Find Maclaurin series for this f ( x ) = x 3 tan − 1 ⁡ ( 2 x ) ; | x | &lt; 1 2

Question

Find Maclaurin series for this  f  (  x  )  =      x    3        tan          −      1        ⁡  (  2  x  )  ;        |    x      |    &amp;lt;      1    2

Kelsie Marks · Accepted Answer

Hint: The Taylor series for       ∑          n      =      0              ∞        (  −  1      )          n                  x              2        n        +        1                    2      n      +      1       when       |    x      |    ≤      1  .

Nica2t · Accepted Answer

For the geometric series with common ratio

- x^{2}

we have:

\sum_{n = 0}^{\infty} (- x^{2})^{n} = 1 - x^{2} + x^{4} - x^{6} + \dots = \frac{1}{1 + x^{2}}

for

| x | < 1.

. Integrate

C + \sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{2 n + 1}}{2 n + 1} = \tan^{- 1} x .

Plug in

x = 0

to discover that

C = 0.

. Replace

x

by

2 x

to get the series for

\tan^{- 1} 2 x

then multiply the whole lot by

x^{3}

.

Find Maclaurin series for this f(x)=x^(3)tan^(−1)(2x); |x|<1/2

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Answer & Explanation

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