1
$\begingroup$

Define

$$V_n(\theta) = \sum_{k=-n-1}^{n+1} e^{in\theta} + \sum_{k=n+2}^{2n+1}\frac{2n+2-k}{n+1}(e^{ik\theta}+e^{-ik\theta})$$

I want to show that $$V_n(\theta) = 2K_{2n+1}(\theta) - K_n(\theta)$$ where $K_n(\theta)$ is the Fejer Kernel given by

$$K_n(\theta) = \sum_{k=-n}^{n}\left(1 - \frac{|k|}{n+1}\right)e^{ik\theta}$$

Expanding the definition of $K_n$, $$2K_{2n+1}(\theta) - K_n(\theta) = 2\sum_{k=-2n-1}^{2n+1}(1 - \frac{|k|}{2n+2})e^{ik\theta} - \sum_{k=-n}^n(1-\frac{|k|}{n+1})e^{ik\theta}$$

Then simplify the coefficients by finding a common denominator,

$$ \sum_{k=-2n-1}^{2n+1}\frac{2n+2-|k|}{n+1}e^{ik\theta} - \sum_{k=-n}^n\frac{n+1-|k|}{n+1}e^{ik\theta} $$

I want to get some $(e^{ik\theta}+e^{-ik\theta})$ terms from the first summation,

$$ \frac{2n+2-0}{n+1}e^0 + \sum_{k=1}^{2n+1}\frac{2n+2-k}{n+1}(e^{ik\theta} + e^{-ik\theta}) - \sum_{k=-n}^n\frac{n+1-|k|}{n+1}e^{ik\theta}$$

Then I can break apart the first summation so that the indices match with $V_n$,

$$ 2 + \sum_{k=1}^{n+1}\frac{2n+2-k}{n+1}e^{ik\theta} + \sum_{k=n+2}^{2n+1}\frac{2n+2-k}{n+1}(e^{ik\theta} + e^{-ik\theta}) - \sum_{k=-n}^n\frac{n+1-|k|}{n+1}e^{ik\theta}$$

Great, the middle summation matches a term from $V_n$. Now my task seems easier, I just need to prove

$$\sum_{k=-n-1}^{n+1}e^{ik\theta} = 2 + \sum_{k=1}^{n+1}\frac{2n+2-k}{n+1}e^{ik\theta} - \sum_{k=-n}^n\frac{n+1-|k|}{n+1}e^{ik\theta}$$

and the "proof" is complete.

Unfortunately, this is where I get stuck. Futhermore, I'm not sure how to get rid of the $2$ term and this leads me to believe I've made an error somewhere. Can someone point out my error and put me or the right track?

$\endgroup$
2
  • $\begingroup$ $K_n(\theta)$ is nothing more than a polynomial in $e^{i \theta}$, you need some help for computing the difference of two polynomials ? $\endgroup$ – reuns Jan 27 '16 at 22:19
  • $\begingroup$ @user1952009 yes, in fact. Otherwise I wouldn't have posted this question... I've since solved this problem, I'll post an answer shortly. $\endgroup$ – cdk Jan 27 '16 at 22:29
0
$\begingroup$

You made a factor of 2 error in the step of finding a common denominator: The denominator of one term needs to be $2(n+1)$.

Following the calculation you did correcting for that error, I still don't get a proof of what you are trying to prove...

$\endgroup$
1
  • $\begingroup$ Yes, it's $2(n+1)$. But the summation for $K_{2n+1}$ has a constant $2$ in front of it, which cancels. $\endgroup$ – cdk Jan 27 '16 at 21:34

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.