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

User Mvvijesh
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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}
User GNG
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