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Christian - I Still Love H.E.R.
Lv 6
Christian - I Still Love H.E.R. asked in Science & MathematicsMathematics · 6 years ago

Find the Maclaurin series and radius of convergence for f(x) = 1/(16 - x)^1/4?

Not sure how to start this problem. Can anyone help me out?

Maclaurin series for:

f(x) = 1/(16 - x)^1/4

Thanks!

1 Answer

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  • kb
    Lv 7
    6 years ago

    Using the Binomial series:

    f(x) = (16 - x)^(-1/4)

    ......= (16 (1 - x/16))^(-1/4)

    ......= 16^(-1/4) (1 - x/16))^(-1/4)

    ......= (1/2) (1 + (-x/16))^(-1/4)

    ......= (1/2) [1 + Σ(n = 1 to ∞) ((-1/4)(-1/4 - 1)...(-1/4 - n + 1)/n!) (-x/16)^n]

    This series converges (save endpoints) when |-x/16| < 1

    ==> |x| < 16.

    So, the radius of convergence equals 16.

    I hope this helps!

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