TITLE: Prove that $\int_{0}^{t} H^{2}_{u} \, du$ is $\mathcal{F}_{t}$-measurable
QUESTION [0 upvotes]: This question is from a step in a proof in the book "Stochastic Calculus Applied to Finance" by D. Lamberton and B. Lapeyre, page 41.
Given an $\mathcal{F}_{t}$-adapted, mean square integrable process $H_{t}$, I want to prove that  $\int_{0}^{t} H^{2}_{u} \, du$ is $\mathcal{F}_{t}$-measurable. The argument in the book is the following: "this result is true if $H$ is a simple process and, by density, it is true if $H$ is mean-square integrable".
I am stuck with the "by density" part. Indeed, given that the simple processes are dense in the space of mean square integrable processes with the norm $\|X\| = E[\int_{0}^{t} X^{2}_{u} \, du]$, choose a sequence $H^{n}$ of processes converging to $H$ in this space. One would like to prove that $\int_{0}^{t} (H^{n})^{2}_{u} \, du$, or any subsequence, converges to $\int_{0}^{t} H^{2}_{u} \, du$.
I have only figured out that  $\mathbb{E}\big[\int_{0}^{t} (H_{s}^{n})^{2}  \,\mathrm{d}s \big] - \mathbb{E}\big[\int_{0}^{t} H_{s}^{2} \, \mathrm{d}s \big] \xrightarrow[n\to \infty]{} 0$, but I need a convergence in $L^1$ of $\int_{0}^{t} (H_{s}^{n})^{2} \, \mathrm{d}s$ to $\int_{0}^{t} H_{s}^{2} \, \mathrm{d}s$ to finish the proof.

REPLY [2 votes]: Since
$$|x^2-y^2| = |x+y| \cdot |x-y| \leq (|x|+|y|) |x-y|$$
for any $x,y \in \mathbb{R}$, we have
$$\begin{align*} \mathbb{E} \left| \int_0^t (H_s^n)^2 \, ds - \int_0^t H_s^2 \, ds \right| &= \mathbb{E} \int_0^t |(H_s^n)^2-H_s^2| \, ds \\ &\leq \mathbb{E} \left( \int_0^t |H_s-H_s^n| (|H_s| + |H_s^n|) \, ds  \right). \end{align*}$$
Applying the Cauchy Schwarz inequality, we get
$$\begin{align*} \mathbb{E} \left| \int_0^t (H_s^n)^2 \, ds - \int_0^t H_s^2 \, ds \right| &\leq \sqrt{\mathbb{E} \left( \int_0^t |H_s-H_s^n|^2 \, ds \right)} \sqrt{\mathbb{E} \left( \int_0^t (|H_s|+|H_s^n|)^2 \, ds \right)} \\ &\leq \sqrt{2} \sqrt{\mathbb{E} \left( \int_0^t |H_s-H_s^n|^2 \, ds \right)} \sqrt{\mathbb{E} \left( \int_0^t (|H_s|^2+|H_s^n|^2) \, ds \right)} \end{align*}$$
where we have used that $$(x+y)^2 \leq 2x^2+2y^2.$$
As $H^n \to H$ in $L^2(\mathbb{P} \otimes \lambda|_{[0,t]})$, the right-hand side converges to $0$ as $n \to \infty$.


TITLE: Is the convolution well defined in that case?
QUESTION [0 upvotes]: Let $f , g : {\mathbb{R}}^n \to \mathbb{R}$ (or $\mathbb{C}$) two measurable functions. We suppose that $f \in L^{\infty}({\mathbb{R}}^n)$, $\mbox{supp} f$ is a compact set in ${\mathbb{R}}^n$ and $g \in L^1(K)$ for all compact set $K$ in ${\mathbb{R}}^n$. Prove that, given $x \in {\mathbb{R}}^n$,
$$
(f*g)(x) = \int_{{\mathbb{R}}^n} f(x - y) g(y) dy
$$
is well defined, that is, $f$ and $g$ are non-negative or non-positive, or $\varphi \in L^1({\mathbb{R}}^n)$, where $\varphi : y \mapsto f(x - y) g(y)$ for all $y \in {\mathbb{R}}^n$. My idea is the next: we know that exists $M \in [0 , \infty)$ such that $|f(z)| \leq M$ for almost every $z \in {\mathbb{R}}^n$, because $f \in L^{\infty}({\mathbb{R}}^n)$, so, using properties about the integral,
$$
|(f*g)(x)| \leq \int_{{\mathbb{R}}^n} |f(x - y)| |g(y)| dy \leq M \int_{{\mathbb{R}}^n} |g(y)| dy = M \int_{\mbox{supp} f} |g(y)| dy + M \int_{{(\mbox{supp} f)}^c} |g(y)| dy\mbox{.}
$$
Now, we also know that $g \in L^1(\mbox{supp} f)$ because $\mbox{supp} f$ is a compact set in ${\mathbb{R}}^n$ and $g \in L^1(K)$ for all compact set $K$ in ${\mathbb{R}}^n$. Then
$$
M \int_{\mbox{supp} f} |g(y)| dy < \infty\mbox{,}
$$
so we only should prove that
$$
\int_{{(\mbox{supp} f)}^c} |g(y)| dy < \infty
$$
but I have no idea. Thank you very much.

REPLY [0 votes]: I believe Folland does it this way:
$1).$ Define $\tau_y f(x)=f(x-y)$ and observe that $\left \| \tau_y f \right \|_{\infty }=\left \| f \right \|_{\infty }.$ Then,
$2).\  \left \| f* g \right \|_{1}=\left \| \int f(\cdot -y)\cdot g(y)dy \right \|_1=\left \| \int \tau_y f(\cdot)\cdot g(y)dy \right \|_1\le \left \| f \right \|_{\infty }\int |g(y)|dy=\left \| f \right \|_{\infty }\left \| g \right \|_1$


TITLE: How to solve $x^2=2^x$ algebraically?
QUESTION [1 upvotes]: I've been trying to solve this very basic looking equation algebraically and can't find a way. Is there one?
$$2^x=x^2$$
So far I have:
$$\ln 2^x=\ln x^2$$
$$x\ln 2=2\ln x$$
$$\frac{x}{\ln x}=\frac{2}{\ln 2}$$
But now I'm stuck. Any hints appreciated!

REPLY [1 votes]: We can solve equations of the form $x^\alpha=\beta^x$ for $x\in\Bbb C$ analytically using the special function known as the Lambert W function $W(z)$, defined as the inverse function of $ze^z$. It has infinitely many branches (like how $\log z$ is only defined up to an integer multiple of $2\pi i$), so there will in general by an infinite number of complex solutions.
Here's how. To start off with, we can absorb the constants together and introduce $\exp$:
$$x^\alpha=\beta^x\iff x=\beta^{x/\alpha}=e^{\gamma x} \qquad{\rm where}~\gamma=\frac{\ln\beta}{\alpha}$$
The way forward from here goes as follows:
$$\begin{array}{ll} x=e^{\gamma x} & \iff xe^{-\gamma x}=1 \\ & \iff -\gamma x e^{-\gamma x}=-\gamma \\ & \iff -\gamma x=W(-\gamma) \\ & \iff x=\frac{W(-\gamma)}{-\gamma}=-\frac{\alpha}{\ln\beta}W(-\frac{\ln\beta}{\alpha}).\end{array}$$
Of course this doesn't help us find $x=2,4$ in this case, but the theory can help us numerically approximate solutions in general where there are no easy symbolic solutions.


