Question

The taylor series about x=5 for a certain function converges

Answer 1

The evaluation of the function $f(x)=e^x$ at a=0 and x=5 by the taylor series will have a value 65.37.

What is taylor's series?

We know the fromula of taylor series is given as:

$f(x)=f(a)+\dfrac{f'((a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+\dfrac{f'''(a)}{3!}(x-a)^3.................$

Now for the function $f(x)=e^x$ the taylor's series will become.

$f(x)=e^a+e^ax+\dfrac{e^ax^2}{2!}+\dfrac{e^ax^3}{3!}+\dfrac{e^ax^4}{4!}$

$f(x)=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}$

So at x=5 the values will become

$f(x)=1+5+12.5+20.83+26.04=65.37$

hence the evaluation of the function $f(x)=e^x$ at a=0 and x=5 by the taylor series will have a value 65.37.

To know more about Taylor series follow

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