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Thread: Geometric progression?

  1. #1
    Super Member fardeen_gen's Avatar
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    Geometric progression?

    Find the sum of the terms of the G.P., $\displaystyle a + ar + ar^2 + ...\ \infty$ where $\displaystyle a$ is the value of $\displaystyle x$ for which the function $\displaystyle 7 + 2x\log_{e}25 - 5^{x - 1} - 5^{2 - x}$ has the greatest value and $\displaystyle r$ is the limit $\displaystyle \lim_{x\rightarrow 0}\int_0^x \frac{t^2}{x^2\tan(\pi + x)} dt$
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    Find the sum of the terms of the G.P., $\displaystyle a + ar + ar^2 + ...\ \infty$ where $\displaystyle a$ is the value of $\displaystyle x$ for which the function $\displaystyle 7 + 2x\log_{e}25 - 5^{x - 1} - 5^{2 - x}$ has the greatest value and $\displaystyle r$ is the limit $\displaystyle \lim_{x\rightarrow 0}\int_0^x \frac{t^2}{x^2\tan(\pi + x)} dt$
    First

    $\displaystyle r = \lim_{x\rightarrow 0}\int_0^x \frac{t^2}{x^2\tan(\pi + x)} dt = \lim_{x \to 0} \frac{\frac{1}{3}x^3}{x^2\tan(\pi + x)} = \frac{1}{3} \lim_{x \to 0} \frac{x}{\tan(\pi + x)} = \frac{1}{3} $ by L'Hopitals rule.

    If we let $\displaystyle y = 7 + 4x\ln 5 - 5^{x - 1} - 5^{2 - x}$ then using calculus we find the max is located at $\displaystyle x = 2$ giving a maximum of

    $\displaystyle
    a = 8 \ln 5 + 1
    $

    Since $\displaystyle r < $1, then the sum of the series is $\displaystyle S_\infty = \frac{a}{1-r} = \frac{3}{2} (8 \ln 5 + 1) $
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  3. #3
    Super Member fardeen_gen's Avatar
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    Answer is 3 according to the problem book. Have they taken any assumptions?
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  4. #4
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    Quote Originally Posted by danny arrigo View Post
    First

    $\displaystyle r = \lim_{x\rightarrow 0}\int_0^x \frac{t^2}{x^2\tan(\pi + x)} dt = \lim_{x \to 0} \frac{\frac{1}{3}x^3}{x^2\tan(\pi + x)} = \frac{1}{3} \lim_{x \to 0} \frac{x}{\tan(\pi + x)} = \frac{1}{3} $ by L'Hopitals rule.

    If we let $\displaystyle y = 7 + 4x\ln 5 - 5^{x - 1} - 5^{2 - x}$ then using calculus we find the max is located at $\displaystyle x = 2$ giving a maximum of

    $\displaystyle
    a = 8 \ln 5 + 1
    $

    Since $\displaystyle r < $1, then the sum of the series is $\displaystyle S_\infty = \frac{a}{1-r} = \frac{3}{2} (8 \ln 5 + 1) $
    I miss read the question, $\displaystyle a = 2$, the $\displaystyle x$ value of the maximum. I originally used the actual maximum value. So

    $\displaystyle
    S = \frac{a}{1-r} = \frac{2}{1 - \frac{1}{3}} = 3
    $
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