Question

If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Answer 1

Given that
`f^((n))(0)=(n+1)!`, we have for
`f(x)` the Taylor series expansion about 0 as


`f(x)=\displaystyle\sum_(n=0)^\infty((n+1)!)/(n!)x^n=\sum_(n=0)^\infty(n+1)x^n`

Replace
`n+1` with
`n`, so that the series is equivalent to


`f(x)=\displaystyle\sum_(n=1)^\infty nx^(n-1)`

and notice that


`\displaystyle(\mathrm d)/(\mathrm dx)\sum_(n=0)^\infty x^n=\sum_(n=1)^\infty nx^(n-1)`

Recall that for
`|x|<1`, we have


`\displaystyle\sum_(n=0)^\infty x^n=\frac1{1-x}`

which means


`f(x)=\displaystyle\sum_(n=1)^\infty nx^(n-1)=(\mathrm d)/(\mathrm dx)\frac1{1-x}`

`\implies f(x)=\frac1{(1-x)^2}`