1. ## Ominous Sum:

Question: Prove for some general function f(x) for which $\forall a \epsilon \mathbb{R} : x\geqslant a \cap$ f(x) is uniformly continuous on [a,b],
$f(x)=\Sigma_{k=1}^n(\frac{1}{n}(f(\frac{kx-(n-k)a}{n})+\frac{k(x-a)}{n}f'(\frac{kx+(n-k)a}{n})))$

-i invented the problem via FTC following these steps
$f(x)=\frac{d}{dx}\int_{a}^{x}f(t)dt=\frac{d}{dx}(\ lim_{n->\infty}\Sigma_{k=1}^n(\frac{x-a}{n}f(\frac{kx+(n-k)a}{n})))$ $=\lim_{n->\infty}\Sigma_{k=1}^n\frac{d}{dx}((\frac{x-a}{n}f(\frac{kx+(n-k)a}{n}))$
$=\lim_{n->\infty}(\Sigma_{k=1}^n(\frac{1}{n}(f(\frac{kx+( n-k)a}{n})+\frac{k(x-a)}{n}f'(\frac{kx+(n-k)a}{n}))))$

particularly letting a=x gives $\lim_{n->\infty}(\Sigma_{k=1}^n(\frac{1}{n}(f(\frac{kx+( n-k)x}{n}))))$= $\lim_{n->\infty}(\Sigma_{k=1}^n(\frac{1}{n}(f(x))=f(x)$

2. Originally Posted by JeffN12345
Question: Prove $\forall a \epsilon \mathbb{R} : x\geqslant a$, $f(x)=\Sigma_{k=1}^n(\frac{1}{n}(f(\frac{kx-(n-k)a}{n})+\frac{k(x-a)}{n}f'(\frac{kx+(n-k)a}{n})))$
WITHOUTusing The Fundamental Theorem Of Calculus

-i invented the problem via FTC following these steps
$f(x)=\frac{d}{dx}\int_{a}^{x}f(x)dx=\frac{d}{dx}(\ lim_{n->\infty}\Sigma_{k=1}^n(\frac{x-a}{n}f(\frac{kx+(n-k)a}{n})))$ $=\lim_{n->\infty}\Sigma_{k=1}^n\frac{d}{dx}((\frac{x-a}{n}f(\frac{kx+(n-k)a}{n}))$
$=\lim_{n->\infty}(\Sigma_{k=1}^n(\frac{1}{n}(f(\frac{kx+( n-k)a}{n})+\frac{k(x-a)}{n}f'(\frac{kx+(n-k)a}{n}))))$

particularly letting a=x gives $\lim_{n->\infty}(\Sigma_{k=1}^n(\frac{1}{n}(f(\frac{kx+( n-k)x}{n})+\frac{k(x-a)}{n}f'(\frac{kx+(n-k)x}{n}))))$= $\lim_{n->\infty}(\Sigma_{k=1}^n(\frac{1}{n}(f(x)+0*f'(x))) )=f(x)$
You preformed some unjustified madness in there, Jeff.

3. Is the question sensical now

4. the proposition is false!
further, your solution has some flaw, too.
pay attention to the oder of derivative and limit!

5. Originally Posted by Shanks
the proposition is false!
further, your solution has some flaw, too.
pay attention to the oder of derivative and limit!
I was told differentials and limits and sums are all linear operators so their order can be interchanged...

6. No! This is not always the case.
That is why I said "pay attention to the interchange of oder".