How do you find the power series for #f'(x)# and #int f(t)dt# from [0,x] given the function #f(x)=Sigma n^-nx^n# from #n=[1,oo)#?

Question

1490 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-30T12:39:56

#f'(x) = sum_(n=1)^oo (x/n)^(n-1)#

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

and both series converge absolutely for every #x in RR#

Write the series as:

#sum_(n=1)^oo n^-nx^n = sum_(n=1)^oo (x/n)^n#

For any #x in RR# we can see that choosing #N > 2x# we have:

#x/n < 1/2# for #n > N#

and then:

#(x/n)^n < (1/2)^n# for #n > N#

so that the series is convergent. As the radius of convergence is #R=oo#, we can differentiate and integrate the series term by term for any #x#:

#f'(x) = sum_(n=1)^oo d/dx ((x/n)^n) = sum_(n=1)^oo (nx^(n-1))/n^n = sum_(n=1)^oo (x/n)^(n-1)#

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