Approximating an integral

  • Thread starter Zaare
  • Start date
  • Tags
    Integral
In summary, the problem is that there exist positive constants c_1 and c_2 such that c_1 \log n \le \int\limits_{ - \pi }^\pi {\left| {D_n \left( t \right)} \right|dt} \le c_2 \log n for n=2,3,4,....
  • #1
54
0
I'm stuck trying to solve the following problem:

If [tex]D_n[/tex] is the Dirichlet kernel, show that there exist positive constants [tex]c_1[/tex] and [tex]c_2[/tex] such that
[tex]
c_1 \log n \le \int\limits_{ - \pi }^\pi {\left| {D_n \left( t \right)} \right|dt} \le c_2 \log n ,
[/tex]​
for [tex]n=2,3,4,...[/tex]. By [tex]\log[/tex] they mean the natural logarithm.

I know that
[tex]
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)}}
[/tex]​
And it's easy to see that
[tex]
\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)
[/tex]​
But that is all I can do. I have no idea on how [tex]\log[/tex] comes into the picture. Any help, suggestions or tips would be much appreciated.
 
Physics news on Phys.org
  • #2
Zaare, isn't [itex]|D_n(t)|[/itex] always greater than [itex]\frac{1}{2\pi}[/itex]? Thus the integral of that over -pi to pi is 1. Thus that integral is always greater than 1. So what does [itex]c_1[/itex] (in terms of n) have to be to make the expression hold? Same dif for the other side right since (according to your results:

[tex]|D_n(t)|\leq \frac{1}{\pi}(\frac{1}{2}+n)[/tex]

So the integral is always less than 1+2n. Now, what does [itex] c_2 [/itex] have to be to satisfy this one?
 
  • #3
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.
 
  • #4
HackaB said:
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:


[tex]|D_n(t)|=|\frac{1}{2\pi}+\frac{1}{\pi}\sum_N^n Cos(Nt)|[/tex]
 
  • #5
Hi, notice the Dirichlet kernel is an even function, so it's enough to consider the integral on 0 to pi.

Using the

[tex]\frac{{\sin \left( {nt + \frac{t}{2}} \right)}}{{2\pi \sin \left( {\frac{t}{2}} \right)}}[/tex]

version, you see it changes sign whenever [itex]\sin \left( {nt + \frac{t}{2}} \right)=0[/itex], 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 [tex]1/\sin(t/2)[/tex] 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 [tex]\sin{x}\leq x[/tex].

One more thing, you'll have to treat the interval containing 0 in a different way, but I'll leave that to you.
 
  • #6
Finally, I solved this a couple of days ago.
Schmoe, I followed your "recipe". It was of great help.
Thank you all.
 

1. What is the purpose of approximating an integral?

The purpose of approximating an integral is to estimate the value of a definite integral when the exact value cannot be determined. This is commonly done in situations where the integral is difficult or impossible to solve analytically.

2. How is an integral approximated?

An integral can be approximated using numerical methods such as the trapezoidal rule, Simpson's rule, or the midpoint rule. These methods involve dividing the integral into smaller parts and using the values of the function at specific points to calculate an estimate of the integral.

3. What are the limitations of approximating an integral?

One limitation is that the accuracy of the approximation depends on the number of intervals used in the numerical method. Using a small number of intervals may result in a less accurate approximation. Additionally, approximating some integrals may require a large number of intervals, making the calculation time-consuming.

4. Can an integral be approximated for any function?

No, not all integrals can be approximated using numerical methods. Some functions may be too complex or may not have a known equation, making it impossible to use traditional numerical methods. In these cases, other techniques such as Monte Carlo integration may be used to approximate the integral.

5. How do I know if the approximation is accurate enough?

The accuracy of an approximation can be evaluated by comparing it to the exact value of the integral, if it is known. Alternatively, the approximation can be compared to other approximations using a different numerical method. It is also important to consider the context of the problem and determine if the level of accuracy is sufficient for the intended use of the approximation.

Suggested for: Approximating an integral

Back
Top