@Jonas...perhaps i just don't like inequalities. it's not that analysis doesn't seem to have some first-rate ideas...it just doesn't move me emotionally.
No, I was not born that way. I actually tried to learn this by myself, not postpone everything until the last moment and then claim I don't understand. Of course you don't understand if you don't work hard.
@DavidWheeler There are not much inequalities in operator theory!
I have never said how good I am at math or how easy it is, you're just saying that to shift the blame to me, while you are the one that doesn't seem to work...
But in that place says: "Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be the function defined, for all x in [a, b], by
well! if i'd know it would have garnered me your e-mail, i'd have behaved more badly in the past. is it too late for me to start proclaiming: i suck at math?
@robjohn If that's the case, we are done! No Lebesgue is needed. Can you link a proof of that without the assumption of continuity of $f$. Is it standar? I do not remember
@leo separate out the set where $f$ is not continuous into a subset of the partition whose total length is less than $\epsilon$, then do the proof there.
For example take the constant function $1$ over $[0,1]$. It is Riemann integrable. Then restrict it to $\Bbb Q$ (a null set) then you can't use Riemann
@Ilya I would say that $n$ is fixed inside the summation on the RHS and so the limit is meaningless or the $n$ in the limit is different from the $n$ in the summation.
@Jordan: one martial artist once said to me, never be afraid to fail -- be afraid "not to fail". The Samurai principle works also in Math -- details matter -- do not care about the failing comments at all. Just push on.
I am trying to find this by using integration by parts but I am not sure how to do it.
$$\int_0^{\pi/2} sinx^7cosx^5$$
I tried rewriting as
$\int_0^{\pi/2} sinx*sinx^6*cosx^5$ = $sinx(1-cosx^3)*cosx^5$
but that seems to only give me a very, very long loop that doesn't help me at all. How do...
@Jordan: the problem is that if you are traumatized now with unthinkable comments, you pretty much sacrify your greatest asset i.e. "your ability to think". There is nothing bad with failing but learn to fail, do not run away.
@hhh: You can't get away without evaluating an integral here or there. Instead of consistently substituting (only the first sub is necessary), just evaluate the integral
what i'm trying to get at, is if we define: $\langle f,g \rangle = \int_{-\pi}^{\pi} f(x)\overline{g(x)} dx$ then the formula for the $c_n$ follows from the orthogonality of the basis functions.
in much the same way as if we define a vector in the standard basis $u = a_1e_1+\dots+a_ne_n$ then the j-th coordinate is $\langle u,e_j \rangle$ where we take the "usual" inner product (aka dot product)
(It is a saying in Dutch, so probably something got lost in translation. It means something like if it is silent then don't get something that makes it noisy).
And also, if I mishbehaved: Sorry. I don't really like that attitude of doing nothing and then complaining that you don't understand (after trying the same thing several times...). I'll shut up next time.
@hhh Okay, I've checked your handwriting. You're messing it up. See the thing as an inner product, not a integral and the same thing with the basis vectors.
@mixedmath $\int_{-\pi}^{\pi} \left( \sum_{m=-\infty}^{\infty} c_m e^{imx-inx}\right) dx$ where $c_m=c_{m}(x)$. The hard point is $c_{m}(x):=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x) e^{-inx} dx$ -- now getting integral inside integral and $f(x)$ is the sum with integral....I cannot see how you evaluate this...
(I did only one substitution but I think I need some clever way to go around the $c_m$...)
@ZhenLin Hey you know how in ravi vakil's notes he gives the result on the bijection between homogeneous prime ideals in $A$ and prime ideals in $A_0$?
It is well-known that if $S$ is a graded ring, and $f$ is a homogeneous element of positive degree, then there is a bijection between the homogeneous prime ideals of the localisation $S_f$ and the prime ideals of $S_{(f)}$, the subring of $S_f$ comprising the homogeneous elements of degree $0$. T...
I am trying to find this by using integration by parts but I am not sure how to do it.
$$\int_0^{\pi/2} sinx^7cosx^5$$
I tried rewriting as
$\int_0^{\pi/2} sinx*sinx^6*cosx^5$ = $sinx(1-cosx^3)*cosx^5$
but that seems to only give me a very, very long loop that doesn't help me at all. How do...
