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geduiwelh

Answered question

2021-02-23

Find a power series representation centered at 0 for the given function using known power series.

f (x) = \ln \sqrt{1 - x^{5}}

lamanocornudaW

Skilled2021-02-24Added 85 answers

Consider the function

f (x) = \ln \sqrt{1 - x^{5}}

We have to find the power series representation centered at 0 for the given function

f (x) = \ln \sqrt{1 - x^{5}}

Use property

{\ln a}^{b} = b \ln a

Hence,

f (x) = \ln \sqrt{1 - x^{5}}

= {\ln (1 - x^{5})}^{\frac{1}{2}}

= \frac{1}{2} \ln (1 - x^{5})

Hence

f (x) = \frac{1}{2} \ln (1 - x^{5})

First differentiate

f (x) = \frac{1}{2} \ln (1 - x^{5})

with respect to x

f^{'} (x) = \frac{d}{d x} (\frac{1}{2} \ln (1 - x^{5}))

= \frac{1}{2} \frac{d}{d x} (\ln (1 - x^{5}))

= \frac{1}{2} \cdot \frac{1}{1 - x^{5}} \times (- 5 x^{4})

= - \frac{5}{2} \cdot \frac{x^{4}}{1 - x^{5}}

Hence,

f^{'} (x) = - \frac{5}{2} \cdot \frac{x^{4}}{1 - x^{5}}

Now, use the series expansion centered at 0 of

\frac{1}{1 - x}

, which is given below

\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n}

Hence,

f^{'} (x) = - \frac{5 x^{4}}{2} \frac{1}{1 - x^{5}}

= - \frac{5 x^{4}}{2} \sum_{n = 0}^{\infty} {(x^{5})}^{n}

= - \frac{5 x^{4}}{2} \sum_{n = 0}^{\infty} x^{5 n}

= - \frac{5}{2} \sum_{n = 0}^{\infty} x^{5 n + 4}

Hence,

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