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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