldots]$ ($k\in \mathbb{N}$) of a semicirculant matrix
$A=[a_{0}, a_{1}, a_{2},\ldots]$ is presented. More precisely the sequence $\{a_{m}(k)\}_{m\geq0}$ is obtained using a recursive method. To do this, we have proved that for all $k\in \mathbb{N}$,
\begin{equation}\label{equ 13}
\begin{array}{ccc}
a_{i}(k) &=& L(A)(i,0)a_{0}^{k}
\binom{k}{0}+\cdots+L(A)(i,j) a_{0}^{k-j}
\binom{k}{j}+\cdots+L(A)(i,i) a_{0}^{k-i}
\binom{k}{i},
\end{array}
\end{equation}
where we adopt the convention that for any element $a\in R$ and any nonnegative integers $k\leq p$, $a^{k-p}\binom{k}{p}=\delta_{k,p}$.
\\The double sequence $\{L(A)(n,m)\}_{n,m}$ of elements of $R$ is defined by
\begin{eqnarray}\label{eq: 4001}
L(A)(n,m)=\displaystyle\sum_{\substack{\Delta(n,m)}}\binom{m}{k_{1},\ldots,k_{n}}a_{1}^{k_{1}}\cdots a_{n}^{k_{n}},
\end{eqnarray}
where $\Delta(n,m)$, $m\leqslant n$ are integers, is the solution set of the following system of equations
\begin{eqnarray*}
\begin{cases}
(k_{1},\ldots,k_{n})\in \mathbb{N}^{n}\\
k_{1}+\cdots+k_{n}=m\\
k_{1}+2k_{2}+ \cdots + nk_{n}=n.
\end{cases}
\end{eqnarray*}
We note that Formula \eqref{equ 13} expresses $a_{i}(k)$ in closed-form that is not easy to use. Here we provide an easy closed-form formula in terms of determinants. As a consequence, in the last section we give a closed-form formula for the kth power of $r$-circulant matrices and we describe a method for finding the solution set of the well known Diophantine equation
\begin{eqnarray*}
k_{1}+2k_{2}+ \cdots + nk_{n}=n.
\end{eqnarray*}
Throughout this paper $R$ will denote an arbitrary commutative ring with identity.

