Why does e^{-0.5} equal \sum_{n=0}^\infty \frac{(-1)^n}{n!}(0.5)^n?

[nhrock3]

i need to calculate $$e^{-0.5}$$

why the solution develops $$e^{-x}$$ and puts 0.5
and not $$e^{+x}$$ and putting -0.5
?

 

[HallsofIvy]

I have no idea what you are talking about. What do you mean by "develops $e^{-x}$? Writing as a Taylor's series? Approximating by the tangent line?

I suspect that both them method your book gives and your method would give the same answer. Have you tried it?

 

[nhrock3]

yes developing in taylor series
but in the first we have libnits series and on the other not
so its not the same

why they are not the same?

 

[HallsofIvy]

If $f(x)= e^{-x}$ then f(0)= 1, $f'= -e^{-x}$ so f'(0)= -1, $f"(0)= e^{-x}$ so f"(0)= 1, etc. The "nth" derivative, evaluated at x= 0, is 1 if n is even, -1 if n is odd. The Taylor's series, about x= 0, for $e^{-x}$ is
[tex]\sum_{n=0}^\infty \frac{(-1)^n}{n!}x^n[/itex].<br /> <br /> In particular, <br /> [tex]e^{-0.5}= \sum_{n=0}^\infty \frac{(-1)^n}{n!}(0.5)^n[/itex]<br /> <br /> The usual Taylor's series for $e^x$ is, of course, <br /> [tex]\sum_{n=0}^\infty \frac{1}{n!}x^n[/itex] <br /> <br /> and now<br /> $$e^{-0.5)= \sum_{n=0}^\infty \frac{1}{n!}(-0.5)^n= \sum_{n=0}^\infty \frac{1}{n!}(-1)^n(0.5)^n$$<br /> $$= \sum_{n=0}^\infty \frac{(-1)^n}{n!}(0.5)^n$$<br /> <br /> They are exactly the same.[/tex][/tex][/tex]