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...
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...
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}##
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...
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
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...
$$\int_{0}^{a} f(x) dx = -\int_{-a}^{0} f(x) dx$$