\section{Sequences $L(A)(n,m)$ and $u_{n}(A)$ corresponding to a semicirculant matrix $A$}
Let $A=[a_{0}, a_{1}, a_{2},\ldots]$ be a semicirculant matrix over $R$. Then The sequence $(L(A)(i,j))_{i,j}$ is independent on the coefficient $a_{0}$ and it is uniquely determined by the following recursive formula (see Lemma 2.2 of \cite{Mou}):
\begin{equation}\label{eq:SD2}
\left\{
\begin{array}{llllll}
L(A)(0,0)&=&1&\\
L(A)(i,0)&=&0& \quad \text{for } & i\neq 0\\
L(A)(i,j)&=&0& \quad \text{for } & i< j\\
L(A)(i,j+1)&=&a_{1}L(A)(i-1,j)+\cdots+a_{i-j}L(A)(j,j)
\end{array}
\right.
\end{equation}
Let $L(A)$ be the matrix $(L(A)(i,j))_{0\leqslant i,j}$. Then $L(A)$ is the following lower triangular matrix
$$ \begin{tikzpicture}[baseline=(math-axis),every right delimiter/.style={xshift=-3pt},every left delimiter/.style={xshift=3pt}]
\matrix [matrix of math nodes] (matrix)
{
|(m11)| 1 &|(m12)| & |(m13)|& |(m14)|&|(m15)| &|(m16)|&|(m17)|&|(m18)|&|(m19)|\\
 |(m21)| 0  & |(m22)| a_{1} &|(m23)| &|(m24)| & |(m25)| &|(m26)| & |(m27)|&|(m28)| &|(m29)| \\
|(m31)| 0  & |(m32)| a_{2} & |(m33)| a_{1}^{2} & |(m34)|&|(m35)|     &|(m36)| &|(m37)| &|(m38)| &|(m39)| \\
|(m41)| 0  & |(m42)| a_{3} & |(m43)| 2a_{1}a_{2} & |(m44)| a_{1}^{3} & |(m45)| & |(m46)|   & |(m47)|& |(m48)|&|(m49)| \\
  |(m51)| 0  & |(m52)| a_{4} & |(m53)| 2a_{1}a_{3}+a_{2}^{2} & |(m54)| 3a_{1}^{2}a_{2}  & |(m55)| a_{1}^{4}    &|(m56)|   &|(m57)| &|(m58)|&|(m59)| \\
  |(m61)| 0  & |(m62)| a_{5} & |(m63)|2a_{1}a_{4}+2a_{2}a_{3} & |(m64)| 3a_{1}^{2}a_{3}+3a_{1}a_{2}^{2}  & |(m65)| 4a_{1}^{3}a_{2} & |(m66)|a_{1}^{5} & |(m67)|&|(m68)| &|(m69)| \\
  |(m71)|0  & |(m72)| a_{6} & |(m73)| 2a_{1}a_{5}+2a_{2}a_{4}+a_{3}^{2} & |(m74)|3a_{1}^{2}a_{4}+6a_{1}a_{2}a_{3}+a_{2}^{3} & |(m75)| 4a_{1}^{3}a_{3}+6a_{1}^{2}a_{2}^{2} & |(m76)| 5a_{1}^{4}a_{2} & |(m77)|a_{1}^{6} &|(m78)| &|(m79)|\\
  |(m81)|\vdots  &|(m82)|\vdots     & |(m83)|  \vdots & |(m84)|\vdots &|(m85)| \vdots
  &|(m86)| \vdots & |(m87)|\vdots       & |(m88)|\ddots & |(m89)|\\
};
\node[draw,dashed,inner sep=0.2pt,fit=(m33) (m63) (m33)] {};
\node[draw,dashed,inner sep=0.2pt,fit=(m74) (m74)] {};
\draw[ dashed ,->] (-2.6,0.2)--(-1.7,-1.3);
\coordinate (math-axis) at ($(matrix.center)+(0em,-0.25em)$);
\end{tikzpicture}$$
Each colomn of this matrix can be deduced from the precedent one.
For example $$\big<(a_{1}, a_{2}, a_{3}, a_{4}),(2a_{1}a_{4}+2a_{2}a_{3}, 2a_{1}a_{3}+a_{2}^{2}, 2a_{1}a_{2}, a_{1}^{2})\big>=3a_{1}^{2}a_{4}+6a_{1}a_{2}a_{3}+a_{2}^{3}.$$
Let $L_{i}(A)$ be the $i$th row of the matrix $L(A)$ and let $S_{i}(A)$ be the right shift of $L_{i}(A)$ by one position. Then formula~\eqref{eq:SD2} implies that
\begin{equation}\label{eq:SD3}
S_{n}(A)=a_{1}L_{n-1}(A)+\cdots+a_{n}L_{0}(A).
\end{equation}
We now consider the sequence
\begin{equation}\label{eq:SD4}
u_{n}(A)=\left\{
\begin{array}{llllll}
\sum_{i=1}^{n}L(A)(n,i)\\
u_{0}(A)=L(A)(0,0)=1
\end{array}\right.
\end{equation}
It is clear that $u_{n}$ is the sum of all elements of the ($n+1$)th row of $L(A)$. From formula~\eqref{eq:SD3} it follows that
\begin{equation}\label{eq:SD5}
\left\{
\begin{array}{llllll}
u_{n}(A)=a_{1}u_{n-1}(A)+a_{2}u_{n-2}(A)+\cdots+a_{n}u_{0}(A)\\
u_{0}(A)=1
\end{array}\right.
\end{equation}
Clearly, the sequence $u_{n}(A)$ is uniquely determined by the relation \eqref{eq:SD5}.
\\Let us now consider the following sets:
\\$\mathcal{SC}$ the set of all infinite semicirculant matrices with entries in the ring $R$,
\\$\mathcal{D}$ the subset of $\mathcal{SC}$ consisting of all diagonal matrices,
\\$\mathcal{U}$ the set of all sequences satisfying the recurrence relation~\eqref{eq:SD5} for some sequences $\{a_{n}\}_{n\geq1}$ of  the elements of $R$,
\\$\mathcal{V}$ the set of all double sequences satisfying the recurrence relation~\eqref{eq:SD2} for some sequences $\{a_{n}\}_{n\geq1}$ of the elements of $R$.\\
The following result shows the existence of a biunivoque correspondence between any two of the sets: $\mathcal{U}$, $\mathcal{V}$ and $\mathcal{SC}/\!\raisebox{-.65ex}{\ensuremath{\mathcal{D}}}$.
\begin{prop}\label{prop 18}
There is a commutative diagram of bijections
$$\begin{tikzcd}[row sep=2.5em]
 &\mathcal{SC}/\!\raisebox{-.65ex}{\ensuremath{\mathcal{D}}} \arrow{dr}{\varrho} \arrow{ld}[swap]{\psi}\\
 \mathcal{V}  \arrow{rr}{\varphi} && \mathcal{U}
\end{tikzcd}$$
where the maps $\rho$, $\psi$ and $\varphi$ are defined as follows
\begin{eqnarray*}
\varrho(\overline{A})&=&\{u_{n}(A)\}_{n\geq0}, \,\, \overline{A} \;\text{denotes the equivalence class of}\; A\in \mathcal{SC} \,\text{modulo}\, \mathcal{D}.\\
\psi(\overline{A})&=&\{L(A)(n,m)\}_{n,m},\\
\varphi(\{\phi(n,m)\}_{n,m})&=&\{\sum_{i=0}^{n}\phi(n,i)\}_{n\geq0}.
\end{eqnarray*}
These maps have respectively as inverses the following
\begin{eqnarray*}
\psi^{-1}(\{\phi(n,m)\}_{n,m})&=&\overline{[0,\phi(1,1),\ldots,\phi(n,1),\ldots]},\\
\varrho^{-1}(\{u_{n}\}_{n\geq0})&=&\overline{[0,a_{1},\ldots,a_{n},\ldots]},\\
\varphi^{-1}(\{u_{n}\}_{n\geq0})&=&\{\phi(n,m)\}_{n,m},
\end{eqnarray*}
where $\phi(n,m)=\displaystyle\sum_{\substack{\Delta(n,m)}}\binom{m}{k_{1},\ldots,k_{n}}a_{1}^{k_{1}}\cdots a_{n}^{k_{n}}$
and
$\{a_{n}\}_{n\geq1}$ is the sequence of elements of $R$ corresponding to $\{u_{n}\}_{n\geq0}$.
\end{prop}
\proof
Because sequences of $\mathcal{U}$ are uniquely determined by~\eqref{eq:SD5}, the map $\varrho$ is well defined. Let now $A=[a_{0},a_{1},a_{2},\ldots]$ and $B=[b_{0},b_{1},b_{2},\ldots]$ such that $u_{n}(A)=u_{n}(B),$ for all positive integer $n$. Then $u_{1}(A)=u_{1}(B)$ and hence $a_{1}=b_{1}$. Furthermore, the relation~\eqref{eq:SD5} implies that $a_{n}=a_{n}u_{0}(A)=u_{n}(A)-a_{1}u_{n-1}(A)-\cdots-a_{n-1}u_{1}(A)$. It then follows by an easy induction that $a_{n}=b_{n}$ for all positive integer $n$, i.e., $\overline{A}=\overline{B}$. Hence the map $\varrho$ is injective. To show that $\varrho$ is surjective, consider an element $(u_{n})_{n\in \mathbb{N}}$ of $\mathcal{U}$ and let $(a_{n})_{n\in \mathbb{N^{\ast}}}$ the corresponding sequence of $R$. It is clear that $\varrho(\overline{[0,a_{1},\ldots,a_{n},\ldots]})=\{u_{n}\}_{n\geq0}$, and then the map $\varrho$ is bijective.
\\The same argument as for $\varrho$ applies again to $\psi$.
\\Now let $\{\phi(n,m)\}_{n,m}\in \mathcal{V}$ and consider the matrix $A=[0,\phi(1,1),\ldots,\phi(n,1),\ldots]$. It is clear that $\varphi(\{\phi(n,m)\}_{n,m})=\{u_{n}(A)\}_{n\geq0}$ and then $\varphi$ is bijective.\\
The last assertion follows immediately from the fact that $\varphi\circ\psi= \rho$.
\endproof
\begin{prop}\label{prop 19} Let $a_{0},a_{1},a_{2},\ldots\in R$ and let $\delta(a_{0}, a_{1}, \ldots, a_{n})$ be the following determinant
\begin{equation}
\delta(a_{0}, a_{1}, \ldots, a_{n})=
\left|\begin{array}{cccccc}
a_{1}   & a_{2}        &\cdots  &\cdots   & \cdots  &a_{n}\\
a_{0}       &     \ddots    &\ddots  &         &        &\vdots      \\
0       &    \ddots     & \ddots &  \ddots &        &\vdots\\
\vdots  &    \ddots     & \ddots &\ddots   &  \ddots      &\vdots\\
\vdots  &               & \ddots &\ddots   & \ddots & a_{2}\\
0       & \cdots        & \cdots &   0     & a_{0}      &a_{1}
\end{array}\right|.
\end{equation}
Then we have
\begin{eqnarray}
\delta(a_{0}, a_{1}, \ldots, a_{n})&=&\delta(1, a_{1}, \ldots, a_{i}a_{0}^{i-1}, \ldots, a_{n}a_{0}^{n-1}) \label{eq:55}\\
                                &=&\delta(-1, a_{1}, \ldots, a_{i}(-a_{0})^{i-1}, \ldots,a_{n}(-a_{0})^{n-1}) \label{eq:56}
                                \end{eqnarray}
for all $n\in \mathbb{N^{\ast}}$.
\end{prop}
\proof The claim is trivially true for $a_{0}=0$. Suppose $a_{0}\neq 0$. For the proof of the first equality it suffices to multiply every column $C_{i}$ of the determinant $\delta(a_{0}, a_{1}, \ldots, a_{n})$ by $a_{0}^{i-1}$, and then multiply every row $R_{i}$ of the resulting determinant by $a_{0}^{-i+1}$.\\
To obtain the last equality, it suffices to multiply each of the even rows of the determinant $\delta(1, \ldots,a_{i}a_{0}^{i-1}, \ldots, a_{n}a_{0}^{n-1})$ by $-1$, and then multiply each of the even columns of the resulting determinant by $-1$.
\endproof
\begin{prop}\label{prop 20} Let $A=[a_{0},a_{1},a_{2},\ldots]$ be a semicirculant matrix over $R$, and let $\{u_{n}(A)\}_{n\geq0}$ be the sequence of elements of $R$ associated to $A$ given in~\eqref{eq:SD4}. Then we have
\begin{equation}\label{eq:54}
u_{n}(A)=\delta(-1, a_{1}, \ldots, a_{n})
\end{equation}
for all $n\in \mathbb{N^{\ast}}$.
\end{prop}
\proof
By~\eqref{eq:55} we know that $\delta(-1, a_{1},\ldots, a_{n})=\delta(1, a_{1}, \ldots, a_{i}(-1)^{i-1}, \ldots, a_{n}(-1)^{n-1})$. Assume that $n\geqslant1$, and let us expand the determinant $\delta(1, a_{1}, \ldots, a_{i}(-1)^{i-1}, \ldots, a_{n}(-1)^{n-1})$
along the first row. Then we get
\begin{eqnarray*}
\delta(-1, a_{1}, \ldots, a_{n})=\sum_{i=1}^{n-1}a_{i}\delta_{i},
\end{eqnarray*}
where $\delta_{i}$ is the cofactor associated with the entry $(-1)^{i-1}a_{i}$ of the matrix $\delta(1, a_{1}, \ldots, a_{i}(-1)^{i-1}, \ldots, a_{n}(-1)^{n-1})$.\\
It can be easily seen that the cofactor $\delta_{i}$ has the form
   \begin{eqnarray*}
   \delta_{i}=
   \left|
    \begin{array}{ccc;{6pt/6pt}ccclc}
     1& & \star & \multicolumn{4}{c}{\multirow{3}{0cm}{\huge$\star$}}&\\
      & \rdots& &&&&&\\
     0&&1&&&&&\\
     \hdashline[6pt/6pt]