TITLE: Is my process of proving inequalities using Mathematical Induction correct?
QUESTION [0 upvotes]: $$ 2^n<(n+1)!   \quad  \text{for all integers} \quad  n \ge 2$$
Base Case: $$ 2^2< (2+1)! = 4< 6 $$
Correct
Assumption: $$ P(k): 2^k<(k+1)!$$
$$P(k+1)= 2^{k+1}<(k+2)!$$
Now we start with the Assumption : $$ 2^k<(k+1)!$$
$$ \text{Multiply 2 on both sides} \quad 2^{k+1}<2(k+1)! $$
$$ \text{Now we can say} \quad  (k+2)!> 2(k+1)! \quad \text{to prove this inequality}$$
$$ (k+2)(k+1)! > 2(k+1)! $$
So now all we must prove is that $$ k+2 > 2 \quad \text{Which is True for} \quad k\ge 2$$

REPLY [1 votes]: In addition to the good advice already in an answer, I recommend more care in how you use symbols, especially the $=$ symbol. Yes, it does mean the thing on the left is the same as the thing on the right, but sometimes one can disagree on exactly what each "thing" is.
For example, many (most?) people reading $2^2<(2+1)!=4<6$ would interpret it as
\begin{array}{ccccc}
2^2<(2+1)! & \quad\text{and}\quad & (2+1)!=4 & \quad\text{and}\quad & 4<6
 \\[1ex]
\text{(true)} && \text{(oops!)} && \text{(true)}
\end{array}
You obviously did not mean to write the equation with "oops!" below it,
and I think most people will realize that, but they'll have to think about it
harder than they should have to and they may hold that against you.
You could write something like "$2^2<(2+1)!$ because $4<6$", or if you want to keep it in one big math expression,
$$2^2=4<6=(2+1)!.$$
People will still read the $<$ in the middle as applying only to the
numeric expressions immediately to its left and right, but now that
gives you $4 < 6$, which is the essence of this part of the proof.
Also, people are used to seeing "chains" of inequalities like
$$ a \ \Box\ b \ \Box\ c \ \Box\ d \ \Box\  x \ \Box\  y \ \Box\  z$$
where each of the $\Box$ symbols is replaced by $=$, $<$, or $\leq$,
and they know that this chain implies that $a \leq z$
(and if any of the relationships is $<$, then $a < z$).
For similar reasons, $P(k): 2^k<(k+1)!$ is fine but
$P(k+1)= 2^{k+1}<(k+2)!$ is confusing.
You could make an argument for writing $P(k+1)= \left(2^{k+1}<(k+2)!\right)$,
but I still prefer "$P(k):$".
Also, once you choose a style, use it consistently.


TITLE: Krull dimension of Nagata rings
QUESTION [4 upvotes]: I want any help concerning proof of the following theorem of Nagata:
Let $S$ be the set of all $f\in R[x]$ with the property that the coefficients of $f$ generate the unit ideal. Then
(a) $S$ is a multiplicative set consisting entirely of non-zero-divisors;
(b) $\dim S^{-1}R[x]=\dim R$, where $\dim$ stands for the Krull dimension of $R$.
Thanks in advance!

REPLY [2 votes]: Denote by $c(f)$ the ideal generated by all the coefficients of $f\in R[X]$.
(a) Hint. $c(f)=R$ iff $\bar f\ne\bar 0$ in $(R/\mathfrak m)[X]$ for any maximal ideal $\mathfrak m$ of $R$.
(b) (Here you missed the condition $R$ noetherian, otherwise the result can be wrong.)
Clearly $\dim S^{-1}R[X]\ge\dim R$.
Now let $P$ be a prime ideal of $R[X]$ maximal with the property $P\cap S=\emptyset$. It's obvious that $S=R[X]\setminus\bigcup_{\mathfrak m\in\operatorname{Max}R}\mathfrak m[X]$. From $P\cap S=\emptyset$ we get $P\subseteq\bigcup_{\mathfrak m\in\operatorname{Max}R}\mathfrak m[X]$. Now we claim that there is $\mathfrak n\in\operatorname{Max}R$ with the property $P\subseteq\mathfrak n[X]$. In order to prove this let's consider $\mathfrak a$ the set of all elements of $R$ which appear as a coefficient of a polynomial from $P$. This is an ideal of $R$, and $\mathfrak a\ne R$. Then there is $\mathfrak n\in\operatorname{Max}R$ with the property $\mathfrak a\subseteq\mathfrak n$, and therefore $P\subseteq\mathfrak n[X]$. In particular, $\operatorname{ht} P\le\operatorname{ht}\mathfrak n[X]=\operatorname{ht}\mathfrak n$, and we are done.


TITLE: Confusion regarding a proof: Finite etale coverings
QUESTION [0 upvotes]: I've a confusion regarding Corollary 2.4. It says that the graphs $\Gamma_f$, $\Gamma_g$ gives us sections which agree at $y'$. I don't understand what section it is talking or it is a section of which map? Also, why does that imply $\Gamma_f$, $\Gamma_g$ are isomorphic?

REPLY [1 votes]: It means that $\Gamma_f,\Gamma_g: Y'\to Y'\times_X Y$ are sections of $Y'\times_X Y\to Y'$, and hence thanks to Lemma 2.3 (you need $Y'$ connected, which is not explicit in the hypothesis) they give you isomorphisms between $Y'$ and connected components $U_f, U_g$ of $Y'\times_X Y$. But, since $f(y')=g(y')$ and the two maps $k(y)\to k(y')$ agree, this tells you that the two maps ${\rm Spec}~k(y')\to Y$ agree and hence, by the usual property of the fiber product, also the two maps ${\rm Spec}~ k(y')\to Y'\times_X Y$ agree and hence $\Gamma_f(y')=\Gamma_f(y)$, which implies $U_f=U_g=U$.
Now, $f$ is equal to the composition $Y'\to U_f\to Y'\times_X Y\to Y$, and the same holds for $g$. Hence, to show that $f=g$, you need to show that the two isomorphisms $Y'\to U_f=U$ and $Y'\to U_g=U$ are the same. But this is obvious, since they are both inverses of the projection $U\subseteq Y'\times_X Y\to Y'$.


