- #1
gty656
- 2
- 1
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?
$$T_{n}\left( x\right)= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}{n \choose 2k}x^{n-2k}\left( x^2-1\right)^{k}$$
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?