TITLE: Linear Second order differential equation, special method?
QUESTION [0 upvotes]: The differential equation is as follows:
$(x+1)y''-(2x+3)y'+(x+2)y=0$ where $y(x)$ is a funtion on $\mathbb{R}$. First, I had to show that $y_1(x)=e^x$ is a solution for the DV. The second question was to find the general solution of the DV. I instantly started on the series of solution solving method, but I think I have to use the 'hint'  $y_1(x)=e^x$ is a solution to the DV, but how to do this?

REPLY [2 votes]: Let
$$y(x) = y_1(x) \, z(x) = e^x \, z(x),$$
where $z(x)$ is your new unknown function. Then $y'(x) = e^x \, z(x) + e^x \, z'(x)$ and $y''(x) = \cdots$, and when you substitute this into your ODE you get something with a $z''$ term and a $z'$ term, but no $z$ term, so if you let $w(x)=z'(x)$ you get a first-order linear ODE for $w$, and such equations can always be solved in principle, using an integrating factor.
(This procedure is called reduction of order.)


TITLE: Arc length of $y^3 = x^2$
QUESTION [1 upvotes]: I am trying to find the arc length of $y^3 = x^2$ and I am suppose to use two formulas, one for in terms of x and one for in terms of y.
At first I need to find (0,0) (1,1) and I start with in terms of x
$$y\prime = \frac{2}{3}x^\frac{-1}{3}$$
$$\left(\frac{2}{3}x^\frac{-1}{3}\right)^2 = \frac{4}{9x^\frac{2}{3}}$$
$$\int_0^1 \sqrt{1 + \left( \frac{2}{3}x^\frac{-1}{3}\right)^2}\, dx$$
$$\int_0^1 \sqrt{1 + \frac{4}{9x^\frac{2}{3}}}\, dx$$
This is undefined at 0 so it is an improper function. I do not think I can continue.
Now in terms of y.
$$x = y^\frac{3}{2}$$
$$x \prime = \frac{3\sqrt{y}}{2}$$
$$ \left(\frac{3\sqrt{y}}{2}\right)^2 = \frac{9y}{4}$$
$$\int_0^1 \sqrt{1+ \frac{9y}{4}}\, dy$$
I have no idea how to factor that out, I have tried many ways but I can not get it.

REPLY [3 votes]: Neater is to do it parametrically, setting $x=t^2$ and $y=t^3$, and using the formula
$$
s = \int_0^1\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt\\
=\int_0^1t\sqrt{9t^2+4}dt\,.
$$
Now you set $u=9t^2+4$, getting $\int_4^{13}u^{1/2}du/18$, evaluating to $(13^{3/2}-8)/27$.


TITLE: Capillary action in conical capillary tube
QUESTION [0 upvotes]: Hi! I want to clear my concepts regarding surface tension . This approach works well in case of a cylindrical tube but fails when the tube is conical. Why so? I tried this question by balancing the force of surface tension with the weight of water in the tube:
$$S(2πb)cos(θ+\frac{1}{2}α)=ρ(\frac{1}{3}πb²h)g$$
But my answer is wrong by using this approach. I know another approach to solve this question by balancing pressure. But I want to know what am I doing wrong in this, so that I can clear my concepts. Please help me.

REPLY [1 votes]: You have ignored the normal forces by the capillary tube, which balance the weight too alongside surface tension.
Hence you must first balance the pressure either to solve it directly or to  find these normal forces


text:
\begin{document}
\centerline{\bf\large Solution of Polynomial Equations with Nested Radicals}\vskip .2in

\centerline{Nikos Bagis}

\centerline{Stenimahou 5 Edessa}
\centerline{Pellas 58200, Greece}
\centerline{bagkis@hotmail.com}

\[
\]

\centerline{\bf Abstract}

\begin{quote}
In this article we present solutions of arbitrary polynomial equations in nested periodic radicals
\end{quote}

\textbf{Keywords}: Nested Radicals; Quintic; Sextic; Septic;  Equations; Polynomials; Solutions; Higher functions;

\section{A general type of equations}

Consider the following general type of equations
\begin{equation}
ax^{2\mu}+bx^{\mu}+c=x^{\nu}.
\end{equation}
Then (1) can be written in the form
\begin{equation}
a\left(x^{\mu}+\frac{b}{2a}\right)^2-\frac{\Delta^2}{4a}=x^{\nu},
\end{equation}
where $\Delta=\sqrt{b^2-4ac}$. Hence
\begin{equation}
x^{\mu}=\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+\frac{1}{a}x^{\nu}}
\end{equation}
Set now $\sqrt[d]{x}=x^{1/d}$, $d\in\textbf{Q}_{+}-\{0\}$, then
\begin{equation}
x=\sqrt[\mu]{\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+\frac{1}{a}x^{\nu}}}
\end{equation}
hence
\begin{equation}
x^{\nu}=\sqrt[\mu/\nu]{\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+\frac{1}{a}x^{\nu}}}
\end{equation}
\textbf{Theorem 1.}\\
The solution of (1) is
\begin{equation}
x=\sqrt[\mu]{\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+\frac{1}{a}\sqrt[\mu/\nu]{\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+\frac{1}{a}\sqrt[\mu/\nu]{\frac{-b}{2a}+\sqrt{\frac{\Delta^2}{4a^2}+...}}}}}}
\end{equation}
\\
\textbf{Proof.}\\
Use relation (4) and repeat it infinite times.\\
\\
\textbf{Corollary 1.}\\
The equation
\begin{equation}
ax^6+bx^5+cx^4+dx^3+ex^2+fx+g=0
\end{equation}
is always solvable with nested radicals.\\
\\
\textbf{Proof.}\\
We know (see [1]) that all sextic equations, of the most general form (7) are equivalent by means of a Tchirnhausen
transform
\begin{equation}
y=kx^4+lx^3+mx^2+nx+s
\end{equation}
to the form
\begin{equation}
y^6+e_1y^2+f_1y+g_1=0
\end{equation}
But the last equation is of the form (1) with $\mu=1$ and $\nu=6$ and have solution
\begin{equation}
y=\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}-\frac{1}{e_1}\sqrt[1/6]{\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}-\frac{1}{e_1}\sqrt[1/6]{\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}+...}}}}}
\end{equation}
where $\Delta_1=\sqrt{f_1^2-4e_1g_1}$. Hence knowing $y$ we find $x$ from (8) and get the solvability of (7) in nested radicals. More precisely it holds
$$
kx^4+lx^3+mx^2+nx+s=
$$
\begin{equation}
=\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}-\frac{1}{e_1}\sqrt[1/6]{\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}-\frac{1}{e_1}\sqrt[1/6]{\frac{-f_1}{2e_1}+\sqrt{\frac{\Delta_1^2}{4e_1^2}+...}}}}}
\end{equation}
\\

