
#19
Apr505, 05:12 PM

P: 998

Well, swapping them is ok if you have uniform convergence of the differentiated series (which he does~).
ie. if [tex]f(x) = \sum_{n=0}^\infty f_n(x)[/tex] on a closed interval [itex]I[/itex] and [tex] \sum_{n=0}^\infty f^\prime_n(x) [/tex] is uniformly convergent on [itex]I[/itex] and each [itex]f^\prime_n[/itex] is continuous on [itex]I[/itex] then [tex]f^\prime(x) = \sum_{n=0}^\infty f^\prime_n(x)[/tex] on [itex]I[/itex]. Of course, that doesn't mean that the function isn't differentiable outside of the interval. 



#20
Apr505, 05:15 PM

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Only when AB < 1, though. The fact the series diverges almost everywhere for AB > 1 doesn't guarantee the derivative doesn't exist.




#21
Apr505, 06:21 PM

P: 998

yep, that's definitely true~




#22
Apr505, 07:27 PM

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Thanks guys. I need to review some Analysis. So, as I see it I can swap the order of summation and differentiation as long as the series converges uniformly to a function which it does if AB<1. This I can prove via the Weierstrass Mtest for both the function and the sum of the derivatives of [itex]f_n(x)[/itex].
So, however, when AB becomes greater than 1 this no longer applies and another approach must be used. Is that correct? I found a good reference proving "a" function exists which is nowhere differentiable which is under the title of "The Weierstrass Function": Link to proof Think I'll get a hardcopy and spend some time going over this proof. I suspect it can be applied to prove nowhere differentiable of the cosine series above. 



#23
Apr505, 08:13 PM

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#24
Apr605, 06:22 PM

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In an effort to approach this problem, I've chosen to fall back to an easier one first:
[tex]f(x)=\sum_{n=0}^\infty \frac{GetDistance[4^n x]}{4^n}[/tex] Where the GetDistance[x] function returns the distance from x to the integer nearest x. This is the sawtooth function commonly used to demonstrate continuous and nowhere differentiable. I've attached a plot superimposing [itex]f_0[/itex],[itex]f_0+f_1[/itex] and[itex]f_0+f_1+f_2[/itex]. My Analysis book has a proof I can follow of it's nowhere differentiable behavior. Essentially, we seek to show: [itex]\lim_{n\rightarrow\infty}{} \frac{f(a+h)f(a)}{h}[/itex] does not have a limit. My interpretation of this is that it has a "corner" at all points in its domain. Next, I'll pursue the following: [tex]f(x)=\sum_{n=0}^\infty \frac{Sin[(n!)^2 t]}{n!}[/tex] I'm told that T.W. Koerner in his book "Fourier Analysis" gives a detailed proof that this function is nowhere differentiable. Again, this involves an analysis of the difference quotient. Apparently using the same technique, one can prove the nondifferentiability of the Weierstrass function. So, I'll work with them a bit first myself and if nothing pans out, well, it's a pleasant ride to the University library from my home. 



#25
Apr705, 08:28 PM

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I've retrieved the reference cited above from T.W.Koerner's book, "Fourier Analysis". I wish to focus on reviewing the proof here of nowhere differentiable for the function below (reviewing/reporting here will help me better understand it and perhaps will help others who also wish to learn more about these functions). I'll report it in parts. I should add that he states after thoroughly investigating this proof, one should be able to apply it to prove nowhere differentiable of the cosine series of the original post which is my ultimate goal.
[tex]f(t)=\sum_{r=0}^\infty \frac{1}{r!}Sin[(r!)^2 t][/tex] In general, we seek to investigate the difference quotient: [tex]\mathop {\lim }\limits_{x_n \to x} \frac{f(x)f(x_n)}{xx_n}[/tex] In order to investigate this limit, Koerner breaks the sum into three parts which turn out to be the key to the proof: [tex] h_n(t)=\sum_{r=0}^{n1}\frac{1}{r!}Sin[(r!)^2 t][/tex] [tex] k_n(t)=\frac{Sin[(n!)^2t]}{n!}[/tex] [tex] I_n(t)=\sum_{r=n+1}^{\infty}\frac{1}{r!}Sin[(r!)^2 t][/tex] And thus: [tex] f(t)=h_n(t)+k_n(t)+I_n(t)[/tex] These function components will be used to evaluate the difference quotient. However, some Lemmas are needed in order to set bounds on the various quantities. That's next. 



