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Thread: Approximating an infinite series

  1. #1
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    Approximating an infinite series

    Approximate a sum for the infinite series:
    $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^4}$
    with a maximum error of $\displaystyle 0.02$

    Answer:
    $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^4}\approx 1.0748$

    My attempt:
    For every $\displaystyle N\in\mathbb{N}$:
    $\displaystyle \int\limits_{N+1}^t\frac{1}{x^4}\, dx$$\displaystyle =\left [ \frac{1}{3x^3} \right ]_{N+1}^t=\frac{1}{3(N+1)^3}-\frac{1}{3t^3}$

    For $\displaystyle t\rightarrow \infty$:
    $\displaystyle \int\limits_{N+1}^t\frac{1}{x^4}\, dx$$\displaystyle =\frac{1}{3(N+1)^3}$

    $\displaystyle f(x)=\frac{1}{x^4}$
    $\displaystyle \int\limits_{N+1}^{\infty}f(x)\, dx$$\displaystyle =\int\limits_{N+1}^{\infty}\frac{1}{x^4}\, dx$$\displaystyle =\frac{1}{3(N+1)^3}$
    $\displaystyle f(N+1)=\frac{1}{(N+1)^4}$

    $\displaystyle \int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)=\frac{1}{3(N+1)^3}+\frac{1}{(N+1)^4}=\frac{ 1}{3}\frac{N+4}{(N+1)^4}$

    Expanding we get:
    $\displaystyle \int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)=\frac{1}{3}\frac{N+4}{n^4+4N^3+6N^2+4N+1}$$\displaystyle =\frac{N+4}{N(3N^3+12N^2+18N+12)+3}$

    Since:
    $\displaystyle N(3N^3+12N^2+18N+12)+3\ge N(3N^3+12N^2+18N+12)$

    We get that:
    $\displaystyle \int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)$$\displaystyle \le \frac{N+4}{N(3N^3+12N^2+18N+12)}\le \frac{1}{N(3N^3+12N^2+18+\frac{1}{3})}$

    Which shows us that:
    $\displaystyle \frac{1}{N(3N^3+12N^2+18+\frac{1}{3})}\le 0.02$

    However when I try to solve for $\displaystyle N$ I get a negative $\displaystyle N$-value which doesn't make sense. Can somebody help me get the correct $\displaystyle N$
    Last edited by MathIsOhSoHard; Oct 8th 2012 at 06:45 PM.
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  2. #2
    Behold, the power of SARDINES!
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    Re: Approximating an infinite series

    Quote Originally Posted by MathIsOhSoHard View Post
    Approximate a sum for the infinite series:
    $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^4}$
    with a maximum error of $\displaystyle 0.02$

    Answer:
    $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^4}\approx 1.0748$

    My attempt:
    For every $\displaystyle N\in\mathbb{N}$:
    $\displaystyle \int\limits_{N+1}^t\frac{1}{x^4}\, dx$$\displaystyle =\left [ \frac{1}{3x^3} \right ]_{N+1}^t=\frac{1}{3(N+1)^3}-\frac{1}{3t^3}$

    For $\displaystyle t\rightarrow \infty$:
    $\displaystyle \int\limits_{N+1}^t\frac{1}{x^4}\, dx$$\displaystyle =\frac{1}{3(N+1)^3}$

    $\displaystyle f(x)=\frac{1}{x^4}$
    $\displaystyle \int\limits_{N+1}^{\infty}f(x)\, dx$$\displaystyle =\int\limits_{N+1}^{\infty}\frac{1}{x^4}\, dx$$\displaystyle =\frac{1}{3(N+1)^3}$
    $\displaystyle f(N+1)=\frac{1}{(N+1)^4}$

    $\displaystyle \int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)=\frac{1}{3(N+1)^3}+\frac{1}{(N+1)^4}=\frac{ 1}{3}\frac{N+4}{(N+1)^4}$

    Expanding we get:
    $\displaystyle \int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)=\frac{1}{3}\frac{N+4}{n^4+4N^3+6N^2+4N+1}$$\displaystyle =\frac{N+4}{N(3N^3+12N^2+18N+12)+3}$

    Since:
    $\displaystyle N(3N^3+12N^2+18N+12)+3\ge N(3N^3+12N^2+18N+12)$

    We get that:
    $\displaystyle \int\limits_{N+1}^{\infty}f(x)\, dx + f(N+1)$$\displaystyle \le \frac{N+4}{N(3N^3+12N^2+18N+12)}\le \frac{1}{N(3N^3+12N^2+18+\frac{1}{3})}$

    Which shows us that:
    $\displaystyle \frac{1}{N(3N^3+12N^2+18+\frac{1}{3})}\le 0.02$

    However when I try to solve for $\displaystyle N$ I get a negative $\displaystyle N$-value which doesn't make sense. Can somebody help me get the correct $\displaystyle N$
    Solving that 4th order equation for N is a nightmare. You can get an approximate value

    $\displaystyle 3N^4< N(3N^3+12N^2+18+\frac{1}{3}) \iff \frac{1}{N(3N^3+12N^2+18+\frac{1}{3})} < \frac{1}{3N^4}$

    This much better to solve

    $\displaystyle \frac{1}{3N^4} < \frac{2}{100} \iff 100 < 6n^4 $

    You can see that $\displaystyle n=3$ works
    Thanks from MathIsOhSoHard
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  3. #3
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Approximating an infinite series

    $\displaystyle \sum _{k=1}^{\infty } \frac{1}{k^4}$=$\displaystyle 1+\frac{1}{2^4}+\frac{1}{3^4}$

    $\displaystyle \frac{1}{3^4}<.02$
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  4. #4
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    Re: Approximating an infinite series

    Hey MathIsOhSoHard.

    Can you show us how you solved the roots of the equation N(3N^3 + 12N^2) + 18 + 1/3)*0.02 - 1 = 0? (Knowing the roots means you know which values of N will be on either side of the inequality).
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