5
$\begingroup$

So I got to the infamous "the proof is left to you as an exercise" of the book when I tried to look up how to get the Lagrange form of the remainder for a Taylor polynomial. Is this right?

Given

$R_{n}(x)=\frac{1}{n!}\int_{0}^{x}f^{n+1}(t)(x-t)^{n}dt$,

find out why

$R_{n}(x)=\frac{1}{(n+1)!}f^{n+1}(c)x^{n+1}$ for some $c\in [0,x]$

According to FTC,

\int_{0}^{x}f'(t)dt = f(x) - f(0)

Also, according to the Mean Value Theorem, there exists a $c$ such that

f'(c)(x-0)=f(x)-f(0)

so

\int_{0}^{x}f'(t)dt = f'(c)(x-0)

finding the derivative of both sides with respect to $x$:

f'(x) = f'(c)

so

$f^{n+1}(x) = f^{n+1}(c)$

Going back to the integral form of the remainder:

$R_{n}(x)=\frac{1}{n!}\int_{0}^{x}f^{n+1}(t)(x-t)^{n}dt$,

I replace $f^{n+1}(x)$ with $f^{n+1}(c)$ (This is the step I am most unsure of)

Since f'(c) is a constant, I pull it out of the integral and integrate what's left under the integral, giving me

$R_{n}(x)=\frac{1}{(n+1)!}f^{n+1}(c)x^{n+1}$ for some $c\in [0,x]$

If this is right, then does it mean that f'(c) is the average value of f'(x) from $0$ to $x$?

Sorry if my LaTeX/wording/proof is off. I'd appreciate any corrections/answers to be as simple (notation-wise) as possible please - 1st year undergrad here...

1 Answers 1

2

No, it is not right.

f'(c) (x-0) = f(x) - f(0) is true, but $c$ depends on $x$.

So it is something like

f'(c_x) (x-0) = f(x) - f(0)

So derivative of f'(c) x is not really f'(c).

And the step from

f'(x) = f'(c) to $f^{n+1}(x) = f^{n+1}(c)$ is also wrong. If $c$ is constant (according to your proof), then the derivative on the right side becomes $0$.

  • 0
    @G.P. If$x$is fixed, what would one mean by derivative with respect to x?2011-01-14