The Chebyshev polynomials can be defined recursively as:

$T_0(x)=1$; $T_1(x)=x$;

$T_{n+1}(x)=2xT_n(x) + T_{n-1}(x)$

The coefficients of these polynomails for a function, $\space f(x)$, under certain conditions can be obtained by the following integral:


Fixing some integer $N$, the zeros of $T_{N+1}(x)$ are :

$$x_j=\cos\frac{\pi(i+\frac12)}{n+1}, \space j=0, 1, 2, ...,N $$

The coefficients can then be calculated to be given by:

$$a_n=\frac{2}{n+1}\sum_{j=0}^{N} f(x_j)T_k(x_j)$$

Can I get any help regarding how to calculate these coefficients for the function:

$f(x)=\large\frac{1}{e^\frac{x-\alpha}{\beta} \space +1}$. Where $\alpha$ and $\beta$ are constants.

  • $\begingroup$ Do you need step by step calculation or just numerical values of coefficients? $\endgroup$ – Ömer Oct 19 '13 at 14:06
  • $\begingroup$ step by step. I want to understand the process. $\endgroup$ – Hasan Oct 19 '13 at 14:20
  • $\begingroup$ @user1772257, could you post your values of coefficients as well? $\endgroup$ – Hasan Oct 20 '13 at 22:05
  • $\begingroup$ I didn't calculate them but i can recommend you www2.maths.ox.ac.uk/chebfun/examples/approx/html/… $\endgroup$ – Ömer Oct 21 '13 at 12:11
  • 1
    $\begingroup$ what the hell are "certain conditions"? $\endgroup$ – Quonux Feb 13 '14 at 1:03
  • (I) Notice that the Chebyshev coefficient of $f(x)$ are the Fourier coefficient of $f(\cos x)$, so the problem is equivalent to find the Fourier series of $\frac{1}{e^{\cos x}+1}$;
  • (II) Notice that $\frac{1}{e^x+1}$ can be expressed as a power series in terms of Bernoulli numbers;
  • (III) Notice that $(\cos\theta)^n$ can be expressed as a combination of $\cos(n\theta),\cos((n-2)\theta)\ldots$ in virtue of the De Moivre identity: $$(\cos\theta)^n = \left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^n = \frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}e^{(n-2k)i\theta}=\frac{1}{2^{n-1}}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{k}\cos((n-2k)\theta);$$
  • (IV) Use (II) and (III) in order to write $a_n$ as a sum of (Bernoulli numbers)$\cdot$(binomial coefficients).
  • $\begingroup$ Could you explain more of step (III). How does $(cos(x))^n$ fit in? $T_n cos(x)=cos(nx)$ for chebyshev polynomials. Should just a trigonometric identity be used? $\endgroup$ – Hasan Oct 19 '13 at 16:44
  • $\begingroup$ I made it more explicit. You have $\frac{1}{e^{\cos x}+1}=\sum B_n (\cos x)^n$, then apply (III) in order to extract from the last identity the coefficient of $\cos(nx)$. $\endgroup$ – Jack D'Aurizio Oct 19 '13 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.