Theorem 1 is a generalization of a theorem of Euler (see [4] pg.306-307):\\
\\
\textbf{Theorem 2.}\\
If the root of
\begin{equation}
aqx^p+x^q=1
\end{equation}
is $x$, then
\begin{equation}
x^n=\frac{n}{q}\sum^{\infty}_{k=0}\frac{\Gamma(\{n+pk\}/q)(-q a)^k}{\Gamma(\{n+pk\}/q-k+1)k!}\textrm{, }n=1,2,3,\ldots
\end{equation}
Where $\Gamma(x)$ is Euler's the Gamma function.\\
Moreover the solution $x$ of (12) can given with nested radicals:
\begin{equation}
x=\sqrt[q]{1-aq\sqrt[q/p]{1-aq\sqrt[q/p]{1-aq\sqrt[q/p]{1-\ldots}}}}
\end{equation}
\\

The general quintic
\begin{equation}
ax^5+bx^4+cx^3+dx^2+ex+f=0
\end{equation}
can reduced also by means of a Tchirnhausen transform into
\begin{equation}
x^5+Ax+B=0
\end{equation}
Define now the hypergeometric function
\begin{equation}
\textrm{BR}(t)=-t\cdot{}_4F_3\left[\left\{\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}\right\};\left\{\frac{1}{2},\frac{3}{4},\frac{5}{4}\right\};-\frac{3125 t^4}{256}\right]
\end{equation}
\\
\textbf{Theorem 3.}\\
The solution of (16) is
\begin{equation}
x=\sqrt[4]{\frac{-A}{5}}\textrm{BR}\left(-\frac{\sqrt[4]{\frac{5^5}{-A^5}}}{4}B\right)
\end{equation}
and also holds
\begin{equation}
x=\sqrt[5]{-B-A\sqrt[5]{-B-A\sqrt[5]{-B-A\sqrt[5]{-B-\ldots}}}}
\end{equation}
\\
\textbf{Application.}\\
If $Im(\tau)>0$ and $j_{\tau}$ denotes the $j$-invariant, then
$$
\frac{1}{j_{\tau}^{5/3}}\left(R\left(e^{2\pi i \tau}\right)^{-5}-11-R\left(e^{2\pi i\tau}\right)^5\right)^{5/3}=
$$
\begin{equation}
=\sqrt[3/5]{\frac{-125}{j_{\tau}}+\sqrt{\frac{12500}{j^2_{\tau}}+\sqrt[3/5]{\frac{-125}{j_{\tau}}+\sqrt{\frac{12500}{j_{\tau}^2}+\ldots}}}},
\end{equation}
where $R(q)$ is the Rogers-Ramanujan continued fraction.\\
\\
\textbf{Proof.}\\
Equation (1) for $a=1/j_{\tau}^{1/3}$, $b=250/j_{\tau}^{1/3}$, $c=3125/j_{\tau}^{1/3}$ and $\mu=1$, $\nu=5/3$ takes the form
\begin{equation}
x^{2}+250x+3125=j_r^{1/3}x^{5/3}
\end{equation}
Which is a simplified form of Klein's equation for the icosahedral equation [3] and have solution $x=Y_{\tau}=R(q^2)^{-5}-11-R(q^2)^5$. By Theorem 1 we can express $Y_{\tau}$ in nested-periodical radicals . This completes the proof.\\
\\
For more details about equation $ax^2+bx+\frac{b^2}{20a}=C_1x^{5/3}$ one can see [2].

\section{The reduction of a given polynomial}

As someone can see if we assume the equation
\begin{equation}
aH_7^3+bH_7^2+cH_7+d=u,
\end{equation}
then we can rewrite (22) in the form
\begin{equation}
H_7^3=-\frac{b}{a}H_7^2-\frac{c}{a}H_7-\frac{d}{a}+\frac{u}{a}
\end{equation}
Hence
\begin{equation}
H_7=\sqrt[3]{\frac{u}{a}-\frac{\Delta_3}{4ba}-\frac{b}{a}\sqrt[1/2]{\frac{c}{2b}+H_7}}
\end{equation}
where $\Delta_3=4bd-c^2$.\\
From the above a nested radical solution of (22) is
$$
H=H_7=H_7(a,b,c,d,u)=
$$
\begin{equation}
=\sqrt[3]{\frac{u}{a}-\frac{\Delta_3}{4ba}-\frac{b}{a}\sqrt[1/2]{\frac{c}{2b}+\sqrt[3]{\frac{u}{a}-\frac{\Delta_3}{4ba}-\frac{b}{a}\sqrt[1/2]{\frac{c}{2b}+...}}}}
\end{equation}
Let also the equation
\begin{equation}
ax^{3\mu}+bx^{2\mu}+cx^{\mu}+d=x^{\nu}.
\end{equation}
Clearly if we drop the $x^{\nu}$ term of (23), then becomes a 3$rd$ degree equation in $x^{\mu}$ and can be written under certain conditions as
\begin{equation}
x^{\mu}=H_7\left(a,b,c,d,x^{\nu}\right)
\end{equation}
or equivalently
\begin{equation}
x=\sqrt[\mu]{H_7(a,b,c,d,x^{\nu})}
\end{equation}
or equivalently
\begin{equation}
x^{\nu}=\sqrt[\mu/\nu]{H_7(a,b,c,d,x^{\nu})}
\end{equation}
Hence\\
\\
\textbf{Theorem 4.}\\
If $H_7$ is the function defined in (22), then equation (26) have solution in nested radicals
\begin{equation}
x=\sqrt[\mu]{H_7\left(a,b,c,d,\sqrt[\mu/\nu]{H_7\left(a,b,c,d,\sqrt[\mu/\nu]{H_7\left(a,b,c,d,\ldots\right)}\right)}\right)}.
\end{equation}
\\
\textbf{Note.}\\ Here we must mention that we don't interested about the conditions, since we want to establish formulas that agreed numerically. One can say that without these conditions our results are conjectures.\\
\\
\textbf{Theorem 5.}\\
Given a general septic equation
\begin{equation}
\sum^{7}_{k=0}a_kx^k=0,
\end{equation}
after using Tchirnhausen transform can written in the form
\begin{equation}
y^7=ay^3+by^2+cy+d,
\end{equation}
which has solution
\begin{equation}
y=H_7\left(a,b,c,d,\sqrt[1/7]{H_7\left(a,b,c,d,\sqrt[1/7]{H_7\left(a,b,c,d,
\ldots\right)}\right)}\right)
\end{equation}
\\
\textbf{Proof.}\\Taking $\nu=7$ and $\mu=1$ in Theorem 4 we get the desired result.\\
\\
\textbf{Example.}\\
Consider the equation
\begin{equation}
x^7=x^3+x^2-1/2
\end{equation}
We want to find a solution of (34). For this we find $H_7$, which is solution of
\begin{equation}
H_7^3+H_7^2-1/2=u.
\end{equation}
Solving the above equation with expansion (25) we get
\begin{equation}
H_7(u)=\sqrt[3]{u+\frac{1}{8}-\sqrt[3/2]{u+\frac{1}{8}-\sqrt[3/2]{u+\frac{1}{8}-...}}}
\end{equation}
Hence a solution of (34) is
\begin{equation}
x=H_7\left(\sqrt[1/7]{H_7\left(\sqrt[1/7]{H_7\left(\sqrt[1/7]{H_7\left(\ldots\right)}\right)}\right)}\right)
\end{equation}
Or if someone preferes a root of
\begin{equation}
x=\sqrt[3]{x^7+\frac{1}{8}-\sqrt[3/2]{x^7+\frac{1}{8}-\sqrt[3/2]{x^7+\frac{1}{8}-...}}}
\end{equation}
\\