TITLE: Zeros of non-trivial solution of linear ODE of second order
QUESTION [1 upvotes]: The exercise is as follows: Prove that an arbitrary non-trivial solution of $x''+tx=0$ on interval $[-25,25]$ has at least $25$ and at most $28$ zeros/roots. How am I supposed to do this? I tried splitting the interval and using standard criteria but I get to many solutions.
I tried the following: I know that the distance between two consecutive zeros $t_1,t_2, t_2>t_1$ of $x''+q(t)x=0$ on interval $\mathbb{I}$ is $$\frac{\pi}{\sqrt{M}}<t_2-t_1<\frac{\pi}{\sqrt{m}},$$
where $M:=\sup_{\mathbb{I}}q(t)$ and $m:=\inf_{\mathbb{I}}q(t)$. And $q(t)>0$ on $\mathbb{I}$.
On the other hand we know that $x''+q(t)x=0$ has at most one zero on interval $\mathbb{I}$ if $q(t)<0$.
Knowing this I split the interval into $[-25,-1], (-1,1)$ and $ [1,25]$. Calculating the distance between zeros I got that on the first interval we have one zero at most. On the second I have no clue, but on the third there are at most like $38.$ However, I am not sure whether these are all possible zeros or just of one solution and if it is just of one solution I don't know whether that solution with that many zeros is a solution on the whole interval.

REPLY [1 votes]: This is an algorithmic solution, that does not give exactly the solution provided by the OP.
Assuming a zero in $t_k$, the previous zero $t_{k-1}<t_k$ should satisfy on the interval $[t_{k-1},t_k]$
$$
\frac{\pi}{\sqrt{t_k}}<t_k-t_{k-1}<\frac{\pi}{\sqrt{t_{k-1}}}
$$
so the minimum distance is when
$$
t_{k-1}=t_k-\frac{\pi}{\sqrt{t_k}}\quad\land\quad t_{k-1}>0\tag1
$$
while the maximum distance is when
$$
t_k-t_{k-1}<\frac{\pi}{\sqrt{t_{k-1}}}
$$
i.e. $t_{k-1}$ is a solution of
$$
t_{k-1}(t_k-t_{k-1})^2=\pi^2\quad\land\quad0<t_{k-1}<t_k\tag2
$$
By calculating the sequence of values of $(1)$ starting with $t_k=25$ we get
$$
\left(
\begin{array}{cc}
 1 & 24.3717 \\
 2 & 23.7353 \\
 3 & 23.0905 \\
 4 & 22.4367 \\
 5 & 21.7735 \\
 6 & 21.1002 \\
 7 & 20.4163 \\
 8 & 19.721 \\
 9 & 19.0135 \\
 10 & 18.2931 \\
 11 & 17.5585 \\
 12 & 16.8088 \\
 13 & 16.0425 \\
 14 & 15.2582 \\
 15 & 14.4539 \\
 16 & 13.6276 \\
 17 & 12.7766 \\
 18 & 11.8977 \\
 19 & 10.9869 \\
 20 & 10.0391 \\
 21 & 9.04756 \\
 22 & 8.00312 \\
 23 & 6.89261 \\
 24 & 5.69599 \\
 25 & 4.37966 \\
 26 & 2.87849 \\
 27 & 1.0268 \\
\end{array}
\right)
$$
while for the sequence $(2)$ we get
$$
\left(
\begin{array}{cc}
 1 & 24.3635 \\
 2 & 23.7185 \\
 3 & 23.0643 \\
 4 & 22.4005 \\
 5 & 21.7265 \\
 6 & 21.0417 \\
 7 & 20.3452 \\
 8 & 19.6362 \\
 9 & 18.9138 \\
 10 & 18.177 \\
 11 & 17.4244 \\
 12 & 16.6545 \\
 13 & 15.8658 \\
 14 & 15.0562 \\
 15 & 14.2232 \\
 16 & 13.3638 \\
 17 & 12.4743 \\
 18 & 11.5499 \\
 19 & 10.5843 \\
 20 & 9.56866 \\
 21 & 8.4905 \\
 22 & 7.33014 \\
 23 & 6.05324 \\
 24 & 4.58627 \\
 25 & 2.66006 \\
\end{array}
\right)
$$
In the first case, in which we are maximising the number of zeroes, we should take into account a possible zero in $25$ and another in $t<0$, reaching a maximum of $29$ zeroes. In the second case we have nothing to add, and obtain a minimum of $25$ zeroes.


TITLE: Show that $F$ is a Galois extension on $\mathbb{Z}_p$ and Show that for each $a \in F$ then $a = x^2 + y^2$ with $x,y \in F$
QUESTION [0 upvotes]: Problem: Let $F$ be a finite field

Show that characteristic of $F$ is a prime number $p>0$ and $|F| = p^n, n \in \mathbb{Z}^+$.
Show that $F$ is a Galois extension on $\mathbb{Z}_p$.
Show that for each $a \in F$ then $a = x^2 + y^2$ with $x,y \in F$.

My attempt:

$F$ is a finite field implies $\text{char} (F)$ and $|F| = p^n$. If $\text{char} (F) = 0$ then $\mathbb{Q}$ is the prime subfield of $F$. Then $F$ infinite. Hence $\text{char} (F) = p>0$ and $\mathbb{Z}_p$ is the prime subfield of $F$. Consider the vector space $F$ on $\mathbb{Z}_p$. We have $\dim_{\mathbb{Z}_p} F = n = \dim_{\mathbb{Z}_p} \mathbb{Z}_p^n$. Hence, $F \cong \mathbb{Z}_p \times \dots \times \mathbb{Z}_p \Rightarrow |F| = p^n$

I need help for the question number $2$ and $3$. Thank all!

REPLY [1 votes]: If $\mathbb{F}$ is a field with  $p^n$, then try to show it is the splitting field of $x^{p^{n}}-x$ over $\mathbb{F}_{p}$.(You might want to recall the fact that $\mathbb{F}-\lbrace 0 \rbrace$ is a multiplicative group, and use Lagrange's theorem!).
As for part 3, if $p=2$, consider the map $\tau: \mathbb{F}^{\star} \to \mathbb{F}^{\star}$, $\tau(x)=x^2$. Notice that $\tau$ is a homomorphism of groups, that is injective (check this!), and hence surjective, and hence bijective. Every element in $\mathbb{F}$ is hence a square.
Now, if $p$ is odd, consider the same map, $\tau: \mathbb{F}^{\star} \to \mathbb{F}^{\star}$. It is also a group homomorphism, whose kernel has two distinct elements. This means that the image of $\tau$, has $\frac{p^{n}-1}{2}$ elements, which means that there are $\frac{p^{n}-1}{2}$ distinct non zero squares, and hence $\frac{p^{n}+1}{2}$ squares in $\mathbb{F}$, including $0 \\$.
Let $a \in \mathbb{F}$. Consider the sets $S= \lbrace s^2, s \in \mathbb{F} \rbrace$ and the set $H= \lbrace a-y^2, y \in \mathbb{F} \rbrace$. These are subsets of $F$ with $\frac{p^{n}+1}{2}$ elements each. They must intersect (if they did not, then what would the cardinality of the union be?). And hence... finish it!


