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Math  /  Calculus

QuestionUse term-by-term differentiation or integration to find a power series centered at x=0x=0 for: f(x)=tan⁡−1(x9)=∑n=0∞f(x)=\tan ^{-1}\left(x^{9}\right)=\sum_{n=0}^{\infty} □\square
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Studdy Solution

STEP 1

1. We need to find a power series representation for f(x)=tan⁡−1(x9) f(x) = \tan^{-1}(x^9) .
2. The power series should be centered at x=0 x = 0 .
3. We can use known power series expansions and manipulate them using differentiation or integration.

STEP 2

1. Recall the power series for tan⁡−1(x) \tan^{-1}(x) .
2. Substitute x9 x^9 into the series.
3. Simplify the resulting series.

STEP 3

Recall the power series expansion for tan⁡−1(x) \tan^{-1}(x) centered at x=0 x = 0 :
tan⁡−1(x)=∑n=0∞(−1)nx2n+12n+1 \tan^{-1}(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}
This series is valid for ∣x∣≤1 |x| \leq 1 .

STEP 4

Substitute x9 x^9 into the power series for tan⁡−1(x) \tan^{-1}(x) :
tan⁡−1(x9)=∑n=0∞(−1)n(x9)2n+12n+1 \tan^{-1}(x^9) = \sum_{n=0}^{\infty} (-1)^n \frac{(x^9)^{2n+1}}{2n+1}
This simplifies to:
tan⁡−1(x9)=∑n=0∞(−1)nx18n+92n+1 \tan^{-1}(x^9) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{18n+9}}{2n+1}

STEP 5

The resulting power series for f(x)=tan⁡−1(x9) f(x) = \tan^{-1}(x^9) centered at x=0 x = 0 is:
f(x)=∑n=0∞(−1)nx18n+92n+1 f(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{18n+9}}{2n+1}
This series represents the function tan⁡−1(x9) \tan^{-1}(x^9) for ∣x9∣≤1 |x^9| \leq 1 , which simplifies to ∣x∣≤1 |x| \leq 1 .
The power series centered at x=0 x = 0 for f(x)=tan⁡−1(x9) f(x) = \tan^{-1}(x^9) is:
∑n=0∞(−1)nx18n+92n+1 \boxed{\sum_{n=0}^{\infty} (-1)^n \frac{x^{18n+9}}{2n+1}}

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