- #1

gty656

- 2

- 1

$$T_{n}\left( x\right)= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}{n \choose 2k}x^{n-2k}\left( x^2-1\right)^{k}$$

How to extract the coefficients of this polynomial of degree n ?

I tried using Newton's binomial but got a double sum

$$\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}{n \choose 2k}x^{n-2k}\left( \sum_{m=0}^{k} {k \choose m} x^{2m}\left( -1\right)^{k-m} \right) \\

\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\sum_{m=0}^{k}\left( -1\right)^{k-m} {n \choose 2k} \cdot {k \choose m} x^{n+2m-2k}\\

$$

Now how to continue counting this sum ?

What would it look like to change the order of summation and would it do anything ?What else did I try ?

Well, I worked out the sum

$$\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}{n \choose 2k}x^{n-2k}\left( x^2-1\right)^{k} $$

for n=8

and I hypothesized that

$$T_{n}\left( x\right)= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \sum_{m=k}^{\lfloor \frac{n}{2} \rfloor}\left( -1\right)^{k}{n \choose 2m} \cdot {m \choose k} x^{n-2k} $$

However, it would be useful to demonstrate the correctness of this hypothesis and count the sum of

$$\sum_{m=k}^{\lfloor \frac{n}{2} \rfloor}{n \choose 2m} \cdot {m \choose k}$$

Here I would like to point out that Wolfram Alpha counts this sum incorrectly

$$\sum_{m=k}^{\lfloor \frac{n}{2}\rfloor}{{n \choose 2m} \cdot {m \choose k}} = \frac{n}{2n-2k} \cdot 2^{n-2k} \cdot {n - k \choose k}$$

And it would even be a nice result but

first of all it is not quite correct ( Have you noticed why ?)

and secondly it comes from a hypothesis I made after dissecting the formula for $n=8$.

It seems to me that this hypothesis of mine would be enough to prove by induction after n but how would it look?