The next case is the octic equation which is equivalent to
\begin{equation}
x^8=ax^4+bx^3+cx^2+dx+e
\end{equation}
and with above method is again solvable in nested radicals, since the $H=H_8$ is of degree 4 and hence solvable in radicals. But, instead solving a 4-th degree equation we can use the function $H_7(u)$, since making the change of variable $x=H_7(u)$ in (39) we have
\begin{equation}
H_7^8(u)=aH_7^4(u)+u
\end{equation}
By this way
\begin{equation}
H_8(u)=H_7^8(u)-aH_7^4(u),
\end{equation}
with $H_7(u)=H_7(b,c,d,e;u)$ is the expansion (25) and
\begin{equation}
H_8(\xi)=\xi,\textrm{ and }x=H_7(\xi)
\end{equation}
\\

The 9$th$ degree polynomial equation have $H=H_9$ and $deg(H_9)=5$. In this case we can use the hypergeometric function solution (18) or the radical solution (19).\\
The desired equation we want to solve is
\begin{equation}
\sum^{9}_{k=0}a_kx^k=0
\end{equation}
This reduces to
\begin{equation}
ay^5+by^4+cy^3+dy^2+ey+f=y^9,
\end{equation}
after a Tchirnhausen transform
\begin{equation}
y=kx^4+lx^3+mx^2+nx+s
\end{equation}
Hence we concern to construct the function $H_9=H_9(u)$ such that
\begin{equation}
aH_9(u)^5+bH_9(u)^4+cH_9(u)^3+dH_9(u)^2+eH_9(u)+f=u
\end{equation}
For to construct $H_9$ we use the Tchirnhausen transform again
\begin{equation}
G_9(u)=k'H_9(u)^4+l'H_9(u)^3+m'H_9(u)^2+n'H_9(u)+s',
\end{equation}
where $(k',l',m',n',s')$ are depending from $(a,b,c,d,e,u)$.
By this method we arrive to
\begin{equation}
G_9(u)^5+A G_9(u)+B=0
\end{equation}
Hence $G_9(u)$ is that of (18), with $A$,$B$ functions of $(a,b,c,d,e,u)$, or denoting simply $A=A(u)$ and $B=B(u)$:
\begin{equation}
G_9(u)=\sqrt[4]{\frac{-A(u)}{5}}\textrm{BR}\left(-\frac{\sqrt[4]{\frac{5^5}{-A(u)^5}}}{4}B(u)\right)
\end{equation}
Finally we substitute (49) to (47) and solve with respect to $H_9(u)$.
Hence with the above notation the next theorem holds\\
\\
\textbf{Theorem 6.}\\
i) The equation
\begin{equation}
ax^{5\mu}+bx^{4\mu}+cx^{3\mu}+dx^{2\mu}+ex^{\mu}+f=x^{\nu}
\end{equation}
have solution in periodical nested radicals
\begin{equation}
x=\sqrt[\mu]{H_9\left(\sqrt[\mu/\nu]{H_9\left(\sqrt[\mu/\nu]{H_9\left(
\ldots\right)}\right)}\right)},
\end{equation}
with $H_9$ expressible by means of (18) (or if preferred by (19)).\\
ii) A root of (44) is
\begin{equation}
kx^4+lx^3+mx^2+nx+s=H_9\left(\sqrt[1/9]{H_9\left(\sqrt[1/9]{H_9\left(
\ldots\right)}\right)}\right).
\end{equation}
\\

By this method one can see that we can solve any polynomial equation with nested radicals.

\newpage

\[
\]

\centerline{\bf References}\vskip .2in

\noindent

[1]: J.V. Armitage and W.F. Eberlein. 'Elliptic Functions'. Cambridge University Press. (2006)

[2]: N.D. Bagis. 'On a general sextic equation solved by the Rogers-Ramanujan continued fraction'. arXiv:1111.6023v2[math.GM]. (2012)

[3]: W. Duke. 'Continued fractions and Modular functions'. Bull. Amer. Math. Soc. (N.S.), 42. (2005), 137-162.

[4]: B.C. Berndt. 'Ramanujan`s Notebooks Part I'. Springer Verlag, New York. (1985).

[5]: M.L. Glasser. 'The Quadratic Formula Made Hard or A Less Radical Approach to Solving Equations'. arXiv:math/94112244v1 [math.CA] 16 Nov (1994).

\end{document}

config: arxiv


TITLE: How does plasm affect the speed of light?
QUESTION [1 upvotes]: 99% of the universe is made of plasma according to Imperial College London. I understand that the plasma is made of charged particles and photons are easily absorbed and re-emit photons. So how does the plasma affect the speed of light?

