Let's consider such polynomial: $$ W_n(z)=\sum_{k=1}^n kz^{k-1} $$ Tell whether this statement is true for any $n \in \mathbb{N}$: $$W_n(z)=0 \Rightarrow |z|&lt;1$$  Here is what I have evaluated:   It is obvious that $z$ can't be a real number bigger or equal to $0$.  Also, if $n$ is even, then $z$ can't be a real number smaller than $-1$, because if it was, we would have: $$W_n(z)=\sum_{k=1}^n kz^{k-1}=\sum_{k=1}^{\frac{n}{2}}\Big((2k+1)z^{2k}+2kz^{2k-1}\Big)=\sum_{k=1}^{\frac{n}{2}}z^{2k-1}\Big(2kz+2k+z\Big)$$ Both $z^{2k-1}$ and $2kz+2k+z$ are smaller than $-1$ so each term of the last sum is bigger than $1$ so the final sum is bigger than $0$.
122769
This is a homework problem that I wasn't able to solve, and I feel a little silly about it (the assignment has been submitted). The problem:     Let $P(z) = 1 + 2z + 3z^2 + \cdots + nz^{n-1}$. By considering $(1-z)P(z)$, show that zeros of $P(z)$ are contained inside the unit disc.   This is from Chapter 1 of Bak's Complex Analysis. A solution should only use algebraic manipulations of complex numbers (in other words, this problem is very easy to solve by employing the Gauss-Lucas theorem, but that's not "on the table" -- complex differentiation hasn't been defined in the scope of what's permissible).  What I've done:  $$\begin{align*} (1-z)P(z) &amp;= 1+z+z^2+\cdots+z^{n-1}-nz^n \\  &amp;= \frac{z^n-1}{z-1}-nz^n. \end{align*}$$ Let $z$ be a zero of $P(z)$; then it is a zero of $(1-z)P(z)$. Consequently, $$\begin{align*} nz^n &amp;= \frac{z^n-1}{z-1} \\ nz^n(z-1) &amp;= z^n-1 \\ nz^n(z-1)+1 &amp;= z^n \end{align*} $$  Then $$|z^n| = |z|^n = |nz^n(z-1)+1|.$$ From the triangle inequality, $$ \begin{align*} |z|^n &amp;\le |nz^n(z-1)| + |1| \\ |z|^n &amp;\le n|z^n||z-1| + |1| \\ n|z|^n(|z-1|) &amp;\le 1 \\ |z|^n &amp;\le \frac{1}{n|z-1|} \end{align*} $$  At this point, I am stuck. I feel like I went down the wrong path, or maybe I just made an arithmetical error, but I can't find where.
60278
We know that for every polynomial of $n$.order Fundamental Theorem of Algebra guarantees $n$ complex roots.  Lets consider the complex exponential function $f(z)=\exp(z)$. As $f(z)$ is holomorphic, we are allowed to evaluate the power sum of $f(z)$.  $$f(z)=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$   We also know that   $$f(z)=\exp(z)=\exp(x+iy)=\exp(x)\exp(iy)=\exp(x)\cdot(\cos(y)+i\sin(y))$$  A zero of this expression would only be possible if $\exp(x)$ or $\cos(y)+i\sin(y)$ or both would vanish. But $\exp(x)&gt;0$ and $|\cos(y)+i\sin(y)|=1$, hence $\exp(z)$ doesn't vanish. As the exponential is fully defined by the powersum with infinite radius of convergence, we can conclude that the powersum, despite of beeing of "polynomial like type" does not have any zeros.  I would be glad if someone could give me an intuitive explanation why the fundamental theorem of algebra is only true for finite polynomials.
52254
Let $f$ be an entire function and let for each $a\in \mathbb R$, there exists at least one coefficient $c_n$ in $f(z)=\sum\limits_{n=0}^{\infty}c_n(z-a)^n$, which is zero. Then    $f^{(n)}(0)=0$ for infinitely many $n\geq 0$ $f^{(n)}(0)=0$ for every $n\geq 0$ $f^{(2n+1)}(0)=0$ for every $n\geq 0$ There exists $k\geq 0$ such that $f^{(n)}(0)=0$ for all $n\geq k$   We know that $c_n=\frac{f^{(n)}(a)}{n!}$ for all $n\in \{0,1,2\ldots\}$. Thus for $a=0$, $c_n=\frac{f^{(n)}(0)}{n!}$. By hypothersis, atleast one $c_n=0$. After that I could not do anything. Please help!