TITLE: Doubt in problem regarding resistance a of cylinder
QUESTION [0 upvotes]: I've been working on some exercises and I'm in doubt if my procedure with this one is correct. We have a hollow cylinder with internal radius $r_a$, external radius $r_b$, resistivity $\rho$ and length $L$. A difference of potential $V$ is then applied on the extremes paralel to the axis of the cylinder.
First I'm asked to find the resitance. Well, for this one I've just used that $R=\rho L/A$ and calculated $A=\pi(r_b^2-r_a^2)$. Second, I'm asked to find the current density when $V$ is applied. Well, what I did was suppose that the cylinder is an ohmic resistor so that $V=Ri$ holds, and then I've got $i=V/R$. Then the current density should be $J=i/A$ and so we have $J=V/RA$ which gives $J=V/\rho L$. Is this correct? Can I assume that the cylinder is ohmic?
Third I'm asked to find the electric field. Now I've used that $\rho = E/J$ so that we must have $E = \rho J$ so that $E = V/L$, this says tha that the field is uniform, but I'm a little unsure about it. Finally the fourth one asks to find the resitance again if now the current flows radially from inside to ouside. In this final case I think that it'll change just the cross sectional area, but I didn't find some easy way to write this down.
Can someone give a little help with this?

REPLY [1 votes]: Your calculations are correct, provided the cylinder is indeed ohmic. The constant $E$ you're getting is the difference in electric field between both terminals.
As for the current flowing from inside to outside, as you said the cross sectional area will be different, and so will the length. The length $L=r_b-r_a$, but the cross sectional area is not uniform, because at the beginning of the wire (the interior), $A=2\pi r_aL$, and at the "end" of the wire (the exterior), $A=2\pi r_bL$. So you'll have to treat each portion of the wire as its own infinitesimal resistor $dR$, and the total resistance is the series combination of them:
$$dR=\rho\frac{dl}{2\pi lL}$$
$$R=\int_{r_a}^{r_b}\rho\frac{1}{2\pi lL}dl$$


TITLE: Can there exist a $C^2$ function $f: \mathbb R^3 \to \mathbb R$ with $f_x = 3x - 2y + 4z.
QUESTION [1 upvotes]: Can there exist a $C^2$ function $f: \mathbb R^3 \to \mathbb R$ with $f_x = 3x - 2y + 4z, f_y = -2x+3y +z $ and $f_z = 3x + y - 5z$. Explain your answer.
$\int f_udu = \frac{3}{2}x^2 - 2xy + 4zu + f(y, z) = f(x, y, z)$
$\int f_y dy = -2xy + zy + \frac{3}{2}y^2 + f(x, z) = f(x, y, z)$
$\int f_z dz = 3xz + yz - \frac{5z^2}{2} + f(x, y) = f(x, y, z)$
since $\frac{d}{dz} (\int f_u du) \neq f_z$ hence does not exist any for which the partial derivatives are given
is this correct?

REPLY [0 votes]: There's a few different approaches. Yours is fine.
Hans Engler has a really good one. Here's a third, somewhere in between.
You can make a vector $\vec{v}=<f_x,f_y,f_z>$
Then if $\nabla \times \vec{v}=0$ then such a function exists.
$\nabla \times \vec{v}\implies v_i=\sum_{j=1}^3\sum_{k=1}^3 \epsilon_{ijk}\frac{\partial}{\partial x^j}\frac{\partial f}{\partial x_k}$
Where $\epsilon_{ijk}$ is the Levi-Civita symbol. $x_1=x, x_2=y, x_3=z.$
You recover Mr. Engler's with perhaps a bit more of work.
Through abuse of notation you can find $\nabla \times \vec{v}$, the curl of $\vec{v}$ with a matrix.


TITLE: Probability for #suits in 5 card poker hand
QUESTION [4 upvotes]: Given a 5 card poker hand from a standard deck, I'm looking to calculate the probability of getting: all 1 suit, 2 different suits, 3 different suits or 4 different suits.
All one suit is straight-forward -
$\frac{\binom{13}{5}*\binom{4}{1}}{\binom{52}{5}}$- pick five different ranks, each from the same suit.
Likewise, 4 seems fairly simple:
$\frac{\binom{4}{1}\binom{13}{2}\binom{13}{1}^3}{\binom{52}{5}}$ - pick one suit to grab two cards from, then pick one card from each other suit.
Its on 2 and 3 that I get kind of stuck - I'm not sure how to set them up! I don't see why something along the lines of
$\frac{\binom{4}{2}*\binom{26}{6} - \binom{13}{5}*\binom{4}{1}}{\binom{52}{5}}$ doesn't work for 2 suits; i.e. picking 2 suits, choose 5 cards, subtracting off the ways in which you could end up with one suit.
Similarly, for 3 I would expect
$\frac{\binom{4}{3}*\binom{39}{5}-\binom{4}{2}*\binom{26}{5}}{\binom{52}{5}}$ to give the answer (picking 5 cards from the group containing 3 suits, subtracting off those hands with fewer than 3 suits), but if I sum the probabilities it comes out incorrectly.
Thank you very much for your help!

REPLY [3 votes]: As an example of another technique, which is useful for counting poker hands, particularly two pairs, we count the $3$-suit hands.
There are two types of $3$-suit hand: $3$-$1$-$1$ and $2$-$2$-$1$.
We count the $3$-$1$-$1$ hands. The suit we have $3$ of can be chosen in $\dbinom{4}{1}$ ways. For each such way, the actual cards can be chosen in $\dbinom{13}{3}$ ways. Now the suits we have $1$ of can be chosen in $\dbinom{3}{2}$ ways, and the actual cards in $\dbinom{13}{1}^2$ ways, for a total of
$$\binom{4}{1}\binom{13}{3}\binom{3}{2}\binom{13}{1}^2.$$
Next we count the $2$-$2$-$1$ hands. The suit in which we have a singleton can be chosen in $\dbinom{4}{1}$ ways. For each such choice, the actual card can be chosen in $\dbinom{13}{1}$ ways. Now the suits we have doubletons in can be chosen in $\dbinom{3}{2}$ ways, and the actual cards in $\dbinom{13}{2}^2$ ways, for a total of
$$\binom{4}{1}\binom{13}{1}\binom{3}{2}\binom{13}{2}^2.$$


TITLE: Similarities between particular solutions of differential equations and systems of linear equations
QUESTION [1 upvotes]: I know from linear algebra that for a non-homogeneous consistent linear system that has an infinite number of solutions, if the last row of the augmented matrix has all zeros, then we have an infinite number of particular solutions.
I was wondering if there is a similar parallel with differential equations, namely say I have a differential equation of the form
$y''(x)(Ax+B)^2 + y(x) =$ some cubic, and I want to find a particular solution of this. Then I make an $ansatz$ that my $y(x)$ is some cubic as well, of the form $y(x) = a_0 x + a_1 x + ... a_3 x^3$
When comparing coeffcients on the left and right hand sides, I have come across cases where by the coeffcients $a_0$, ... $a_1$ on the L.H.S. do not have particular values, but are instead expressed as expressions involving one of the other coefficients as a free variable.
Is it possible that differential equations too may have an infinite number of particular solutions?
Ben

