A type of Combinatorial equality：$\sum_{k=0}^{n}\binom{n}{k} \cos\frac{k}{2}\pi=2^{\frac{n}{2}}\cos\frac{n}{4}\pi.$

Question

When computing the Taylor series of the function $f(z)=e^z\cos z,$ I use two methods:

On the one hand, using Cauchy product, \begin{align*} e^z\cos z &=\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right) \left(\sum_{n=0}^{\infty}\frac{\cos\frac{n}{2}\pi}{n!}z^n\right)\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\frac{1}{(n-k)!} \frac{\cos\frac{k}{2}\pi}{k!}\right)z^n\ (\text{Cauchy Product})\\[3pt] &=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\sum_{k=0}^{n}\binom{n}{k}\cos\frac{k}{2}\pi \right)z^n,\ z\in\mathbb{C}; \end{align*} On the other hand, \begin{align*} e^z\cos z &=e^z\cdot\frac{e^{i z}+e^{-i z}}{2}\\ &=\frac{e^{(1+i)z}+e^{(1-i)z}}{2}\\[3pt] &=\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{(1+i)^n}{n!}z^n +\sum_{n=0}^{\infty}\frac{(1-i)^n}{n!}z^n\right)\\ &=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(1+i)^n+(1-i)^n}{n!}z^n\\ &=\frac{1}{2}\sum_{n=0}^{\infty} \frac{2^{\frac{n}{2}}\left(e^{\frac{n}{4}\pi i}+e^{-\frac{n}{4}\pi i}\right)}{n!}z^n\\ &=\sum_{n=0}^{\infty}\left(\frac{2^{\frac{n}{2}}}{n!}\cos\frac{n}{4}\pi\right)z^n,\ z\in\mathbb{C}. \end{align*} So Compare the corresponding coefficients, we get the following Combinatorial equality: $$\sum_{k=0}^{n}\binom{n}{k} \cos\frac{k}{2}\pi=2^{\frac{n}{2}}\cos\frac{n}{4}\pi.$$ What I want to konw: is there an elementary method or constructive method ( which is suitable for high school student!) to prove this Combinatorial equality? Any help and hint will welcome!

Answer 1

There is a fundamental way to prove this. I used a simple binomial expansion and Euler's identity to solve this... Here's the solution:

$$ \begin{aligned} &\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) \cos \frac{k \pi}{2}=?\\ &\text { Let } T_{k}=\left(\begin{array}{l} n \\ k \end{array}\right)\left\{\cos \left(\frac{k \pi}{2}\right)+i \sin \left(\frac{k \pi}{2}\right)\right\}\\ &=\left(\begin{array}{l} n \\ k \end{array}\right) e^{\frac{i k \pi}{2}} \quad[\text { Euler's identity }] \end{aligned} $$

$$ \begin{aligned} \sum_{k=0}^{n} T_{k} &=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) e^{\frac{i k \pi}{2}} \\ &=\left(e^{i \pi / 2}+1\right)^{n} \quad[\text { Binomial expansion }] \\ &=(1+i)^{n} \\ &=2^{n / 2}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^{n} \\ &=2^{n / 2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{n} \\ S_{k} &=2^{n / 2} e^{\frac{i n \pi}{4}} \end{aligned} $$

$$ \text { But, } \operatorname{Im}\left(T_{k}\right)=\left(\begin{array}{l} n \\ k \end{array}\right) \cos \frac{k \pi}{2} $$

$$ \operatorname{Im}\left(S_{k}\right)=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) \cos \frac{k \pi}{2}=2^{n / 2} \cos \frac{n \pi}{4} $$

Answer 2

Here are some more general results. Let $\theta$ and $\phi$ be arbitrary complex numbers. For any integer $n\geq 0$, we have $$\sum_{k=0}^n\,\binom{n}{k}\,\cos(\theta+k\phi)=2^n\,\cos^n\left(\frac{\phi}{2}\right)\,\cos\left(\theta+\frac{n\phi}{2}\right)\,;\tag{*}$$ $$\sum_{k=0}^n\,(-1)^k\,\binom{n}{k}\,\cos(\theta+k\phi)=\left\{\begin{array}{ll}(-1)^{\frac{n}{2}}2^n\,\sin^n\left(\frac{\phi}{2}\right)\,\cos\left(\theta+\frac{n\phi}{2}\right)&\text{if $n$ is even}\,,\\ (-1)^{\frac{n-1}{2}}2^n\,\sin^n\left(\frac{\phi}{2}\right)\,\sin\left(\theta+\frac{n\phi}{2}\right)&\text{if $n$ is odd}\,;\end{array}\right.\tag{#}$$ $$\sum_{k=0}^n\,\binom{n}{k}\,\sin(\theta+k\phi)=2^k\,\cos^k\left(\frac{\phi}{2}\right)\,\sin\left(\theta+\frac{k\phi}{2}\right)\,;\tag{$\star$}$$ and $$\sum_{k=0}^n\,(-1)^k\,\binom{n}{k}\,\sin(\theta+k\phi)=\left\{\begin{array}{ll} (-1)^{\frac{n}{2}}2^n\,\sin^n\left(\frac{\phi}{2}\right)\,\sin\left(\theta+\frac{n\phi}{2}\right)&\text{if $n$ is even}\,,\\(-1)^{\frac{n+1}{2}}2^n\,\sin^n\left(\frac{\phi}{2}\right)\,\cos\left(\theta+\frac{n\phi}{2}\right)&\text{if $n$ is odd}\,.\end{array}\right.\tag{\$}$$ A way to prove these identities is just as in P. Patil's answer. Here is another way.