     &&&a_{1}&-a_{2}&\cdots&a_{n-i}(-1)^{n+1-i}&\\
     &\multicolumn{1}{c}{\multirow{2}{0.5cm}{\huge$0$}}&&1&\ddots&&\vdots&\\
     &&& &\ddots&\ddots&\vdots&\\
     &&&0&&1&a_{1}&
    \end{array}
    \right|
   \end{eqnarray*}
   Thus for $i\geqslant1$ we have $\delta_{i}=\delta(-1,a_{1},\ldots,a_{n-i-1})$.
Hence
\begin{eqnarray*}
\delta(-1,a_{1},\ldots,a_{n})=\sum_{i=1}^{n-1}a_{i}\delta(-1,a_{1},\ldots,a_{n-i-1}).
 \end{eqnarray*}
It follows that the sequence $\delta(-1,a_{1},\ldots,a_{n})$ satisfies the recurrence relation~\eqref{eq:SD5}. But since $\delta(-1,a_{1})=a_{1}$, we have $\delta(-1,a_{1},\ldots,a_{n})=u_{n}$ for all $n\in \mathbb{N^{\ast}}$.
\endproof
The following proposition is useful for the proof of the main result.
\begin{prop}\label{prop 201} Let $X$ be an indeterminate over $R$ and let $\{a_{n}\}_{n\geq1}$ be a sequence of the elements of $R$. Let $\mathcal{X}_{n}(X)$ be the polynomial sequence defined by
\begin{equation}\label{eq:201}
\mathcal{X}_{n}(X)=\delta(-X,a_{1},\ldots,a_{n}).
\end{equation}
 Then one has
\begin{equation}\label{eq: 1201}
\mathcal{X}_{n}(X)=\displaystyle\sum_{i=1}^{n}L(A)(n,i)X^{n-i},
\end{equation}
where $A=[0,a_{1},a_{2},\ldots]$.
\end{prop}
\proof
Let $A=[0,a_{1},a_{2},\ldots]$ and consider the semicirculant matrix $B=[0,a_{1},a_{2}X,\ldots,a_{n}X^{n-1},\ldots]$ over $R[X]$. Then
\begin{eqnarray*}
L(B)(n,m)&=&\sum_{\substack{\Delta(n,m)}}\binom{m}{k_{1},\ldots,k_{n}}a_{1}^{k_{1}}(a_{2}X)^{k_{2}}\cdots (a_{n}X^{n-1})^{k_{n}}\\
&=&\sum_{\substack{\Delta(n,m)}}\binom{m}{k_{1},\ldots,k_{n}}a_{1}^{k_{1}}\cdots a_{n}^{k_{n}}X^{k_{2}+\cdots+(n-1)k_{n}},
\end{eqnarray*}
and since $k_{2}+\cdots+(n-1)k_{n}=(k_{1}+2k_{2}+\cdots+nk_{n})-(k_{1}+k_{2}+\cdots+k_{n})$ it follows that $k_{2}+\cdots+(n-1)k_{n}=n-m$, and $L(B)(n,m)=L(A)(n,m)X^{n-m}$. On the other hand, Formula~\eqref{eq:56} together with Formula~\eqref{eq:54} yield $u_{n}(B)=\mathcal{X}_{n}(X)$. As a consequence of Proposition~\ref{prop 18}, we have $\mathcal{X}_{n}(X)=\displaystyle\sum_{i=1}^{n}L(B)(n,i)=\sum_{i=1}^{n}L(A)(n,i)X^{n-i}$.
\endproof
Let $k$ and $n$ be nonnegative integers and let $[.]_{k}^{n}: R[X]\longrightarrow R[X]$ be the linear map defined by
\begin{eqnarray*}
[X^{i}]_{k}^{n}=X^{k-(n-i)}\binom{k}{n-i},
\end{eqnarray*}
where we use the convention that for $k\leq p$,
\begin{equation}\label{eq,,1}
X^{k-p}\binom{k}{p}=\delta_{k,p},
\end{equation}
the Kronecker delta. Then we have The following result which gives a closed-form expression of the kth power of semicirculant matrices.
\begin{thm}\label{Thmmm 1}
Let $\{a_{n}\}_{n\in \mathbb{N}}$ be a sequence of the elements of $R$. For all nonnegative integer $k$, the kth power of the semicirculant matrix $A=[a_{0},a_{1},a_{2},\ldots]$ are given as follows:
\begin{eqnarray}
A^{k}=[[\mathcal{X}_{0}]_{k}^{0}(a_{0}), [\mathcal{X}_{1}]_{k}^{1}(a_{0}), [\mathcal{X}_{2}]_{k}^{2}(a_{0}), \ldots]
\end{eqnarray}
where $\mathcal{X}_{n}$ is the polynomial $\delta(-X,a_{1},\ldots,a_{n})$ given in~\eqref{eq:201} and $\mathcal{X}_{0}=1$.
\end{thm}
\proof
Follows immediately from Proposition~\ref{prop 201} and Formula~\eqref{equ 13}.
\endproof
\begin{ex}
Consider a semicirculant matrix $A=[3, 2, 2, -3, 3]$. Let $k$ be any nonnegative integer.
We have
\begin{eqnarray*}
\mathcal{X}_{3}(X)= \begin{vmatrix} 2& 2& -3\\-X&2& 2\\0&-X&2\end{vmatrix}=-3X^{2}+8X+8.
\end{eqnarray*}
Then the $(1,4)$ entry of matrix $A^{k}$ is
$-3\times 3^{k-1}\binom{k}{1}+8\times 3^{k-2}\binom{k}{2}+8\times 3^{k-3}\binom{k}{3}$.
\end{ex}
\begin{ex}
Consider another semicirculant matrix $B=[0, 2, 1, 0,3]$. Let $k$ be any nonnegative integer.
We have
\begin{eqnarray*}
\mathcal{X}_{4}(X)= \begin{vmatrix} 2& 1&0&3\\-X&2&1&0\\0&-X&2&1\\0&0&-X&2\end{vmatrix}=3x^{3} + x^{2} + 12x + 16.
\end{eqnarray*}
Then the $(1,5)$ entry of matrix $B^{k}$ is $3\binom{k}{1}0^{k-1} + \binom{k}{2}0^{k-2} + 12\binom{k}{3}0^{k-3} + 16\binom{k}{4}0^{k-4}$.
\end{ex}
\section{Applications}
\subsection{A closed-form expression for the kth power of $C_{n,r}$}
\begin{thm}\label{Thm 10} Let $r, c_{0},\ldots,c_{n-1}\in R$ and let $k$ be any nonnegative integer. Consider the $r$-circulant matrix
\begin{eqnarray*}
C_{n,r}=\text{circ}_{n,r}(c_{0},\ldots,c_{n-1}).
\end{eqnarray*}
Then the pth strip of $C_{n,r}^{k}$ is
\begin{eqnarray*}
(C_{n,r}^{k})_{p}=\sum_{m\equiv p\;(\bmod\, n)}c_{0}^{k-m}[\mathcal{X}_{m}]_{k}^{m}(c_{0})r^{\floor*{\frac{m}{n}}},
\end{eqnarray*}
where
\begin{eqnarray*}
\mathcal{X}_{m}(X)=
\left\{
\begin{array}{lll}
\delta(-X,c_{1},\ldots,c_{m})& \text{if} &m\geq 1\\
1& \text{if}& m=0,
\end{array}
\right.
\end{eqnarray*}
and $\floor*{x}$ denotes the greatest integer less than or equal to $x$.
\end{thm}
\proof
Follows immediately from Theorem~\ref{Thmmm 1} and Theorem 4.1 of~\cite{Mou}.
\endproof
\begin{rmk}
The sequence $r^{\floor*{\frac{m}{n}}}$, appearing in the pth strip
\begin{eqnarray*}
(C_{n,r}^{k})_{p}=\sum_{m\equiv p}c_{m}(k)r^{\floor*{\frac{m}{n}}}
\end{eqnarray*}
of the $r$-circulant matrix $C_{n,r}^{k}$, is nothing but the geometric sequence with
common ratio $r$.
\end{rmk}
\begin{ex}
Let $C=circ_{5,2}(5, 4, 3, 2, 1)$ and let $[C]=[5, 4, 3, 2, 1,0,0,\ldots]$ be the associated infinite semicirculant matrix. Put
\begin{eqnarray*}
[C]^{k}=[c_{0}(k), c_{1}(k), c_{2}(k),\ldots].
\end{eqnarray*}
Using the method provided in Theorem~\ref{Thmmm 1}, we obtain
\begin{eqnarray*}
c_{s}(3)&=&0 \quad\text{for all}\quad s\geq (3\times4)+1=13;\\
\mathcal{X}_{12}(X)&=&x^{9} + \cdots, \quad c_{12}(3)=5^{9}\times5^{-9}\binom{3}{3}=1;\\
\mathcal{X}_{11}(X)&=&6x^{8} + \cdots, \quad c_{11}(3)=6\times 5^{0}\binom{3}{3}=6;\\
\mathcal{X}_{10}(X)&=&21X^{7} + \cdots,\quad c_{10}(3)=21\times 5^{0}\binom{3}{3}=21;\\
\mathcal{X}_{9}(X)&=&56X^{6} + \cdots,\quad c_{9}(3)=56\times5^{0}\binom{3}{3}=56;\\
\mathcal{X}_{8}(X)&=&x^{6} + 111x^{5} + \cdots,\quad c_{8}(3)=5^{1}\binom{3}{2}+111\times5^{0}\binom{3}{3}=126;\\
\mathcal{X}_{7}(X)&=&4X^{5} + 174X^{4} + \cdots,\quad c_{7}(3)=4\times5^{1}\binom{3}{2}+174\times5^{0}\binom{3}{3}=234;\\
\mathcal{X}_{6}(X)&=&10X^{4} + 219X^{3} + \cdots,\quad c_{6}(3)=10\times5^{1}\binom{3}{2}+219\times5^{0}\binom{3}{3}=369;\\
\mathcal{X}_{5}(X)&=&20X^{3} + 204X^{2} + \cdots,\quad c_{5}(3)=20\times5^{1}\binom{3}{2}+204\times5^{0}\binom{3}{3}=504;\\
\mathcal{X}_{4}(X)&=&x^{3} + 25x^{2} + 144x +\cdots,\quad c_{4}(3)=5^{2}\binom{3}{1}+25\times5^{1}\binom{3}{2}+144\times5^{0}\binom{3}{3}=594;\\
\mathcal{X}_{3}(X)&=&2X^{2} + 24X + 64,\quad c_{3}(3)=2\times5^{2}\binom{3}{1}+24\times5^{1}\binom{3}{2}+64\times5^{0}\binom{3}{3}=574;\\
\mathcal{X}_{2}(X)&=&3X + 16,\quad c_{2}(3)=3\times5^{2}\binom{3}{1}+16\times5^{1}\binom{3}{2})=465;\\
\mathcal{X}_{1}(X)&=&4,\quad c_{1}(3)=4\times5^{2}\binom{3}{1}=300;\\
c_{0}(3)=5^{3}&=&125.
\end{eqnarray*}
Hence \\$[C]^{3}=[125, 300, 465, 574, 594, 504, 369, 234, 126, 56, 21, 6, 1,0,0,\ldots]$.
\\Therefore,
\begin{eqnarray*}
C_{0}^{3}&=&125\times2^{0}+504\times2^{1}+21\times2^{2}=1217\\
C_{1}^{3}&=&300\times2^{0}+369\times2^{1}+6\times2^{2}=1062\\
C_{2}^{3}&=&465\times2^{0}+234\times2^{1}+2^{2}=937\\
C_{3}^{3}&=&574\times2^{0}+126\times2^{1}=826\\
C_{4}^{3}&=&594\times2^{0}+56\times2^{1}=706.
\end{eqnarray*}
Thus
\begin{eqnarray*}
C^{3}=circ_{5,2}(1217, 1062, 937, 826, 706).
\end{eqnarray*}
\end{ex}
\subsection{The calculation of $\Delta(n)$}
Consider the solution set $\Delta(n,p)$ of the following system of equations
\begin{eqnarray*}
\begin{cases}
(k_{1},\ldots,k_{n})\in \mathbb{N}^{n}\\
k_{1}+\cdots+k_{n}=p\\
k_{1}+2k_{2}+ \cdots + nk_{n}=n.
\end{cases}
\end{eqnarray*}
Here, $p$ and $n$ are positive integers such that $p\leqslant n$.
\\Let $\Delta(n)$ be the solution set of the diophantine equation
\begin{eqnarray*}
k_{1}+2k_{2}+ \cdots + nk_{n}=n.
\end{eqnarray*}
We have clearly
\begin{eqnarray*}
\Delta(n)=\bigcup_{1\leq i\leq n} \Delta(n,i).
\end{eqnarray*}
Let $X,x_{1},\ldots,x_{n}$ be independent indeterminates over $R$. From \eqref{eq: 1201} we have
\begin{eqnarray*}
\mathcal{X}_{n}(X)=\displaystyle\sum_{i=1}^{n}L(A)(n,i)X^{n-i},
\end{eqnarray*}
and from~\eqref{eq: 4001} we have
\begin{eqnarray*}
L(A)(n,i)=\displaystyle\sum_{\substack{\Delta(n,i)}}\binom{i}{k_{1},\ldots,k_{n}}x_{1}^{k_{1}}\cdots x_{n}^{k_{n}},
\end{eqnarray*}
where $A=[0,x_{1},\ldots,x_{n}]$.
\\We observe that we can determine the set $\Delta(n)$ by computing the polynomial $\mathcal{X}_{n}(X)$, and then Comparing the polynomial $L(A)(n,i), 1\leq i\leq n,$ and the term of degree $n-i$ of $\mathcal{X}_{n}(X)$.
\begin{ex}
We have $\mathcal{X}_{4}(X)=x_{4}X^{3}+(2x_{1}x_{3}+x_{2}^{2})X^{2}+3x_{1}^{2}x_{2}X+x_{1}^{4}$. Then
\begin{eqnarray*}
\Delta(4,1)&=&\{(0,0,0,1)\},\\
\Delta(4,2)&=&\{(1,0,1,0),(0,2,0,0)\},\\
\Delta(4,3)&=&\{(2,1,0,0)\},\\
\Delta(4,4)&=&\{(4,0,0,0)\}
\end{eqnarray*}
\end{ex}
\begin{ex}
We have $\mathcal{X}_{6}(X)=x_{6} X^{5} + (2x_{1}x_{5} + 2x_{2}x_{4} + x_{3}^{2})X^4 + (3x_{1}^{2}x_{4} + 6x_{1}x_{2}x_{3} + x_{2}^{3})X^{3}+ (4x_{1}^{3}x_{3} + 6x_{1}^{2}x_{2}^{2})X^{2} + 5x_{1}^{4}x_{2}X + x_{1}^{6}$. Then
\begin{eqnarray*}
\Delta(6,1)&=&\{(0,0,0,0,0,1)\},\\
\Delta(6,2)&=&\{(1,0,0,0,1,0),(0,1,0,1,0,0),(0,0,2,0,0,0)\},\\
\Delta(6,3)&=&\{(2,0,0,1,0,0),(1,1,1,0,0,0),(0,3,0,0,0,0)\},\\
\Delta(6,4)&=&\{(3,0,1,0,0,0),(2,2,0,0,0,0)\},\\
\Delta(6,5)&=&\{(4,1,0,0,0,0)\},\\
 \Delta(6,6)&=&\{(6,0,0,0,0,0)\}
\end{eqnarray*}
\end{ex}

