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Math Help - problem of Integral proof

  1. #1
    kin
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    problem of Integral proof



    how to solve (b) ......?
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    MHF Contributor chisigma's Avatar
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    You have to take into account that for |x|<1 is...

    \frac{x^{4}\cdot (1-x)^{4}}{1-x^{2}}= (1-x)^{4}\cdot (x^{4} + x^{6} + x^{8} + \dots)

    Kind regards

    \chi \sigma
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  3. #3
    kin
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    Quote Originally Posted by chisigma View Post
    You have to take into account that for |x|<1 is...

    \frac{x^{4}\cdot (1-x)^{4}}{1-x^{2}}= (1-x)^{4}\cdot (x^{4} + x^{6} + x^{8} + \dots)

    Kind regards

    \chi \sigma

    it's \frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}} not (1-x),,,,,,,,,
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by kin View Post
    it's \frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}} not (1-x)...
    Very sorry!...

    ... in such a case is...

    \frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}}= (1-x)^{4}\cdot (x^{4} - x^{6} + x^{8} -...), |x|<1

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by kin View Post

    how to solve (b) ......?
    It seems that you're not expected to use a) to evaluate the integral. The simplest way anyway is probably to expand the numerator, and perform the Euclidian division of X^4(1-X)^4 by 1+X^2 (like with integers, you know?), and get something like X^4(1-X)^4=Q(X)(1+X^2)+aX+b hence your integral equals \int_0^1 Q(x)dx + a\int_0^1 \frac{x}{1+x^2}dx + b\int_0^1\frac{dx}{1+x^2}. The first term is easy, the second term has antiderivative \frac{1}{2}\log(1+x^2), and the third one relates to arctan.

    For the second part, you probably have to write 1\leq 1+x^2\leq 2 hence \frac{1}{2}x^4(1-x)^4\leq \frac{x^4(1-x)^4}{1+x^2}\leq x^4(1-x)^4 and integrate.
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  6. #6
    kin
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    Quote Originally Posted by chisigma View Post
    Very sorry!...

    ... in such a case is...

    \frac{x^{4}\cdot (1-x)^{4}}{1+x^{2}}= (1-x)^{4}\cdot (x^{4} - x^{6} + x^{8} -...), |x|<1

    Kind regards

    \chi \sigma
    but i still dont understand your point...
    how does it help to solve the integral..?
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  7. #7
    kin
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    Quote Originally Posted by Laurent View Post
    It seems that you're not expected to use a) to evaluate the integral. The simplest way anyway is probably to expand the numerator, and perform the Euclidian division of X^4(1-X)^4 by 1+X^2 (like with integers, you know?), and get something like X^4(1-X)^4=Q(X)(1+X^2)+aX+b hence your integral equals \int_0^1 Q(x)dx + a\int_0^1 \frac{x}{1+x^2}dx + b\int_0^1\frac{dx}{1+x^2}. The first term is easy, the second term has antiderivative \frac{1}{2}\log(1+x^2), and the third one relates to arctan.

    For the second part, you probably have to write 1\leq 1+x^2\leq 2 hence \frac{1}{2}x^4(1-x)^4\leq \frac{x^4(1-x)^4}{1+x^2}\leq x^4(1-x)^4 and integrate.
    thank u very much ,
    this is a straightforward method....
    are there any ways of combining result of (a) to solve it quickly...?
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  8. #8
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    which book are these problems from?
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  9. #9
    kin
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    Quote Originally Posted by debanik1 View Post
    which book are these problems from?
    they are not from books....just from my tutorial notes...

    anyway, i have solved it.

    thanks for help , guys...
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