Maximum value of Coefficient of Fibonacci Polynomial

[Bling Fizikst]

Homework Statement

Denote by $F_0(x), F_1(x), \ldots$ the sequence of Fibonacci polynomials, which satisfy the recurrence $F_0(x)=1, F_1(x)=x,$ and $$F_n(x)=xF_{n-1}(x)+F_{n-2}(x)$$ for all $n\geq 2$. It is given that there exist unique integers $\lambda_0, \lambda_1, \ldots, \lambda_{1000}$ such that$$x^{1000}=\sum_{i=0}^{1000}\lambda_iF_i(x)$$ for all real $x$. For which integer $k$ is $|\lambda_k|$ maximized?

Relevant Equations

ok

Tried to find patterns on the coefficients of the polynomials . But could only go as far as : $$x: \frac{i+1}{2}$$ $$x^2: \frac{i(i+2)}{8}$$ for the $i$th fibonacci polynomial . Quite stuck on this one for a while , so , not sure if i have to change routes . But seems like if i find the coefficients of the polynomial , i can set them all to zero except the one for the $1000$ th power . It seems i can only start thinking about the maximising part after finding the coefficents and setting them to zero .

 

[fresh_42]

This becomes complicated quickly. E.g. see the formulas here:
https://arxiv.org/pdf/1505.06697

It says that
$$
F_{n-1}(x)=\dfrac{(x+\sqrt{x^2+4})^n-(x-\sqrt{x^2+4})^n}{2^n\sqrt{x^2+4}}
$$
and
$$
F_{n-1}(x)=\sum_{k=0}^{\lfloor \frac{n-1}{2}\rfloor}\binom{n-k-1}{k}x^{n-2k-1}
$$
which at least gives us a hint how to solve it. And Wikipedia says

The Fibonacci polynomials form a sequence of orthogonal polynomials.

but I haven't figured out what the weight function and the integral domain are.

 

[WWGD]

This is maybe related to a Generating Function.

 

[Gavran]

There is a relation between the Fibonacci polynomials and the standard basis polynomials
$$ x^n=F_n(x)+\sum_{j=0}^{\left\lfloor \frac{n-1}{2} \right\rfloor}(-1)^{j+1}(\binom{n}{j+1}-\binom{n}{j})F_{n-2j-2}(x) $$ where $n\gt0$ and $F_{-1}(x)=0$.