- #1

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show that if x [itex]\in[/itex] [-1/2 , 1/2] then

|f[itex]^{n+1}[/itex](x)| <= 2[itex]^{n + 1}[/itex] * n!

I am having a hard time seeing how 2[itex]^{n + 1}[/itex] * n! comes into play.

I have that the taylor series for f is [itex]\Sigma[/itex] [itex]\frac{x^n}{n}[/itex]

If a take a derivative it becomes x^(n-1) and if I plug anything on the interval it is less than one. I am thinking that I did this wrong because of how big that upper bound is/.