@AlexYoucis "@Baby Dragon I certainly hope not. I'd like to hold lot until a professional weighs in. If they don't, maybe I'll change it to an answer – Alex Youcis Oct 26 '13 at 2:41"
Does the following hold $$ \int_{-\infty}^{\infty} \frac{P(x)}{G(x)} e^{iax} \mathrm{d}x = \sum_{k=1}^m \operatorname{Res}\left[ \frac{P(x)}{G(x)} e^{iax} , z_k\right] $$ where $a \in \mathbb{R}$ ? Where $z_k$ are the poles of the function.
I think this holds as one can choose a half circle contour, and from jordans lemma then the arc will tend to zero as $R \to \infty$. And since the denominator is larger than the nominator + 2, the real integral converges.
I was under the impression that a function was analytic iff it was differentiable, however, damtp.cam.ac.uk/user/examples/B7a.pdf question 1ii seems to think differently. Am I misinterpreting the question?
@N3buchadnezzar Firstly, a factor of $2\pi i$ is missing. Secondly, if $G$ has any zeros on $\mathbb{R}$, things get a little more complicated. Thirdly, if $G$ has no zeros on $\mathbb{R}$, you sum only the residues in one half-plane. Fourthly, if $a < 0$, you need the lower half-plane, and thus a factor of $-1$ for the orientation of the contour.
@Alyosha "analytic" means representable by a power series. It turns out that a function is (complex-) analytic on an open set $U\subset\mathbb{C}$ if and only if it is complex differentiable on $U$, but $f$ being complex differentiable at one point $z_0$ does not imply that $f$ is analytic at that point (that needs complex differentiability on a neighbourhood of $z_0$).
@DanielFischer I think my problems may stem from never having seen the proof of 'it turns out that...'. Does that theorem have a name/ can you reference a proof of it?
@Alyosha It's a direct consequence of Cauchy's integral formula. In $$f(z) = \frac{1}{2\pi i}\int_{\lvert\zeta-z_0\rvert = r} \frac{f(\zeta)}{\zeta-z}\,d\zeta$$ expand $\frac{1}{\zeta-z}$ into a geometric series $$\sum_{\nu=0}^\infty \frac{(z-z_0)^\nu}{(\zeta-z_0)^{\nu+1}}.$$
I just forgot about the factor $2\pi i$ I know that is needed, I was also thinking about the particluar case where $G(z)$ has no zeros on the real line. I am just refreshing now, but why would you need the lower plane contour if $a<0$ ?
@N3buchadnezzar Because $$\left\lvert e^{iaz}\right\rvert = e^{-a\operatorname{Im} z}$$ becomes large in the upper half-plane for $a < 0$, so you can't deduce that the integral over the semicircle tends to $0$ by the standard estimate or Jordan's lemma.
@Alyosha A lot of the topological (and thus analytical) background is usually given within the course....however: the motivation of the definitions may seem unmotivated.
If $A$ is finite set of linearly independent vectors then the dimension of the subspace spanned by $A$ is equal to the number of vectors in $A$.
This is obviously true. Since $A$ is a finite set of linearly independent vectors and spans a subspace, $A$ is a basis for that subspace spanned by $A...
Ahh! India just lost a thriller. Very disappointed :( @DavidWheeler I am struggling with the Leibniz notation. I am currently learning the u substitution, though I have no idea how people do by assuming something as $u$ and do something with $du$. I do it my way, identify $f$ and $g$ use the chain rule in reverse. I have till now no idea what $dx$ and $dy$ is, how the ratio becomes a deriviative. Can you help?
@Matthew Ah, me too! You saw it [?], it was a real thriller. 10 was needed in the last over with 2 wickets in hand for Pakistan. The last over went like this: W,0,6,6 :(
Example (sorry for being late): Integrate $x\cdot e^{x^2}$. I note that that the derivative of x^2 is 2x. Therefore, it is of the form $F'(g(x))g'(x)$, where F=e^x, g=x^2. So, integral is F(g(x) and then multiply by 1/2, which I disregarded at the beginning.
