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

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?

  • $\begingroup$ $K_n(\theta)$ 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

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...

  • $\begingroup$ 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

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