REPLY [3 votes]: There is in fact a very strong connection between (certain kinds of) differential equations and systems of linear equations. They are both instances of linear transformations.
Consider the collection of all infinitely differentiable functions. There is a function $D$ from this set to itself that maps a function $f(x)$ to
$$D(f) = f''(x)(Ax+B)^2 + f(x).$$
This map is linear: $D(f+g) = D(f)+D(g)$, and $D(\alpha f) = \alpha D(f)$ for all infinitely differentiable functions $f$ and $g$, and all real numbers $\alpha$.
In particular, fix a function $k_0(x)$, and suppose that $D(f) = k_0(x)$ (that is, $f$ is a particular solution to $y''(x)(Ax+B)^2 + y(x) = k_0(x)$). Then:

If $g(x)$ is any solution to $D(g)=0$, that is, $g''(x)(Ax+B)^2 + g(x)=0$, then $f(x)+g(x)$ is also a solution to $D(y)=k_0(x)$. This because $D(f+g) = D(f)+D(g) = k_0(x)+0 = k_0(x)$.
If $h$ is any other solution to $D(y)=k_0(x)$, then there exists a solution $g$ to $D(g)=0$ such that $h(x) = f(x)+g(x)$. That is, all solutions to $D(y)=k_0(x)$ are obtained by taking a single particular solution and addding solutions to $D(y)=0$.
To see this, notice that if $D(f)=D(h)=k_0(x)$, then $D(h-f) = D(h)-D(f) = k_0(x)+k_0(x) = 0$. So $h-f$ is a solution to $D(y)=0$. And $h=f+(h-f)$, so $h$ can be obtained by adding a solution to $D(y)=0$ to $f$.

In particular, the differential equation
$$y''(x)(Ax+B)^2 + y(x) = \text{some cubic}$$
that has at least one particular solution has exactly as many particular solutions as there are solutions to
$$y''(x)(Ax+B)^2 + y(x) = 0.$$
So if the homogeneous equation has infinitely many (essentially distinct) solutions, and your original equation has at least one particular solution, then the original equation will in fact have infinitely many (essentially distinct) solutions as well.
This is in complete analogy to the case of a system of linear equations, which can be seen as given by the matrix equation $A\mathbf{x}=\mathbf{b}$, where $A$ is the matrix of coefficients of the system: given any particular solution $\mathbf{x}_0$, you get all solutions by taking $x_0+\mathbf{z}$, where $\mathbf{z}$ is any solution to $A\mathbf{z}=\mathbf{0}$.
This is all part of linear algebra. Both systems of linear equations and linear differential equations can be studied together, and many general results apply to both, so it's no surprise you are seeing a lot of similarities: in fact, there are a lot of similarities.


TITLE: is $\det(A^2 + I)$ always non negative?
QUESTION [10 upvotes]: Obviously $\det(A^2)$ is (casework), but is the above matrix non-negative? $\det(A)\det(A) \geq 0$ as $\det(A) > 0$ or $\det(A) < 0$ yields positive when squared. However, I am not sure that when adding the identity matrix that it is also positive.

REPLY [2 votes]: It is possible also to use eigenvalues to prove the claim.
Determinant is the product of eigenvalues and eigenvalues of the polynomial are polynomials of eigenvalues.
Denote $B=A^2+I$ and eigenvalues of $A$ as $r_{Aj}$ when they are real, and $c_{Ak}=a_k+b_ki$ when they are complex with non-zero imaginary part.
As we know for real matrices complex eigenvalues come in conjugate pairs $c_{Ak}=a_k+b_ki,c_{Ak*}=a_k-b_ki$.
Then eigenvalues of $B$ are $r_{Bj}^2+1$ ( which are obviously positive, even $\ge 1$)  for real eigenvalues, and $c_{Bk}=(a_k+b_ki)^2+1,c_{Bk*}=(a_k-b_ki)^2+1$ for complex ones.
$$c_{Bk}=a_k^2-b_k^2+1 + 2a_kb_ki$$
$$c_{Bk*}=a_k^2-b_k^2+1 - 2a_kb_ki$$
We have received once again a pair of conjugated complex numbers.
The product of conjugated complex numbers is non-negative number ( $cc^*=\vert c\vert^2$) and the whole product is non-negative.
Generally  we can say even more: the determinant is mostly positive,  except the case when matrix $A$ has $\pm i \ \ $ eigenvalues (in that case determinant is $0$, $B=A^2+I$ is singular).


TITLE: Residual vectors and Range
QUESTION [0 upvotes]: I wanted to understand the question here a little better: Let $r = b - Ax$ be the corresponding residual vector. Which of the following three vectors is a possible value for $r$? Why?
If I understand correctly, the hint posted by the one answerer there implies that I should find the range of $A$ and then check whether the vectors from (a), (b), and (c) are orthogonal to see if they are truly residuals in this case. Lets call these vectors $r_1, r_2, r_3$ respectively.
I know the range of $A$ can be defined as the span of its columns. My question here is, should I start finding all possible dot products of $r_1, r_2, r_3$ with the columns of $A$, and check if any of them come out to zero? I'm not sure that would work. How do I check if  $r_1, r_2, r_3$ is orthogonal to the range?

REPLY [2 votes]: While it's true that the residual vector needs to be orthogonal to every linear combination of the columns (i.e. every vector in the columnspace), you don't need to compute the dot product with every such vector. It suffices to just check that the possible residual vector is orthogonal with a basis (or even just a spanning set) for the columnspace.
Since every column is in the columnspace, certainly every residual vector must be perpendicular to every column. Conversely, if $r$ is perpendicular to every column, then by linearity of the map $x \mapsto x \cdot r$, then $x$ is perpendicular to every linear combination of columns.
So, you really don't have to worry about checking every vector in the columnspace.


TITLE: Why does secant method converge
QUESTION [1 upvotes]: Assume $f$ is continuous and twice differentiable on $[a,b]$ such that $f'(x)>0$ and $f''(x)>0$, $x \in [a,b]$.
If $f(b)>0$ and $f(a)<0$ and I choose $x_0=a$,why are we gauraunteed convergence with the iterations:
$$x_n=x_{n-1}-f(x_{n-1})\frac{b-x_{n-1}}{f(b)-f(x_{n-1})}$$
I made some progress, but it relies on an assumption that I don't know how to prove.
I assume that $x_k$ is bounded above by the root. Meaning if $f(\xi)=0$ then $\xi>x_k$ for all $k$.
What I did:
$x_{k+1}=x_k-f(x_k)\frac{b-x_k}{f(b)-f(x_k)}$
Notice that our initial $x_0$ was $a$, and $f(a) <0$. we also assumed $x_k < \xi$ and that $f$ is continuous and $f'(x)>0$, it follows that $f(x_k)<f(\xi)=0$. We also know that $b-x_k >0$ (again, from the assumption on $\xi$) and that $f(b)-f(x_k)>0$. This is because $f$ is monotonic rising and continuous on $[a,b]$ and $f(b)>0$. It follows that $f(b)>f(x)$ for all $x \in [a,b]$.
So overall, the term $-f(x_k)\frac{b-x_k}{f(b)-f(x_k)}$ is positive. and so $x_{k+1}=x_k+\alpha$ where $\alpha>0$ and so $x_{k+1}>x_k$ and the sequence is monotonic rising.
Since we also assumed it is bounded by $\xi$, it converges. We still need to prove it converges to $\xi$:
Let's call the limit $L$:
$$\lim_{k \to \infty}x_{k}=\lim_{k \to \infty}x_{k+1}=L$$
$L=L-f(L)\frac{b-L}{f(b)-f(L)}$, it follows that $f(L)\frac{b-L}{f(b)-f(L)}=0$
We assumed $x_k$ is bounded by $\xi$. This means that $L \leq \xi <b$. So $L \neq b$.
This indeed implies $f(L)=0$, since there is only one root, this means $L=\xi$.
End of proof.
All of this fails because we rely on the assumption $x_k \leq \xi$.
How do I prove this assumption?

