Continuous but Not Differentiable?Date: 12/27/2003 at 02:47:40 From: Amir Subject: Weierstrauss I have heard that Weierstrauss found a function, f(x) = sigma(b^n*cos(a^n*pi*x)) which is continuous everywhere, but differentiable nowhere. How is that possible? Date: 12/29/2003 at 14:27:07 From: Doctor Fenton Subject: Re: Weierstrauss Hi Amir, Thanks for writing to Dr. Math. Weierstrass's example is quite complicated, and although searching the web with the phrase "nowhere differentiable" found a number of sites which state examples or give graphs or animations, I found no site which actually proves the statement. T. W. Koerner, in his book _Fourier Analysis_, gives a detailed proof of the non-differentiability of the function oo --- \ sin((k!)^2 t) f(t) = / ------------- --- k! . k=0 Analyzing the difference quotient oo --- f(t+h) - f(t) \ 1 [ sin((k!)^2(t+h)) - sin((k!)^2 t) ] ------------- = / - [ -------------------------------- ] h --- h [ k! ] , k=0 there is a single term in the series whose value dominates the other terms in the series. However, this is a fairly complicated argument, and I would refer you to Koerner's book. He says that using the same technique, one can prove the non-differentiability of the Weierstrass function. If you just want an example of a non-differentiable function, van der Waerden's example is much simpler to analyze. Let {x} denote the "distance to the nearest integer function": on [0,1), { x 0 <= x <= .5 {x} = { { 1 - x .5 < x < 1 , and {x} is periodic with period 1 for other values of x. Then define oo --- \ {10^m x} g(x) = / -------- --- 10^m . m=0 Because of the periodicity of {x}, we need consider only points in [0,1). Let a be a point in [0,1), with decimal expansion a = 0.a(1)a(2)a(3)... . Define a sequence h(n) by { 10^(-n) if a(n) is not 4 or 9; h(n) = { { -10^(-n) if a(n) is 4 or 9 . [For example, if a = 0.1497... , then a(1) = 1, so h(1) = .1 ; a(2) = 4, so h(2) = -.01 ; a(3) = 9, so h(3) = -.001; a(4) = 7, so h(4) = .0001; etc. ] In the difference quotient oo --- g(a+h(n)) - g(a) 1 \ {10^m (a+h(n))} - {10^m a} ---------------- = ---- / -------------------------- h(n) h(n) --- 10^m m=0 the right side reduces to the sum of n terms, each of which is +1 or -1, so it is an integer of the same parity as n. As n->oo, the difference quotients are integers which change between even and odd with each increase of 1 in n, so they cannot possibly converge, and g is not differentiable at a. If you have any questions, please write back and I will try to explain further. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ Date: 12/29/2003 at 23:42:18 From: Amir Subject: Thank you (Weierstrauss) Thank you for your help! I did not think that you would answer my question so immediately. Your site will be very good for me and many others. Once again, thanks for your help. |
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