## exponent error question

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
?
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 Recognitions: Gold Member Science Advisor Staff Emeritus 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?
 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?

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## exponent error question

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
$$\sum_{n=0}^\infty \frac{(-1)^n}{n!}x^n[/itex]. In particular, [tex]e^{-0.5}= \sum_{n=0}^\infty \frac{(-1)^n}{n!}(0.5)^n[/itex] The usual Taylor's series for $e^x$ is, of course, [tex]\sum_{n=0}^\infty \frac{1}{n!}x^n[/itex] and now [tex]e^{-0.5)= \sum_{n=0}^\infty \frac{1}{n!}(-0.5)^n= \sum_{n=0}^\infty \frac{1}{n!}(-1)^n(0.5)^n$$
$$= \sum_{n=0}^\infty \frac{(-1)^n}{n!}(0.5)^n$$

They are exactly the same.