REPLY [1 votes]: Your first guess is $x_0=a$ and $f(a)<0$. Because $f$ is convex
$$\frac{f(x_1)-f(a)}{x_1-a}< \frac{f(b)-f(a)}{b-a}.$$
With this you can prove $f(x_1)<0$ and $x_1 < \xi$.
Note that
$$f(x_1) < f(a) + \frac{x_1-a}{b-a}[f(b)-f(a)]= f(a)-\frac{f(a)}{f(b)-f(a)}[f(b)-f(a)]=0$$
On each subsequent iteration you start with $f(x_i) <0$ and consequently you get $f(x_{i+1})<0$ and $x_{i+1}< \xi$.


TITLE: Decide about convergence of the series for $a>0$ and $q\in \mathbb R$
QUESTION [2 upvotes]: My series is: $$\sum_{n=1}^{+ \infty } \frac{q^{n}n^{2}cos( \frac{n\pi}{5} )}{ \frac{n^{3}}{1444}+1001n^{2}+1444an }$$
for $q \in \mathbb{R}$ and $a > 1$.
I did this task for $|q|<1$ and $|q|>1$ using the Cauchy criterion and for $q=-1$ using the Leibnitz criterion. However I cannot do this task for $q=1$. My idea is to use Raabe test but there are big number and it is difficult to use it so I think exist easier way.

REPLY [2 votes]: You can prove it converges using the Dirichlet Test. https://en.wikipedia.org/wiki/Dirichlet%27s_test
When $q=1$ this sum will be less than
$$1444\sum_{n=1}^\infty \frac{\cos{(\frac{n\pi}{5})}}{n}$$
$|\sum_{n=1}^N \cos{(\frac{n\pi}{5})}| \le 10$ for every integer $N$, this is easy to see because the sum repeats every $20$, and the sum at $20$ is $0$ (Note I choose $10$ randomly, you could get a way better bound if you wanted). And $\frac1n$ is decreasing and goes to 0, so the sum converges.


TITLE: Uniform continuity implies existence of limit of integral
QUESTION [3 upvotes]: Let $f: (0,1) \rightarrow \mathbb{R} $ be uniformly continuous. Prove that $$ \lim_{\epsilon \to 0} \int^{1-\epsilon}_{\epsilon}\!\!f(t)dt \in \mathbb{R}$$ Any ideas?? $f$ can be extended to a continuous function $\hat{f}$ in $[0,1]$, so an integrable one. My best guess is the integral will equal $\int^{1}_{0}\hat{f}(t)\,dt$. Am I correct?? A rigorous proof will be very much appreciated. Thanks in advance.

REPLY [1 votes]: Indeed, $f$ can be extended to $\hat f$ on $[0,1]$.  Let $\epsilon > 0$.  Then \begin{align} \bigg| \int_{0}^1 \hat fdx - \int_{\epsilon}^{1- \epsilon} \hat f dx\bigg| &= \bigg| \int_{0}^{\epsilon} \hat f dx + \int_{1-\epsilon}^{\epsilon} \hat f dx \bigg|
\\
&\leq 2\epsilon ||\hat f||_{\infty}\end{align}
where $||\hat f||_{\infty}$ is the supremum of $\hat f$ over $[0,1]$ and is finite by uniform continuity on a compact set.


TITLE: Sum of the areas of all n-sided polygons inscribed within each other
QUESTION [6 upvotes]: Imagine you take an $n$-sided regular polygon with side length $x$, and connect the midpoints of each of its sides to construct a smaller but identical $n$-sided regular polygon within it.
Now perform the same process with the smaller polygon, and then with the next one, and the next and so on.
If you continue this, endlessly inscribing smaller and smaller regular $n$-sided polygons, what is the sum of all of their individual areas in terms of $x$ and $n$?
Below is an image describing this question when n = 6:

I know there must be a common ratio between each of the areas, so i'm thinking about turning this into a geometric series of the form $\frac{a}{1-r}$ (since $r$ < 1), where $a$ = the area of the first polygon, however I do not know how to find the specific values for $a$ and $r$ in terms of $x$ and $n$.
Any help with this question would be greatly appreciated!

REPLY [2 votes]: Let the side length of the outermost polygon and side length of the first constructed polygon be $x_0$ and $x_1$ respectively. The external angles of the polygons will each be $\frac{2\pi}{n}$, so the internal angles will each be $\pi - \frac{2\pi}{n}$. Consider the triangle shown in the diagram below:

Applying the cosine law, we get
$$\begin{align}
(x_1)^2 &= \left( \frac{x_0}{2} \right)^2 + \left( \frac{x_0}{2} \right)^2 - 2\left( \frac{x_0}{2} \right)\left( \frac{x_0}{2} \right) \cos \left( \pi - \frac{2\pi}{n} \right) \\
\\
&= \frac{(x_0)^2}{2} + \frac{(x_0)^2}{2} \cos \left( \frac{2\pi}{n} \right) \\
\\
&= \frac{(x_0)^2}{2} \left[ 1 + \cos \left( \frac{2\pi}{n} \right) \right] \\
\\
&= \frac{(x_0)^2}{2} \left[ 2 \cos^2 \left( \frac{\pi}{n} \right) \right] \\
\\
&= (x_0)^2 \cos^2 \left( \frac{\pi}{n} \right) \\
\\
x_1 &= x_0 \cos\left( \frac{\pi}{n} \right)
\end{align}$$
Thus the common ratio of the sides is $\cos \left( \frac{\pi}{n} \right)$, so the common ratio of the areas will be $\boxed{\cos^2 \left( \frac{\pi}{n} \right)}$


TITLE: Almost sure convergence of eigenvectors under nice assumptions
QUESTION [2 upvotes]: I have that a sequence of symmetric, real, positive semi-definite random matrices, $M_n$, converges almost surely to a real-valued positive semi-definite diagonal matrix, $D$, with at least one diagonal entry equal to zero. How can I go about showing that the smallest eigenvector of $M_n$ converges almost surely to the smallest eigenvector of $D$, or that $v_i$ converges almost surely to $0$ if $D_{i,i}\neq 0$, where $v$ is the smallest eigenvector? Note that each $M_n$ is not necessarily diagonal.

