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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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elena-14-01-66 [18.8K]
First, you should solve for f(2x), which equals 2*(2x)=4x.  Now, solve the integral of f(2x)=2*(2x)=4x, to get that\int\ {(f(2x)=4x)} \, dx= 2x^2.  You can check this by taking the integral of what you got.  Now by the Fundamental Theorem\int\limits^2_0 {4x} \, dx=[2x^2] ^{2}_{0}=2(2)^{2}-2(0)^2=8.

This should be the answer to your question, if I understood what you were asking correctly. 
8 0
3 years ago
Which one is it I need to know quick and I’ll mark you Brainly
QveST [7]

Answer: i think

the corrcect answer is 0.5*10^-1

Step-by-step explanation:

hope this helps

5 0
2 years ago
1( below is the pre image dilate by a factor of 2
bonufazy [111]

Before:It's too short. Write at least 20 characters to explain it well.

After:It's too short. Write at least 20 characters to explain it well.

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3 years ago
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Step-by-step explanation:

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6 0
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Answer:

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