Consider the operator $\hat{\gamma}_\phi$ defined by $$(\hat{\gamma}_\phi\,f)(\theta):=2\,\cos\left(\frac{\phi}{2}\right)\,f\left(\theta+\frac{\phi}{2}\right)$$ for all $f:\mathbb{C}\to\mathbb{C}$ and $\theta\in\mathbb{C}$. Observe that $$(\hat{\gamma}_\phi\,\cos)(\theta)=\cos\left(\theta\right)+\cos\left(\theta+\phi\right)$$ for all $\theta\in\mathbb{C}$. By induction, we can easily see that $$2^n\,\cos^n\left(\frac{\phi}{2}\right)\,\cos\left(\theta+\frac{n\phi}{2}\right)=(\hat{\gamma}_\phi^n\,\cos)(\theta)=\sum_{k=0}^n\,\binom{n}{k}\,\cos(\theta+k\phi)\,.$$ This proves (*). Similarly, we can prove ($\star$) by observing that $$(\hat{\gamma}_\phi\,\sin)(\theta)=\sin(\theta)+\sin(\theta+\phi)\,.$$ We can also take the derivative of (*) with respect to $\theta$ to get ($\star$).

For (#), if we replace $\phi$ by $\phi+\pi$ in (*), then the left-hand side becomes $$\sum_{k=0}^n\,\binom{n}{k}\,\cos(\theta+k\phi+k\pi)=\sum_{k=0}^n\,(-1)^k\,\binom{n}{k}\,\cos(\theta+k\phi)\,.$$ The right-hand side is as given in (#) since $\cos\left(\dfrac{\phi+\pi}{2}\right)=-\sin\left(\dfrac{\phi}{2}\right)$ and $$\cos\left(\theta+\frac{n\phi+n\pi}{2}\right)=\left\{\begin{array}{ll}(-1)^{\frac{n}{2}}\,\cos\left(\theta+\frac{n\phi}{2}\right)&\text{if $n$ is even}\,,\\ (-1)^{\frac{n+1}{2}}\,\sin\left(\theta+\frac{n\phi}{2}\right)&\text{if $n$ is odd}\,.\end{array}\right.$$

Likewise, if we replace $\phi$ by $\phi+\pi$ in ($\star$), the left-hand side becomes $$\sum_{k=0}^n\,\binom{n}{k}\,\sin(\theta+k\phi+k\pi)=\sum_{k=0}^n\,(-1)^k\,\binom{n}{k}\,\sin(\theta+k\phi)\,.$$ The right-hand side is as given in (\$) since $\cos\left(\dfrac{\phi+\pi}{2}\right)=-\sin\left(\dfrac{\phi}{2}\right)$ and $$\sin\left(\theta+\frac{n\phi+n\pi}{2}\right)=\left\{\begin{array}{ll}(-1)^{\frac{n}{2}}\,\sin\left(\theta+\frac{n\phi}{2}\right)&\text{if $n$ is even}\,,\\ (-1)^{\frac{n-1}{2}}\,\cos\left(\theta+\frac{n\phi}{2}\right)&\text{if $n$ is odd}\,.\end{array}\right.$$ Alternatively, we can simply take the derivative of (#) with respect to $\theta$.

Another way to deal with (#) and (\$) is to define the operator $\hat{\sigma}_\phi$ as follows. For $f:\mathbb{C}\to\mathbb{C}$, let $$(\hat{\sigma}_\phi\,f)(\theta):=2\,\sin\left(\frac{\phi}{2}\right)\,f\left(\theta+\frac{\phi}{2}\right)$$ for all $\theta\in\mathbb{C}$. Show that $$(\hat{\sigma}_\phi\,\cos)(\theta)=-\sin(\theta)+\sin(\theta+\phi)$$ and $$(\hat{\sigma}_\phi\,\sin)(\theta)=+\cos(\theta)-\cos(\theta+\phi)$$ for all $\theta\in\mathbb{C}$. Then, perform induction to get (#) and (\$) (both identities should be proven within the same induction procedure).