REPLY [1 votes]: By smallest eigenvector of $M_n$, I assume that you consider an eigenvector associated to a smallest eigenvalue of $M_n$. Anyway, we consider only eigenvectors with norm $1$. Let $S_p$ be the set of symmetric real matrices of dimension $p$.
EDIT. Clearly you are not interested in my first answer; I rewrite my post and then, I leave the party. More I read your question, less it seems to me clear. If I correctly understand:
One has for a.e. $t\in \Omega$, $M_n(t)$ converges to $D(t)$.  Let $\lambda_n(t)$ be the smallest eigenvalue of $M_n(t)$; we assume that the $M_n[i,j]$ follow a probability law with a $C^{\infty}$ density function.
It is known that if $(n,t)\rightarrow M_n(t)$ is continuous in $(+\infty,0)$, then $(t,n)\rightarrow \lambda_n(t)$ too; thus $\lambda_n$ converges a.s. to $0$. Unfortunately a unit eigenvector associated to $\lambda_n(t)$ is not necessarily continuous and you have no results about its convergence.
Yet, if $D$ has a sole zero on the diagonal and if the $M_n[i,j]$ are $C^{\infty}$, then
the convergence of the previous eigenvector works.


TITLE: Express a formula in terms of trigonometric expressions
QUESTION [1 upvotes]: I am studying Kerr Black holes using Hobson's General relativity an introduction for physicists book.
In order to find circular radius for photons, two conditions need to be satisfied:
$$r_c=3\mu\frac{b-a}{b+a}$$ and
$$(b+a)^3=27\mu^2(b-a)$$
According to the book: The equations may be solved by setting y=a+b in the second condition and substituting the resulting value of b into the first. This is what I did and I obtained:
$$r_c=\frac{\mu\alpha^{1/3}(3\mu^2+3\alpha^{1/3}-2a)}{\mu^2+\alpha^{2/3}}$$
where for simplification I defined $\alpha=\sqrt{a^2\mu^4-\mu^6}-a\mu^2$.
However, the book further says that one can easily simplify $r_c$ as:
$$r_c=2\mu\Big(1+\cos\Big[\frac{2}{3}\cos^{-1}\Big(\pm\frac{a}{\mu}\Big)\Big]\Big)$$
and $$b=3\sqrt{\mu{r_c}}-a$$
I am stuck and don't know how is it possible to simplify my expression for $r_c$ into the elegant expression as given by the book. This seems to be a physics question but I am stuck with the algebraic manipulation.

REPLY [1 votes]: \begin{align}
r_c&=3\mu\frac{b-a}{b+a} \tag{1}\label{1}
\\
(b+a)^3&=27\mu^2(b-a) \tag{2}\label{2}
\end{align}
To get the expression for b from \eqref{1},
\begin{align}
b-a&=\frac{r_c}{3\mu}(b+a)
,
\end{align}
combined with \eqref{2},
\begin{align}
 (b+a)^3&=27\mu^2\frac{r_c}{3\mu}(b+a)
 ,\\
 (b+a)^2&=9\mu{r_c}
 ,\\
 b&=3\sqrt{\mu{r_c}}-a
 .
\end{align}
From \eqref{2}
\begin{align}
3\,\mu\,\frac{b-a}{b+a}
&=\frac{(b+a)^2}{9\,\mu}=r_c
\tag{3}\label{3}
.
\end{align}
Now we have two expressions for $r_c$.
One has a factor $\frac{\mu}{(b+a)}$,
the other has its reciprocal $\frac{(b+a)}{\mu}$,
and they both are begging to be canceled.
When we multiply them, we'll get a nice simplified
expression for $r_c^2$:
\begin{align}
r_c^2&=
\tfrac13\,(b+a)(b-a)
,\\
r_c^2&=\sqrt{\mu\,r_c}(3\,\sqrt{\mu\,r_c}-2\,a)
,\\
\left(\frac{r_c}{\mu}\right)^2
&=
\sqrt{\frac{r_c}{\mu}}
\left(
3\,\sqrt{\frac{r_c}{\mu}}-\frac{2\,a}{\mu}
\right)
,\\
\left(\sqrt{\frac{r_c}{\mu}}\right)^3
-
3\,\sqrt{\frac{r_c}{\mu}}
&=
-\frac{2\,a}{\mu}
,\\
4\,\left(\tfrac12\sqrt{\frac{r_c}{\mu}}\right)^3
-
3\,\left(\tfrac12\sqrt{\frac{r_c}{\mu}}\right)
&=
-\frac{a}{\mu}
\end{align}
Recall that
\begin{align}
4\,\cos^3 x-3\,\cos x=\cos3x.
\end{align}
So, we have
\begin{align}
\cos3x&=-\frac{a}\mu
;\\
3x&=\arccos\left(-\frac{a}\mu\right)+2\,\pi k,\quad k=0,1,2
;\\
x&=\tfrac13\arccos\left(-\frac{a}\mu\right)+\tfrac23\,\pi k,\quad k=0,1,2
;\\
\end{align}
Hence
\begin{align}
\cos x=
\tfrac12\sqrt{\frac{r_c}{\mu}}
&=
\cos\left(
\tfrac13\,\arccos\left( -\frac{a}{\mu} \right)
+\tfrac23\,\pi\,k
\right)
,\quad k=0,1,2
;\\
\tfrac12\frac{r_c}{\mu}
&=
2\,
\cos^2\left(
\tfrac13\,\arccos\left( -\frac{a}{\mu} \right)
+\tfrac23\,\pi\,k
\right)
,\quad k=0,1,2
;\\
r_c&=2\,\mu
\left(
1+\cos\left(
\tfrac23\,\arccos\left( -\frac{a}{\mu} \right)
+\tfrac43\,\pi\,k
\right)
\right)
,\quad k=0,1,2
.
\end{align}


TITLE: limit involving two variables with $\ln(x)$
QUESTION [2 upvotes]: $$ \lim_{(x,y) \to (1,0)} \frac{(x-1)^2\ln(x)}{(x-1)^2 + y^2}$$
I tried L' Hospitals Rule and got now where. Then I tried using $x = r\cos(\theta)$ and $y = r\sin(\theta)$, but no help. How would I approach this? T

REPLY [0 votes]: $$|\frac{(x-1)^2\ln(x)}{(x-1)^2 + y^2}| \leq |\frac{(x-1)^2 \ln (x)}{(x-1)^2}| \stackrel{x \neq 1}{=} |\ln x|$$
But $\ln x \to 0$ as $x \to 1$.
Hence, by squeeze theorem:
$$ \lim_{(x,y) \to (1,0)} \frac{(x-1)^2\ln(x)}{(x-1)^2 + y^2}=0$$


TITLE: Simultaneous small fractional parts of polynomials
QUESTION [7 upvotes]: Fix $\epsilon>0$. For a finite set of arbitrary-degree polynomials with integer constant term, $p_1(x), ..., p_m(x)\in \mathbb{R}[x]$ is it possible to find an $n\in \mathbb{N}$ such that $$\max_{i=1,...,m}||p_i(n)||<\epsilon$$ where $||\cdot||$ denotes the distance to the nearest integer?
In the description of the book Small Fractional Parts of Polynomials a problem of this type is listed as an open question for degree greater than 4, but the book is from 1977. Research has continued since then on small fractional parts of polynomials. Has progress improved on this problem, or is it still open?
Updates: Discussion in the comments indicates that this can be proved by Weyl's equidistribution criterion. While this proves the result for $\mathbb{Q}$-linearly independent collections $\{p_1(n),...,p_k(n), 1, n, n^2,...\}$ it does not prove it when the collection is linearly dependent, since equidistribution no longer holds in that case. How can we generalize to when the polynomials may be $\mathbb{Q}$-linearly dependent?

