0
votes
1answer
22 views

Use Leibniz' formula to show that the $(2n)$th derivative of $(2x^2 + 3x +1)sinx$ is $(-1)^n(2x^2+3x-8n^2+4n+1)sinx+(-1)^{n+1}(8nx+6n)cosx$ wrt $x$

If I let $f=f(x)=sinx$ and $g=g(x)=2x^2+3x+1$ and $D=$ First derivative wrt $x$, $D^2=$ Second derivative wrt $x$ and $D^n=$ $nth$ derivative wrt $x$ then, Leibniz' formula states that $\displaystyle ...
1
vote
2answers
106 views

Proof that $\dfrac{1}{e^x}=e^{-x}$ without converting it to $e^{x}e^{-x}=1$.

I want to show that $\dfrac{1}{e^x} = e^{-x}$ from the Taylor expansion of $e^x$. To express $\dfrac{1}{e^x}$ as a power series, I let: $$ \left(\dfrac{1}{0!}x^0 + \dfrac{1}{1!}x^1 + ...
6
votes
2answers
189 views

Prove $(1-x)^{2k+1} \sum\limits_{n\ge 0}\binom{n+k-1}{k}\binom{n+k}{k} x^n = {\sum\limits_{j\ge 0} \binom{k-1}{j-1}\binom{k+1}{j} x^j} $

I stumbled upon the identity $$(1-x)^{2k+1} \sum\limits_{n\ge 0}\binom{n+k-1}{k}\binom{n+k}{k} x^n = {\sum\limits_{j\ge 0} \binom{k-1}{j-1}\binom{k+1}{j} x^j}. $$ The right-hand side is a polynomial. ...
12
votes
1answer
255 views

Prove that $\exp(\log(\frac{1}{1-x})) = \frac{1}{1-x}$

I am trying to prove this directly by comparing the coefficients in the two series rather than using formal calculus. Here is what I have so far, but I think I made a mistake: \begin{align*} ...
2
votes
2answers
86 views

Sum of products of binomial coefficient $-1/2 \choose x$

I am having trouble with showing that $$\sum_{m=0}^n (-1)^n {-1/2 \choose m} {-1/2 \choose n-m}=1$$ I know that this relation can be shown by comparing the coefficients of $x^2$ in the power series ...
4
votes
2answers
115 views

Series of inverses of binomial coefficients

Can you think of a simple way of proving that $$ \sum_{n=k+1}^\infty \frac{1}{n \choose k} $$ is rational for any $k \geq 2$? Here's the background. Consider a series: $$ \sum_{n=1}^\infty ...
2
votes
2answers
572 views

Power series for the Bessel function using Laplace transforms?

The Bessel's function of the first kind of order zero, $J_0$ is the solution to $$ty''+y'+ty=0$$ which satisfies $J_0(0)=1$ The Laplace transform of this equation gives ...
1
vote
1answer
66 views

Simplification of Binomial Expansion.

How $$(x+h)^n-x^n=nhx^{(n-1)}\text{ ?}$$ My attempt : $$ \begin{align} (x+h)^n-x^n & =nhx^{(n-1)} \\[8pt] & =\left[\sum_{k=0}^{n}\binom{n}{k}x^{(n-k)}h^k\right]-x^n \\[8pt] & = ...
1
vote
1answer
116 views

Sum of power series with inverse coefficients

Are there any relations between the sums of $\displaystyle\sum_{n=0}^{\infty} a_n x^n$ and $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{a_n}$? For example, I know ...
1
vote
0answers
90 views

$\sum_{k=0}^{\infty}\frac1{4^k(2k+1)}\binom{2k}{k}x^{2k+1}\sum_{k=0}^{\infty}(-1)^k\frac1{4^k}\binom{2k}{k}x^k =$

Prove that for $|x|<1$, $\sum_{k=0}^{\infty}\frac1{4^k(2k+1)}\binom{2k}{k}x^{2k+1}\sum_{k=0}^{\infty}(-1)^k\frac1{4^k}\binom{2k}{k}(-x^2)^k = ...
1
vote
2answers
59 views

What is the function given by $\sum_{n=0}^\infty \binom{b+2n}{b+n} x^n$, where $b\ge 0$, $|x| <1$

For a nonnegative integer $b$, and $|x|<1$, what is the function given by the power series $$ \sum_{n=0}^\infty \binom{b+2n}{b+n} x^n. $$ For $b=0$, this post shows $$ \sum_{n=0}^\infty ...
4
votes
2answers
239 views

Finding a closed form expression for this sum [duplicate]

For non-negative $n$, find $$ \sum_{k=0}^n \binom{2k}{k}\binom{2n-2k}{n-k}. $$ I can't figure this out. Any ideas?
0
votes
1answer
202 views

Using the general binomial theorem to find a series-like expression for $\sqrt 2$

How do I use the general binomial theorem (i.e. the series expansion of ${(1+x)^\alpha}$ for $ |x|<1$) to show the following? $$\sqrt 2=1+\frac 1{2^2}+\frac{1\cdot3}{2!\cdot{2^4}} ...
1
vote
2answers
51 views

Integer sequence comparison, binomials Vs power

I need to know which sequence grows faster with n: $$ f(n) = \sum_0^{floor(n/3)} {n \choose 3*i+1} $$ compared to $$ g(n) = 2^n -1 $$ it seems f(10)>5000 is greater than g(10)=1023 but I would ...
3
votes
1answer
297 views

A sum involving permutation

Does there exist a nice closed form formula for the sum $$\sum_{k=0}^m P(m,k)x^k$$ where $P(m,k)=C(m,k)*k!$, $C(m,k)$ being the "m choose k" number. Formula given by Maple 11 is complicated. I ...
5
votes
2answers
921 views

Show $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$

How do you prove that $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$? I tried to identify the sum as a binomial series, but the $4$ and the $-1/2$ puzzle me. (This series arises in ...