Question

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)`?

Answer 1

`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`

Answer 2

`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.