Solving for Constants in Dirichlet Kernel Integral Approximations

[Zaare]

I'm stuck trying to solve the following problem:

If D_n is the Dirichlet kernel, show that there exist positive constants c_1 and c_2 such that

<br /> c_1 \log n \le \int\limits_{ - \pi }^\pi {\left| {D_n \left( t \right)} \right|dt} \le c_2 \log n ,<br />​

for n=2,3,4,.... By \log they mean the natural logarithm.

I know that

<br /> D_n \left( t \right) = \frac{1}{\pi }\left( {\frac{1}{2} + \sum\limits_{N = 1}^n {\cos \left( {Nt} \right)} } \right) = \frac{1}{{2\pi }}\sum\limits_{N = - n}^n {e^{iNt} } = \frac{{\sin \left( {nt + \frac{t}{2}} \right)}}{{2\pi \sin \left( {\frac{t}{2}} \right)}}<br />​

And it's easy to see that

<br /> \left| {D_n \left( t \right)} \right| = \frac{1}{\pi }\left( {\frac{1}{2} + \left| {\sum\limits_{N = 1}^n {\cos \left( {Nt} \right)} } \right|} \right) \le \frac{1}{\pi }\left( {\frac{1}{2} + n} \right)<br />​

But that is all I can do. I have no idea on how \log comes into the picture. Any help, suggestions or tips would be much appreciated.

 

[saltydog]

Zaare, isn't |D_n(t)| always greater than \frac{1}{2\pi}? Thus the integral of that over -pi to pi is 1. Thus that integral is always greater than 1. So what does c_1 (in terms of n) have to be to make the expression hold? Same dif for the other side right since (according to your results:

|D_n(t)|\leq \frac{1}{\pi}(\frac{1}{2}+n)

So the integral is always less than 1+2n. Now, what does c_2 have to be to satisfy this one?

 

[HackaB]

D_2(t) (according to the forumula involving sines) has a root at t = 2 pi/5. So its absolute value isn't always greater than any positive number.

 

[saltydog]

D_2(t) (according to the forumula involving sines) has a root at t = 2 pi/5. So its absolute value isn't always greater than any positive number.

Thanks for pointing that out HackaB. I see now that the last inequality should be:


|D_n(t)|=|\frac{1}{2\pi}+\frac{1}{\pi}\sum_N^n Cos(Nt)|

 

[shmoe]

Hi, notice the Dirichlet kernel is an even function, so it's enough to consider the integral on 0 to pi.

Using the

\frac{{\sin \left( {nt + \frac{t}{2}} \right)}}{{2\pi \sin \left( {\frac{t}{2}} \right)}}

version, you see it changes sign whenever \sin \left( {nt + \frac{t}{2}} \right)=0, find all these points and treat seperately the integral on each of these intervals. This removes the absolute value problem.

Now finding the integral on each of these intervals isn't easy, but fortunately a good enough approximation isn't difficult. Find (constant) bounds for 1/\sin(t/2) seperately on each of these smaller intervals to remove it from the integrals, then integrate, sum and you're nearly done. Nothing fancy for the bounds is needed, a partial hint remember \sin{x}\leq x.

One more thing, you'll have to treat the interval containing 0 in a different way, but I'll leave that to you.

 

[Zaare]

Finally, I solved this a couple of days ago.
Schmoe, I followed your "recipe". It was of great help.
Thank you all.