121582
Fix a nonnegative integer $n$. Let $P(x)$ be a polynomial of degree $n$ (over the real numbers) such that for all $k\in\left\{0,1,\ldots,n\right\}$, we have $P(k)=\dfrac{k}{k+1}$. Find $P(n+1)$.   My attempt:  I consider a new polynomial $Q(x)=(x+1)P(x)-x$. This $Q$ is a polynomial of degree $\leq n+1$ satisfying: $Q(k)=0\quad\forall k\in\{0,1,2,\ldots,n\}$. Therefore $Q(x)=\lambda x(x-1)\cdots(x-n)$ for a certain real $\lambda$. Hence, $Q(-1)=\lambda \cdot (-1)^{n+1}(n+1)!$, but also $Q(-1)=(-1+1)P(-1)+1=1$. Comparing these, we get $\lambda \cdot (-1)^{n+1} (n+1)! = 1$, thus $\lambda=\dfrac{(-1)^{n+1}}{(n+1)!}$. Since $Q(n+1)=\lambda(n+1)!$ then $Q(n+1)=(-1)^{n+1} $, and from $P(x) = \dfrac{Q(x)+x}{x+1}$ we deduce: $$P(n+1)=\dfrac{(-1)^{n+1}+n+1}{n+2}.$$ I doubted that the answer were $P(n+1)=\frac{n+1}{n+2}$ but the term $(-1)^{n+1}$ wasn’t expected. Is there any mistake here? Does someone have an alternative proof?
90968
Must show:     $\sum_{i=0}^n i{n \choose i} = n2^{n-1}$ using a combinatorial argument. I have tried picking the problem apart by determining that the summation is equal to 1/2 of the size of all subsets of a set of size n times n.  But I am having trouble relating that to what each term of the summation represents.
97681
I'm trying to prove that  $$\sum_{k=0}^n {n \choose k} (-1)^k \frac{1}{k+1} = \frac{1}{n+1}$$  So far I've tried induction (which doesn't really work at all), using well known facts such as  $$\sum_{k=0}^n {n \choose k} (-1)^k = 0$$  and trying to apply identities like  $${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$$  Would anyone be able to point me towards the right method? Should I be looking to apply an identity or is there a method I'm missing?
103852
Let $\{f_n(x)\}_{n\in \Bbb N}$ be a sequence of polynomials defined inductively as   $$\begin{matrix} f_1(x) &amp; = &amp; (x-2)^2 &amp; \\ f_{n+1}(x)&amp; = &amp; (f_n(x)-2)^2, &amp;\text{ for all }n\ge 1.\end{matrix}$$      Let $a_n$ and $b_n$ respectively denote the constant term and the coefficient of $x$ in $f_n(x)$. Then         $(\text A)\,\, a_n=4, b_n=-4^n$   $(\text B)\,\,a_n=4, b_n=-4n^2$   $(\text C)\,\, a_n=4^{(n-1)!}, b_n=-4^n$   $(\text C)\,\, a_n=4^{(n-1)!}, b_n=-4n^2$      Please help me how to solve this problem. I am not able to solve it.  The problem can be found on this .
88033
I'm going to university in October and thought I'd have a go at a few questions from one of their past papers. I have completed the majority of this question but I'm stuck on the very last part. In honesty I've been working on this paper a while now and I'm a bit tired so I'm probably giving up earlier than I usually would.  I won't write out the full question, only the last part:     Let $$S_r(n) = \sum_{m=0}^n\ (-1)^m m^r {n \choose m}$$ where r is a   non-negative integer . Show that $S_r(n)=0$ for $r&lt;n$. Evaluate   $S_n(n)$.   I have shown that $S_r(n)=0$ for $r&lt;n$ by taking $(1+z)^n= \sum_{m=0}^n\ z^m {n \choose m}$, letting $D_r(f(z))=z\frac d{dz}(z\frac d{dz}...(\frac d{dz}(f(z)))...)$ where $z\frac d{dz}$ is applied $r$ times and applied it to both sides.  The left hand side gives a polynomial, degree n which has factor $(1+z)$ for all $r&lt;n$ and the right hand side gives $\sum_{m=0}^n\ z^m (m)^r {n \choose m}$. Setting $z=-1$ yields the required result.   There is some build up to this, so I'm fairly certain that this was the intended method.  I'm stuck however on the very last part. I have tried finding a form for $D_r((1+z)^n)$, but I'm fairly sure that this isn't the correct approach, as the wording of the question implies that $S_n(n)$ needs to be considered separately.  I'm surprised that I didn't find that this question had already been asked, so apologies if it has been.  Thank you.
