Winter 2004#5

Problem Statement: Let f_M(x)=\lim\limits_{N\rightarrow\infty}[\cos(M!\pi x)]^{2N} be defined on [a,b]. (a) Find g(x)=\lim\limits_{M\rightarrow\infty}f_M(x). (b) State a definition of Riemann Integrable and use it to prove or disprove g(x) is Riemann Integrable on [a,b].

(a) g(x)=\lim\limits_{M\rightarrow\infty}\lim\limits_{N\rightarrow\infty}[\cos(M!\pi x)]^{2N}

Note that -1\leq \cos x\leq 1 for x\in [a,b] and that 0\leq \cos^2x\leq 1 for x\in [a,b]. We know that \cos^2 x=1 if and only if x=k\pi for some k\in\mathbb{Z}. With this in mind we may re-write the above limit as g(x)=\lim\limits_{M\rightarrow\infty}\lim\limits_{N\rightarrow\infty}[\cos^2 (M!\pi x)]^N. If M!x\in\mathbb{Z} then it follows that M!x=k for some k\in\mathbb{Z} and so we may write x=\dfrac{k}{M!}\in\mathbb{Q}.

So, if x\notin\mathbb{Q} it follows that 0\leq \cos^2(M!\pi x)<1 and thus, \lim\limits_{N\rightarrow\infty}[\cos^2(M!\pi x)]^{N}=0. If x\in\mathbb{Q} then for some values of M, f_M(x)\neq 1 but as M\rightarrow\infty all of these functions go to 1. This gives us that g(x)=0 if x\notin\mathbb{Q} and g(x)=1 if x\in\mathbb{Q}.

(b) Claim: f_M(x) is not Riemann Integrable on [a,b]. A function f(x) is Riemann Integrable on [a,b] if for every \varepsilon>0 there exists a \delta>0 such that if \pi_1=\{x_0,x_1,\dots,x_n\} and \pi_2=\{y_0,y_1,\dots,y_m\} are partitions of [a,b] such that ||\pi_1||<\delta and ||\pi_2||<\delta then it follows that |\sum\limits_{k=1}^{n}f(\xi_k)(x_k-x_{k-1})-\sum\limits_{j=1}^{m}f(\zeta_j)(y_j-y_{j-1})|<\varepsilon. Note that \xi_k is the associated point in [x_{k-1},x_k] and that \zeta_k is the associated point in [y_{k-1},y_k].

Proof:

Assume that f(x) is Riemann Integrable on [a,b]. Let \varepsilon=\dfrac{b-a}{5}, then there exists a \delta>0 such that |\sum\limits_{k=1}^{n}f(\xi_k)(x_k-x_{k-1})-\sum\limits_{j=1}^{m}f(\zeta_j)(y_j-y_{j-1})|<\varepsilon. Since this must be true for any associated points in \pi_1 and \pi_2 we may choose our associated points.

Pick associated points \xi_k\in\mathbb{Q} for each 0\leq k\leq n and \zeta_j\notin\mathbb{Q} for each 0\leq j\leq m. We may do this since \mathbb{Q} is dense in \mathbb{R}. Since f(x) is Riemann Integrable this implies that

|\sum\limits_{k=1}^{n}f(\xi_k)(x_k-x_{k-1})-\sum\limits_{j=1}^{m}f(\zeta_j)(y_j-y_{j-1})|<\varepsilon

|\sum\limits_{k=1}^n(x_k-x_{k-1})-\sum\limits_{j=1}^m 0|<\dfrac{b-a}{5}

|(b-a)-0|<\dfrac{b-a}{5}

b-a<\dfrac{b-a}{5}

We know that b-a>0 since b>a and so the above statement is a contradiction. Thus, our assumption that f(x) is Riemann Integrable must be incorrect.

\Box

Reflection: Wow, what a problem. The qual writers had a lot of fun writing this one. They gave us this horribly, nasty looking thing that reduced down to the Dirichlet Function. Wow. Ok, enough about the problem itself, now to talk about the math. The hard part of this problem was realizing that g(x) was the Dirichlet function. Once we got to there showing it was not Riemann Integrable was a piece of cake. Finding g(x) really rested on the fact that we could rewrite (\cos M!\pi x)^{2N}=(\cos^2 M!\pi x)^N. Since we can bound \cos^2 M!\pi x between 0 and 1 we know that it will go to 0 if \cos M!\pi x\neq 1, otherwise, \cos M!\pi x will always equal 1 and so it’s limit will also be 1. This trick with this problem is to recognize which values of x cause the function to go to 0 and which values cause the function to go to 1. After recognizing that this depends on if x is an integer or not we are able to get the Dirichlet Function and the rest is a piece of cake.

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This entry was posted in Analysis, Math, Riemann Integrable, Sequence of Functions. Bookmark the permalink.

2 Responses to Winter 2004#5

  1. Morgan Sherman says:

    Nice solution! One correction I would make: if x \in \mathbb{Q} then f_M(x) *does* depend on M. For example if M=1 and x=1/2 then f_M(x) \neq 1. However, for this particular value of x, you’ll see that f_M(x) =1 for every M > 1….

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