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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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Answer:

The value of x = 9

Step-by-step explanation:

Given the angles of the triangle

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We know that the sum of angles of a triangle is 180°.

Therefore,

(13x+2)° + (5x-7)° + (3x-4)° = 180°

13x+2 + 5x-7 + 3x-4 = 180°

21x + 2 - 11 = 180°

21x - 9 = 180°

21x = 180° + 9

21x = 189

divide both side sby 21

21x/21 = 189/21

x = 9

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cricket20 [7]

Line CB touches circle A only at one point (point B), hence CB is a tangent.

<h3>What is a circle?</h3>

A circle is the locus of a point such that its distance from a fixed point (center) is always constant.

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