Unlucky Polynomial

Note $f(2(x)^2-1)=f(2(-x)^2-1)$ . Hence $f(x)^2=f(-x)^2$

Since $f(x)$ is a non zero polynomial, $f(x) = f(-x)$ for all $x$ or $f(x)=-f(-x)$ for all $x$ . Since the degree of $f(x)$ is $13$ , $f(x) > 0$ , $f(-x) < 0$ for large $x$ . Hence $f(x)$ is an odd function.

Let $f(x) = ax^{13}+bx^{11}+cx^{9}+dx^{7}+ex^{5}+fx^{3}+gx$

Comparing coefficient of $x^26$ , $2a^2-8192a=0$ . Hence $a = 2^12$ Comparing coefficient of $x^24$ , $4ab+53248a=0$ . Hence $b = -2^10*13$ Comparing coefficient of $x^22$ , $4ac-159744a+2b^2-2048b=0$ . Hence $c = 2^8*13*5$ Comparing coefficient of $x^20$ , $4ad+292864a+4bc+11264b=0$ . Hence $d = -2^6*13*12$ Comparing coefficient of $x^18$ , $4ae-366080a+4bd-28160b+2 c^2-512c=0$ . Hence $e = 2^4*13*14=2912$

Note that if $g(x) = 2x^2 - 1$ , then we must have $f(g(x)) = g(f(x))$ . Since $g(x) = \cos(2 \arccos x)$ is the 2nd Chebyshev polynomial, we see that a possible solution is the 13th Chebyshev polynomial $f(x) = cos(13 arccos x)$ , which easily commutes with $g$ .

Let's first show that this solution is unique. We first note that $f(-x)^2 = f(x)^2$ by the functional equation. The rational function $f(-x)/f(x)$ takes only values +1 or -1, and therefore is constant. Thus $f$ is either odd or even, and must be odd since it has degree 13. It follows that $f(x) = Ax^{13} + Bx^{11} + Cx^9 + Dx^7 + Ex^5 + Fx^3 + Gx$ . The $x^{26}$ term of $f(g(x))$ is $(2^{13})A$ , while the corresponding term of $g(f(x))$ is $2A^2$ . This uniquely determines $A = 2^{12}$ . Now consider the $x^{24}$ terms. By binomial expansion, $f(g(x))$ has $-13(2^{12}) A$ , while $g(f(x))$ has $4AB$ . This uniquely determines B after dividing through by A. Similarly, $x^{22}$ yields a linear equation with 4AC on the right-hand and a combination of A and B on the left. Since A is non-zero, this determines C. This pattern continues, determining all coefficients uniquely (note that this analysis only shows uniqueness, not existence).

[Note: Knowing that we want it to be the Chebyshev polynomial, a quicker approach will be to conisder $g (x) = x^{13} f ( \frac {x+x^{-1}}{2})$ which transforms the functional equation into $g(x^2) = 2g(x)^2 - x^{26}$ . Then show that $g(x) = \frac {x^{26}+1}{2}$ . - Calvin]

With uniqueness in hand, we conclude that $f(x)$ must be the 13th Chebyshev polynomial. The explicit formula $T_n(x) = \frac12 [ (x+\sqrt{x^2-1})^n + (x-\sqrt{x^2-1})^n ]$ shows that the $x^5$ coefficient is given by that of $\binom{13}{1} x^1 (x^2-1)^6 + \binom{13}{3} x^3 (x^2-1)^5 + \binom{13}{5} x^5 (x^2-1)^4,$ which is $\binom{13}{1}\binom{6}{2} + \binom{13}{3} \binom{5}{1} + \binom{13}{5} = 13( 15 + 22\cdot 5 + 99) = 13(224) = 2912$ .

First, let $f(x)=\sum^{13}_{i=0}a_ix^i$ . Expand both sides using the Binomial Theorem. We claim that all $a_{2k}, k$ an integer are 0. This is because $a_{13} \neq 0$ , but if we compare the coefficients of $x^{25}$ , $4a_{13}a_{12}=0\Rightarrow a_{12}=0$ . Similarly when we compare coefficients of $x^{23}, x^{21}, ... , x$ , we will get all $a_{2k}, k$ an integer are 0. Now first we compare the coefficients of $x^{26}$ , we get $4096a_{13}=a_{13}^2\Rightarrow a_13=4096$ . Then compare the coefficients of $x^{24}$ , we get $-4096 \times 13a_{13}=4a_{11}a_{13}\Rightarrow a_{11}=-13 \times 2^{10}$ . Then compare the coefficients of $x^{22}$ , $2048(78a_{13}+a_{11})=2(a_{11}^2+2a_9a_{13})\Rightarrow a_9=16640$ . Compare the coefficients of $x^{20}$ , we get $-1024(286a_{13}+11a_{11})=4(a_9a_{11}+a_7a_{13})\Rightarrow a_7=-9984$ . Lastly, compare the coefficients of $x^{18}$ , we have $512(715a_{13}+55a_{11}+a_9)=2(a^2_9+2a_7a_{11}+2a_5a_{13})\Rightarrow a_5=2912$ , so the answer is 912.

Notice that $\cos(2x)=2\cos ^2 x -1.$ We will first prove that $f$ is the $13$ th Chebyshev polynomial, that is the polynomial that expresses $\cos (13t)$ in terms of $\cos t$ .

To do this, consider polynomial $g(x)=x^{13}f\left(\frac{x+x^{-1}}{2}\right)$ . One can check that the functional equation for $f$ gives the following functional equation for $g$ : $g(x^2)=2(g(x))^2-x^{26}$ Note that the degree of $g$ , from its definition, equals $26$ . Suppose $g(x)=ax^{26}+h(x),$ where $\deg h(x) \leq 25.$

Looking at the highest degree of the functional equation for $g,$ we obtain $a=\frac{1}{2}$ . Then the functional equation for $h$ becomes $h(x^2)=2x^{26}h(x)+2(h(x))^2-x^{26}.$ If $h$ is not a constant, then the degree of $2x^{26}h(x)$ is strictly greater than the degrees of $h(x^2),$ $2(h(x))^2$ and $x^{26}$ . This is impossible, so $h(x)$ is a constant. It is easy to see that this constant is $\frac{1}{2}.$ Therefore, $g(x)=\frac{x^{26}+1}{2}$ . This implies that $f(\frac{x+x^{-1}}{2})=\frac{x^{13}+x^{-13}}{2}$ for all $x$ , in particular for $x=e^{2\pi i t}$ for real $t$ . For these $t$ we get $f(\cos t)=\cos (13t)$ .

To find the coefficient for $x^5$ , consider De Moivre's formula: $(\cos t +i\sin t)^{13}= \sum \limits_{k=0}^{13} \frac{13!}{k!(13-k)!} i^{13-k} \cos ^kt \sin ^kt.$ Replacing for even $n-k$ the expression $\sin^{n-k}t$ by $(1-\cos ^2t)^{\frac{n-k}{2}}$ , and taking the real part, we get a formula for $f(x)$ . Note that the coefficient for $x^5$ comes from the sum of three terms: $\cos t \frac{13!}{1!12!} i^{12} (1-\cos^2 t)^{6} + \cos^3 t \frac{13!}{3!10!} i^{10} (1-\cos^2 t)^{5} + \cos^5 t \frac{13!}{5!8!} i^{8} (1-\cos^2 t)^{4}.$ Thus the coefficient is $\frac{13!}{1!12!}\frac{6!}{2!4!} + \frac{13!}{3!10!}\frac{5!}{1!4!} + \frac{13!}{5!8!}\frac{4!}{0!4!}= 2912,$ so the answer is $912.$