Toss the coin 2017 times

Let's generalize it a little bit.

We consider a binomial experiment where probability of success changes after every trial.

Let us denote probability of success at $i$ th trial to be $p(i)$ and that of failure at $i$ th trial to be $q(i) = 1 -p(i)$

So, what's the probability of $1$ success in a sequence of $n$ trials.

$P(1 success) = p(1) q(2) q(3) \cdots q(n) + q(1) p(2) q(3) \cdots q(n) + \cdots + q(1) q(2) q(3) \cdots p(n)$

Obviously in layman's term - when $p(1)$ comes, $q(1)$ does not.

So, how can we make a model for that given that we'd have to find for $m$ successes?

We use generating functions.

$P(\text{1 success}) = [x^1] { (p(1) x + q(1)) (p(2) x + q(2)) (p(3) x + q(3)) \cdots (p(n) x + q(n)) }$

where $[x^k] { f(x)}$ means coefficient of $x^k$ in $f(x)$

What about $\text{2 success}$ ? We want $2 p(i)$ 's and rest $q$ 's such that "when $p(1)$ comes, $q(1)$ does not."

So, very clearly, it is coefficient of $x^2$ in our given generating function.

Therefore, $P(\text{m successes}) = a_m$ , where $f(x) = (p(1) x + q(1)) (p(2) x + q(2)) (p(3) x + q(3)) \cdots (p(n) x + q(n)) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$

So that's our new binomial distribution and we can closed forms of problems like odd number of successes etc

We would need to find expected number of successes .

$\displaystyle E(X) = \dfrac{\sum_{k=0}^{n}{P(\text{m successes}) m}}{\sum_{k=0}^{n}{P(\text{m successes})}}$

$\displaystyle = \dfrac{a_0 0 + a_1 1 + a_2 2 + \cdots + a_n n}{a_0 + a_1 + a_2 + \cdots + a_n}$

$\displaystyle = \dfrac{f'(1)}{f(1)}$

$\displaystyle = \dfrac{d}{dx} (\log(f(x))) |_{x=1}$

$\displaystyle = \sum_{i=1}^n {\frac{d}{dx} (p(i) x + q(i)) } |_{x=1}$

$\displaystyle \large \boxed{\displaystyle E(X) = \sum_{i=1}^n {p(i)}}$

$\displaystyle \text{Variance} = \sigma^2 = \dfrac{a_0 0^2 + a_1 1^2 + a_2 2^2 + \cdots + a_n n^2}{a_0 + a_1 + \cdots + a_n} - E^2$

$\displaystyle = \frac{(xf'(x))'}{x} |_{x=1} - E^2$

$\displaystyle = \frac{xf''(x)}{f(x)} |_{x=1} + \frac{f'(x)}{f(x)} |_{x=1} - E^2$

It can be shown that $f''(1) = (p(1) + p(2) + \cdots + p(n))^2 - p(1)^2 - p(2)^2 - \cdots - p(n)^2$ .

$f(1) = 1, f'(1) = p(1) + p(2) + \cdots + p(n)$ .

$\displaystyle = E^2 - \sum_{i=1}^n{p(i)^2} + \sum_{i=1}^n {p(i)} - E^2$

$\displaystyle \large \boxed{\displaystyle \sigma^2 = \sum_{i=1}^n{p(i)} - \sum_{i=1}^n {(p(i))^2}}$

Check for normal binomial distribution!

For our case, $\text{Variance} = \displaystyle \sigma^2 = \sum_{i=1}^{2017}{\sin^2\left(\frac{i\pi}{2017}\right)} - \sum_{i=1}^n {\sin^4\left(\frac{i\pi}{2017}\right)}$

$\displaystyle \sum_{i=1}^n{\sin^2\left(\frac{i\pi}{n}\right)} = \sum_{i=1}^n{\frac{1}{2}\left(1 - \cos\left(\frac{2i\pi}{n}\right)\right)} = \frac{n}{2}$

$\displaystyle \sum_{i=1}^n{\sin^4\left(\frac{i\pi}{n}\right)} = \sum_{i=1}^n{\frac{1}{4}{\left(1 - \cos\left(\frac{2i\pi}{n}\right)\right)}^2}$

$\displaystyle = \sum_{i=1}^n{\frac{1}{4}\left(1 + \frac{1}{2}\left(1 + \cos\left(\frac{4i\pi}{n}\right)\right)\right)} = \frac{n}{4} + \frac{n}{8} = \frac{3n}{8}$

Here I have repeatedly used $\cos(2x) = 2\cos^2(x) - 1 = 1 - 2\sin^2(x)$ and $\cos(\alpha) + \cos(\alpha + \beta) + \cdots + \cos(\alpha + (n-1)\beta) = \frac{\sin\left(\frac{n\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} \cos\left(\alpha + \frac{(n-1)\beta}{2}\right)$ .

This yields our answer as $\frac{n}{8} = \frac{2017}{8}$

This solution is incomplete.

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Essentially, we want to evaluate $\text{Var}(x) = E(X^2) - (E(X))^2$ , where $E(X^n)$ denotes the $n^\text{th}$ moment, $\displaystyle E(X^n) = \sum_{j=1}^{2017} \left (\sin^2 \dfrac{j \pi}{2017} \right)^n$ .

So this boils down to a trigonometric expression.

$\begin{aligned} \text{Var}(x) &=& E(X^2) - (E(X))^2 \\ &=& \sum_{j=1}^{2017} \left (\sin^2 \dfrac{j \pi}{2017} \right)^2 - \sum_{j=1}^{2017} \left( \left (\sin^2 \dfrac{j \pi}{2017} \right) \right)^2 \\ &=& \sum_{j=1}^{2017} \left (\sin^2 \dfrac{j \pi}{2017} \right) \left( 1 - \sin^2 \dfrac{j \pi}{2017}\right) \\ &=& \sum_{j=1}^{2017} \left (\sin^2 \dfrac{j \pi}{2017} \cos^2 \dfrac{j \pi}{2017}\right) \\ &=& \dfrac14 \sum_{j=1}^{2017} 4 \left (\sin^2 \dfrac{j \pi}{2017} \cos^2 \dfrac{j \pi}{2017}\right) \\ &=& \dfrac14 \sum_{j=1}^{2017} \sin^2 \dfrac{2 j \pi}{2017} \qquad \qquad \qquad \qquad \text{ because } \sin(2A) = 2\sin A \cos A \\ &=& \dfrac14 \sum_{j=1}^{2017} \dfrac{1 - \cos(4j\pi/2017)}{2}\qquad \qquad \qquad \text{ because } \sin^2(A) = \frac12 (1-\cos(2A)) \\ &=& \dfrac18 \sum_{j=1}^{2017} 1 -\dfrac18 \sum_{j=1}^{2017} \cos \left( \dfrac{4\pi j}{2017} \right) \\ \end{aligned}$

What's left to do is to show that $\displaystyle \sum_{j=1}^{2017} \cos \left( \dfrac{4\pi j}{2017} \right) = 0$ , which can be solved using Chebyshev polynomial of the first kind or roots of unity . I'll finish this whenever I feel like it.