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So I have this eqn.

$$ f(x)= \frac {e^ \frac{-x^2}{2}} {\sqrt{2\pi}} $$

I need to find:

$$ \int\limits_{-1}^1 f(x)dx $$

So I want to use this series to integrate. I know that:

$$ e^x = \sum\limits_{n=0}^\infty \frac {x^n}{n!} $$

So without considering the coefficient of the function, I can build my eqn into this series as such:

$$ e^{\frac{-x^2}{2}} =\sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n}}{2^nn!} $$

This is where I get lost. What do I do with the coefficient of e. How do I build this into a series and include the coefficient of e:

$$ \frac {1}{\sqrt{2\pi}}\ $$

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  • $\begingroup$ you know that total probability is one and standard normal distribution ranges from -1 to 1 so your integral final value is 1 $\endgroup$
    – MRK
    Sep 22 '13 at 23:38
  • $\begingroup$ you cannot use elementary methods to integrate normal distribution and even if we use special function that will give the answer as an infinite series $\endgroup$
    – MRK
    Sep 22 '13 at 23:39
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    $\begingroup$ ^ what!? the domain of standard normal distribution is not [-1,1]. $\endgroup$
    – user81327
    Sep 23 '13 at 0:07
  • $\begingroup$ Use linearity of summation. $\endgroup$
    – Tunococ
    Sep 23 '13 at 0:50
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    $\begingroup$ lol domain for standard normal distribution is $$ (-\infty, \infty) $$ $\endgroup$ Sep 23 '13 at 6:29
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Ok I figured it out.

$$ f(x)= \frac {e^ \frac{-x^2}{2}} {\sqrt{2\pi}} $$

I need to find:

$$ \int\limits_{-1}^1 f(x)dx $$

So I want to use this series to integrate. I know that:

$$ e^x = \sum\limits_{n=0}^\infty \frac {x^n}{n!} $$

So without considering the coefficient of the function, I can build my eqn into this series as such:

$$ e^{\frac{-x^2}{2}}=\sum\limits_{n=0}^\infty \frac{(-1)^nx^{2n}}{2^nn!} $$

So we can just build up and get the origial.

$$ \frac {1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}=\frac {1}{\sqrt{2\pi}} \sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n}}{2^nn!} $$

Then all we need to do is expand out the series to a few values of n, then we can anti-differentiate the series, and then we can solve using the FTC, where the integral from [a,b] = F(b) - F(a).

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    $\begingroup$ Almost! Note that when you replace $x$ with $-x^2/2$ in the Taylor series, you should get $$ e^{\frac{-x^2}{2}}=\sum\limits_{n=0}^\infty \frac{(-1)^nx^{2n}}{2^nn!} $$ which for odd $n$ will be negative, but for even $n$ will be positive. So $$ \mathrm{e}^{-x^2} \sim 1-\frac{x^2}{2}+\frac{x^4}{8}+\mathcal{O}(x^6), $$ which should converge when $x \in (-1,1)$. $\endgroup$ Sep 23 '13 at 0:51
  • $\begingroup$ Yeah my algebra was off in the OP, but I ended up fixing it in my answer there. I ended up with exactly that series. And it did converge. And it ends up as 68%, which was kinda funny. And I ended up using the -1^n in my notes. It's a much clearer notation I agree. $\endgroup$ Sep 23 '13 at 6:13
  • $\begingroup$ Why is 68% funny? How many terms is this? I think the 'true' value is roughly 0.68, which is good, this is the area under the bell curve from -1 std deviation to +1 std deviation. $\endgroup$ Sep 23 '13 at 14:02
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    $\begingroup$ Sorry, I didn't mean funny as in incorrect. I like math, so I get a kick out of things like this. $\endgroup$ Sep 23 '13 at 15:14
  • $\begingroup$ Yep, consistency is the beautiful thing about maths :) $\endgroup$ Sep 24 '13 at 1:22
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Hint: You cannot find an explicit answer for your integration. However the integration of normal distribution can be stated in terms of Q Function. $$ Q(x)=\int_{x}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-t^2}{2}}dt $$ There are various estimates for this function which may be useful. For instance for $x>0$: $$ \frac{x}{(x^2+1)\sqrt{2\pi}}e^{\frac{-x^2}{2}} \leq Q(t)\leq \frac{1}{x\sqrt{2\pi}}e^{\frac{-x^2}{2}} $$

Remark: If you want to continue working with series, it is enough to take the coefficient of the exponential function out of the integration.

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  • $\begingroup$ Thanks for the input. There is actually a very easy way to approximate this integral. Note that, since I'm working on an approximation, we don't need to use an infinite series. Rather, we can consider that n goes to 5 or some arbitrary number, and get an approximation of the integral of f(x). And as you said, you can actually take the coefficient right out. I will answer below. $\endgroup$ Sep 23 '13 at 0:18

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