taylor series: n-th der. at x = 5 is ((-1)^n n!)0/(2^n(n+2)

kapplayer

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Feb 18, 2008
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The problem is ths one

The Taylor series about x=5 for a certain function f converges to f(x) for all x in the interval of convergence. The nth derivative of f at x = 5 is given by f^(n)(5) = ((-1)^n(n!)0/(2^n(n+2), and f(5)=1/2.

Write the third-degree Taylor polynomial for f about x = 5.

Find the radius of convergence of the Taylor series for f about x = 5.

Show that the sixth-degree Taylor polynomial for f about x = 5 approximates f(6) with error less than 1/1000

So far i have for part a) f(5)+f'(5)(x-5)+(1/2!)f''(5)(x-5)^2+(1/3!)f'''(5)(x-5)^3
how do i find my f'(5), f''(5)......

Im not sure about the rest of the problem
 
Re: taylor series

kapplayer said:
The problem is ths one

The Taylor series about x=5 for a certain function f converges to f(x) for all x in the interval of convergence. The nth derivative of f at x = 5 is given by f^(n)(5) = ((-1)^n(n!)0/(2^n(n+2), and f(5)=1/2.

Write the third-degree Taylor polynomial for f about x = 5.

Find the radius of convergence of the Taylor series for f about x = 5.

Show that the sixth-degree Taylor polynomial for f about x = 5 approximates f(6) with error less than 1/1000

So far i have for part a) f(5)+f'(5)(x-5)+(1/2!)f''(5)(x-5)^2+(1/3!)f'''(5)(x-5)^3
how do i find my f'(5), f''(5)...... given f^(n)(5) = ((-1)^n(n!)0/(2^n(n+2) here f[sup:33uzw5e3]1[/sup:33uzw5e3](x) = df(x)/dx
Im not sure about the rest of the problem
 
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