config: arxiv


TITLE: Visualization of the real projective plane
QUESTION [5 upvotes]: Consider a closed (compact and without boundary) and non-orientable 2-manifold $M$. By Whitney embedding theorem, one can embed $M$ in $\mathbb{R}^4$. $M$ cannot be embeded in $\mathbb{R}^3$ and just can be immersed in $\mathbb{R}^3$.
We can not imagine a four-dimensional object. Can we use different figures of immersed $M$ in $\mathbb{R}^3$ and obtain imagination of $M$ in $\mathbb{R}^4$ (even a bit)? Can knowing of different figures of immersed $M$ help to imagine embedded $M$ in higher dimension? For example, the real projective plane is a closed and non-orientable 2-manifold, so it can be embedded in $\mathbb{R}^4$ and we cannot imagine this. But different figures of immersed real projective plane in $\mathbb{R}^3$ are within our reach. (following figures, from left to right respectively: cross-cap, Roman's surface, Boy's surface)

REPLY [6 votes]: You could slice the 4D object with a 3D 3-flat that varies in the 4th dimension.
Here are snapshots of such a slicing of the hypercube down a diagonal:

     (Image above from Fleischfilm.)

Here is a video of a (much!) more complex object, a "4D quaternion Julia set," being sliced by a moving 3-flat: video link. And here is a still from the animation:

(Image above from Creative Applications Network.)


TITLE: Maple: assign derivative to function
QUESTION [4 upvotes]: This is probably a basic Maple question.  I'm trying to introduce $g$ as the derivative of $f$:

Somewhat puzzling, Maple now says $g$ is two times the function $x()$.
I've tried g := x -> diff(f(x),x) and g := diff(f,x), but no luck. Google tells me I can use subs(x=3,g) to evaluate the derivative at $x=3$, but that's not very practical.
Is there a way to define a Maple function as the derivative of another Maple function?

