Question

(Cauchy Mean Value Theorem) Suppose the function \[f:(-1,1)\to\mathbb{R}\] has n derivatives and that its nth derivative \[f^{(n)}:(-1, 1)\to\mathbb{R} \]is bounded. Also assume that
\[f(0) = f'(0) = \dotso = f^{(n-1)}(0) = 0.\]
Prove \[\exists M>0 : |f(x)| \leq M|x|^{n} \\\forall x \in (-1, 1).\]

Probably more effort to LaTeX that than figure it out, oh well...

Answer 1

i have an idea not sure if it is correct

Answer 2

no it isn't let me try again

Answer 3

The solution almost definitely involves a watered down version of the Lagrange Remainder Theorem that they give before defining Taylor polynomials (since Lagrange remainders are ultimately derived by Cauchy Mean Value), which I can type up in a second...

Answer 4

Theorem: Let I be an open interval and n be a natural number and suppose that the function \[f:I\to\mathbb{R}\] has n derivatives. Suppose also that at \[x_0 \in I, f^{(k)}(x_0) = 0; 0\leq k\leq n-1.\]Then \[\forall x\neq x_0, x\in I \]There exists a z strictly between x, x0 such that
\[f(x) = \frac{f^{(n)}(z)}{n!}(x-x_0)^n.
\]

Answer 5

So here we're given just such an f that that final form can be used, I guess we can set
\[M = \frac{f^{(n)}(z)}{n!}\]And then monkey with it somehow...

Answer 6

i am no help, sorry

Answer 7

No worries!

Answer 8

Yeah I was right. Given that our function is n-times differentiable on an open interval etc etc and given that we have
\[f^{(k)}(0) = 0, 0\leq k \leq n-1 \] it's permissable to set x0 = 0 and the theorem becomes
\[f(x) = \frac{f^{(n)}(z)}{n!}(x)^n, 0<|z|\leq 1
\]Thus,
\[|f(x)| = \bigg|\frac{f^{(n)}(z)}{n!}x^n\bigg| = \bigg|\frac{f^{(n)}(z)}{n!}\bigg||x|^n
\]And therefore,
\[|f(x) \leq M|x|^n
\]for
\[M\geq \bigg|\frac{f^{(n)}(z)}{n!}\bigg|\quad\Box
\]