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(2n 1)2nn! 2 V2T n=0 Since this is an alternating series (because the parity on the power on x means that will always have th    #1 please and the answer should be in the form of a piecewise function

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Answer #1

1.

The probability density function must be positive between -1 and 1.

Also, since it represents picking a number at random, its value must be the same for all numbers

between -1 and 1, then the function is constant in the interval [-1,1]

Therefore the function is of the form

if -1 f(x) kif 1 1 if 1 0

where k is a constant.

Now, recall that the integral of a PDF over all reals must be equal to 1.

Hence

roo kdr 1 f(x)dx

Then

(1)

k(1 (1)) 1

2

Thus, the PDF is

if r 1 if 1 1 if r 0 f (x) 2 0 1

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