Graduate Coefficients of Chebyshev polynomials

[gty656]

Not long ago, I derived the formula for Chebyshev polynomials

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

 

[DrClaude]

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

This appears to be a correct representation of $T_{n}\left( x\right)$, checked by explicit calculation of the factors for various values of $n$.

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

Mathematica gives a complicated answer in terms of the gamma and hypergeometric functions, so there doesn't appear to be a simple closed formula.

 

[gty656]

I checked for myself in Wolfram some initial values for n
and it was ok however I would like to see a full proof with induction being enough for me
(When proving the induction step, you will probably need to refer to the definition of recursion
except that here we have second-order linear recursion)As for this sum, in my case Wolfram calculated it as follows

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

However, I noticed that the problem would be, for example, n=0.
I thought it was possible to find a similar formula to the above however also correct for n=0.

I would also like to add that there is a slight error in my post namely in the second line it should look like this in sequence:

I tried using Newton's binomial but got a double sum
$\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}{n \choose 2k}x^{n-2k}\left( \sum_{m=0}^{k} {k \choose m} x^{2m}\left( -1\right)^{k-m} \right) \\$​

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