#26
Apr805, 07:07 PM

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In order to prove nowhere differentiable, we must show that the difference quotient diverges at all x. Thus we seek to analyze:
[tex]\mathop {\lim }\limits_{a \to x}\frac{f(x)f(a)}{xa}[/tex] If we can show that for any such x, this limit increases without bound as additional elements of the function sequence are added, then we will have shown the derivative is undefined at x. It suffices therefore, to consider the absolute value of the quotient: [tex]\mathop{\lim}\limits_{a\to x}\frac{f(x)f(a)}{xa}[/tex] In order to evaluate this limit, f(x) is decomposed into the three functions stated earlier which are dependent on the summation index n. Later, we will evaluate the limit as n approaches infinity. Thus we analyze: [tex]k_n(x)k_n(a)[/tex] [tex]h_n(x)h_n(a)[/tex] [tex]L_n(x)L_n(a)[/tex] Lemma A: If[itex] K\geq 3[/itex] is an integer and [itex]x\in D_f[/itex] (x is in the domain of the function), then there exists an [itex]a\in D_f[/itex], such that: [tex]\frac{\pi}{K}<xa\leq\frac{3\pi}{K}[/tex] and: [tex]Sin[Kx]Sin[Ka]\geq 1[/tex] This is the central idea in the whole proof and needs to be fully understood in order to proceed further. Proof: Consider the plot of the function Sin[Kx] and an arbitrary point, x, along it as well as the following interval I'll call I: [tex]I=(x+\frac{\pi}{K},x+\frac{3\pi}{K}][/tex] An attached plot shows this setup for K=6 and x=1.2. Note that within I, Sin[Kx] takes on all values between 1 and 1. Any value, a, within this interval will be in absolute value: [tex]\frac{\pi}{K}<xa\leq\frac{3\pi}{K}[/tex] But since Sin[Kx] takes on all values between 1 and 1 in this interval, I can choose a such that Sin[Ka] is either 1 or 1 so that: [tex]Sin[Kx]Sin[Ka]\geq 1[/tex] QED Using this Lemma, we can now evaluate [itex]k_n(x) [/itex] which I'll do next. 



#27
Apr905, 07:01 PM

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Using Lemma A, for any [itex]x\in D_f[/itex], we can find an "a" in the neighborhood of x such that:
[tex]\frac{\pi}{(n!)^2}<xa\leq\frac{3\pi}{(n!)^2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)[/tex] Note that as n goes to infinity, a approaches x. Thus, the final limit will be evaluated as n goes to infinity. Using the definition of k(x) above, we have: [tex]k_n(x)k_n(a)=\frac{1}{n!}Sin[(n!)^2 x]Sin[(n!)^2 a][/tex] which, according to Lemma A is bounded from below, thus giving:: [tex]k_n(x)k_n(a)\geq\frac{1}{n!}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)[/tex] The determination of a bound for h(x) is a very interesting problem and deserves its own post: 



