# Trying to expand (1+x)^n using Taylor Expansion

1. Jun 23, 2012

### Rococo

1. The problem statement, all variables and given/known data

I need to use Taylor Expansion to show that:

(1+x)^n = 1 + nx + n(n-1)(x^2)/2! + ...

2. Relevant equations

y(x0 + dx) = y(x0) + dx(dy/dx) + [(dx)^2/2!](d^2y/dx^2) + ...

3. The attempt at a solution

I've only just begun Taylor Expansion, according to my textbook I need the above equation

(1+x)^n
So: x0 = 1
and dx = x

y(1+x) = (1+x)^n
So: y(x) = x^n
dy/dx = nx^n-1
d^2y/dx^2 = (n)(n-1)x^n-2

However putting all of this into the equation I get:

y(1+x) = y(1) + (x)(nx^n-1) + [x^2/2!][(n)(n-1)x^n-2] + ...

(1+x)^n = 1^n + (nx)(x^n-1) + (n)(n-1)[x^2/2!](x^n-2) + ...

(1+x)^n = 1 + (nx)(x^n-1) + (n)(n-1)[x^2/2!](x^n-2) + ...

Which I get as my final answer.

As you can see, there is a (x^n-1) in the second term, and a (x^n-2) in the third term, that shouldn't be there.

2. Jun 23, 2012

### Dick

The function you want to Taylor expand is y(x)=(1+x)^n. Then your Taylor expansion is y(x)=y(0)+y'(0)x+y''(0)x^2/2!+... Start by finding the derivatives of y evaluated at 0. What are y(0), y'(0), y''(0) etc etc?

3. Jun 24, 2012

### Rococo

Thanks, here's what I get now:

y(0) = 1^n

y'(0) = n(1^n-1) = n

y''(0) = (n-1)(n)(1^n-2) = (n)(n-1)

So putting that into the equation I now get:

(1+x)^n = 1^n + (n)(x) + (n)(n-1)(x^2)/2! + ...

Is that the right method?

4. Jun 24, 2012

### Dick

Ok, that works. Try using more parentheses in expressions. (1^n-1) is unclear. I thought it meant (1-1)=0. It should be (1^(n-1)).

5. Jun 24, 2012

### HallsofIvy

Staff Emeritus
By the way:
$$1= \begin{pmatrix}n \\ 0 \end{pmatrix}$$
$$n= \begin{pmatrix}n \\ 1 \end{pmatrix}$$
$$n(n-1)= \begin{pmatrix}n \\ 2 \end{pmatrix}$$
etc.

6. Dec 20, 2012

### Shurok

HallsofIvy, you forget multiply the left side by 1/2 in last formula:

n choose 2 = n(n-1)/2

7. Dec 20, 2012

### HallsofIvy

Staff Emeritus
Yes, thanks.