Apostol   5.9.35   Given $P_0(x) = 1,\; P_n'(x) = nP_{n-1}(x),\; \int_0^1{P_n(x) \ dx} = 0 \enspace \forall \:n\geq 1$  (f)Prove that   $$ P_n(1-x) = {(-1)}^n P_n(x)$$  if $n\geq 1$
80337
Consider the Bernoulli polynomials $B_n(x)$ given by the expansion  $$\frac{te^{xt}}{e^t-1} = \sum\limits_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}.$$  I want to prove the identity $$B_n(1-x)=(-1)^nB_n(x).$$  This formula is given in a lot of books, but unfortunately without any proof. Therefore I'm wondering how to prove it. Do you have any idea?  I tried for instance:  $\sum\limits_{n=0}^{\infty}B_n(1-x)\frac{t^n}{n!}=\frac{te^{-xt}}{e^t-1}e^t=\left(\sum\limits_{n=0}^{\infty}B_n(-x)\frac{t^n}{n!}\right)\sum\limits_{n=0}^{\infty}\frac{t^n}{n!}=\sum\limits_{n=0}^{\infty}\left(\sum\limits_{l=0}^n\binom{n}{l}B_{l}(-x)  \right)\frac{t^n}{n!}$.  Hence  $B_n(1-x)=\sum\limits_{l=0}^n\binom{n}{l}B_{l}(-x)$.  But I do not know how to proceed.  On the other hand I know that $B_n(x+1)-B_n(x)=n x^{n-1}$. Hence  $B_{n}(1-x)=B_{n}(-x)-n(-x)^{n-1}$. But again, I do not know how I can go further.   Hopefully someone can help me.  Best regards
88366
Here $n$ belongs to natural numbers. Firstly, I proved the relation by putting $n = 1$ . Then, taking $$|\sin(mx)| \leq m|\sin x|$$ true, I had to prove $$|\sin(m + 1)x| \leq (m + 1)|\sin x|$$  Now, here I got stuck. How to prove it?? Please help.
49133
Show that: $$P(N(x)&gt;n)= \frac{x^n}{n!}$$ I was thinking that we can take $X_1+...+X_n$ to be a single random random variable $S_n$ and that the $n$ in the $P(N(x)&gt;n)$ is not the same as the $n$ in the definition of $N(x$). But unsure how to actually start proceeding to prove this. Any help appreciated, thank you.
116500
$\textbf{QUESTION-}$ Let $P$ be a p-group with $|P:Z(P)|\leq p^n$. Show that $|P'| \leq p^{n(n-1)/2}$.  If $P=Z(P)$ it is true. Now let $n &gt; 1$, then  If I see $P$ as a nilpotent group and construct its upper central series, it will end , so let it be,  $e=Z_0&lt;Z_1&lt;Z_2&lt;......&lt;Z_r=P$   Now as $Z_{i+1}/Z_i=Z(P/Z_i)$, so if if I take some $x\in Z_2$\ $Z_1$ then $N$={$[x,y]|y\in P$} $\leq Z_1(P)$ and $N \triangleleft P $, so $P/N$ is a group with order $\leq p^{n-1}$.  Now if I let $H=P/N$ then obviously |$H/Z(H)$|$\leq p^{n-1}$.  Now $H'\cong P'N/N \cong P'/(P' \cap N)$ so from here I could finally bring $P'$ atleast into the picture, now |$P'$|=$|H'|.|P'\cap N|$ so $|P'|\leq |H'||N|$.   This is where I am $\textbf{STUCK}$  Now , from here how can I calculate or find some power $p$ bounds on $|H'|$ and $|N|$ so i could get my result.
74620
Okay, maybe I'm missing something obvious but consider $h_n(x) = x^{ 1+ \frac{1}{2n-1}}$ on the set $[-1,1]$. Now consider a fixed $x \in [-1,1]$, we have:  $$\lim_{n\rightarrow \infty} h_n(x) = x \lim_{n \rightarrow \infty} x^\frac{1}{2n-1} = x$$  But my book says it's $|x|$? Why?
60543
I'm not entirely sure if I'm going about proving $n^2+n$ is even for all the natural numbers correctly.  $P(n): = n^2+n$     $P(1) = 1^2+1 = 2 = 0$ (mod $2$), true for $P(1)$   Inductive step for $P(n+1)$:     $\begin{align}P(n+1) &amp;=&amp; (n+1)^2+(n+1)\\ &amp;=&amp;n^2+2n+1+n+1\\ &amp;=&amp;n^2+n+2(n+1)\end{align}$   Does this prove $n^2+n$ is even as it's divisible by $2$? Thanks!
11017
Let $p(z)=z^{n}+a_{n-1}z^{n-1}+...+a_{0}$ be a polynomial of degree $n\geq1$.  How can you prove that either $p(z)=z^{n}$ or there exists $z'$ with $|z'|=1$ such that $|p(z')|&gt;1$.  Maybe we can use the maximum modulus principle and consider $q(z)=z^{n}p(1/z)$.