REPLY [1 votes]: It's not that simple. Plasma affects things, but the effect on the speed of light can be ignored to some extent.
The intergalactic plasma density average is about 1 particle per cubic meter, and most of the cosmic microwave radiation CMB has come through space with no interaction with it. The last time the CMB was scattered was at the surface of last scattering, when the electrons and protons recombined Into hydrogen atoms and made the universe mostly transparent to photons,about 380,000 years after the Big Bang. All we have seen of the universe so far is from after that.
The main effect on those photons has been due to cosmological expansion and thus the redshift of about a factor of 1000. The speed has been c, locally. For the most part the major effect on cosmological distance that changes that is from gravitational lensing, with the Wolfe-Sachs effect.
The astrophysical effects from plasma interactions have been relevant but different, and mostly from regions or areas where the plasma density is higher, such as near stars, in their coronas or atmospheres, and in hot interstellar or galactic regions where collisions or explosions have heated up the gases. In those case it is important, and not a simple single effect. See the different effects summarized at
https://en.m.wikipedia.org/wiki/Plasma_(physics), and much more detailed at
http://www.sp.ph.imperial.ac.uk/~sjs/APmaster.pdf as well as
some in https://en.m.wikipedia.org/wiki/Astrophysical_plasma.
And how it can have some effect in disturbing slightly the CMB spectrum, at https://en.m.wikipedia.org/wiki/Sunyaev–Zel'dovich_effect
But it does have effects in determining luminosities from galactic clusters and nebula at https://arxiv.org/abs/1605.08333, and can have a small effect in the determinations of the cosmic distance scale. Also redshift relationships such as in https://arxiv.org/abs/1404.0870.
So, mostly the issue has been how does the plasma absorption of some of the EM radiation or how much does it emit so one can determine distances. I.e., luminosities, spectra of absorption or emission, thus some density or distance measurements using those luminosities or spectra. Since those bodies of plasma overall are a small part of the distances travelled the speed of light can be taken to be c.
The detailed research continues, but it's just one aspect of the astrophysics and cosmology of the universe.


TITLE: Exercise on Radon–Nikodym derivative
QUESTION [0 upvotes]: Consider the measure space $(\mathbb{N}, P(\mathbb{N}))$  and two measures $\mu$ and $\nu$, where $\mu$ is the counting measure and $\nu$ is defined as
$$ \nu(E) = \sum_{n \in E}(n+1)$$
Compute $d\mu/d\nu$ if its possible.
--
How would you proceed?
Thanks in advice!
Julian

REPLY [3 votes]: By definition the Radon-Nikodym derivative is a function $f$ such that
$$\nu (E)=\int _Efd\mu,$$ in the counting measure this means
$$\nu (E)=\int _Efd\mu =\sum _{n\in E}f(n),$$
so is not $f(n)=n+1.$
Edit: Taking the inverse problem, $d\mu /d\nu =\frac{1}{n+1}.$


TITLE: Heisenberg's Uncertainty Principle - finding uncertainty in wavelength
QUESTION [2 upvotes]: I am confused about this problem:

I needed to find the uncertainty of a wavelength using Heisenberg's uncertainty principle. In the solution, they differentiated $λ=c/f$ with respect to frequency to get $\frac{-c}{f^{2}}$.  λ is then substituted back in to get $\frac{- λ^{2}}{c}$. $Δf$ had already been found, so the final equation to find $Δλ$ was:
$$Δλ = \frac{- λ^{2}}{c}*Δf$$
Which honestly makes no sense to me. Why is it necessary to differentiate λ? I was thinking that the final equation would be something like:
$$Δλ = \frac{c}{Δf}$$
since $λ = \frac{c}{f}$. Why wouldn't my method work?

REPLY [2 votes]: As to why differentiation, we are talking about basic calculus.  You want to find out how $\lambda$ varies with $f$.
$\lambda =\frac{c}{f}$
$\frac{d\lambda}{df}=-\frac{c}{f^2}=-\frac{\lambda ^2}{c}$
$d\lambda=-\frac{\lambda ^2}{c}\ df$
Change the differentials to $\Delta$'s and you get what they have.  Your way is incorrect in that
$d\lambda\neq \frac{c}{df}$


TITLE: Proof formula with induction
QUESTION [0 upvotes]: How can I prove by induction that $\forall n\in\mathbb{N}: \ 3^{2n} - 1$ is divisible by $8$.
Proof for $n=1$: $\displaystyle3^{2\cdot1} - 1 = 8$ which is divisible by $8$.
How can I prove it for $n+1$? this is where I got so far: $3^{2(n+1)} - 1 = 3^{2n+2} -1$

REPLY [1 votes]: Assume $3^{2n}-1$ is divisible by $8$.
$$3^{2(n+1)}-1=3^{2n}3^{2}-1=9*3^{2n}-1=9*3^{2n}-9+8=9\big(3^{2n}-1\big)+8$$
$3^{2n}-1$ and $8$ are both divisible by $8$, therefore the sum is also divisible by $8$. This along with the proof that the statement holds true for $n=1$ proves this statement by induction.


TITLE: Proving that $x^{2}+2$, $x^{2}-x+4$, and $x^{3}+3x-1$ are irreducible over $\mathbb{Q}$
QUESTION [4 upvotes]: Let $f$, $g$ and $h$ be the polynomials given by:
  $$f(x)=x^{2}+2$$
  $$g(x)=x^{2}-x+4$$
  $$h(x)=x^{3}+3x-1$$
  Show that $f$, $g$ and $h$ are irreducible over $\mathbb{Q}$.

I do this:
$$x^{2}+2=(\sqrt{2}-i x) (\sqrt{2}+i x)$$
then it is irreducible over $\mathbb{Q}$.
$$x^{2}-x+4=4+(-1+x)(x)$$
then it is irreducible over $\mathbb{Q}$.
$$x^{3}+3x-1=-1+x (3+x^2)$$
then it is irreducible over $\mathbb{Q}$.
Is this true?
Thanks very much.

REPLY [3 votes]: Any non-trivial factorization of those polynomials would contain a linear factor, implying a rational root. The rational root test easily shows there are no rational roots.


TITLE: Is there an expression for the ith term of this sequence $1,1,2,1,2,3,1,2,3,4,1,2,3,4,5...?$
QUESTION [3 upvotes]: I'm trying to do some work in Excel and if I found a formula for this sequence it would help a lot. I don't particularly need to know why the formula works.
I have found the sequence here .
But there in no nice form for it.
Thanks!

