Quick! Where's the Chebyshev Table?

I came up with a way to determine the coefficient of $\cos^3(x)$ for $\cos((4k+3)x)$ .

We apply the standard method of using De Moivres theorem.

$cos((4k+3)x)=\Re[(cos(x)+isin(x))^{4k+3}]$

Expanding the RHS we find the coefficient of $\cos^3(x)$ can be found.

$\binom{4k+3}{1}\cos(x)i^{4k+2}\sin^{4k+2}+\binom{4k+3}{3}\cos^3(x)i^{4k}\sin^{4k}(x)+...$

With a little bit of simplifying this implies the coefficient of $\cos^3(x)$ is

$C_3=\frac{4(4k+3)(2k+1)(k+1)}{3}$

Let's try for a few known values that we can test from wiki to make sure.

for $k=0$ , $C_3=4$

for $k=1$ , $C_3=56$

for $k=2$ , $C_3=220$

To find it for $\cos(2015x)$ we use $k=503$

$\implies Y=2015\times676704$

$\displaystyle \cos(nx) = \frac{{e}^{inx} + {e}^{-inx}}{2}$

We want that in terms of $\cos(x)$ .

$\displaystyle \cos(nx) = \frac{{cos(x) + i\sqrt{1 - \cos^2(x)}}^{n} + {cos(x) - i\sqrt{1 - \cos^2(x)}}^{n}}{2}$

Let's change some variables

$\displaystyle T_n(z) = \frac{{z + \sqrt{z^2 -1}}^{n} + {z - \sqrt{z^2 - 1}}^{n}}{2}$

Now, we will use binomial theorem,

$\displaystyle T_n(z) = \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor}{\binom{n}{2k} {(z^2 -1)}^{k} {z}^{n-2k}}$ [I guess that's quite simple to get. Try yourself.]

$\displaystyle T_n(z) = \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor}{\binom{n}{2k} {(1-{z}^{-2})}^{k} {z}^{n}}$

$\displaystyle T_n(z) = \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor}{\binom{n}{2k} \sum_{m=0}^{k}{\binom{k}{m} (-1)^m {z}^{-2m}}{z}^{n}}$

Now, changing order of summation,

$\displaystyle T_n(z) = \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor}{\frac{n}{2} \frac{(n-k-1)!}{(n-2k)! k!} {(-1)}^{k} {(2z)}^{n-2k}}$

Now, the problem can easily be done since we know the series and hence, the coefficients.

I will give a much more detailed proof of "changing order of summation" part tomorrow. It's late now.

I solved the problem by observation.

• Let's call each chebyshev polynomial $\cos nx$ $P_{n}$ My observations are the following:

• Only odd numbers of $n$ yield chebyshev polynomials with $x^{3}$

• The abs. value of coefficients of $x^{3}$ for n=3, 5, 7, 9 are: 4, 20, 56, and 120. These can be written as 4 x 1, 4 x 5, 4 x 14, and 4 x 30. Notice: 1, 5, 14, 30 are in the form $\displaystyle \sum_{x=1}^{\frac{n}{2}-0.5} x^{2}$ . So, for $P_{2015}$ , our abs. value of the coefficient will be in the form $4 \displaystyle \sum_{x=1}^{1007} x^{2}$ . Use the formula $\dfrac{1}{6} z(z+1)(2z+1)$ to compute the power sum with exponent of 2 to get $\dfrac{1}{6} (1007)(1008)(2015)$ and multiply that by 4 to obey the rule.

• $P_{3, 7, 11...}$ have a positive value of the coefficient for $x^{3}$ (this is an arithmetic progression in the form $T_{y} = n= 3+4(y-1)$ and $P_{5, 9, 13...}$ have negative value of coefficient for $x^{3}$ (this is an arithmetic progression in the form $T_{y}= n = 5+4(y-1)$ . So, $2015 = 3+4(y-1)$ gives a whole number result, no decimals, so our coefficient is positive.

The answer calls for dividing the coefficient of $x^{3}$ in $P_{2015}$ by $2015$ so we have $\dfrac{(4)(1007)(1008)(2015)}{(6)(2015)}$ which is equal to $\dfrac{(4)(1007)(1008)}{(6)}$ , and the answer is $\boxed{676704}$ .