The limit of the k-times convolution of the function \( f(x) = 1 + \cos(2\pi x) \) as \( k \) approaches infinity is determined using Fourier analysis. The Fourier coefficients of \( f \) are \( \hat{f}(n) = \frac{1}{2} \) for \( n = \pm1 \) and \( \hat{f}(n) = 1 \) for \( n = 0 \). The convolution in the Fourier domain results in \( \hat{f_k}(n) = 2^{-k} \) for \( n = \pm1 \) and \( \hat{f_k}(n) = 1 \) for \( n = 0 \). Consequently, the limit \( \lim_{k \to \infty} f_k(1/2) \) evaluates to 1.
PREREQUISITESMathematicians, signal processing engineers, and students studying Fourier analysis and convolution techniques.
Hi Sonifa, and welcome to MHB!Sonifa said: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 said: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 $f$.
Opalg said: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 $$\lim_{k\to\infty}f_k(1/2).$$