Differentiation with sigma notation I want to solve one function that contains sigma notation. 
$f(x)=1- e^{(-x/a)}\sum_{i=0}^{m-1}\frac{(\frac{x}{a})^i}{i!}$ 
 A: Well, let's tackle it step by step so you see what is happening. Let $g : \mathbb{R} \to \mathbb{R}$ be given by:
$$g(x)=\sum_{i=0}^{m-1}\frac{1}{i!}\left(\frac{x}{a}\right)^i$$
And let $h : \mathbb{R} \to \mathbb{R}$ be another function given by $h(x)=e^{-x/a}$. So your function is in reallity $f:\mathbb{R} \to \mathbb{R}$ given by $f = 1 - hg$. This implies by the properties of the derivative that we have:
$$f'=-(h'g+hg')$$
So we need to compute just the derivatives of $h$ and of $g$. Indeed $h$ is very simple, if you know about the chain rule and about differentiating exponentials you'll trivially get
$$h'(x)=\frac{-e^{-x/a}}{a}$$
Now the derivative of $g$ requires some thought. Just to make the steps clearer I'll use Liebniz' notation. We want to calculate
$$\frac{d}{dx}\sum_{i=0}^{m-1}\frac{1}{i!}\left(\frac{x}{a}\right)^i$$
Now, the derivative is linear, so that the derivative of a sum is the sum of the derivatives, which allows putting the derivative inside the sum. Also linearity says that the derivative of the product of a constant by a function is the constant times the derivative of the function. This allows to write the following:
$$\frac{d}{dx}g(x)=\sum_{i=0}^{m-1}\frac{1}{i!}\frac{d}{dx}\left(\frac{x}{a}\right)^i$$
Now we use the chain rule. We know how to differentiate $x^i$ but we don't know how to differentiate $(x/a)^i$. So we use the chain rule, differentiate outside "with respect to $x/a$" and multiply by the derivative of what's inside with respect to $x$:
$$\frac{d}{dx}g(x)=\sum_{i=0}^{m-1}\frac{1}{i!}i\left(\frac{x}{a}\right)^{i-1}\frac{1}{a}$$
This simplifies to the following:
$$\frac{d}{dx}g(x)=\sum_{i=0}^{m-1}\frac{1}{a(i-1)!}\left(\frac{x}{a}\right)^{i-1}$$
Now that we have both $h'$ and $g'$ we just substitute them back:
$$f'(x)=-\left( \frac{-e^{-x/a}}{a}\sum_{i=0}^{m-1}\frac{1}{i!}\left(\frac{x}{a}\right)^i + e^{-x/a}\sum_{i=0}^{m-1}\frac{1}{a(i-1)!}\left(\frac{x}{a}\right)^{i-1}\right)$$
We can simplify it by taking $e^{-x/a}$ outside:
$$f'(x)=-e^{-x/a}\left( \frac{-1}{a}\sum_{i=0}^{m-1}\frac{1}{i!}\left(\frac{x}{a}\right)^i + \sum_{i=0}^{m-1}\frac{1}{a(i-1)!}\left(\frac{x}{a}\right)^{i-1}\right)$$
Now we can put the $-1/a$ inside the first sum and combine both sums getting:
$$f'(x)=-e^{-x/a}\sum_{i=0}^{m-1}\left( \frac{-1}{a i!}\left(\frac{x}{a}\right)^i + \frac{1}{a(i-1)!}\left(\frac{x}{a}\right)^{i-1}\right)$$
And this can also be written as
$$f'(x)=-\frac{e^{-x/a}}{a}\sum_{i=0}^{m-1}\left( \frac{-(i-1)!}{i!(i-1)!}\left(\frac{x}{a}\right)^i + \frac{i(i-1)!}{i!(i-1)!}\left(\frac{x}{a}\right)^{i-1}\right)$$
And finally we get a less intimidating version of the derivative of this function:
$$f'(x)=-\frac{e^{-x/a}}{a}\sum_{i=0}^{m-1}\left(\frac{1}{i!}\left(\frac{x}{a}\right)^{i-1}\left(i-\frac{x}{a}\right)\right)$$
I hope it was that you wanted. Good luck with your studies.
A: The sum is the leading terms in $e^{\frac xa}$ so as $m \to \infty, f(x) \to 0$  As each term in the sum is smaller than the last (if $\frac xa$ is small enough-how small does it need to be?) you can approximate the sum as $e^{\frac xa}-(\frac xa)^m\frac 1{m!}$ so $f(x) \approx e^{-\frac xa}(\frac xa)^m\frac 1{m!}$
