How do you find the power series for #f'(x)# and #int f(t)dt# from [0,x] given the function #f(x)=Sigma 10^nx^n# from #n=[0,oo)#?

1 Answer
Steve M
Feb 5, 2017

# f'(x) = sum_(n=0)^oo \ n10^n x^(n-1) #

# int_0^x \ f(t) \ dt=sum_(n=0)^oo \ 10^n / (n+1) \ x^(n+1) #

Explanation:

We have:

# f(x) = sum_(n=0)^oo \ 10^nx^n #

And so:

# f'(x) = d/dx sum_(n=0)^oo \ 10^nx^n #
# " " = sum_(n=0)^oo \ d/dx10^nx^n #
# " " = sum_(n=0)^oo \ 10^nd/dxx^n #
# " " = sum_(n=0)^oo \ 10^n nx^(n-1) #
# " " = sum_(n=0)^oo \ n10^n x^(n-1) #

And:

# int_0^x \ f(t) \ dt= int_0^x \ sum_(n=0)^oo \ 10^nt^n \ dt#
# " "= sum_(n=0)^oo \ int_0^x \ 10^nt^n \ dt#
# " "= sum_(n=0)^oo \ 10^n \ int_0^x \ t^n \ dt#
# " "= sum_(n=0)^oo \ 10^n \ [ t^(n+1)/(n+1) ]_0^x #

# " "= sum_(n=0)^oo \ 10^n \ { x^(n+1)/(n+1) - 0 } #
# " "= sum_(n=0)^oo \ 10^n \ x^(n+1)/(n+1) #
# " "= sum_(n=0)^oo \ 10^n / (n+1) \ x^(n+1) #

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