Lim of Convolution: Fourier Analysis Solution

[Sonifa]

Let f(x)=1+cos 2\pix and let fk=f*...*f (k-times convolution)

what is the value of lim fk(1/2) when k tends to infinity

Should use something about the Fourier Analysis, Could someone help me how to solve this problem?

 

[Opalg]

Let f(x)=1+cos 2\pix and let fk=f*...*f (k-times convolution)

what is the value of lim fk(1/2) when k tends to infinity

Should use something about the Fourier Analysis, Could someone help me how to solve this problem?

Hi Sonifa, and welcome to MHB!

If you want to get some replies to your post, I think you need to explain the problem a bit more clearly. In particular,
(1) the definition of $f(x)$ is unclear. Is it meant to be $f(x) = 1 + \cos\bigl(\frac2\pi x\bigr)$?
(2) what is the domain of the function? It could be the unit interval, the whole real line, or something in between. That will affect the definition of convolution, because $f*f$ is defined as $$(f*f)(x) = \int f(y)f(x-y)\,dy,$$ where the integral is taken over the domain of $f$.

 

[Sonifa]

Hi Sonifa, and welcome to MHB!

If you want to get some replies to your post, I think you need to explain the problem a bit more clearly. In particular,
(1) the definition of $f(x)$ is unclear. Is it meant to be $f(x) = 1 + \cos\bigl(\frac2\pi x\bigr)$?
(2) what is the domain of the function? It could be the unit interval, the whole real line, or something in between. That will affect the definition of convolution, because $f*f$ is defined as $$(f*f)(x) = \int f(y)f(x-y)\,dy,$$ where the integral is taken over the domain of $f$.

The function should be $f(x) = 1 + \cos\bigl(2\pi x\bigr)$ and it is defined on the unict circle R/Z

 

[Opalg]

If $f(x) = 1 + \cos(2\pi x) = 1 + \frac12e^{2\pi ix} + \frac12e^{-2\pi ix}$ then the (complex) Fourier coefficients are given by $$\hat{f}(n) = \begin{cases}\frac12&(n = \pm1), \\ 1&(n=0, \\ 0&(\text{otherwise}). \end{cases}$$ The Fourier transform converts convolution products to pointwise products, so it follows that $$\hat{f_k}(n) = \begin{cases}2^{-k}&(n = \pm1), \\ 1&(n=0, \\ 0&(\text{otherwise}). \end{cases}$$ Therefore $f_k(x) = 1 + 2^{-k}\bigl(e^{2\pi ix} + e^{-2\pi ix}\bigr) = 1 + 2^{1-k}\!\cos(2\pi x).$ From that, you should easily be able to find $$\lim_{k\to\infty}f_k(1/2).$$

 

[Sonifa]

If $f(x) = 1 + \cos(2\pi x) = 1 + \frac12e^{2\pi ix} + \frac12e^{-2\pi ix}$ then the (complex) Fourier coefficients are given by $$\hat{f}(n) = \begin{cases}\frac12&(n = \pm1), \\ 1&(n=0, \\ 0&(\text{otherwise}). \end{cases}$$ The Fourier transform converts convolution products to pointwise products, so it follows that $$\hat{f_k}(n) = \begin{cases}2^{-k}&(n = \pm1), \\ 1&(n=0, \\ 0&(\text{otherwise}). \end{cases}$$ Therefore $f_k(x) = 1 + 2^{-k}\bigl(e^{2\pi ix} + e^{-2\pi ix}\bigr) = 1 + 2^{1-k}\!\cos(2\pi x).$ From that, you should easily be able to find $$\lim_{k\to\infty}f_k(1/2).$$

Finally, I got the same solution as yours. But still many thanks!