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iren [92.7K]
3 years ago
8

find the Taylor Series for tan−1(x) based at 0. Give your answer using summation notation and give the largest open interval on

which the series converges. (If you need to enter [infinity] , use the [infinity] button in CalcPad or type "infinity" in all lower-case.)
Mathematics
1 answer:
enyata [817]3 years ago
6 0

Let f(x)=\tan^{-1}x. Then f'(x)=\frac1{1+x^2}. Note that f(0)=0.

Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{n\ge0}x^n

This means that for |-x^2|=|x|^2, or -1, we have

\displaystyle f'(x)=\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n\ge0}(-x^2)^n=\sum_{n\ge0}(-1)^nx^{2n}

Integrate the series to get

f(x)=f(0)+\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}

\implies\tan^{-1}x=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}

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Part B:

Given that <span>her score is always in between the lowest score and the highest score shown.

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Part C:

Given that <span>her score is always in between the lowest score and the highest score shown.

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Part D:

Given that <span>her score is always in between the lowest score and the highest score shown.

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3 years ago
Simplify this expression
I am Lyosha [343]
Hello there!

Let's put these into simpler terms:
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We are now left with:
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I hope this helps!
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