REPLY [1 votes]: I've pulled this out of a comment, though it doesn't quite answer the question. There are some flexible ways of generating sequences in Excel which don't depend on a closed formula. If you need to reference a sequence within a formula, then you can generate the sequence (eg on another worksheet) and use a "lookup" formula to pick the term you need. There are ways of doing this (using a name for the cells containing the sequence) which can make the final formula easier to read and check.
For example, to generate the sequence you have asked about in this question, put 1 in cells A1 and B1, and use the formula =IF(B1>MAX($A1:A1),1,B1+1) in cell C1. Then copy this formula across the first row. I've left the formula in plain text, so you ought to be able to copy it. Obvious adjustments would apply for this to work in a different row, or in a column.
There are limitations on this technique, of course, but it generates shorter formulae, which can be easier to understand when you come back to them later.


TITLE: Why do water waves with longer wavelengths travel faster?
QUESTION [4 upvotes]: When wind blows over the sea, it excites a range of wavelengths.
Why do longer wavelengths travel faster?

REPLY [3 votes]: Definitions

deep water limit : when the wavelength of the mode/wave is much less than the depth of the water
shallow water limit : when the wavelength of the mode/wave is much larger than the water depth

Answer

Why do longer wavelengths travel faster?

In the deep water limit, the phase speed of, say, a gravity wave is $V_{ph} \propto \sqrt{\lambda}$, where $\lambda$ is the wavelength.  This is because the waves are dispersive, i.e., their frequency depends upon their wavenumber/wavelength.


TITLE: Intuition behind the definition of the d'Alembert operator
QUESTION [0 upvotes]: The d'Alembert operator is defined as $$\square^{2}=\frac{\partial^{2}}{\partial t^{2}}-\nabla^{2}$$
My question is what is the intuition behind this definition? My intuition tells me that we should Instead define such operator like a $$\square^{2}=\frac{\partial^{2}}{\partial t^{2}}+\nabla^{2}=\frac{\partial^{2}}{\partial t^{2}}+\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}$$
Because then we would have space derivatives (Laplace operator) + time derivatives. But with d'Alembert definition we have time derivative - space derivative. Why?

REPLY [3 votes]: The ultimate reason the d'Alembertian is useful is that the metric of spacetime is (+,-,-,-); that is, the invariant interval along a path is $\int (dt^2 - dx^2 - dy^2 - dz^2)$.
Given that metric, the natural extension to space time of
$$
\delta_{ii} \partial_i \partial_i
$$
is
$$
g^{\mu\nu} \partial_\mu \partial_\nu
$$
which is (up to a sign) the d'Alembertian.
The interesting thing is that he came up with this before Minkowski space was a common concept.

REPLY [1 votes]: Because $$\nabla^2 \phi = \frac{\partial^{2}}{\partial t^{2}} \phi$$
is up to scaling the wave-equation. So solving it is equivalent to find members of the kernel of the d'Alembert operator. Thinking about solving PDEs this way is helpful.


TITLE: Logic: some basic plane geometry
QUESTION [1 upvotes]: Suppose you've got the language of some basic plane geometry, i.e. two 1-place relation symbols $P$ and $L$ for point and line and one 2-place relation symbol $I$ for point $x$ lies on line $y$. Now, how is it possible to express the following axiom in this language: for every line $l$ and point $x$ (which is not on $l$), there is a unique line $m$ through point $x$.
This is what i've got: $\forall l,x$ $\neg I(x,l)\rightarrow \exists m$ $I(x,m)$.
Is this a correct sentence to begin with, and how could I express the uniqueness of the line $m$?

REPLY [1 votes]: Given that $\forall x \in X. \Phi$ translates to $\forall x. X(x) \implies \Phi$, and $\exists y \in Y. \Psi$ is $\exists y.\ Y(y) \land \Psi$ we can write
$$
\newcommand{\L}{\mathtt{L}}
\newcommand{\P}{\mathtt{P}}
\newcommand{\I}{\mathtt{I}}
\forall l \in \L.\ \forall x \in \P.\ \I(l,x) \lor \exists m \in \L.\ \I(m,x) \land \forall k \in \L.\ \I(k,x) \implies m = k.$$
In case you would like to express "unique line parallel to $l$" we could translate it to "unique line that doesn't intersect with $l$" or rather "doesn't has common points", for example
\begin{align}
&\forall l \in \L.\ \\
&\forall x \in \P.\ \mathtt{false}\\
&\quad\quad\lor\ \I(l,x) \\
&\quad\quad\lor\ \exists m \in \L.\ \mathtt{true}\\
&\quad\quad\quad\quad \land\ \I(m,x) \\
&\quad\quad\quad\quad \land\ \Big(\forall y \in \P.\ \\
&\quad\quad\quad\quad\quad\quad\quad\ \neg \I(m,y) \lor \neg \I(l,y)\Big) \\
&\quad\quad\quad\quad \land\ \Bigg(\forall k \in \L.\ \mathtt{false}\\
&\quad\quad\quad\quad\quad\quad\quad \lor\ m = k \\
&\quad\quad\quad\quad\quad\quad\quad \lor\ \neg\I(k,x) \\
&\quad\quad\quad\quad\quad\quad\quad \lor\ \neg\Big(\forall z \in \P.\ \\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\ \neg \I(k,z) \lor \neg \I(l,z)\Big)
\Bigg).
\end{align}
Quite long, but I hope this helps $\ddot\smile$


TITLE: Azuma's inequality to McDiarmid's inequality?
QUESTION [17 upvotes]: I was going through some notes on concentration inequalities when I noticed that there are two commonly-cited forms of McDiarmid's inequality. Long story short: I know how to prove the weaker one from Azuma's inequality. I also know how to prove the stronger one not directly from Azuma. But is there a way to prove the strong one from Azuma's inequality?
In what follows I will assume for simplicity that all differences are bounded by 1.

