Let $f'(0)=f''(0)=1$, $f^{(12)}(x)$ is differentiable and $g=f(x^{10})$ What's the value of $g^{(11)}(0)$?

Let $f'(0)=f''(0)=1$, $f^{(12)}(x)$ is differentiable and $g=f(x^{10})$

What's the value of $g^{(11)}(0)$?

I think it is important to use the fact that $f^{(12)}(x)$ is differentiable. However, I don't know hot to use it.

Can anyone help me solve the question?

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by $f^{12}(x)$ do you mean $f^{(12)}(x)$ or $(f(x))^{12}$? – Mercy King Nov 3 '12 at 14:46
First one. 12 order derivative of $f(x)$ – John Hass Nov 3 '12 at 14:47
You can use Faà di Bruno's formula. See en.wikipedia.org/wiki/Faà_di_Bruno%27s_formula – Mercy King Nov 3 '12 at 14:53

From Taylor's Theorem we have $$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+h(x)x^2=f(0)+x+\frac{1}{2}x^2+h(x)x^2, \ x \in \mathbb{R},$$ where $h$ is a function s.t. $\displaystyle{\lim_{x \to 0}h(x)=0}.$ The polynomial $f(0)+x+\frac{1}{2}x^2$ is unique with the above property.

Since $g(x)=f\left(x^{10}\right)$ and $f$ is $13$ times differentiable $\Rightarrow \ \ g$ is $13$ times differentiable. Now $$g(x)=f\left(x^{10}\right)=f(0)+x^{10}+\frac{1}{2}x^{20}+h\left(x^{10}\right)x^{20}=\\ f(0)+x^{10}+x^{13}\left(\frac{1}{2}x^{7}+h\left(x^{10}\right)x^{7}\right), \ x \in \mathbb{R}.$$

Since $\displaystyle{\lim_{x \to 0}\left(\frac{1}{2}x^{7}+h\left(x^{10}\right)x^7\right) =0}$ from uniqueness in Taylor's theorem we conclude that $f(0)+x^{10}=g(0)+g'(0)x+\frac{g''(0)}{2}x^2+\ldots +\frac{g^{(13)}(0)}{13!}x^{13}$.

Therefore $g^{(11)}(0)=0.$

Actually $g^{(k)}(0)=0, \ \ \ \forall k \in \{1,2,\ldots,9,11,12,13\}$ and
$g(0)=f(0), \ g^{(10)}(0)=10!.$

The fact that $f^{(12)}$ is differentiable is important because to apply Taylor's theorem to $g$ we need to know how many times $g$ is differentiable. Of course he could have said that $f^{(10)}$ is differentiable.

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Taylor's Theorem tells us that a $k$-times differentiable function can be written in the form:

$$f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_kx^k + h(x)x^{k}$$

where $h(x) \to 0$ as $x \to 0$. If $f'(0) = f''(0) = 0$ then $a_1 = a_2 = 0.$ Thus:

$$f(x) = a_0 + a_3x^3 + a_4x^4 + \cdots + a_kx^k + h(x)x^{k}.$$

You define $g(x) := f(x^{10})$ and so we have:

$$g(x) = a_0 + a_3x^{30} + a_4x^{40} + \cdots + a_kx^{10k} + h(x^{10})x^{10k}.$$

Clearly $g^{(p)}(0) = 0$ for all $1 \le p \le 29.$ In particular $g^{(11)}(0) = 0.$ Notice we only really needed $f$ to be twice differentiable. Anything else was a bonus.

Without Taylor's Theorem you could apply the chain rule ten times.

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I thing it should be $f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_kx^k + h(x)x^{k}$ and not $f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_kx^k + h(x)x^{k+1}$. For example if $f(x)=1+x+x^2, \ f$ is $1-$ time differentiable function. What is $h$? – P.. Nov 8 '12 at 18:42
Quite right... Thanks for pointing that out. – Fly by Night Nov 8 '12 at 18:47
One more thing: $1\leq p \leq 19, p \neq 10$ – P.. Nov 8 '12 at 19:47
I disagree. The first $29$ derivatives of $a_0 + a_3x^{30} + o(x^{30})$ will vanish when $x=0.$ – Fly by Night Nov 8 '12 at 20:13