REPLY [2 votes]: The answer worked for the OP, but I have had trouble getting "D" to behave the way I want it to.  I found a solution using "unapply".  If the function is named f, then
fp:= unapply(diff(f(x),x),x)

creates a function fp that's the derivative of f.  diff(f(x), x) is an expression, and unapply converts it to a function of x.
This can also be done for functions of several variables (see Maplesoft's documentation for unapply here: https://www.maplesoft.com/support/help/maple/view.aspx?path=unapply)
UPDATE: Hey there, in case anyone else feels compelled to suggest this be deleted, yes I know what "D" does.  It might do what you want most of the time.  But I've written code where "D" didn't behave the way I wanted to.  unapply is a decent workaround.  I see several other Maple users suggesting it in this Maple forum.  One of them wrote "I find that in some situations D seems a little stupid, so often I prefer diff with unapply."
If you think there's something wrong with my answer, then a comment might be nice.  This workaround has worked for me and others when "D" didn't.


TITLE: Do the results of statistical mechanics depend upon the choice of macrostates?
QUESTION [20 upvotes]: Consider a magnet with temperature $T$. We can observe its net magnetization $M$, so we say that a value of $M$ specifies a macrostate. Statistical mechanics tells us which macrostate the magnet is in. To do this, we compute the free energy $F(M)$ and minimize it. The free energy is defined by
$$F = U - TS$$
and hence depends on the entropy $S$. This entropy is determined by the number of microstates that could be compatible with the given macrostate, i.e. the number of spin states that lead to the magnetization we observe.
This procedure has always felt sketchy to me because it seems to rely on some subjective notion of knowledge. The reason that there are many allowed microstates is because we've postulated that we know nothing about the system besides the net magnetization; if we did know something else, it would decrease the number of consistent microstates and change the entropy, which changes $F$.
As such, it looks like the result of the calculation depends on the set of macrostates we use! For example, suppose I somehow attached a measuring device to every single spin. Then in principle, I could specify my macrostates with a long list containing the state of every spin; then there is only one microstate corresponding to each macrostate. In this case, $S = 0$ for each macrostate, $F = U$, and the minimum free energy is attained for minimum $U$.
Then I conclude that the system is always in the ground state!
What's wrong with this reasoning? Is it somehow illegal to make this macrostate choice? Could acquiring all of this information about the spins necessarily change how the magnet behaves, e.g. from something like Landauer's principle? In general, can changing macrostate choice ever change the predictions of statistical mechanics?

REPLY [2 votes]: What's wrong with these reasoning?

First of all let me say what is right: you are right that the definition of macrostates is a choice. In your example, we could split the paramagnet into two equal sections (in our minds) and describe the macrostate in terms of the magnetizations $M_1$ and $M_2$ of each section. If we keep subdividing in this way eventually we end up in the situation you describe, where we consider the magnetization of each spin separately.
What's wrong is your handling of the thermodynamic limit. The statement 'the physical macrostate minimizes free energy' is only true in this limit. In any finite-size system the statistical mechanics gives you a probability distribution over the macrostates of the system, and while the one with minimum free energy has maximum likelihood, there is no reason for the distribution to be sharp. In particular, if you consider each spin seperately then the probability distribution is simply given by the Boltzmann factor, $P(\vec{s}) \sim \exp[\mu \vec{s}\cdot \vec{H}/k T]$.
Now, in order for the distribution to become sharp (and thus the conclusion that the physical macrostate minimizes free energy be valid), what is needed is that the number of microstates per macrostate becomes large. If I consider the macrostate to be described by the total magnetization, $M$, or the magnetizations $(M_1,M_2,\ldots)$ of a fixed finite number of different sections, then as the number of spins $N$ becomes large the number of microstates per macrostate also grows (exponentially in $N$), so the thermodynamic limit works fine. However, if I consider the macrostate as specifying the magnetization of every spin then the number of microstates per macrostate is a constant (equal to one), and we have a problem.
In short, your argument fails because you cannot assume the free energy is minimized as you have defined it.

Is it somehow illegal to make this macrostate choice?

No-one is going to arrest you, but in order for the thermodynamic limit to work the ratio macrostates/microstates must grow (exponentially) with $N$.

Could acquiring all of this information about the spins necessarily change how the magnet behaves, e.g. from something like Landauer's principle?

As I understand it, Landauer's principle implies that there is a minimum entropy cost to aquiring information, but says nothing about where this excess entropy must be held. If you are considering the paramagnet in equilibrium with a thermal bath nothing will change. Of course, in a real paramagnet continuous measurement of each spin certainly would affect how the system behaves.

In general, can changing macrostate choice ever change the predictions of statistical mechanics?

It changes what your model can predict, yes. For example, if I define the macrostate in terms of the two magnetizations ($M_1,M_2$) then I get more information (in principle) than if I define macrostate in terms of the magnetization of the whole system, $M$. There are some cases where this might be significant, for example if the external field has spatial variation. However, the predictions must be compatible with each other, in the sense that $M = M_1 + M_2$ (in the thermodynamic limit) or for a finite system $P(M) = \sum_{M_1+M_2=M}P(M_1,M_2)$.


TITLE: Dirichlet/Chebotarev's Theorem for natural/analytic density
QUESTION [9 upvotes]: A little bird told me that Dirichlet/Chebotarev's Theorem is true not only for Dirichlet density, but also for analytic/natural density. I'm having trouble finding a citation for this. Does anyone know of one?

REPLY [5 votes]: There is a nice treatment of this and discussions of the error term in Serre's book "Lectures on $N_X(p)$". He also gives various applications but I don't think he proves it.
I should note that the natural density version holds over number fields but not over global function fields. Serre discusses this in his book.


TITLE: $\Lambda^k(A)$ is a trace class operator
QUESTION [1 upvotes]: Let $A$ be a trace class operator. I am trying to understand the proof of ($\|\cdot\|_1$ means the trace class norm)
$$\|\Lambda^k(A)\|_1\leq  \frac{\|A\|_1^k}{k!}$$
I have a proof but I think there is a mistake in it.
In the proof is the followig: $|\Lambda^k(A)| =\Lambda^k(|A|)$ with eigenvalues $\mu_{i_1}(A)\cdots \mu_{i_k}(A)$ for $i_1<\cdots <i_k$ so $$\sum\limits_{i_1<\cdots< i_k} \mu_{i_1}(A)\cdots \mu_{i_k}(A)=\frac{1}{k!}\sum\limits_{i_1<\cdots< i_k} \mu_{i_1}(A)\cdots \mu_{i_k}(A)$$
The equality doesn't make sense for me.

REPLY [1 votes]: The second summation should probably be without $i_1<...<i_k$, so that would be $k!$ duplicates.


TITLE: Is a modular element of maximal ideal unique if it exists?
QUESTION [2 upvotes]: Let $A$ be an algebra over the complex numbers (not necessary unital or commutative). An ideal $I$ in $A$ is called modular if there exists a non-zero $e\in A$ such that $ea-a\in I$ and $ae-a\in I$ for all $a\in A$. Equivalently, $e+I$ is a unit for the quotient algebra $A/I$. In this case, we call $e$ a modular element for $I$. Also, it is easy to see that $e\in I$ if and only if $I=A$.
If $A$ is unital, then $1$ is a modular element for every ideal in $A$.
Although I have not seen a counterexample yet, I am quite sure that a modular element is  not unique in general. Namely, if $e,f\in A$ are both modular elements for $I$, then we only know that $e+I=f+I$ by uniqueness of the unit and this does not imply that $e=f$.
Now I was wondering if we can make assumptions on $I$ that give uniqueness of $e$.
My first guess was maximality: Can we conclude that $e=f$ when $I$ is maximal?
(If no, are there other conditions on $I$ that guarantee uniqueness?)
EDIT: My second guess is the assumption that we can write $I=\ker(\phi)$ for some non-zero homomorphism $\phi\colon A\to\mathbb{C}$. Then $I$ is automatically a modular maximal ideal (so this case is included in my other guess). I can prove that $\phi(e)=1=\phi(f)$...

REPLY [1 votes]: Although I have not seen a counterexample yet, I am quite sure that a modular element is not unique in general.

It looks like you didn't notice that if $i$ is any element of $I$ at all,
$(e+i)a-a=(ea-a)+ia\in I$ for every $a\in A$, making every element in the coset $e+I$ a modular element for $I$.
That means there's only one condition to make it unique: $I=\{0\}$.


