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Math Help - taylor approximation near infinity

  1. #1
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    taylor approximation near infinity

    Consider F(x)=[x^4 - x^2 + 1]/[x^4 + 2x^3 + x] for x not = 0

    F(x) tends to 1 as x tends to infinity. When x is large, f(x) is approx but not exactly 1

    Use change of variable y=1/x then do a maclaurin expansion in y to show that f(x) is approx 1-2x for large x and find the third term


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  2. #2
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    Quote Originally Posted by Bryn View Post
    Consider F(x)=[x^4 - x^2 + 1]/[x^4 + 2x^3 + x] for x not = 0

    F(x) tends to 1 as x tends to infinity. When x is large, f(x) is approx but not exactly 1

    Use change of variable y=1/x then do a maclaurin expansion in y to show that f(x) is approx 1-2x for large x and find the third term
    Excellent idea. Have you done that? To start with, what is f(1/x)?


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  3. #3
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    No, I am finding every stage challenging, maybe I think I have been doing a little too much maths and have lost the ability to process, but any suggestions would be appreciated

    thanks again
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  4. #4
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    Quote Originally Posted by Bryn View Post
    Consider F(x)=[x^4 - x^2 + 1]/[x^4 + 2x^3 + x] for x not = 0

    F(x) tends to 1 as x tends to infinity. When x is large, f(x) is approx but not exactly 1

    Use change of variable y=1/x then do a maclaurin expansion in y to show that f(x) is approx 1-2x for large x and find the third term


    Many thanks and happy christmas to anyone celebrating
    Step one: find F(1/x) by replacing each "x" by "1/x".
    F(1/x)= \frac{\frac{1}{x^4}- \frac{1}{x^2}+ 1}{\frac{1}{x^4}+ \frac{2}{x^3}+ \frac{1}{x}}

    Get rid of the "little fractions" by multiplying numerator and denominator by x^4
    F(1/x)= \frac{1- x^2+ x^4}{1+ 2x+ x^3}

    Now find the MacLaurin series for that.
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