Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In answering a question on math.SE, I attempted to find integral of Fejér kernel by using

$$ K_n(t) = \frac{1}{n} U_{n-1}^2\left( \cos \frac{t}{2} \right) $$ where $U_n(z)$ stands for the Chebyshev polynomial of the second kind. Then $$ \frac{1}{2 \pi} \int_{-\pi}^\pi K_n(t) \mathrm{d} t = \frac{1}{\pi n} \int_{-1}^1 \frac{U_{n-1}^2(z)}{\sqrt{1-z^2}} \mathrm{d} z $$ Also it is known that the left-hand-side integral is one, I could not find any neat way of showing that the right-hand-side integral equals one.

Note that, by orthogonality property for $U_n(z)$: $$ \int_{-1}^1 U_{n-1}^2(z) \sqrt{1-z^2} \mathrm{d} z = \frac{\pi}{2} $$

Thanks for reading.

share|improve this question
    
You mean other than using $$K_n(t) = \frac{1}{n} U_{n-1}^2\left( \cos \frac{t}{2} \right)$$ and showing the integral of the Fejer Kernel is $1$? –  robjohn Sep 30 '11 at 21:03
1  
@robjohn Yes, I was hoping to use some properties of Chebyshev polynomials, as in Eric's solution. –  Sasha Sep 30 '11 at 21:17
add comment

2 Answers

up vote 8 down vote accepted

The weight $\frac{1}{\sqrt{1-z^2}}$ may not be the one used for the orthogonality relations of $U_n(x)$, but it is however the weight used for the orthogonality relations of $T_n(x)$. Specifically, we have that for $n,m>0$ $$\int_{-1}^1 T_n(z)T_m(z)\frac{dz}{\sqrt{1-z^2}}=\frac{\pi}{2}\delta_{n,m}.$$ (in the case that $m=n=0$, then the integral equals $\pi$)

Hence we need only find a way to relate $U_n(x)$ to a sum of $T_n(x)$ terms. On this note, we have the identities:

$$n\ \text{odd}\Rightarrow\ \ U_n(x)= 2\sum_{k\leq n,\ k\ \text{odd}} T_k (x)$$

and $$n\ \text{even}\Rightarrow\ \ U_n(x)= -1+2\sum_{k\leq n,\ k\ \text{even}} T_k (x).$$

These both follow from induction and the recurrence relations for $U_n(x),\ T_n(x)$.

If $n$ is odd, then by the above we see that the integral will give $\frac{\pi}{2}$ for each diagonal term, and zero on all the others. Hence we conclude that $$\int_{-1}^1 \frac{U_n(z)U_n(z)}{\sqrt{1-z^2}}dz = 4\sum_{k\leq n,\ k\ \text{odd}} \frac{\pi}{2}= \pi (n+1).$$

If $n$ is even, things follow in a similar manner.

share|improve this answer
add comment

A much simpler approach is to let $x=\cos\theta.$ Then $$\int_{-1}^{1}\frac{U_{n-1}(x)^{2}}{\sqrt{1-x^{2}}}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}U_{n-1}(\cos\theta)^{2}d\theta.$$ Since $U_{n-1}(\cos\theta)=\frac{\sin(n\theta)}{\sin\theta},$ our integral becomes $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin(n\theta)^{2}}{\sin(\theta)^{2}}d\theta,$$ and for $n=2k+1,$ we can use the identity $$\frac{\sin((2k+1)\theta)}{\sin(\theta)}=1+\sum_{j=1}^{k}\cos(2jx)$$ to evaluate the integral, and similarly for $k$ even.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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