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!)/n^nx^n from n=[1,oo)?

2 Answers
Mar 2, 2017

f'(x)=sum_1^(oo) (n!)/n^(n-1)x^(n-1)

int_0^x f(t)dt= sum_1^(oo) (n!)/((n+1)n^n) x^(n+1)

Explanation:

The series for f'(x) is :

f'(x)=sum_1^(oo) d/dx ((n!)/n^nx^n)

=sum_1^(oo) (n!)/n^(n-1)x^(n-1)

From the FTC we know that:

d/dx ( int_0^x f(t)dt) = f(x)

It follows that the series for int_0^x f(t)dt is:

int_0^x f(t)dt=sum_1^(oo)( int (n!)/n^nx^n dx)

=sum_1^(oo) (n!)/((n+1)n^n) x^(n+1) +C

Because int_0^0 f(t)dt = 0, we conclude that C = 0

Mar 2, 2017

f'(x) = sum_(n=1)^oo (n!)/n^(n-1) \ x^(n-1)
" " = 1 + x + 2/3x^2 + 3/8x^3 + 24/125x^4 + ...

int_0^x \ f(t) \ dt = sum_(n=1)^oo \ (n!)/n^n \ x^(n+1)/(n+1)
" " = 1/2x^2+ 1/6x^3 +1/18x^4+ 3/360x^5+...

The radii of convergence for the original and subsequent series is

|x| < 1

Explanation:

We have:

f(x) = sum_(n=1)^oo \ (n!)/n^n \ x^n
" " = x + 1/2x^2 + 2/9x^3 + 3/32x^4 + 24/625x^5 + ...

Let us also define the following functions:

D(x) = f'(x) and I(x) = int_0^x \ f(t) \ dt

Then:

D(x) = d/dx sum_(n=1)^oo \ (n!)/n^n \ x^n
" " = sum_(n=1)^oo (n!)/n^n \ d/dx \ x^n
" " = sum_(n=1)^oo (n!)/n^n \ nx^(n-1)
" " = sum_(n=1)^oo (n!)/n^(n-1) \ x^(n-1)
" " = 1 + x + 2/3x^2 + 3/8x^3 + 24/125x^4 + ...

And:

I(x) = int_0^x \ sum_(n=1)^oo \ (n!)/n^n \ t^n \ dt
" " = sum_(n=1)^oo \ (n!)/n^n \ int_0^x \ \ t^n \ dt
" " = sum_(n=1)^oo \ (n!)/n^n \ [t^(n+1)/(n+1)]_0^x

" " = sum_(n=1)^oo \ (n!)/n^n \ x^(n+1)/(n+1)
" " = 1/2x^2+ 1/6x^3 +1/18x^4+ 3/360x^5+...

For the radii of convergence, we can apply the d'Alembert's ratio test:

Suppose that;

S=sum_(r=1)^oo a_n \ \ , and \ \ L=lim_(n rarr oo) |a_(n+1)/a_n|

Then

if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.

For the original series, the test limit is

L = lim_(n rarr oo) | { ((n+1)!)/(n+1)^(n+1) \ x^(n+1) } / {(n!)/n^n \ x^n } |
\ \ = lim_(n rarr oo) | ((n+1)!)/(n+1)^(n+1) * n^n/(n!) * x^(n+1) / x^n |
\ \ = lim_(n rarr oo) | ((n+1))/(n+1)^(n+1) * n^n * x |
\ \ = lim_(n rarr oo) | 1/(n+1)^n * n^n * x |
\ \ = lim_(n rarr oo) | (n/(n+1))^n * x |
\ \ = lim_(n rarr oo) | (n/(n+1) * (1/n)/(1/n))^n * x |
\ \ = lim_(n rarr oo) | (1/(1+1/n))^n * x |
\ \ = | x |

And so the series for f(x) is convergent if |x| < 1

So for f'(x), the test limit is:

L = lim_(n rarr oo) | { ((n+1)!)/(n+1)^n \ x^n } / { (n!)/n^(n-1) \ x^(n-1) } |
\ \ = lim_(n rarr oo) | ((n+1)!)/(n+1)^n * n^(n-1)/(n!) * x^n/x^(n-1) |
\ \ = lim_(n rarr oo) | (n)/(n+1)^n * n^(n-1) * x |
\ \ = lim_(n rarr oo) | (n^n)/(n+1)^n * x |
\ \ = lim_(n rarr oo) | (n/(n+1))^n * x |
\ \ = lim_(n rarr oo) | (n/(n+1) * (1/n)/(1/n))^n * x |
\ \ = lim_(n rarr oo) | (1/(1+1/n))^n * x |
\ \ = | x |

And so the series for f'(x) is convergent if |x| < 1

And finally, for I(x), the test limit is:

L = lim_(n rarr oo) | { ((n+1)!)/ (n+1)^ (n+1) \ x^(n+2)/(n+2)} / {(n!)/n^n x^(n+1)/(n+1)} |
\ \ = lim_(n rarr oo) | ((n+1)!)/ (n+1)^ (n+1) * n^n/(n!) * x^(n+2)/(n+2) *(n+1)/x^(n+1)
\ \ = lim_(n rarr oo) | ((n+1))/ (n+1)^ (n+1) * n^n * x/(n+2) *(n+1)
\ \ = lim_(n rarr oo) | (1)/ (n+1)^n * n^n * x/(n+2) *(n+1) |
\ \ = lim_(n rarr oo) | (n/(n+1))^n * (n+1)/(n+2) * x |
\ \ = lim_(n rarr oo) | (n/(n+1) * (1/n)/(1/n))^n * (n+1)/(n+2) * (1/n)/(1/n)* x |
\ \ = lim_(n rarr oo) | (1/(1+1/n))^n * (1+1/n)/(1+2/n) * x |
\ \ = \ x |

And so the series for I(x) is convergent if |x| < 1

This should not come as any surprise as the radii of convergence is unaffected by differentiation or integration.