Azuma's inequality: Let $S_n = \sum_{i=1}^n X_i$ and $S_0 = 0$. Let the filtration $\{\mathcal{F}_n\}$ be the usual one defined by $\mathcal{F}_n = \sigma(X_1, \ldots, X_n)$. If $\{S_n\}_{n \geq 0}$ is a martingale and $|X_n| \leq 1$ for all $n$, then for $\lambda>0$, $$P(S_n \geq \lambda) \leq \exp \left(-\frac{\lambda^2}{2n} \right)$$

Now here is the "weak" version of McDiarmid's which is an obvious application of Azuma's inequality:

McDiarmid's inequality: Suppose $\xi_1, \ldots, \xi_n$ are independent. Let $Z = f(\xi_1, \ldots, \xi_n)$. Also, assume that: $$
\Big|\; f(\xi_1, \ldots, \xi_k, \ldots, \xi_n) - f(\xi_1, \ldots, \xi_k', \ldots, \xi_n) \;\Big| \leq 1$$ for two realizations of the same random variable $\xi_k$ and $\xi_k'$. Then for $\lambda>0$, $$
P(Z-EZ \geq \lambda) \leq \exp \left(-\frac{\lambda^2}{2n} \right) $$

Now the strong version of McDiarmid's inequality has the same setup except that the "2" in the denominator is moved up to the numerator, so the bound is:
$$P(Z-EZ \geq \lambda) \leq \exp \left(-\frac{2\lambda^2}{n} \right)$$
Clearly this bound is sharper than the one in the version presented above. I know a proof for this bound but it does not explicitly use Azuma's inequality above.
The question: Is it possible to adapt Azuma's inequality to prove this second bound. More specifically, is there a more general form of Azuma's inequality that can be applied to the case of McDiarmid's inequality to yield the tighter bound?
Edit: After pondering it over, I now think that it is not possible. Just intuitively, the bounded martingale difference condition in Azuma's inequality seems somehow "weaker" than the condition in McDiarmid's inequality, which is a Lipschitz-like condition on the function $f$. So one would expect that using Azuma's inequality would not give the sharpest bound.
Edit 2: To shed some more light on this, the proof of the "weak" McDiarmid's using Azuma's inequality goes like this. Define $S_n = E(Z \;|\; \xi_1, \ldots, \xi_m) - EZ$. Then $\{S_m\}_{m \geq 1}$ is a martingale. Also, for $m=n$, $S_m = S_n = Z - EZ$.
To prove that the martingale differences are bounded, let $\xi_1', \ldots, \xi_n'$ be independent copies of the $\xi_1, \ldots, \xi_n$. Then we have:
$$ E\big[f(\xi_1, \ldots, \xi_m, \ldots, \xi_n) \;\big|\; \xi_1, \ldots, \xi_{m-1}\big] = E\big[f(\xi_1, \ldots, \xi_m', \ldots, \xi_n) \;\big|\; \xi_1, \ldots, \xi_m\big] $$
Thus we have:
$$\big|S_m - S_{m-1}\big| \leq E\big[|\, f(\xi_1, \ldots, \xi_m, \ldots, \xi_n) - f(\xi_1, \ldots, \xi_m', \ldots, \xi_n)\,| \;\big|\; \xi_1, \ldots, \xi_m\big]$$
So to satisfy the bounded martingale difference in Azuma's inequality, we only need the expectation of the difference of $f(\cdot) - f(\cdot)$ (conditional on $\xi_1, \ldots, \xi_m$) to be bounded, whereas the assumption in McDiarmid's is slightly stronger. However, the bounded martingale difference must also hold for all $m$, and in particular for the case of conditioning on $\xi_1, \ldots, \xi_n$. In that case it seems to reduce to assumption. So... it does seem like Azuma's inequality (in this form) is not strong enough. Thoughts?
Here is a resource presenting the "standard" proof of the strong version of McDiarmid's inequality: http://empslocal.ex.ac.uk/people/staff/yy267/McDiarmid.pdf

REPLY [4 votes]: The difference is the sharper constants for the sub-Gaussian norm/parameter of the bounded variables. You can still use Azuma's technique to get the sharper bound. Here are some notes I wrote a while ago:
Sub-Gaussian tail bounds

Recall that $\mathbb{E}  e^{\lambda X} \le e^{\sigma^2 \lambda^2/2 }$ implies the following tail bound ($\mathbb{E} X = 0$)
\begin{align*}
 \mathbb{P}( X \ge t) \le \exp\Big( {- \frac{t^2}{2\sigma^2}}\Big)
\end{align*}
Such a random variable is called sub-Gaussian with parameter $\sigma$.
A bounded variable $X \in [a,b]$ has squared sub-G parameter $\sigma^2 = (b-a)^2/4$. This sharp bound requires some work to  show (left as an exercise.)

Azuma-Hoeffding approach
We want to provide concentration bounds for $Z = f(X) = f(X_1,\dots,X_n)$.
Let $\mathbb{E}_i[Z] := \mathbb{E}[Z | X_1,\dots,X_i]$ and $\Delta_i = \mathbb{E}_i[Z] - \mathbb{E}_{i-1}[Z]$. Then, $\{\Delta_i\}$ is a martingale difference sequence: $\mathbb{E}_{i-1}[\Delta_i] = 0$. Let $S_j := \sum_{i=1}^j \Delta_i$ which is only a function of $X_i, i \le j$.
\begin{align*}
 S_n = \sum_{i=1}^n \Delta_i =  Z - \mathbb{E} Z
\end{align*}
Let us assume that (*) $\mathbb{E}_{i-1} [e^{\lambda \Delta_i}] \le e^{\sigma_i^2 \lambda^2/2}$ for all $\lambda \ge 0$, and all $i$, almost surely. Then,
\begin{align*}
 \mathbb{E}_{n-1} [ e^{\lambda S_n}] =  e^{\lambda S_{n-1}}\mathbb{E}_{n-1} [ e^{\lambda \Delta_n}] \le e^{\lambda S_{n-1}} e^{\sigma_n^2 \lambda^2/2}
\end{align*}
Taking $\mathbb{E}_{n-2}$ of both sides:
\begin{align*}
 \mathbb{E}_{n-2} [ e^{\lambda S_n}]  \le e^{\sigma_n^2 \lambda^2/2} \mathbb{E}_{n-2}[ e^{\lambda S_{n-1}} ] \le
 e^{\lambda S_{n-2}} e^{(\sigma_n^2 + \sigma_{n-1}^2)\lambda^2/2}
\end{align*}
Repeating the process, we get $\mathbb{E}_{0} [e^{\lambda S_n}] \le \exp (  (\sum_{i=1}^n \sigma_i^2)\lambda^2/2 ) $. Showing that $S_n$ is sub-G with squared-param. $\sigma^2 = \sum_{i=1}^n \sigma_i^2$.
Bounded difference
Conditional sub-G assumption (*) holds under bounded difference
\begin{align*}
 |f(x_1,\dots,x_{i-1},x_i,x_{i+1},\dots,x_n) - f(x_1,\dots,x_{i-1},x_i',x_{i+1},\dots,x_n)| \le L_i
\end{align*}
Let $g_i(x_1,\dots,x_i) := \mathbb{E}[f(x_1,\dots,x_{i-1},x_i,X_{i+1},\dots,X_n) ]$ so that $\mathbb{E}_i[Z] = g_i(X_1,\dots,X_i)$. This uses independence, i.e. the distribution of the rest does not change by conditioning. It is easy to see that $g_i$ satisfies bounded difference condition with constants $(L_1,\dots,L_i)$, by an application of Jensen.