TITLE: What is the relationship between the 1 and 2D wave equation and the elastic wave equation
QUESTION [4 upvotes]: The elastic wave equation or momentum equation is given by:
$$\rho \frac {\partial^2 u_i} {\partial t^2} = \ \frac {\partial \sigma_{ij}}{\partial x_j}$$
This equation has remarkable similarities with the 1D and 2D wave equations which are given by
$$\frac {\partial^2 u} {\partial x^2} = \frac 1 {c^2} \frac {\partial^2 u} {\partial t^2}$$
and
$$\frac {\partial^2 u} {\partial x^2}+\frac {\partial^2 u} {\partial y^2} = \frac 1 {c^2} \frac {\partial^2 u} {\partial t^2}$$
For instance, expanding one term of the momentum equation we obtain the following:
$$\rho \frac {\partial ^2 u_x} {\partial t^2} = \frac {\partial \sigma_{xx}}{\partial x}+\frac {\partial \sigma_{xy}}{\partial y}+\frac {\partial \sigma_{xz}}{\partial z}$$
Substituting in the values of $\sigma$ based on the constitutive elastic matrix and the definitions for strain we obtain:
$$\rho \frac {\partial ^2 u_x} {\partial t^2} = \frac{E}{(1+\nu)(1-2 \nu)} \left( \frac \partial {\partial x}\left[ (1-\nu)\frac {\partial u_x}{\partial x} + \nu \frac {\partial u_y} {\partial y} + \nu \frac {\partial u_z}{\partial z}\right] + \frac {\partial} {\partial y} \left[\frac 1 2 (1-2 \nu) (\frac {\partial u_x}{\partial y} + \frac {\partial u_y} {\partial x})\right]+\frac {\partial} {\partial z} \left[\frac 1 2 (1-2 \nu) (\frac {\partial u_x}{\partial z} + \frac {\partial u_z} {\partial x})\right] \right)$$
Which bears the following similarities with the 1D and 2D wave equations:

The left hand side is the second partial derivative with respect to time.
The right hand side contains several spatial derivatives taken twice.
dividing both sides by $\rho$ we see emerge a coefficient not unlike the $1/c^2$ term in the 1D and 2D wave equations.

QUESTIONS
I have 2 questions:

Does there exist a relationship between the 1D/2D wave equation and the elastic wave equation? Can this relationship be demonstrated?
What analytical methods can handle a PDE with 4 independent variables and 3 dependent solutions (such as the momentum equation)?

REPLY [2 votes]: Let us use the index notation (Einstein convention). In linear elasticity, the momentum equation $\rho \ddot{u}_i = \sigma_{ij,j}$ is combined with the constitutive law $\sigma_{ij} = C_{ijk\ell}u_{k,\ell}$, where the dot denotes $\partial_t$ and $C_{ijk\ell} = \lambda\delta_{ij}\delta_{k\ell}+\mu(\delta_{ik}\delta_{j\ell} + \delta_{i\ell}\delta_{jk})$ is the stiffness tensor of Hooke's law. Thus, the constitutive law reads $\sigma_{ij} = \lambda u_{k,k}\delta_{ij} + \mu (u_{i,j}+u_{j,i})$, where $\lambda$, $\mu$ are the Lamé parameters and $\delta_{ij}$ is the Kronecker delta. The momentum equation rewrites as
$$
\rho \ddot{u}_i = C_{ijk\ell}u_{k,\ell j} = (\lambda+\mu) u_{j,ji} + \mu u_{i,jj} \, ,
$$
or in intrinsic notation,
\begin{aligned}
\rho \ddot{\bf u} &= (\lambda+\mu) \nabla\,(\nabla\cdot {\bf u}) + \mu \nabla^2{\bf u} \, , \\
&= (\lambda+2\mu) \nabla\,(\nabla\cdot {\bf u}) - \mu\nabla\times\nabla\times {\bf u} \, .
\end{aligned}
In linear elasticity, compressional waves (P-waves) with speed $\sqrt{(\lambda+2\mu)/\rho}$ and shear waves (S-waves) with speed $\sqrt{\mu/\rho}$ are decoupled. Their corresponding wave equations are obtained by applying $\nabla\cdot$ and $\nabla\times$ to the previous equations, respectively. Setting $\psi = \nabla\cdot{\bf u}$ and ${\bf a} = \nabla\times {\bf u}$, one has
\begin{aligned}
\rho \ddot\psi - (\lambda+2\mu) \nabla^2\psi &= 0 \, ,\\
\rho\ddot{\bf a} - \mu\nabla^2 {\bf a} &= 0 \, .
\end{aligned}
Dedicated analytical methods are the classical methods for wave equations (Fourier transform, normal modes decomposition, etc.), based on a decomposition in scalar and vector potentials ${\bf u} = \nabla \varphi + \nabla\times {\bf b}$.
The relationship with Young's modulus $E$ and Poisson's ratio $\nu$ in OP is
$$
E = \frac{\mu (3\lambda+2\mu)}{\lambda+\mu}\, , \nu = \frac{\lambda}{2(\lambda+\mu)} \quad\text{or}\quad \lambda = \frac{E\nu}{(1+\nu)(1-2\nu)}\, , \mu = \frac{E}{2(1+\nu)}\, .
$$


TITLE: Rings in which every J-matrix is non-singular
QUESTION [2 upvotes]: Let $R$ be a ring with identity. A matrix $A=[a_{ij}] \in M_n(R)$ is called a J-matrix if for any $i$, $a_{ii} \not \in J(R)$ but for any $i \not = j$, $a_{ij} \in J(R)$. Now suppose that every J-matrix is non-singular. Is it true that the ground ring $R$ is local ? (Note that the converse is clearly true)

REPLY [2 votes]: I might be interpreting your question wrong, but if not then I believe the answer is yes. This boils down to the fact that a ring is local if and only if the Jacobson radical $J(R)$ is maximal if and only if every element $x \notin J(R)$ is a unit. This characterization of local comes from the fact that the Jacobson radical is the intersection of all maximal left ideals. Then the first if and only if statement is clear and the second follows from the fact that any nonunit is contained in a maximal ideal.
Now, for the proof of your statement. Suppose for some $n$, every $J$-matrix in $M_n(R)$ is nonsingular. Then for any $a \notin J(R)$, consider the matrix $A = \text{diag}(a,1,\ldots,1) \in M_n(R)$. By construction, $A$ is a $J$-matrix so $A$ is nonsingular. However, $A$ has an inverse if and only if $a$ is invertible. Thus, any $a \notin J(R)$ is invertible so $J(R)$ is maximal and $R$ is a local ring.


TITLE: The bizarre behaviour of superfluids! Climbing up walls and geting out of glass beakers
QUESTION [8 upvotes]: SUPERFLUIDITY:
Superfluidity is one of the most amazing quantum phenomena, which we can sit and watch them happening before our eyes!  Watch this video, it is only a couple of minutes long, which will bring you face-to-face with quantum mechanics in all its beauty. Then take a look at the questions below.
http://www.youtube.com/watch?v=9FudzqfpLLs
QUESTIONS:
A superfluid has the amazing ability to climb up the walls of the glass container that it is in: Does it defy gravity?
A superfluid can get out through the walls of the glass container: Quantum tunnelling in action before our eyes?

REPLY [7 votes]: When the superfluid helium leaks out of the container, that's not tunnelling. A few drops of superfluid helium contain many many atoms and for all of them to tunnel through would be extremely improbably as to pretty much not happen. So what makes the helium flow out?
If you notice carefully, the container is not a normal glass mug, but it's bottom is made of some other material (I think it's some kind of ceramic with tiny pores). As the scientist in the video says, the only reason "normal" liquid helium (above the critical temperature) did not flow out through those pores, was that viscous drag kept it from dripping out. After the phase transition to a superfluid, the *super*fluid doesn't have any viscosity and  no longer resists flow. So it flows out of the container.
As far as the wall-climbing goes: It should somehow be that (loosely) it is energetically/probabilistically preferable for the fluid to climb up the walls and then flow out rather than stay in. I don't really understand why exactly it climbs the walls of a container and a cursory search seems to indicate that though there are explanations based on viscosity for this too, there isn't any clear consensus which is perfectly validated by experiment.
I think this video (with annotations) is more informative: https://www.youtube.com/watch?v=2Z6UJbwxBZI


TITLE: Finding determinant for matrix using upper triangle method
QUESTION [1 upvotes]: Here is an example of a matrix, and I'm trying to evaluate its determinant:
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
2 & 5 & -2 & 9 \\
3 & 7 & 0 & 1 \\
\end{pmatrix}
$$
When applying first row operation i get:
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & -1 & -6 & 7 \\
0 & -2 & -6 & -2 \\
\end{pmatrix}
$$
Now, if I continue doing row operations until i get the upper triangle, the determinant will be 14 (which is said to be the correct one).
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & 0 & 2 & 3 \\
0 & 0 & 0 & -7 \\
\end{pmatrix}
$$
However, if I instead apply this certain operation, R4 --> (1/-2)R4...
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & -1 & -6 & 7 \\
0 & 1 & 3 & 1 \\
\end{pmatrix}
$$
...and then carry on with operations, I get a different final answer: The determinant will be 7 in this case!
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & 0 & -1 & 5 \\
0 & 0 & 0 & -7 \\
\end{pmatrix}
$$
Could someone explain that to me - is this operation illegal?
R4 --> (1/-2)R4
How so? Because i always tend to use it, just to simply things a little.

