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
\sum_{n=0}^\infty \frac{(-1)^n}{n!}x^n[/itex].<br /> <br /> In particular, <br /> e^{-0.5}= \sum_{n=0}^\infty \frac{(-1)^n}{n!}(0.5)^n[/itex]&lt;br /&gt; &lt;br /&gt; The usual Taylor&amp;#039;s series for e^x is, of course, &lt;br /&gt; \sum_{n=0}^\infty \frac{1}{n!}x^n[/itex] &amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; and now&amp;lt;br /&amp;gt; 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&amp;lt;br /&amp;gt; = \sum_{n=0}^\infty \frac{(-1)^n}{n!}(0.5)^n&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; They are exactly the same.