@DavidWheeler Yes, another point I was about to make. And I hate that long S sign too. Wonder what Leibniz thought when ''he spent days fixing the notation, unlike Newton''.
@DavidWheeler Wat?? If the limit goes both to the numerator and denominator [quotient property?], it become 0/0, nothing else, and it is not allowed inm the first place.
@Sawarnik well, it turns out that the infinitesimals, and the transfinite numbers (plus the real numbers) still form a field, so we can take reciprocals and stuff
Given $c \in (0,1)$ and $b \in (0,1)$ how would one find the (2-dimensional) lebesgue measure (i.e. the area) of the set $\{(x,y): 0 \leq c x (1-c)y \leq b\}$?
sorry, that set should be $\{(x,y): 0\leq cx + (1-c)y \leq b\}$
@DavidWheeler No, not the derivative of the composition, of $g$. What I am doing is recognizing the $F$ and $g$ in $F'(g(x))g'(x)$. Is the $u$ more than recognizing $g$ in this case?
Didnt you guys had problems studying with Lebniz, early on? Am I a odd case?
@DavidWheeler So that is a waste of space and ink? Instead of a $'$? Why will I make a operator so big, if I wanted to have the $x$ somewhere, why not something like your big D?
It confuses some people, others are quite used to it. My physics instructor always used Newtonian notation (with the dots over a variable to indicate differentiation).
@DavidWheeler I am using Stewart, and he too doesn't use it mostly, apart from writing the theorems in Leibniz as alternate forms. Until the substitutions.
@robjohn Three logicians walk into a bar. A waiter asks, do you all want a drink? The first replies, I don't know, the second replies, I don't know, the third replies, yes.
the "big D" notation is usually used for considering operators, or "functions of "functions"in analogy with using capital letters for linear operators in linear algebra
@Chris'ssis hey, have you seen any presentation constructing Fourier series as extensions of Taylor series? This is the most helpful one I found online
@Chris'ssis If you name your profile as Chris sis, it shows your .., you must be a good sister. And the otherway round too .. Unlike me and my sister, always keep fighting.
What theory or algorithm would I need to research to solve equations such as $\sin(t+am)=a$ (knowing that $-1 \le m \le 1$) for the value of $a$?
My equations may become more complex but have similar properties (and always the constraint on the range of the $m$ variables), such as $\sin(t+\sin(t...
@IanMateus There will probably not be a similar error term since Fourier series are developed in a different manner from Taylor series. Fourier series are developed as a Hilbert space approximation. Taylor series are usually developed as an exact formula with an integral as the remainder.
@Sawarnik Chris says that if you wanna be good at math then you have to be crazy, to go into the matter and fight with the worst problems, and fight a lot. I mean to work day and night.
@Sawarnik No, I'm not.
@robjohn Yes, ill. There is a kind a bad cold here.
@robjohn ok, a bit of speculation now. Is this Hilbert space that relates the so called "orthogonality" to "being able to approximate a function using linear combinations of blah"? I have seen some other functions with this property (Legendre polynomials), so there must be "Legendre series" and so on, right?
@Sawarnik I also wanna try to study $$\sum_{n=0}^{\infty} \frac{1}{\displaystyle \prod_{i=2}^{k} \displaystyle \binom{in}{n}}$$ that I feel it has nice closed forms.
I think that people have the magical ability to find a closed form solution for anything they can perceive, even without conscious knowledge of the un-invented functions making it possible
Otherwise the little "glitches" of everyday life where people somehow just know too much wouldn't add up for me
@IanMateus Note that also in that example, although there is a formula in terms of the hypergeometric functions, the first few cases presented have very nice closed forms.
@IanMateus Mathematica gives me only results in terms of the hypergeometric functions for the last one.
@IanMateus $(14)$ $(15)$ $(16)$ are simply amazingly beautiful.