REPLY [0 votes]: If we  multiply a certain row or column of a matrix $A$ by some scalar $\lambda$ then determinant of $A$ changes to $\lambda|A|.$


TITLE: Any Cauchy net in metric space is convergent?
QUESTION [0 upvotes]: Any Cauchy net in a metric space is convergent?
I know that if (X,d) is metric space COMPLETE then cauchy net in X is convergent in X.
But, only (X,d) is metric space (not complete) It is true?
pd:
As additional doubt. In general, any succession of Cauchy in a set X  is always converges in (probability another) some set in somewhere?

REPLY [0 votes]: No: in particular, the statement is false for sequences, and all sequences are nets (they're those nets for which the directed set is the natural numbers with the obvious order relation). For an explicit example, take any Cauchy sequence of rationals that converges in the reals to $\pi$. In the metric space $\mathbb{Q}$, this sequence does not converge.
I'm not at all sure what you mean by your edit, but I think that you're asking if, for every metric space $X$, there is some complete metric space that includes both $X$ and all limits of $X$. If that's correct, then the answer is yes: you simply take all of those things and throw them together, and you're done: this is called the completion of $X$, and is, for example, one of the standard definitions/constructions of $\mathbb{R}$: it is precisely the completion of the rationals.


TITLE: Number of values of x
QUESTION [2 upvotes]: $$a\dfrac{(x-b)(x-c)}{(a-b)(a-c)}+b\dfrac{(x-c)(x-a)}{(b-c)(b-a)}+c\dfrac{(x-a)(x-b)}{(c-a)(c-b)}=x$$

How many values of $x$ satisfy this equation? It is clear that x=a, x=b, x=c do satisfy the equation, but are those the only three possible solutions?

REPLY [6 votes]: This has to be true for all $x$.
Firstly, there are at least $3$ solutions, assuming $a,b,c$ are distinct.
Secondly, this is a polynomial equation of degree $2$ in one variable $x$. In other words, you can rewrite the equation as $A x^2 + B x + C = 0$. But this can only have at most $2$ solutions, unless $A=B=C=0$.
Combining the two facts, the above equation holds for all $x$. Thus, if you multiply things out, it should come out as an identity.

REPLY [3 votes]: If you move $x$ to the left side, you have a quadratic polynomial in $x$, that has the three zeroes $a,b,c$ that you have already found.  However the only quadratic with more than two roots is the zero polynomial; hence this holds for all $x$.

REPLY [3 votes]: Another way to see this: the expression on the left is the Lagrange polynomial of degree 2 passing through $(a,a)$, $(b,b)$, and $(c,c)$; in other words, it describes the line $y=x$. This value is always equal to $x$, which corresponds with the right-hand side.


TITLE: On a relativistic merry go round, who ages faster — rider or observer?
QUESTION [0 upvotes]: Consider a merry go round with a radius of 1 light year and a tangential velocity of 0.9c. The rider is in a capsule on the edge of the merry go round. The observer is just a guy who is the same age as the rider but just floating in space. The merry go round start's spinning. The rider is in a couch in the capsule and at 0.9 c experiences a very tolerable 7.6 m/s^2 centrifugal acceleration.
From the rider's point of view, the observer is like an orbiting planet with an eccentric orbit, albeit at relativistic speed. So the rider will calculate that the observer will age more slowly.
But the observer will see the rider rotating very fast and will calculate the she will age more slowly.
When the merry go round runs for a few days and then stops, who is the one who experienced time dilation?
I read these recent threads:
How to tell who is experiencing time slower and who faster when travelling at different speed?
On The Twin Paradox The Symmetry Remains
Clearly, we humans continue to be puzzled at this hundred year old story. From reading the threads above, PM2ring's comment is enlightening:
"In Special Relativity, speed causes time dilation, but with constant speed the situation is symmetrical. If observers A & B have a constant relative velocity then A measures B's clock to be running slow by a factor of γ, and B measures A's clock to be running slow by a factor of γ.
To break the symmetry, (at least) one of the observers needs to make one or more changes of reference frame. It's not so much that the acceleration causes time dilation, it's merely the mechanism whereby the reference frame is changed."
This tells me that in my model above, the rider will age more slowly since she is not in an inertial frame of reference. But consider if the observer is on an identical merry go round, just shifted a little in the z direction. If both merry go rounds are spinning in the same direction, they are covariant. Each rider will see the other as stationary. Each will calculate that Neither rider will age slower.
But if the merry go rounds spin in opposite directions, each rider will see the other moving very fast and will calculate that the other should age more slowly. When the merry go rounds stop, they will be surprised to see they are the same age, even though each saw the other whipping by.
These two situations put the riders in symmetrical reference frames. Yet in one direction they see each other age similarly, and when spinning opposite, they see each other age differently.
And it doesn't require a merry go round. If both riders are on rocket sleds light years apart and travel towards each other at 0.9c, each will see the other age more slowly. Yet when they meet, they should be the same age since their reference frames are symmetrical.
I guess it would be nice to have a mechanistic explanation of what is going on. How is it that traveling fast changes the fundamental nature of aging.

REPLY [0 votes]: The choice of merry-go-rounds complicates the issues somewhat and masks the underlying principle, so consider the simpler case you mention of two people approaching each other.
Let's make the two people twins of the same age who start the experiment a distance apart and stationary relative to each other. As they travel toward each other, they will each consider the other to be ageing more slowly, yet when they meet they will find they are still the same age as each other. The paradox is resolved when you remember to take in to account the relativity of simultaneity- each twin will believe they started their journey earlier than the other did. When you factor-in the perceived difference in the start times, they restore the overall symmetry.
To take an extreme example to illustrate the point. Suppose you and I are walking toward each other. I set off on Monday and arrive at our meeting on Thursday, having aged three days. You walk towards me, leaving at what you consider to be Monday and meeting me on Thursday after ageing three days.
Owing to time dilation, from my perspective, you only aged two days while you were walking, not the three you experienced yourself, but owing to the relativity of simultaneity, I consider that you didn't start walking until Tuesday instead of the Monday departure you experienced in your reference frame. So the reduction in ageing which I observed during your walk is balanced by your later start date- you have still aged by three days when you meet me. You see the reciprocal effects.


TITLE: Let $f :\mathbb{R}→ \mathbb{R}$ be a function such that $f^2$ and $f^3$ are differentiable. Is $f$ differentiable?
QUESTION [11 upvotes]: Let $f :\mathbb{R}→ \mathbb{R}$ be a function such that $f^2$ and $f^3$ are differentiable. Is $f$ differentiable?
Similarly, let $f :\mathbb{C}→ \mathbb{C}$ be a function such that $f^2$ and $f^3$ are analytic. Is $f$ analytic?

REPLY [4 votes]: If you do want $f^{p}$ to represent the $p^{\text{th}}$ iterate, then you can let $f$ denote the characteristic function of the irrational numbers.  Then $f^2$ and $f^3$ are both identically zero, yet $f$ is nowhere continuous, let alone differentiable.


TITLE: Number of different quadratic functions mod 12.
QUESTION [1 upvotes]: How many different quadratic functions are there of form:

$x \mapsto ax^2+bx+c \pmod{12}$

All I could come up with is an upper bound of $11*12*12$. This is a puzzle from a YouTube video by Socratica.