27799
Can someone tell me how to get a closed form for  $$\sum_{k=1}^n k^p$$  For $p = 1$, it's just the classic $\frac{n(n+1)}2$.   What is it for $p &gt; 1$?
92609
Usually, the Stirling numbers of the first kind are defined as the coefficients of the rising factorial: $(*) \prod_{i=0}^{n-1}(x+i) = \sum_{i=0}^{n} S(n,i) x^i$.   With this definition, a recursive relation of $S(n,i)$ be derived, and it can be shown that is coincides with the recursive relation on the number of permutation on an n-set with i cycles and they have the same initial conditions, hence they coincide.   1) Is there any possibility to do it the other way around, i.e., define $S(n,i)$ combinatorially and then to show that $\prod_{i=0}^{n-1}(x+i) = \sum_{i=0}^{n} S(n,i) x^i$ holds for $x \in \mathbb{N}$ by some combinatorial argument, and thus it is a polynomial identity?  (For Stirling numbers of the second kind, it is possible: it can be shown that $n^k = \sum_{i=0}^{k} \binom{n}{i} i! S_2(k,i)$ combinatorially ($n^k$ counts function from $[k]$ to $[n]$).)  2) Additionally: equating coefficients in $(*)$ shows that $S(n,i)$ is the elementary symmetric polynomial on $n$ variables of degree $n-i$ evaluated on $(0,1,\cdots ,n-1)$. Is there a combinatorial interpretation of this?
33723
Question:  $$\sum_{k=1}^{n}k\binom{2n}{n+k}=\frac n2\binom{2n}{n}$$ is true for every $n\in \mathbb N$? If this is true, then how can we prove this?   When I was playing with numbers, I conjectured this. Though I found that this holds for many $n$s, I haven't found any proof. Can anyone help?
56917
For a polynomial $P(x)$ of degree $n$, $P(k) = 2^k$ for $k = 0, 1, 2, . . . , n$. Find $P(n+1)$.   If $n=1$, $P(x)=x+1$ and $P(2)=3$.  If $n=2$, $P(x)=0.5x^2+0.5x+1$ and $P(3)=7$.  How to approach further cases? I am stuck.
117373
Any hint to prove this? I tried with properties of binomial coefficient but I can't get anything. $$\sum_{n=k}^{\infty}{{n}\choose{k}}x^n=\dfrac{x^k}{(1-x)^{k+1}}$$
10198
Prove if $|z| &lt; 1$ and  $ |w| &lt; 1$, then $|1-zw^*| \neq 0$ and $| {{z-w} \over {1-zw^*}}| &lt; 1$Given that $|1-zw^*|^2 - |z-w|^2 = (1-|z|^2)(1-|w|^2)$I think the first part can be proven by saying $|1-zw^*| = 0$ if and only if $zw^*$ = 1.  And given the conditions that cannot be true. However I don't know if this part is right.
14225
Let $\ell$ be an even positive integer.  Express $$\sum_{k=0}^n\sum_{i=0}^\ell (-1)^i\binom{n}{k}^2\binom{2k}{i}\binom{2n-2k}{\ell-i}$$in closed form.   Is anybody to give a insight, please? [A Path to Combinatorics for Undergraduates] p.192 problem 8.9
55711
Why is $\displaystyle\sum\limits_{k=0}^{n}(-1)^k\binom{n}{k}^2=(-1)^{n/2}\binom{n}{n/2}$ if $n$ is even ?   The case if $n$ is odd, is clear, since $\displaystyle(-1)^k\binom{n}{k}+(-1)^{n-k}\binom{n}{n-k}=0$ (we have $(n+1)/2$ such pairs.)  but if $n$ is even we have no symmetry, I tried to consider the odd and even terms separately but with no success.   Do you have an idea ? Thanks in advance.
6711
Let $f(z)$ be an entire function s.t $|f(z)|\leq M|z|$ ; $M&gt;0$ is a constant then $f(z)$ Is polynomial of degree 1   My try $f(z)$ is entire so it can be written as a $\sum a_n z^n$  $$|\sum a_n z^n|\leq \sum |a_n||z|^n\leq M|z|$$  So $n=1$  Seems that the move $\sum |a_n||z|^n\leq M|z|$ is problematic