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?



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



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

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

$f'(x) = f'(c)$


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

Going back to the integral form of the remainder:


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...


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 $0$.

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  • $\begingroup$ But if I were to fix $x$ to a particular number, the point I am trying to approximate/find-the-remainder-of, would't it be valid then? $\endgroup$ – G.P. Burdell Jan 14 '11 at 23:07
  • $\begingroup$ @G.P. If x is fixed, what would one mean by derivative with respect to x? $\endgroup$ – Aryabhata Jan 14 '11 at 23:12

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