# Exponent error question

1. Apr 2, 2010

### 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
?

2. Apr 2, 2010

### 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?

3. Apr 2, 2010

### 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?

4. Apr 2, 2010

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

They are exactly the same.