# Steps to differentiate a geometric sum

1. Sep 5, 2014

### Crystals

Can someone guide me with the steps to differentiate a geometric sum, x?

$^{n}_{i=0}$$\sum$x[STRIKE]$^{i}$[/STRIKE]=$\frac{1-x^{n+i}}{1-x}$

If i'm not wrong, the summation means:

$= x^0 + x^1 + x^2 + x^3 + ..... + n^i$

Problem is:
I have basic knowledge on differentiating a normal numbers but how do I apply differentiate on a geometric sum?

Last edited: Sep 5, 2014
2. Sep 5, 2014

### Mogarrr

Its been a while since I looked at these kinds of problems, so I checked out Paul's Online Calculus Notes, which is a good reference.

I would think that...

$\frac {d}{dx} \sum_{n=0}^{i} x^{i} = \sum_{n=0}^{i} \frac {d}{dx} x^{i}$.

From there, I would see the what the series actually looks like, then express the summation as a formula.

3. Sep 5, 2014

### Mogarrr

And I think that this is valid for the derivative of a Power Series.

4. Sep 5, 2014

### HallsofIvy

$$\frac{d}{dx}(\sum a_n(x))= \sum \frac{da_n(x)}{dx}$$

if and only if the initial sum is "uniformly convergent". All power series are uniformly convergent inside the radius of convergence.

5. Sep 5, 2014

### Ray Vickson

You are wrong: it means
$$x^0 + x^1 + x^2 + \cdots + x^n = 1 + x + x^2 + \cdots + x^n.$$
Note how the last term differs from yours. Also, this is equal to
$$\frac{1-x^{n+1}}{1-x}$$
if $x \neq 1$. You wrote something different, with an $i$ in it (but there is no "$i$" outside of the summation---that is, after you do the sum, the "$i$' no longer 'exists').

Anyway, your last sentence makes no sense: you cannot differentiate a normal number (unless you want to get 0 always), but you can differentiate functions of x, like your series.

There are two ways to proceed here: (1) differentiate the series directly; and (2) differentiate the formula you have obtained for the sum. In other words:
$$(1) \;\text{derivative } = \frac{d}{dx} (1 + x + x^2 + \cdots + x^n) = 1 + 2x + 3x^2 + \cdots + n x^{n-1}\\ (2)\; \text{derivative } = \frac{d}{dx} \frac{1-x^{n+1}}{1-x} = \cdots \cdots$$
I will let you do the derivative here.

6. Sep 5, 2014

### Crystals

I did the (2) derivative by Ray Vickson and this is the draft solution I got.

Is this correct to say that the derivatives both sides of the geometric sum , with respect of x is:

$\frac{d}{dx}$$\frac{1-x^{n+1}}{1-x}$
=$\frac{x^{n}(n(x-1)-1)+1}{(x-1)(x-1)}$
=$\frac{x^{n}(n(x-1)-1)+1}{(x-1)^2}$ (simplified final answer)

I got confused with the question saying "derivatives both sides of the geometric sum". I thought I will need to do the derivatives for both LHS and RHS?

Last edited: Sep 6, 2014