Naive bound: We have
\begin{align*}
 \Delta_i = \mathbb{E}_i[Z] - \mathbb{E}_{i-1}[\mathbb{E}_i[Z]] = g_i(X_1,\dots,X_i) - \mathbb{E}_{i-1}[g_i(X_1,\dots,X_i)]
 \end{align*}
Conditioned on $X_1,\dots,X_{i-1}$, we are effectively looking at ($X'_i$ is an independent copy of $X_i$)
\begin{align*}
 g_i(x_1,\dots,x_{i-1},X_i)- \mathbb{E}[g_i(x_1,\dots,x_{i-1},X'_i)]
 \end{align*}
due to independence of $\{X_i\}$. Thus, $|\Delta_i| \le L_i$ condition on $X_1,\dots,X_{i-1}$. That is, $\mathbb{E}_{i-1} [e^{\lambda \Delta_i}] \le e^{\sigma_i^2 \lambda^2/2}$ where $\sigma_i^2 = (2L_i)^2/4 = L_i^2$.
Better bound: We can show that $\Delta_i \in I_i$ where $|I_i| \le L_i$, improving the constant by $4$: Conditioned on $X_1,\dots, X_i$, we are effectively looking at
\begin{align*}
 \Delta_i = g_i(x_1,\dots,x_{i-1},X_i) - \mu_i
 \end{align*}
where $\mu_i$ is a constant (only a function of $x_1,\dots,x_{i-1}$). Then, $\Delta_i + \mu_i \in [a_i,b_i]$ where $a_i = \inf_x  g_i(x_1,\dots,x_{i-1},x)$ and $b_i = \sup_x g_i(x_1,\dots,x_{i-1},x)$. We have
\begin{align*}
 b_i - a_i = \sup_{x,y} \big[ g_i(x_1,\dots,x_{i-1},x) - g_i(x_1,\dots,x_{i-1},y) \big] \le L_i
 \end{align*}
Thus, we have $\mathbb{E}_{i-1}[e^{\lambda \Delta_i}] \le e^{\sigma^2 \lambda^2/2}$ where $\sigma_i^2 = (b_i-a_i)^2/4 \le L_i^2/4$.


TITLE: Why the preimage of a function between two complex Riemann surface is discrete?
QUESTION [1 upvotes]: Let $X,Y$ be two complex Riemann surface and $f:X\rightarrow Y$ be a nonconstant holomorphic map.
I was told that for any $y\in Y$, $f^{-1}(y)$ is discrete in $X$ because $f$ is nonconstant and $X$ is closed.
I wonder why this statement is true? Particularly, I wonder what "$X$ is closed" contributes here?

REPLY [2 votes]: If $f^{-1}(y)$ was not discrete, what could you say about the holomorphic function $g(x) = f(x)-y$?  (I'm assuming $X$ is connected.)


TITLE: Existence of vector space complement and axiom of choice
QUESTION [6 upvotes]: Let say we live in the category of vector spaces over $\mathbb{R}$ or $\mathbb{C}.$ Here are three sentences:

Axiom of choice
Every vector space has a base.
For every vector space $V$ and its subspace $E\subset V$ there is a subspace $F\subset V$ such that $V=E\oplus F.$

I know how to prove that (1)->(2)->(3). How about the inverese? Do (2)->(1) and (3)->(2) hold?
If this is not the case, then is there some weaker version of AC which imply (3)?

REPLY [10 votes]: No, there is no weaker choice principle implying (3). It was shown that (3) implies the axiom of choice in $\sf ZF$.
The proof is via an equivalent of the axiom of choice called "The Axiom of Multiple Choice". You can find the details in Rubin & Rubin's "Equivalents of the Axiom of Choice II" as Theorem 6.35 (pp. 119-120 and 122).
The proof is due to Bleicher from 1964

M. N. Bleicher, Some theorems on vector spaces and the axiom of choice, Fund. Math. 54 (1964), 95--107.

It is interesting to note that in a more relaxed setting where there might be atoms (non-set objects) or that the axiom of regularity fails, it is not known whether or not (3) implies the axiom of choice.


TITLE: antiderivative of $f(x)=(3x^2-1)^2$
QUESTION [0 upvotes]: It is easy to calculate $\int(3x^2-1)^2dx$: $\int(3x^2-1)^2dx=9x^4-6x^2+1dx=\frac{9}{5}x^5-2x^3+x+c$ $c\in\mathbb{R}$.
Now, one of my friends calculated the antiderivative of f with a similar method to subsitution, but which I have never seen before and which I don't understand, he named this method "backwards subsitution".
He get $F(x)=\frac{1}{3}(3x^2-1)^3\frac{1}{6x}$ as an antiderivative of f.
It is $F'=f$, but x has to be nonzero. It seems to be a different result as $\frac{9}{5}x^5-2x^3+x$, but I don't understand what is wrong. Is $F(x)=\frac{1}{3}(3x^2-1)^3\frac{1}{6x}$ an antiderivative of f?
Regards

REPLY [3 votes]: One fundamental principle in calculus is that you don't (or your friend doesn't) get to make up new rules and expect them to work. His method is incorrect and doesn't follow at all from the rule of substitution.


TITLE: Show $e^{D}(f(x)) = f(x+1)$ where $D$ is the derivative operator
QUESTION [8 upvotes]: I would appreciate help showing $e^{D}(f(x)) = f(x+1)$
Where $D$ is the linear operator $D: \mathbb{C}[x] \rightarrow \mathbb{C}[x]$ where (in the context where this statement arose) $x \in \mathbb{N}$;
$f(x) \mapsto \frac{d}{dx} f(x)$
By the Taylor series expansion $e^{D} = \sum_{n=0}^{\infty} \frac{D^n}{n!}$
$(1)$ Then $e^{D} (f(x)) = f(x) + f'(x) + \frac{f''(x)}{2!} +\dots$
I would appreciate help showing that the above line $(1)$ is equal to $f(x +1)$
This is what I have tried, but it feels forced:
for a Taylor series representation of $f(x +1)$ near $x$ I could write
$f(x+1) = f(x) + f'(x)(x+1 - x) + \frac{f''(x)}{2!}(x+1 -x)^2 \dots$
This then is equal to the RHS of line $(1)$.
Could this line of thinking be correct? Thanks very much.

