Recent content by Euklidian-Space

well since ##m_{k}## and ##m_{k}^{'}## are the infs they are definitely bounded by ##\epsilon## if ##|f_n(x) - f(x)|<\varepsilon## for all ##x##; which is true because we have uniform convergence

what exactly is less than ##\epsilon/2##? I was thinking something like... $$|L(f_{n},P) - L(f,p)| = |\sum_{k = 1}^{n} m_{k}^{'}(x_{k} - x_{k - 1}) - \sum_{k = 1}^{n} m_{k} (x_{k} - x_{k - 1}|$$ we then have $$\sum_{k = 1}^{n} m_{k}^{'} - m_{k} (x_{k} - x_{k - 1})$$ now if i can bound...

oh my bad. L(f,p) is the lower sum of the Reimann integral $$L(f,P) = \sum_{k = 1}^{n} m_{k} (x_{k} - x_{k - 1})$$

Homework Statement Suppose that ##f_{n} \rightarrow f## uniformly on [a,b] and that each ##f_{n}## is integrable on [a,b]. Show that given ##\epsilon > 0##, there exists a partition ##P## and a natural number ##N## such that ##\left|L(f_{n}, P) - L(f,P)\right| < \epsilon##. Homework...

wait how do you get \lim_{h \to 0} \frac{f'(a + h) - f'(a) - hf''(a)}{2h} = \frac{f''(a) - f''(a)}{2} = 0

oh i think I see now

Well I tried applying lhopitals again. But I get ##\lim_{n\to0} \frac{f"(a+h) - f"(a-h)}{2}##. Which seems to be a little closer, but not quite.

ok when i apply l'hopital's i get the following $$\lim_{h \rightarrow 0} \frac{f'(a+h) - f'(a - h)}{2h}$$ we differentiate the numerator and denominator with respect to the limit variable right? doesn't seem like it is giving us ##f''(a) = \lim_{h \to 0} \frac{f''(a + h) - f''(a)}{h}##

DEvens, would you use the limit definition for differentiation to prove this? If so how would you resolve the different limit variables?

it seems even with the taylor method you get into the same problem. $$f(a + h) = f(a) + hf'(a) + \frac{1}{2} h^2f''(a) + \epsilon (h)$$ solve for f''(a) and get $$f''(a) = \frac{2f(a + h) - 2f(a) - 2hf'(a) - 2\epsilon (h)}{h^2}$$ so i guess at this point you would take limit of both sides...

ah so use Fund THM of calc?

Again I am sorry man i just don't understand. dy = -dx right? then how does ##\int_{-a}^{0} f(y) dy = \int_{-a}^{0} f(x) dx##? when dy = -dx? I do not see how using a "DUMMY variable" justifies dy = -dx to dy = dx

oh sorry, didnt see that. it is fixed now.

No that's what i meant... Since dy = -dx we get that negative out in front done we?

I am sorry if I am being difficult vela, but I just don't understand why showing $$\int_{0}^{a}f(x)dx = \int_{-a}^{0} f(y) dy$$ shows that $$\int_{0}^{a}f(x)dx = \int_{-a}^{0} f(x) dx$$ if anything i think i have shown... $$\int_{0}^{a} f(x) dx = -\int_{-a}^{0} f(x) dx$$