If your goal is to approximate e^(-1/2) by using the Maclaurin series up to the quadratic term, the best estimate you can achieve for the error is that it must be less than or equal to what?

The interval used will be [-1/2, 0]

My work:
My work so far:
Use e^x maclaurin series, set e^x = e^(-1/2) which means x = -1/2, replace the x in the maclaurin series with -1/2. Not sure what to do after that.

Question

If your goal is to approximate e^(-1/2) by using the Maclaurin series up to the quadratic term, the best estimate you can achieve for the error is that it must be less than or equal to what?

The interval used will be [-1/2, 0]

My work:
My work so far:
Use e^x maclaurin series, set e^x = e^(-1/2) which means x = -1/2, replace the x in the maclaurin series with -1/2.  Not sure what to do after that.

anonymous · Answer

@robtobey

anonymous · Answer

I would approach it in a slightly different way. First consider the Maclaurin series:$$f(x)=f(0)+f'(0)x+\frac{ f"(0)x^{2} }{ 2! }...$$ if we were to apply this to the function e^x it would be like this:$$e^{x}=e^{0}+e^{0}x+\frac{ e^{0}x^{2} }{ 2! }...$$ in fact all the e^0s equals to 1 so the first three terms are: $$e^{x}=1+x+ \frac{ x^{2} }{ 2! }$$ pluging in -1/2 for x and we get 0.625 which is 0.018 from the actual answer (approx. 0.607)