# Taylor expansion help

1. Jun 9, 2014

### V150

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

Suppose that $f(x)=\sum_{n=0}^{\infty}c_{n}x^{n}$for all x.
If $\sum_{n=0}^{\infty}c_{n}x^{n} = 0$, show that $c_{n} = 0$ for all n.

2. Relevant equations
3. The attempt at a solution
I know, by using taylor expansion, $c_{n}=\frac{f^{n}(0)}{n!}$, and because $\sum_{n=0}^{\infty}c_{n}x^{n}$ is zero, $c_{n}$ have to be zero. But, I don't know how to write more logical proof to this.

2. Jun 9, 2014

### AlephZero

That is the right idea. Instead of "using the Taylor expansion", you can just differentiate n times.

$\frac{d^n}{dx^n} f(0) = 0$, and if you differentiate the series term-by-term that gives $c_n = 0$.

To make that idea into a rigorous proof, you have to say why it is valid to differentiate the series term by term. (Hint: convergence).

3. Jun 9, 2014

### V150

@AlephZero
What I want to know is why $c_n = 0$ have to be zero if $\sum_{n=0}^{\infty}c_{n}x^{n} = 0$, not 'IF' $c_n=0$. Would you make this more concrete, please?

4. Jun 9, 2014

### johnqwertyful

$\frac{d}{dx} f(x)=\sum_{n=1}^{\infty} nc_nx^{n-1}$
Then let x=0 to get:
$0=c_1+\sum_{n=2}^{\infty} nc_n 0^{n-1}$

Can you generalize this?

5. Jun 9, 2014

### V150

hmmm. How can I use that to prove $\sum_{n=0}^{\infty}c_{n}x^{n}=0$,then $c_{n}=0$ when x is all real numbers?

Last edited: Jun 10, 2014
6. Jun 9, 2014

### D H

Staff Emeritus
Hint: Use the transitivity of equality. If a=b and b=c then a=c.

What does the above say about $f(x)$ given that $f(x)=\sum_{n=0}^{\infty} c_n x^n$ and that $\sum_{n=0}^{\infty} c_n x^n = 0$ for all x?

7. Jun 10, 2014

### Ray Vickson

The result is false as you have stated it: $\sin(x) = x - x^3/3! + x^5/5! - \ldots$ is equal to 0 for $x = \pm \pi, \pm 2\pi, \ldots$ but its coefficients are not all zero. Did you mean "=0 for all x..."?

8. Jun 10, 2014

### D H

Staff Emeritus
He had to have meant "for all x". Otherwise it's patently false.

9. Jun 10, 2014

### V150

Yes, for all x with the radius of convergence R>0.

10. Jun 10, 2014

### V150

So the question would be, "why $c_{n}$ is 0 for all n when $\sum_{n=0}^{\infty}c_{n}x^{n}=0$ for all $x$ with the radius of convergence R>0?

11. Jun 10, 2014

### HallsofIvy

Staff Emeritus
You have asked that question four times now. Have you given any thought to the responses you have been given? If $f(x)= \sum c_nx^n$, what is f(0)? what is f'(0)? f''(0)?

12. Jun 10, 2014

### V150

Oh, oh. I finally got it. $f'(x)$ is zero because $f(x) = 0$ and $f'(x)$ is like differentiating zero, right?

13. Jun 10, 2014

### D H

Staff Emeritus
Exactly. You never went from $f(x)=\sum_n c_n x^n$ and $\sum_n c_n x^n=0$ to $f(x)=0$ until just now. Once you have $f(x)=0$ for all x, it's a simple matter to show that $f^{(n)}(x) = 0$ for all x and for all n.

You should of course prove that $f^{(n)}(x) = 0$ for all x and for all n.