#28
Apr905, 09:47 PM

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In order to analyze [itex]h_n(x)[/itex], the following Lemma is needed:
Lemma B: For all [itex]x,a\in D_f[/itex]: [tex]Sin[x]Sin[a]\leqxa[/tex] Proof: Using the Mean Value Theorem we have: [tex]Sin[x]Sin[a]=Cos[c](xa)[/tex] Thus: [tex]Sin[x]Sin[a]\leqxa[/tex] QED The following is as messy as the proof gets: Using the definition of [itex]k_n(x)[/itex] above, we have: [tex]h_n(x)h_n(a)\leq\sum_{r=0}^{n1}\frac{1}{r!}Sin[(r!)^2x]Sin[(r!)^2a][/tex] (The triangle inequality) Using Lemma A: [tex]\sum_{r=0}^{n1}\frac{1}{r!}Sin[(r!)^2x]Sin[(r!)^2a]\leq\sum_{r=0}^{n1}\frac{1}{r!}(r!)^2x(r!)^2a[/tex] Now: [tex]\sum_{r=0}^{n1}\frac{1}{r!}(r!)^2x(r!)^2a=\sum_{r=0}^{n1}(r!)xa[/tex] and: [tex]\sum_{r=0}^{n1}(r!)xa=[1+2!+3!+ . . . +(n1)!]xa[/tex] [tex]=[(n1)!+\sum_{r=0}^{n2}r!]xa[/tex] Now: [tex]\sum_{r=0}^{n2}r!\leq\sum_{r=0}^{n2}(n2)![/tex] Thus we have: [tex]h_n(x)h_n(a)\leq[(n1)!+\sum_{n=0}^{n2}(n2)!]xa[/tex] Since: [tex]\sum_{r=0}^{n2}(n1)!=(n1)(n2)![/tex] This reduces to: [tex]h_n(x)h_n(a)\leq 2(n1)!xa[/tex] Since: [tex]xa\leq\frac{3\pi}{(n!)^2}[/tex] Finally then: [tex]h_n(x)h_n(a)\leq\frac{6\pi}{n(n!)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)[/tex] I know, this is messy but it involves a lot of good work with summations, absolute values, and inequalities. The rest is lesscomplicated. 



#29
Apr1005, 05:14 PM

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Lemma C:
[tex]\sum_{r=n}^{\infty}\frac{1}{r!}\leq\frac{2}{n!} [/tex] For [itex]n\geq 2[/itex] Proof: [tex]\sum_{r=n}^{\infty}\frac{1}{r!}=\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{ (n+2)!}+. . .[/tex] [tex]=\frac{1}{n!}[1+\frac{1}{n+1}+\frac{1}{(n+2)(n+1)}+\frac{1}{(n+3)(n+2)(n+1)}+. . .][/tex] [tex]\leq\frac{1}{n!}[1+\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+. . .][/tex] Since the last term is a geometric series, we have: [tex]\sum_{r=n}^{\infty}\frac{1}{r!}\leq\frac{2}{n!}[/tex] QED Analysis of [itex]L_n(x)[/itex] Since: [tex]L_n(x)\leq\sum_{r=n+1}^{\infty}\frac{1}{r!}[/tex] Using Lemma C we have: [tex]L_n(x)\leq\frac{2}{(n+1)!}[/tex] Thus: [tex]L_n(x)L_n(a)\leqL_n(x)+L_n(a)\leq\frac{4}{(n+1)!}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)[/tex] We now have all the expressions needed to analyze the limit of the difference quotient as n goes to infinity. I'll complete the proof next. 



