Proving c_{n} = 0 Using Taylor Expansion: Homework Help

[V150]

Homework Statement

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.

Homework Equations

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.

 

[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).

 

[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<br />, not 'IF' c_n=0. Would you make this more concrete, please?

 

[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?

 

[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?

 

[D H]

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?

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

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.

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..."?

 

[D H]

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

 

[V150]

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

 

[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?

 

[HallsofIvy]

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)?

 

[V150]

Oh, oh. I finally got it. f&#039;(x) is zero because f(x) = 0 and f&#039;(x) is like differentiating zero, right?

 

[D H]

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.