How do you find the power series for #f'(x)# and #int_0^x f(t)dt# given the function #f(x)=sum_(n=0)^oo x^(2n)# ?
1 Answer
# f'(x) = 0 + 2x^1 + 4x^3 + 6x^5 + ... #
# " " = sum_(n=1)^oo (2n)x^(2n-1) #
# int_0^x \ f(t) \ dt = x + x^3/3 + x^5/5 + x^7/7 + ... #
# " " = sum_(n=0)^oo x^(2n+1)/(2n+1) #
Explanation:
We have
# f(x) = sum_(n=0)^oo x^(2n) #
We can expand the fist few terms to get an idea of how the series behaves:
# f(x) = x^0 + x^2 + x^4 + x^6 + ... #
# " " = 1 + x^2 + x^4 + x^6 + ... #
So first we find the derivative by differentiating term by term:
# f'(x) = 0 + 2x^1 + 4x^3 + 6x^5 + ... #
# " " = sum_(n=1)^oo (2n)x^(2n-1) #
And similarly we find the integral by integrating term by term:
# int_0^x \ f(t) \ dt = int_0^x \ 1 + t^2 + t^4 + t^6 + ... \ dt#
# " " = [t + t^3/3 + t^5/5 + t^7/7 + ... ]_0^x #
# " " = {x + x^3/3 + x^5/5 + x^7/7 + ... } - {0} #
# " " = x + x^3/3 + x^5/5 + x^7/7 + ... #
# " " = sum_(n=0)^oo x^(2n+1)/(2n+1) #
