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(a) Find a power series representation for the function

f(x)=x2tan−1(x^3) and determine its radius of convergence.
(b) Find the Taylor series for f(x)=cosx centered at π.

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(a) To find a power series representation for f(x), we can use the formula for the power series representation of tan^-1(x):

tan^-1(x) = ∑ (-1)^n * (x^(2n+1)) / (2n+1)

Substituting x^3 into this formula, we get:

tan^-1(x^3) = ∑ (-1)^n * ((x^3)^(2n+1)) / (2n+1)
= ∑ (-1)^n * (x^(6n+3)) / (2n+1)

Multiplying this by x^2, we get:

x^2 * tan^-1(x^3) = ∑ (-1)^n * (x^(6n+5)) / (2n+1)

So the power series representation for f(x) is:

f(x) = x^2 * tan^-1(x^3) = ∑ (-1)^n * (x^(6n+5)) / (2n+1)

To find the radius of convergence, we can use the ratio test:

lim |(-1)^n * (x^(6(n+1)+5)) / (2(n+1)+1)| / |(-1)^n * (x^(6n+5)) / (2n+1)|
= lim |x^6 / (12n+9)|
= 0

So the radius of convergence is infinity.

(b) To find the Taylor series for f(x) cos(x) centered at x = 1, we can use the formula for the Taylor series of the product of two functions:

(fg)(x) = ∑ [f^(n)(a) / n!] * g(x-a)^n

Taking the derivative of f(x) = x^2 * tan^-1(x^3), we get:

f'(x) = 2x * tan^-1(x^3) + (x^2) * (1 / (1 + x^6))

Evaluating this at x = 1, we get:

f'(1) = 2 * tan^-1(1^3) + (1^2) * (1 / (1 + 1^6))
= 2 * (π/4) + 1/2

Taking the derivative again, we get:

f''(x) =
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