REPLY [5 votes]: The map $$\varphi: (a,b,c)\mapsto \bigl[ x \mapsto ax^2+bx+c \bigr]$$ is a homomorphism from $(\mathbb Z_{12})^3$ to $(\mathbb Z_{12})^{\mathbb Z_{12}}$ as groups under pointwise addition. We want to find the number of elements in the image
$$ \{ \varphi(a,b,c) \mid a\ne 0 \} $$
Fortunately this is the same as the image of $\varphi$ itself. The requirement that $a\ne 0$ doesn't change the image because we always have $\varphi(0,b,c)=\varphi(6,b+6,c)$ since $6x^2+6x\equiv 0\pmod{12}$ for all $x$. (To see this note that $6x^2+6x=6x(x+1)$ and either $x$ or $x+1$ will be even).
The size of the image of a homomorphism between finite groups is the size of the domain divided by the size of the kernel, so we need to count the number of triples $(a,b,c)$ such that
$$ ax^2 + bx + c $$
is the zero function.
First, of course $c$ must be $0$, because $\varphi(a,b,c)(0)=c$.
Next, calculating modulo $3$ we find that $a\equiv b\equiv c\equiv 0\pmod 3$ because $\mathbb Z_3$ is a field and a nonzero (formal) polynomial over a field can only have as many roots as its degree.
So $a,b\in\{0,3,6,9\}$.
Furthermore, plugging in $x=-1$ gives us $a-b\equiv 0$, so $a=b$. So the only triples that could possibly lead to the zero function are
$$ (0,0,0) \qquad (3,3,0) \qquad (6,6,0) \qquad (9,9,0) $$
We already know that $(6,6,0)$ (and of course $(0,0,0)$) is in the kernel, and plugging in $x=1$ elimintates $(3,3,0)$ and $(9,9,0)$.
In total, the kernel of $\varphi$ is exactly $\{(0,0,0),(6,6,0)\}$, so the size of the image of $\varphi$ is
$$ \frac{12^3}{2} = 864 $$


text:
\section{Classification of Groups of Order up to 15}
Tags: Order of Groups

\begin{theorem}
Up to [[Definition:Group Isomorphism|isomorphism]], every [[Definition:Group|group]] of order $\order G \le 15$ is one of the below:
{| class="sortable wikitable"
|- bgcolor="#ececec"
! Order !! Abelian !! Non-Abelian
|-
|1 || $\Z_1$ ||
|-
|2 || $\Z_2$ ||
|-
|3 || $\Z_3$ ||
|-
|4 || $\Z_4, \Z_2 \oplus \Z_2$ ||
|-
|5 || $\Z_5$ ||
|-
|6 || $\Z_6$ || $D_3 = S_3$
|-
|7 || $\Z_7$ ||
|-
|8 || $\Z_8, \Z_4 \oplus \Z_2, \Z_2 \oplus \Z_2 \oplus \Z_2$ || $D_4, \Dic 2$
|-
|9 || $\Z_9, \Z_3 \oplus \Z_3$ ||
|-
|10 || $\Z_{10}$ || $D_5$
|-
|11 || $\Z_{11}$ ||
|-
|12 || $\Z_{12}, \Z_6 \oplus \Z_2$ || $D_6, A_4, \Dic 3$
|-
|13 || $\Z_{13}$ ||
|-
|14 || $\Z_{14}$ || $D_7$
|-
|15 || $\Z_{15}$ ||
|}
where:
: $D_n$ is the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$
: $S_n$ is the [[Definition:Symmetric Group|$n$th symmetric group]]
: $A_n$ is the [[Definition:Alternating Group|alternating group]] on $n$ points
: $\Dic 2$ is the [[Definition:Dicyclic Group|dicyclic group]] of [[Definition:Order of Structure|order]] $4 n$.
\end{theorem}

\begin{proof}
The [[Definition:Abelian Group|Abelian]] cases are the direct result of the [[Fundamental Theorem of Finite Abelian Groups]].
The non-Abelian cases follow from seven separate theorems:
:$(1): \quad$ [[Trivial Group is Cyclic Group]] - determines theorem for order $1$
:$(2): \quad$ [[Prime Group is Cyclic]] - determines theorem for orders $2$, $3$, $5$, $7$, $11$, and $13$
:$(3): \quad$ [[Group of Order Prime Squared is Abelian]] - determines theorem for orders $4$ and $9$
:$(4): \quad$ [[Cyclic Groups of Order p q]] - determines theorem for order $15$
:$(5): \quad$ [[Groups of Order Twice a Prime]] - determines theorem for orders $6$, $10$, $14$
:$(6): \quad$ [[Groups of Order 8]] - determines theorem for order $8$
:$(7): \quad$ [[Groups of Order 12]] - determines theorem for order $12$
{{qed}}
[[Category:Order of Groups]]
pp24q0v3ow83txltn6grzbjuoqymggr
\end{proof}

config: wiki


TITLE: $\exists M >0$ s.d. $ \vert R(z)\vert \leq \frac{M}{x^2}$, calculation of integral with the residue theorem
QUESTION [0 upvotes]: Given a polynomial of the form $R(z):=\frac{P(z)}{Q(z)}$ such that $R(z)$ has no real roots and $deg(Q) \geq deg(P) + 2$, then the integral can be expressed as
$$\int_{-\infty}^{+\infty} R(z)dz=2\pi i\sum_{a \in \Bbb H}{\mathrm{Res}\left ( f;a \right )}$$
where $\Bbb H:=\{z \in \Bbb C: Im(z)>0\}$.
Now for proving this statement, we have first to show that the following limit exists:
$$\int_{-\infty}^{+\infty} R(z)dz=\lim_{r \to -\infty}\int_{r}^{+\infty} R(z)dz+\lim_{r \to +\infty}\int_{-\infty}^{r} R(z)dz$$
The argument they used there, is that $\exists M >0$ s.d. $ \vert R(z)\vert \leq \frac{M}{x^2}$. But I don't see, why it exists such an $M$. Many thanks for some help!

REPLY [1 votes]: $z^{2} \frac {a_0+a_1z+...+a_nz^{n}} {b_0+b_1z+...+b_mz^{m}}$ tends to $0$ if $m >n+2$ and it tends to $\frac {a_n} {b_n}$ if $m=n+2$. Hence $z^{2} \frac {a_0+a_1z+...+a_nz^{n}} {b_0+b_1z+...+b_mz^{m}}$ is a bounded function.
[Since $x^{2}R(x)$ is continuous on the real line and has finite limits at $\pm \infty$ it is a bounded function].

REPLY [1 votes]: The function $x\mapsto x^2R(x)=\frac{x^2P(x)}{Q(x)}$ is a rational function such that the degree of the numerator is smaller than or equal to the degree of the denominator, and therfore the limits $\lim_{x\to\pm\infty}\frac{x^2P(x)}{Q(x)}$ exist and are real numbers. So, that function is bounded and now you can take $M$ such that $(\forall x\in\Bbb R):\left|\frac{x^2P(x)}{Q(x)}\right|\leqslant M$, and then $(\forall x\in\Bbb R):\left|R(x)\right|\leqslant\frac M{x^2}$.


TITLE: Powers of diagonalizable matrices-basic question
QUESTION [0 upvotes]: If $A$ is a diagonalizable matrix so it there is a matrix $P$ such that $A=P^{-1}DP$
So why if we want to calculate $A^n$ we solve $PD^nP^{-1}$ and not $P^{-1}D^nP$?

REPLY [1 votes]: I think you confusing the notations. Let me summarise the facts:
If $P$ is the change of basis matrix from the canonical basis $\mathcal C$ to the new basis $\mathcal B$ (i.e. the columns of $P$ are the coordinates  in $\mathcal C$ of the vectors in the new basis), if $X,Y$,… are the (old) coordinates of some vectors in  $\mathcal C$, $X',Y'$,… the (new) coordinates of the same vectors in basis  $\mathcal B$,

$P$ expresses the old coordinates in function of the new coordinates:
$$ X=PX', \quad Y=PY'.$$
If $A$ is the matrix of a linear map in the canonical basis, $A'$ its matrix in the new basis, then
$$A'= P^{-1}AP\qquad(\text{the $other$ way})$$
The reason for having the formula the other way comes from the relation: $Y=AX$, hence $PY'=APX'$, whence $Y'=P^{-1}APX'$.

Thus, in a basis of eigenvectors, in which the matrix is diagonal, the relation is $D=P^{-1}AP$, whence $A=PDP^{-1}$, and more generally  $A^n=PD^nP^{-1}$.

REPLY [0 votes]: Hello no you calculate $D^n$ by setting the diagonal entries $(.)^n$; then you multiply with $P$ and $P^{-1}$ in the form $P^{-1}D^n P $ and get $A^n = P^{-1}D^n P$.