REPLY [2 votes]: The claim is true as stated, and can be arrived at using Weyl's criterion which was pointed out in the comments. As I post this answer, there is no consensus on the rate of convergence of the $N$.
We proceed by induction on $k$, the number of polynomials. For $k=1$, either $p_1$ has only rational coefficients or it has at least one irrational. If it has an irrational one, Weyl's criterion shows that $p_1(n)$ is an equidistributed sequence, and thus has infinitely many $N$ for which $||p_1(N)||<\epsilon$. If $p_1$ has only rational coefficients, we can simply choose $N$ to be the product of every coefficient's denominator to get $||p_1(N)||=0$.
Now suppose the theorem holds for some fixed $k$. Consider a polynomial list $p_0, p_1, ..., p_k$ of length $k+1$. If $\{p_0(x),p_1(x),...,p_k(x), 1, x, x^2,...\}$ is $\mathbb{Q}$-linearly independent, then Weyl's criterion on the $k+1$ torus applies and equidistribution gives the existence of infinitely many $N$ for which $||p_i(N)||<\epsilon$ for all $i=0,...,k$.
If, on the other hand, the aforementioned collection is instead $\mathbb{Q}$-linearly dependent, Weyl's criterion does not apply. However, $$p_0(x)=q_1p_1(x)+\cdots +q_kp_k(x)+r_0+r_1x+\cdots +r_sx^s$$ for $q_i,r_j\in \mathbb{Q}$. If we let $q$ be the product of all denominators of the $r_j$, we have that at any natural number $n$, $$||p_0(qn)||=||q_1p_1(qn)+\cdots +q_kp_k(qn)||.$$
Apply the inductive hypothesis to $q_1p_1(qx),...,q_kp_k(qx)$. Choose $M$ so that $||q_ip_i(qM)||<\epsilon/2k$ for all $i=1,...,k$. At this small scale, the $||\cdot||$ function is subadditive in $k$ arguments, so $$||p_0(qM)||\leq ||q_1p_1(qM)||+\cdots +||q_kp_k(qM)||<k(\epsilon/2k)<\epsilon.$$
Therefore, at $N=qM$, $||p_i(N)||<\epsilon$ for all $i=0,...,k$. We are finished by induction.


TITLE: Generalizing an euclidean metric problem to singular metrics
QUESTION [0 upvotes]: I have the following problem:
Let $M$ be a symmetric positive semidefinite real matrix of order $n$ and $x,x',y,y' \in \mathbb{R}^n$ such that $(x-x')^TM(x-x') = 0 = (y - y')^TM(y-y')$. Then, $x^TMy = (x')^TM(y')$.
I tried this to solving the problem. We can write
\begin{align}
x^TMy - (x')^TM(y') &= x^TMy - (x')^TMy + (x')^TMy - (x')^TM(y') \\
&= (x-x')^TMy + (x')^TM(y-y'),
\end{align}
so the problem reduces to prove that the last equality is zero. When $M$ is strictly positive definite, then the expresions of the form $u^TMv$ are dot products, and using the Cauchy-Schwarz inequality is easy to see that $(x-x')^TMy = 0 = (x')^TM(y-y')$. But when $M$ is singular I'm not sure if the inequality holds. In fact, the usual proof starts using that $u = 0 \iff \|u\| = 0$, which is not true for the seminorms induced by PSD matrices, so can I do it for the singular case?

REPLY [1 votes]: I think you're going in the right direction, the missing part is to show that a positive semidefinite bilinear form has the Cauchy-Schwarz property. This can be done separating the cases $y^TMy = 0$ and $y^TMy \neq 0$, while letting $x \in \mathbb{R}^n$. http://kaba.hilvi.org/homepage/blog/semi_norms.htm.
The case $y^TMy \neq 0:$
Let $x_{\parallel}=\frac{x^TMy}{y^TMy}y$ and $x_{\perp}=x-x_{\parallel}$. We have
\begin{align}
(x^TMy)^2 &= (x^TMy)^2\frac{(y^TMy)^2}{(y^TMy)^2} = \frac{x^TMy}{y^TMy}y^TMy\frac{x^TMy}{y^TMy}y^TMy = (x_{\parallel}^TMx_{\parallel})(y^TMy),
\end{align}
so if we can show that $x_{\parallel}^TMx_{\parallel} \leq x^TMx$ then $(x^TMy)^2 \leq (x^TMx)(y^TMy)$ i.e. the Cauchy-Schwarz property follows and thus
\begin{align}
x^TMy - (x')^TM(y') &= x^TMy - (x')^TMy + (x')^TMy - (x')^TM(y') \\
&= (x-x')^TMy + (x')^TM(y-y') \\ &\leq (((x-x')^TM(x-x'))(y^TMy))^\frac{1}{2} + (((x')^TMx')((y-y')^TM(y-y')))^\frac{1}{2} \\ &=0,
\end{align}
and we are done. It remains to show $x_{\parallel}^TMx_{\parallel} \leq x^TMx$, using that $x_{\parallel} \cdot x_{\perp} = 0$ and that $M$ is positive semidefinite so $x_{\perp}^TMx_{\perp} \geq 0:$
\begin{align}
x_{\parallel}^TMx_{\parallel} \leq x_{\parallel}^TMx_{\parallel} + x_{\perp}^TMx_{\perp} = x_{\parallel}^TMx_{\parallel} + 2x_{\parallel}^TMx_{\perp} + x_{\perp}^TMx_{\perp} =(x_{\parallel}+x_{\perp})^TM(x_{\parallel}+x_{\perp})=x^TMx
\end{align}


TITLE: Motion on a slope - A Level
QUESTION [1 upvotes]: A particle P of mass 5kg and particle Q of mass 1kg are connected by a light inextensible string. P lies on a slope inclined at a 60 degree angle to the horizontal. The string passes from P parallel to the line of greatest slope, and runs over a light pulley that can freely rotate. Then descends vertically to Q. The coefficient of friction between P and the slope is 0.7.
Find the force of friction acting on P and it’s direction and find the acceleration of P when the system is released from rest.
I was able to find the frictional force 17.15N up the slope) but I can’t find the acceleration. I have tried to use the equation 5gsin(60)-17.15-9.8 = 5a with 9.8 being the weight of Q, but I get the answer of 3.1m/s/s for a, bit the mark scheme says that a (acceleration) is 2.58m/s/s but I don’t see how. Could you please give me some advice on how to get 2.58?

REPLY [1 votes]: You have done all correct, you just messed up in the calculation. The equation
$$5g\sin60° -17.15-9.8=6a$$
has the solution $\boxed{a=2.58\,\text{m}\,\text{s}^{-2}}$ as required.
Hope this helps. Ask anything if not clear :)