10
$\begingroup$

In class, I defined the binomial coefficient using an integral:

$$\binom{n}{k} = \displaystyle \int_0^{2\pi}\dfrac{dt}{2\pi} e^{-ikt}(1+e^{it})^n.$$

I succeeded in demonstrating many standard properties of the binomial coefficient directly using integration: Pascal's identity, Vandermonde identity, Hockey stick. But I could not show that $$\sum_{k=0}^n \binom{n}{k}=2^n.$$

It turns out I have to show the following:

$$\int_{0}^{2\pi}\dfrac{dt}{2\pi}\dfrac{\sin\left(\dfrac{(n+1)t}{2}\right)}{\sin\dfrac{t}2}\cos^n\dfrac{t}{2}=1$$

I do not know how to perform this integration! I need help. It is better if the solution did not involve contour integration.

$\endgroup$
2
$\begingroup$

Hint

Notice that inside the integral we have $$\frac {\sin\left(\frac {(n+1)t}{2}\right)}{\sin\frac t2}$$ which is equal to $D_{\frac n2}$ where $D_n$ denotes the Dirichlet kernel.

Hence using it's properties we have $$\frac {\sin\left(\frac {(n+1)t}{2}\right)}{\sin\frac t2}=D_{\frac n2}=1+2\sum_{r=1}^{\frac n2} \cos (rt) $$

Hope you can continue further

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thanks for the tip. But this was an intermediate step in the proof. Your idea reduces the computation to the integral $\displaystyle \int_0^{2\pi} \cos(rt) \cos^n \frac{t}2 dt/2\pi$, and I didn't see how to do this then. I will try it again. $\endgroup$ – Isomorphism May 21 '18 at 5:21
  • $\begingroup$ @Isomorphism You might have to take cases of whether 'n' is even or odd. For n as odd will be a very simple case and will be solved pretty easily. But the case of n as even will be a bit tedious one $\endgroup$ – Rohan Shinde May 21 '18 at 6:06
-1
$\begingroup$

There is a mistake in my presentation that eludes me, but the essence is the following.

Using $$\sum_{k=1}^{\infty} p^{k} \, \sin(k \, x) = \frac{p \, \sin(x)}{1 - 2 \, p \, \cos(x) + p^{2}}$$ then let $S_{n}$ be the desired summation to be evaluated to obtain: \begin{align} S_{n} &= \sum_{k=0}^{n} \binom{n}{k} \\ &= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n} \, \frac{\sin\left(\frac{(n+1) \, t}{2} \right)}{\sin\left(\frac{t}{2}\right)} \, dt \\ &= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n+1} \, \frac{\sin\left(\frac{(n+1) \, t}{2} \right)}{\sin(t)} \, dt \end{align} Now, \begin{align} \sum_{n=0}^{\infty} S_{n} &= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left[ \sum_{n=0}^{\infty} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n+1} \, \sin\left(\frac{(n+1) \, t}{2} \right)\right] \, \frac{dt}{\sin(t)} \\ &= \frac{1}{2 \pi} \, \int_{0}^{2\pi} \left[ \sum_{n=1}^{\infty} \left(2 \, \cos\left(\frac{t}{2}\right)\right)^{n} \, \sin\left(\frac{n \, t}{2} \right)\right] \, \frac{dt}{\sin(t)} \\ &= \frac{1}{2 \pi} \, \int_{0}^{2 \pi} \frac{2 \cos\left(\frac{t}{2}\right) \sin\left(\frac{t}{2}\right)}{1 - 4 \cos^{2}\left(\frac{t}{2}\right) + 4 \cos^{2}\left(\frac{t}{2}\right)} \, \frac{dt}{\sin(t)} \\ &= \frac{1}{2\pi} \, \int_{0}^{2 \pi} dt = 1 \\ &= - (-1) = - \frac{1}{1 - 2} = - \sum_{n=0}^{\infty} 2^{n}. \end{align}

This yields $$\sum_{n=0}^{\infty} \binom{n}{k} = - 2^{n}.$$

As stated, the mistake that eludes me is the sign error.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't think the second last step is right. The infinite geometric sum can be applied only when the common ratio of the GP lies between $-1$ and $1$ but as per your text it seems that you have taken 2 as the common ratio which I don't think would be proper to continue either $\endgroup$ – Rohan Shinde May 31 '18 at 18:39
-2
$\begingroup$

Using mathematical induction. Base case $n=0$, ok!
Assume that is true for an arbitrary natural number $n$. Since $$ \sum_{k=0}^{n+1} \binom{n+1}{k} = \sum_{k=0}^n \binom{n+1}{k} + \binom{n+1}{n+1} $$ Using your definition, $\binom{n+1}{n+1}$ equals to $1$, then using Pascal's identity $$ \binom{n+1}{k} = \binom{n}{k}+\binom{n}{k-1} \text{ for all } 1\leqslant k\leqslant n $$ You get the following : $$ \sum_{k=0}^{n+1} \binom{n+1}{k} = 2^n + 2^n - 1 +1 = 2^{n+1}$$

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ I know this. I want to prove the statement via the integral. In other words, I want the proof of the integral identity. $\endgroup$ – Isomorphism May 20 '18 at 22:05
  • 2
    $\begingroup$ But you proved Pascal's identity using integral representation... this works fine. $\endgroup$ – Phoenix May 20 '18 at 22:07
  • $\begingroup$ Otherwise, you can perform some contour integration... but you won't... $\endgroup$ – Phoenix May 20 '18 at 22:10
  • $\begingroup$ I don't mind seeing a contour integral proof. I just prefer an elementary integration proof. $\endgroup$ – Isomorphism May 20 '18 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.