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Math Help - Maclaurin series expansion

  1. #1
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    Maclaurin series expansion

    I want to calculate the Maclaurin series for log(1+x^2) by using a previous result, that log(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + ...

    The following is not correct, but I don't know why, nor how to do it the right way

    let u=x^2. Then

    log(1+x^2) = log(1+u) = u - \frac{1}{2}u^2 + \frac{1}{3}u^3 - \frac{1}{4}u^4 + ... = x^2 - \frac{1}{2}x^4 + \frac{1}{3}x^6 - \frac{1}{4}x^8 + ...

    Thanks for any help.
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  2. #2
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    Quote Originally Posted by HD09 View Post
    I want to calculate the Maclaurin series for log(1+x^2) by using a previous result, that log(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + ...

    The following is not correct, but I don't know why, nor how to do it the right way

    let u=x^2. Then

    log(1+x^2) = log(1+u) = u - \frac{1}{2}u^2 + \frac{1}{3}u^3 - \frac{1}{4}u^4 + ... = x^2 - \frac{1}{2}x^4 + \frac{1}{3}x^6 - \frac{1}{4}x^8 + ...

    Thanks for any help.
    Why is that not correct (on [-1,1])?
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  3. #3
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    Because my solutions say so and the long-way calculation I did (which turns out to have a mistake in it now I check again) agreed with those instead of this!! Thanks. I was flabbergasted at how this could possibly be incorrect... what a relief.
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  4. #4
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    Now you've got me curious! What was the "solution sheet" answer that "the long way" agreed with?
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  5. #5
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    It only gave the first two terms but the solutions claim the co-efficient of the second term is 1/4 rather than 1/2
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