Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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. Seeà_di_Bruno%27s_formula – Mercy King Nov 3 '12 at 14:53
up vote 1 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.