REPLY [3 votes]: Use the binomial theorem.
$$
e^D x^n = \sum_{k=0}^\infty \frac{D^k}{k!} x^n.
$$
All of the terms after $k=n$ are $0$, i.e. $D^k x^n=0$ if $k>n$, so this is
\begin{align}
\sum_{k=0}^n \frac{D^k}{k!} x^n & = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} x^{n-k} \\[6pt]
& = \sum_{k=0}^n \binom nk x^{n-k} \\[6pt]
& = (x+1)^n.\quad\longleftarrow\text{By the binomial theorem.}
\end{align}


TITLE: Are there alternative proofs for existence/uniqueness of ODE solutions?
QUESTION [14 upvotes]: Consider the differential equation $\dot x = f(x)$. The standard proofs are

The Picard iteration based proof of existence/uniqueness for Lipschitz $f$.
The Peano existence theorem for continuous $f$.
The Caratheodory existence theorem for measurable $f$.

My question is as follows. Assuming a Lipschitz $f(x)$, are there any other proofs out there for existence of solutions (in some reasonable sense) for ODEs?

REPLY [0 votes]: The theory of Di Perna-Lions, also revisited by Ambrosio, provides existence (and uniqueness, in a suitable sense) results for a.e. initial datum of the ODE $\gamma'_t=v_t(\gamma_t)$ under the assumption that the vector fields $v_t$ are Sobolev/BV and with bounded divergence. Notice that in dimension 1 this latter requirement is equivalent to the $v_t$'s being Lipschitz, but in higher dimensions it is much weaker than that.
The proof uses an argument based on Young measures to reinterpret the (non-linear) ODE in terms of the associated (linear) continuity equation $$\partial_t\mu_t+div(v_t\mu_t)=0$$
The assumptions on the vector fields are used to show that for this latter equation we have existence and uniqueness results in suitable spaces. Then, with what is called `superposition principle' one see that these solutions must be induced by a flow of the given vector fields.
See:
DiPerna, R. J.; Lions, P.-L. Ordinary differential equations, transport theory and Sobolev spaces. Invent. Math. 98 (1989), no. 3, 511–547.
Ambrosio, L. Transport equation and Cauchy problem for BV vector fields. Invent. Math. 158 (2004), no. 2, 227–260.


TITLE: Does $\langle Tv, w \rangle = \langle v, w \rangle$ imply that $T = I$?
QUESTION [1 upvotes]: Suppose $V$ is an inner product space and $T \in \mathcal{L}(V)$. Suppose $\langle Tv, w \rangle = \langle v, w \rangle$ for all $v, w \in V$. Does this imply that $T = I$?
EDIT: Thanks for the replies. In that case, does $\langle Tv, w \rangle = \langle Sv, w \rangle$ imply that $T = S$? Because:
$ \hspace{0.33 in} \langle Tv, w \rangle = \langle Sv, w \rangle$ for all $v, w \in V$
$ \Longrightarrow \langle Tv-Sv, w \rangle = 0$ for all $v, w \in V$
$ \Longrightarrow Tv-Sv = 0$ for all $v \in V$
$ \Longrightarrow T-S = 0$
$ \Longrightarrow T=S$
And the same should hold for $\langle v, Tw \rangle = \langle v, Sw \rangle$, because of additivity in the right slot as well, right?

REPLY [7 votes]: Yes. Just take $w=Tv-v$ to see that $\|Tv-v\|^{2}=0$ so $Tv=v$.

REPLY [4 votes]: $\langle Tv,w\rangle-\langle v,w\rangle=\langle Tv-v,w\rangle=0$ for any $v,w\in V$. Hence $Tv-v=0$ for any $v\in V$ hence $T=Id$.

REPLY [1 votes]: Yes.
$$\forall v,w\quad \langle Tv,w\rangle = \langle v,w\rangle \Leftrightarrow \\ \forall v,w\quad \langle (T-I)v,w\rangle = 0 \Leftrightarrow \\ \forall v\quad (T-I)v = 0 \Leftrightarrow \\ T-I =0$$


text:
\begin{document}

\title[Correction factors for Kac-Moody groups]{Correction factors for Kac-Moody groups and $t$-deformed root multiplicities}

\author{Dinakar Muthiah}
\address{Department of Mathematics and Statistics, University of Massachusetts Amherst, Amherst, MA 01003}
\email{muthiah@math.umass.edu}

\author{Anna Pusk{\'{a}}s}
\address{Department of Mathematics and Statistics, University of Massachusetts Amherst, Amherst, MA 01003}
\email{puskas@math.umass.edu}

\author{Ian Whitehead}
\address{Department of Mathematics, Statistics, and Computer Science, Macalester College, St. Paul, MN 55105}
\email{iwhitehe@macalester.edu}
\maketitle

\begin{abstract}
We study a correction factor for Kac-Moody root systems which arises in the theory of $p$-adic Kac-Moody groups. In affine type, this factor is known, and its explicit computation is the content of the Macdonald constant term conjecture. The data of the correction factor can be encoded as a collection of polynomials $m_\lambda \in \ZZ[t]$ indexed by positive imaginary roots $\lambda$. At $t=0$ these polynomials evaluate to the root multiplicities, so we consider $m_\lambda$ to be a $t$-deformation of $\mult (\lambda)$. We generalize the Peterson algorithm and the Berman-Moody formula for root multiplicities to compute $m_\lambda$. As a consequence we deduce fundamental properties of $m_\lambda$.
\end{abstract}
\section{Introduction}\label{sect:intro}
\input{section-intro}
\section{Preliminaries and Notation}\label{sect:prelim}
\input{section-prelim}
\section{A Formal Definition of the Factor $\m$}\label{sect:thefactorm}
\input{section-definition}
\section{The Affine Case}\label{sect:affinecase}
\input{section-affine-case}
\section{The Generalized Peterson Algorithm}\label{sect:peterson}
\input{section-generalized-peterson-algorithm}
\section{The Generalized Berman-Moody Formula}\label{sect:GenBermanMoody}
\input{section-generalized-berman-moody-formula}
\section{Properties of $\m$}\label{sect:ordertwovanishing}
\input{section-order-two-vanishing}

\bibliographystyle{amsalpha}
\bibliography{papers}

\appendix

\section{Examples of the polynomial $\chi_{\lambda }$}\label{sect:dataandconjectures}

\input{section-short-data}

\end{document}