#30
Apr1005, 06:42 PM

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We now have:
[tex]xa\leq\frac{3\pi}{(n!)^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)[/tex] [tex]k_n(x)k_n(a)\geq\frac{1}{n!}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)[/tex] [tex]h_n(x)h_n(a)\leq\fraq{6\pi}{n(n!)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)[/tex] [tex]L_n(x)L_n(a)\leq\fraq{4){(n+1)!}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)[/tex] Recalling: [tex]f(x)=k(x)+h(x)+L(x)[/tex] and noting that we have expressed each term in terms of n, we wish to evaluate: [tex]\mathop{\lim}\limits_{n\to \infty}\frac{(k_n(x)+h_n(x)+L_n(x))(k_n(a)+h_n(a)+L_n(a))}{xa}[/tex] A short Lemma to assist with evaluating this absolute value: Lemma D: [tex]a+b+c\geqabc[/tex] Proof: [tex]a=(a+b+c)(b+c)\leqa+b+c+b+c[/tex] [tex]ab+c\leqa+b+c[/tex] Since: [tex]b+c\leqb+c[/tex] we have: [tex]abc\leqab+c\leqa+b+c[/tex] QED Thus: [tex]f(x)f(a)\geqk_n(x)k_n(a)h_n(x)h_n(a)L_n(x)L_n(a)[/tex] [tex]\ \ \ \ \ \ \geq\frac{1}{n!}\frac{6\pi}{n(n!)}\frac{4}{(n+1)!}[/tex] Since: [tex]\frac{1}{n!}\frac{6\pi}{n(n!)}\frac{4}{(n+1)!}=\frac{1}{n!}(1\frac{6\pi}{n}\frac{4}{n+1})[/tex] And: [tex]\frac{1}{n!}(1\frac{6\pi}{n}\frac{4}{n+1})\geq\frac{1}{n!}(1\frac{30}{n!})[/tex] Finally, for n>60, [tex]\frac{1}{n!}(1\frac{30}{n!})\geq\frac{1}{2(n!)}[/tex] Substituting this expression and the results from equation (1) above, [tex]\mathop{\lim}\limits_{n\to\infty}\frac{(k_n(x)+h_n(x)+L_n(x))(k_n(a)+h_n(a)+L_n(a))}{xa}\geq\frac{\frac{1}{2(n!)}}{\frac{3\pi}{(n!)^2}}=\frac{n!}{6\pi}[/tex] Thus for [itex]n\geq 60[/itex], the limit goes to infinity proving that the derivative does not exist for arbitrary point x in the domain. QED Initially, I was overwhelmed with this proof and was having a difficult time. The author made an interesting point: "reflection and the passage of time will help a great deal". He was right. Now I understand it fully and am comfortable with its construction. It's now a relatively simple matter to prove the same for the Cosine series which I'll do in a brief summary report next time. 



#31
Apr1205, 06:32 AM

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"relatively simple matter"
. . . . . . yea, right . . . . . I'm wrong and I ain't proud. Can someone show me how to prove: [tex]f(x)=\sum_{r=0}^{\infty} A^r Sin[B^r x][/tex] Is nowhere differentiable. Even with all I did above I can't apply it to this equation with A<1 and B an integer: I end up getting the difference quotient is larger than minus infinity which is meaningless. 



#32
Apr1205, 06:03 PM

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I was incorrectly interpreting the partial sums of a geometric series. It should be:
[tex]\sum_{r=0}^{n1}(\frac{1}{A})^r+(\frac{1}{A})^n+\sum_{r=n+1}^{\infty}(\frac{1}{A})^r=[/tex] [tex]\frac{1(\frac{1}{A})^n}{1\frac{1}{A}}+(\frac{1}{A})^n+\frac{(\frac{1}{A})^{n+1}}{1\frac{1}{A}}[/tex] The important point is to devise sizes for A and B such that the following partial differences are less than 1. The most difficult is for [itex]h_n(x)[/itex]. I rationalized that since factorials squared were being used above, then I should try to make the B term quite large with respect to A. In the following example, I used [itex]B=A^2[/itex]. [tex]k_n(x)k_n(x_0)\geq(\frac{1}{A})^n[/tex] [tex]I_n(x)I_n(x_0)\leq\frac{2}{A1}(\frac{1}{A})^n[/tex] [tex]h_n(x)h_n(x_0)\leq(\frac{A^n1}{A1})\frac{3\pi}{A^nA^n}[/tex] Plugging this into the differential difference quotient leads to the following expression: [tex]\mathop{\lim}\limits_{n\to\infty}\frac{(\frac{1}{A})^n[1\frac{2}{a1}\frac{3\pi(A^n1)}{A^n(A1)}]}{\frac{3\pi}{B^n}}[/tex] With [itex]B=A^2[/itex] This limit tends to infinity if [itex]A\geq 13[/itex] Thus, [tex]f(x)=\sum_{r=0}^\infty(\frac{1}{13})^r Sin[(13)^{2r}x][/tex] is nowhere differentiable. (this is not a proof) 


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