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Math Help - Riemann Integrability

  1. #1
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    Riemann Integrability

    I have no idea how to prove this. Can someone help?
    Given a real number x, denote by [x] the largest integer less than or equal to x. Prove that the function defined
    f(x) = \sum_{n = 0}^{\infty}\frac{nx-[nx]}{2^n} is Riemann integrable on any interval [a,b]. Then compute
    \int_0^1 f(x)dx
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  2. #2
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    Quote Originally Posted by namelessguy View Post
    I have no idea how to prove this. Can someone help?
    Given a real number x, denote by [x] the largest integer less than or equal to x. Prove that the function defined
    f(x) = \sum_{n = 0}^{\infty}\frac{nx-[nx]}{2^n} is Riemann integrable on any interval [a,b]. Then compute
    \int_0^1 f(x)dx
    for n = 0 the term in your series is 0. so i'll assume that n > 0. let f_n(x)=\frac{nx - [nx]}{2^n}, \ n \geq 1.

    each f_n is integrable on any interval [a, b], because it's continuous almost everywhere on the

    interval. let \sigma_n(x)=\sum_{j=1}^n f_j(x). so \sigma_n is integrable on [a, b] for all n.

    since 0 \leq f_n(x) \leq \frac{1}{2^n}, \ \forall x, by Weierstrass test, \{\sigma_n \} is uniformly convergent. so f(x)=\lim_{n\to\infty} \sigma_n(x)

    is integrable on [a, b] and \int_a^b f(x) dx = \lim_{n\to\infty} \int_a^b \sigma_n(x) dx. if a = 0 and b = 1, then this gives us:

    \int_0^1 f(x) dx = \lim_{n\to\infty} \int_0^1 \sigma_n(x) dx = \lim_{n\to\infty} \int_0^1 \sum_{j=1}^n f_j(x) \ dx=\lim_{n\to\infty} \sum_{j=1}^n \int_0^1 f_j(x) dx

    =\lim_{n\to\infty} \sum_{j=1}^n \int_0^1 \frac{jx -[jx]}{2^j}dx=\lim_{n\to\infty} \sum_{j=1}^n \frac{1}{2^{j+1}}=\sum_{j=1}^{\infty} \frac{1}{2^{j+1}}=\frac{1}{2}. \ \ \ \square
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  3. #3
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    Quote Originally Posted by namelessguy View Post
    I have no idea how to prove this. Can someone help?
    Given a real number x, denote by [x] the largest integer less than or equal to x. Prove that the function defined
    f(x) = \sum_{n = 0}^{\infty}\frac{nx-[nx]}{2^n} is Riemann integrable on any interval [a,b]. Then compute
    \int_0^1 f(x)dx
    I will do the special case [a,b]=[0,1]. Define a sequence of functions f_n:[0,1]\mapsto \mathbb{R} as f_n(x) = \tfrac{1}{2^n}(nx-[nx]). Notice that |f_n(x)| \leq \tfrac{1}{2^n}. Since \sum _{n=0}^{\infty}\tfrac{1}{2^n} < \infty it follows by the Weierstrass test that the series of functions \sum_{n=0}^{\infty} f_n(x) converges uniformly to a function f(x). Each function f_n(x) is discontinous at \tfrac{k}{n} for 1\leq k<n. This implies that f is integrable and \int_0^1 f(x) dx = \sum_{n=0}^{\infty}\int_0^1 f_n(x) dx. It remains to compute \int_0^1 f_n(x) dx. The graph of f_n consists of a n congruent right-angles triangles each having height \tfrac{1}{2^n} and with base width 1/n. Thus, \int_0^1 f_n(x) dx = \frac{1}{2} n\cdot \frac{1}{n} \cdot \frac{1}{2^n} = \frac{1}{2^{n+1}}. I might have missed maybe \pm 1 on the exponent, but you get the idea. Now this is a geometric series which sums easily.

    EDIT: Got beaten in the response by a fellow algebraist.
    Last edited by ThePerfectHacker; May 15th 2008 at 04:43 PM.
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