How can we show that $D_n(t)$ has a bounded integral over a certain interval?

  • MHB
  • Thread starter Euge
  • Start date
In summary, $D_n(t)$ is the Dirichlet kernel used in Fourier analysis to approximate periodic functions. Its bounded integral is important because it indicates the accuracy of the approximation and the function's representation by its Fourier series. The boundedness of $D_n(t)$'s integral is determined by evaluating it over the given interval, with a finite result indicating boundedness and an infinite result indicating unboundedness. If the integral is unbounded, it means that the Dirichlet kernel does not accurately approximate the function due to irregularities. The boundedness of the integral can be proven analytically using mathematical techniques, but there are also other approaches such as numerical methods.
  • #1
Euge
Gold Member
MHB
POTW Director
2,073
244
Here is this week's POTW:

-----
Let $D_n(t) = \sum\limits_{\lvert k\rvert \le n} e^{2\pi i kt}$ for $t\in [-.5, .5]$. Show that if $n \ge 2$, there are positive constants $A$ and $B$ independent of $n$, such that

$$A \le \frac{1}{\log n}\int_{-.5}^{.5} |D_n(t)|\, dt \le B$$

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
  • #2
Seeing that users are having some trouble, I’m giving another week for POTW submissions and corrections. You might find it useful to show first that $|D_n(t)|$ is bounded by a constant times the smaller of $n$ and $\lvert t\rvert^{-1}$ for all nonzero $t\in [-.5,.5]$.
 
  • #3
Opalg gets honorable mention for his near perfect solution. Since his overall argument is solid, I’ll post his solution below. Just note that in fact, $D_n(t) = \sin[(2n+1)\pi t]/\sin(\pi t)$.

\(\displaystyle D_n(t) = \sum_{k=-n}^n e^{2\pi ikt}\) is a geometric series with sum $$\frac{e^{-2\pi int}(1 - e^{2\pi i(n+1)t})}{1 - e^{2\pi it}} = e^{-\pi i(2n-1)t}\frac{\sin(n+1)\pi t}{\sin\pi t}.$$ So \(\displaystyle |D_n(t)| = \frac{|\sin(n+1)\pi t|}{|\sin\pi t|}\), and $$\int_{-1/2}^{1/2}|D_n(t)|\,dt = \int_{-1/2}^{1/2}\frac{|\sin(n+1)\pi t|}{|\sin\pi t|}\,dt = 2\int_0^{1/2}\frac{|\sin(n+1)\pi t|}{\sin\pi t}\,dt.$$ On the interval $[0,1/2]$, $2t \leqslant \sin\pi t \leqslant \pi t$. It follows that \(\displaystyle \int_{-1/2}^{1/2}|D_n(t)|\,dt\) lies between two positive multiples of\(\displaystyle \int_0^{1/2}\frac{|\sin(n+1)\pi t|}{ t}\,dt\), so it will be sufficient to estimate the size of that integral. The substitution $x = (n+1)\pi t$ gives $$ \int_0^{1/2}\frac{|\sin(n+1)\pi t|}{ t}\,dt = \int_0^{(n+1)\pi/2}\frac{|\sin x|}{x}\,dx = \sum_{r=0}^n \int_{r\pi/2}^{(r+1)\pi/2}\frac{|\sin x|}{x}\,dx.$$ On the interval $[0,\pi/2]$, $\frac{\sin x}x$ lies between $\frac12$ and $1$.On the interval $[r\pi/2,(r+1)\pi/2]$ for $r\geqslant1$, \(\displaystyle \frac{|\sin x|}{x} \leqslant \frac1{r\pi/2}.\) Therefore \(\displaystyle \int_{r\pi/2}^{(r+1)\pi/2}\frac{|\sin x|}{x}\,dx \leqslant \int_{r\pi/2}^{(r+1)\pi/2}\frac1{r\pi/2}\,dx = \frac1r.\)To get an inequality in the opposite direction, notice that \(\displaystyle |\sin x| \geqslant \frac1{\sqrt2}\) throughout one or other of the subintervals $[r\pi/2,(r+0.5)\pi/2]$, $[(r+0.5)\pi/2,(r+1)\pi/2]$ (each of which has length $\pi/4$). On that subinterval, \(\displaystyle \frac{|\sin x|}x \geqslant \frac{1/\sqrt2}{(r+1)\pi/2} = \frac{\sqrt2}{(r+1)\pi}.\) It follows that \(\displaystyle \int_{r\pi/2}^{(r+1)\pi/2}\frac{|\sin x|}{x}\,dx \geqslant \frac{\sqrt2}{4(r+1)} \geqslant \frac{\sqrt2}{8r}\) (when $r\geqslant1$).Thus \(\displaystyle \int_{r\pi/2}^{(r+1)\pi/2}\frac{|\sin x|}{x}\,dx\) lies between two fixed (independent of $r$) multiples of $\frac1r$, and therefore \(\displaystyle \int_{-1/2}^{1/2}|D_n(t)|\,dt\) lies between two positive multiples of \(\displaystyle \sum_{r=1}^n\frac1r\). But that harmonic sum is asymptotically close to $\ln n$.In conclusion (modulo tidying up the little details of what happens when $r=0$, and how to incorporate the Euler–Mascheroni constant), \(\displaystyle \int_{-1/2}^{1/2}|D_n(t)|\,dt\) lies between two fixed (independent of $n$) positive multiples of $\ln n$.
 

FAQ: How can we show that $D_n(t)$ has a bounded integral over a certain interval?

What is $D_n(t)$ and why is it important to show its bounded integral?

$D_n(t)$ is the Dirichlet kernel, which is used in Fourier analysis to approximate periodic functions. It is important to show that its integral is bounded over a certain interval because this indicates that the approximation is accurate and the function can be well-represented by its Fourier series.

How is the boundedness of $D_n(t)$'s integral determined?

The boundedness of $D_n(t)$'s integral is determined by evaluating the integral itself over the given interval. If the result is a finite value, then the integral is bounded. If the result is infinite, then the integral is unbounded.

What does it mean if $D_n(t)$'s integral is unbounded?

If $D_n(t)$'s integral is unbounded, it means that the Dirichlet kernel does not accurately approximate the given periodic function over the specified interval. This could be due to the function having sharp discontinuities or other irregularities.

Can the boundedness of $D_n(t)$'s integral be proven analytically?

Yes, the boundedness of $D_n(t)$'s integral can be proven analytically by using mathematical techniques such as integration by parts and applying theorems related to Fourier series. However, it may require advanced mathematical knowledge and techniques to do so.

Are there any other approaches to showing the boundedness of $D_n(t)$'s integral?

Yes, there are other approaches to showing the boundedness of $D_n(t)$'s integral, such as using numerical methods to approximate the integral and analyzing the results. This can be useful when dealing with complex or difficult integrals that cannot be easily evaluated analytically.

Back
Top