116589
Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\dots +a_0$ be a real polynomial such that $a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots + \frac{a_n}{n+1}=0$.  Show that $P$ has a root in $[0,1]$. I would like proof verification and alternative proofs.  My solution:      notice that the antiderivative of $P(x)$ is $\frac{a_nx^{n+1}}{n+1}+\frac{a_{n-1}x^n}{n}+\dots + \frac{a_1x^2}{2}+a_0x+C$. From here $\int_{0}^1P(x)dx=a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots + \frac{a_n}{n+1}=0$. If $P$ had no roots then $P$ would be a strictly positive(or negative) continuous function, so its integral would be strictly positive ( or negative), a contradiction.
33610
Let $P(x)$ be a nonconstant polynomial whose coefficients are positive integers. If $P(n)$ divides $P(P(n) - 2015 )$ for every natural number n, then prove that $$P (-2015) = 0$$  I've observed that the proof depends on  $ P(n)$ divides $ P(P(n)-2015)-P(-2015)\,$ but I cannot see how to prove that. Could someone please elaborate on how to proceed.
75589
Given a positive integer $n$. How to prove that $\sum\limits_{k=0}^{n}(-1)^k \dbinom{n}{k} = 0 $ ?   I've tried using mathematical induction, then:  $$p(1)=\sum_{k=0}^{1}(-1)^k \binom {1}{k}=0 $$  And my induction hypothesis is: $$p(n)=\sum_{k=0}^{n}(-1)^k \binom {n}{k}=0 $$ So, i need prove: $$p(n+1)=\sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k}=0 $$ Let $\sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k}$, $$\begin{align} \sum_{k=0}^{n+1}(-1)^{k} \binom {n+1}{k} &amp;= \sum_{k=0}^{n}(-1)^{k} \binom {n}{k}+(-1)^{k} \binom {n+1}{n+1} \\ &amp;=0 + (-1)^{k} \end{align}$$ What am i doing wrong?
67032
How to prove that $p-mq \mid f(m)$ where $m$ is any integer, $$f(x) = A_0 + A_1 x + A_2 x^2 + ... + A_{n-1} x^{n-1} + A_n x^n$$ $$f(x)∈ ℤ[x]$$ $p/q$ is a zero for $f(x)$ and $p$ and $q$ are coprime $⇒ \, ∃u,w ∈ ℤ \mid 1 = up + wq$.
73938
I've got following problem. Let $z$ be a complex number. Let $P(z)$ be a polynomial of degree $n$. Suppose that $|P(z)| \le M$ for $|z| \le 1$. Prove that for $|z| \ge 1$ we have. $|P(z)| \le M \cdot |z|^n$. Can somebody help?
7994
Q. Let $P(x)$ be a non-constant polynomial whose coefficients are   positive integers. If $P(n)$ divides $P(P(n)-2015)$ for every natural   number $n$, prove that $P(-2015)=0$.   In one of the sources, the solution given is as follows: Note that $P(n)-2015-(-2015)=P(n)$ divides $P(P(n)-2015)-P(-2015)$ for every positive integer $n$. But $P(n)$ divides $P(P(n)-2015)$ for every positive integer n.   Therefore, $P(n)$ divides $P(-2015)$ for every positive integer $n$. Hence $P(-2015)=0$.  I am not able to understand that how $P(n)-2015-(-2015)=P(n)$ divides $P(P(n)-2015)-P(-2015)$. Please help me out.
63191
Prove the following in $S_n$:  Let $\alpha = (a_1,a_2,\ldots,a_s)$ be a cycle and let $\pi$ be a permutation in $S_n$. Then $\pi\alpha\pi^{-1}$ is the cycle $(\pi(a_1),\ldots,\pi(a_s)).$     I'm not sure where to even start.
93565
Consider the Fibonacci numbers $F(0) = 0; F(1)=1; F(n) = F(n-1) + F(n-2)$.  Prove by induction that for all $n&gt;0$,  $$F(n-1)\cdot F(n+1)- F(n)^2 = (-1)^n$$  I assume $P(n)$ is true and try to show $P(n+1)$ is true, but I got stuck with the algebra. How do we reach $P(n+1)$ from $P(n)$? Also, strong induction may be used instead.
101987
let $x=(x(1),x(2),\ldots,x(n),\ldots)\in c_0$  So for any $\varepsilon&gt;0$ there exists a $n_0\in\mathbb{N}$ such that $|x(n)|&lt;\varepsilon_2$ for all $n\geq n_0$.  Now for all $n\geq n_0$ ,$||x_n−x||_\infty=\sup{|x(m)|}_{m\geq n_0}\leq \varepsilon_2&lt;\varepsilon$  i have a doubt that regarding the last line. What makes it possible?
114720
I got stuck on this inequality for a day. If $p$ is positive integer, then the problem becomes too easy, but I can't find how we deal with the general case when p can be any positive number. Can someone give me a hint? I really appreciate     Prove that $|x^p - y^p| \le p|x-y|(x^{p-1} + y^{p-1})$ provided that $1 \le p \lt \infty$ and $x, y \ge 0$