Hi I am just wondering,
If I have a signal $f[n]\in \mathbb{C}^L$, i.e. $f$ is $L$-periodic, i can also define $h[n]=f[-n]$.
Is it true that the Fourier transform of $f$, say $\hat{F}$, and the Fourier tranform of $h$, say $\hat{H}$ is related by the equation below?
$$|\hat{F}[k]|=|\hat{H}[k]|...
@skullpatrol Conservation of mass implies that there aren't enough particles in the entire universe to write a comprehensive guide for understanding women
Well, @teadawg, it was only after the publication of one of my best papers in one of the best journals that I found a very basic (multivariable chain rule) error in it. :(
Fortunately, it is just embarrassing; it didn't affect the validity of the result.
@TheArtist In my country, the smartests persons I've ever met it happened to be during the interviews. I'd like to work with some of them and present them the real face of the reality. :-) (well, you only remain with personal satisfaction, nothing more)
I agree, @Ted. A bit too much to present in one course... but I think a few of them can be good motivating examples.
And now the next time someone asks me 'what's this good for?', I can say actual things instead of muttering "Uh... uh... it's used pretty much everywhere"
I was very pleased when I figured out (writing the linear algebra book) a cool argument that the vandermonde matrix is nonsingular by applying the change of basis formula.
Well, you should actually learn some of what you plan to mumble.
@TedShifrin Which is why I posted the question... of course I'm not going to find it sufficient to just glance at the answers if I want to talk about them. :)
(One of the answerers, the one who answered about computer graphics and rotation matrices, is a UCLA first year... that's 4 of us that can be considered vaguely active users now. Almost 20%!)
That's actually one of the fun things about teaching: You get to actually learn new, cool stuff. I only wish my probability students had learned some. And, it turns out, many have no clue how to set up the easiest double-integrals, despite their having had homework on this.
@Chris'ssis Yes I've read the answers in the post, $\frac{ab^2}{6}$ is the method I like, however I don't know how one would come up with that ....I don't understand at all Omrans answer, and it seems to be an interesting answer
@Chris'ssis Omrans answer is mathematics I've never seen. You know, before I joined mathematic stack exchange I never knew such brilliant smart people like you existed.
Things can split even if they don't canonically split.
@BalarkaSen Suggested approach: show that the composite covering map is the same as the composite $\Sigma_{11} \to \Sigma_6 \to \Sigma_2$; consider the degree two deck transformation from the first map as an element of $\text{Aut}(\Sigma_{11},\Sigma_2)$; check how it commutes with the other guy.
(Of course, this requires you write down a homeomorphism between the two ways of looking at $\Sigma_{11}$ we're using.)
Actually I was thinking something along the lines of this : $p_1$ and $p_2$ be two different covers of $\Sigma_{11} \to \Sigma_2$, where $p_2$ is the composite and $p_2$ is the $\Sigma_{11} \to \Sigma_{11}/\Bbb {Z_{10}}$. This means that there is a self-homeo $h$ of $\Sigma_2$ with itself such that $p_1 = h \circ p_2$
@BalarkaSen Let $q:E\to X$ be a covering map, $A\subset X$ locally path-connected. Then the restriction of $q$ to each component of $q^{-1}(A)$ is a covering map onto its image.
Is this an ok proof of $A \backslash (B \cap C) = (A \backslash B) \cup (A \backslash C)$? $$ \begin{aligned} \text{Suppose that } x \in A \backslash (B \cap C) & \iff x \in A \land \lnot ( x \in B \cap C ) \\ & \iff x \in A \land \lnot ( x \in B \land x \in \cap C ) \\ & \iff x \in A \land (x \not\in B \lor x \not\in \cap C ) \\ & \iff ( x \in A \land x \not\in B) \lor (x \in A \land x \not\in \cap C ) \\ & \iff (x \in A \backslash B) \lor (x \in A \backslash C) \\ & \iff x \in (A \backslash B) \cup (A \backslash C) \end{aligned} $$
@TheArtist have you seen this one I created yesterday? $$\int_0^{\infty}\left( \frac{\log ^2(\gamma x+1)}{(\gamma x+1)^{\large \sqrt[3]{\log(2)}}}-\frac{\log ^2(\pi x+1)}{(\pi x+1)^{1+\sqrt[3]{\log(2)}}}\right) \frac{1}{x} \, dx=\frac{2}{\log(2)}$$
Here is an elementary way to evaluate the integral without involving any special functions or advance formulas. Notice that
$$\int_0^\infty e^{-y}\sin(xy)\;\mathrm dy=\frac{x}{1+x^2}$$
Hence we have
\begin{align}
\int_0^\infty \frac{x}{\left(1+x^2\right)\left(e^{2\pi x}+1\right)}\,\mathrm dx&=\in...
In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function:
It is the first of the polygamma functions.
== Relation to harmonic numbers ==
The digamma function, often denoted also as ψ0(x), ψ0(x) or (after the shape of the archaic Greek letter Ϝ digamma), is related to the harmonic numbers in that
where Hn is the n-th harmonic number, and γ is the Euler-Mascheroni constant. For half-integer values, it may be expressed as
== Integral representations ==
If the real part of x is positive then the digamma function has the following integral representation...
Random Set Theory Question : What do you mean by an equinumerosity among sets? Could someone give me an example of $\Bbb A \approx \Bbb B$. From what I understand, if there exists some bijection $f : \Bbb A \to \Bbb B$, then the sets are equinumerous.
@MikeMiller That's pretty vague. There could be an infinite number of bijections between any two sets. Does this mean that all sets are equinumerous to one another or is this property only referenced in light of some specific bijective function.
@MikeMiller That's neat. That brings to my mind a more twisted question. In the sets you've given what would you call $f: B \to A$ where $f(x) = 0$ . Sure, it isn't a bijection but what is the relationship between its domain and codomain called?
Consider the map $Aut(X, Z) \to Aut(Y, Z)$ by taking an element $h : X \to X$ of the former and mapping it to $p \circ h$ where $p$ is the covering map $X \to Y$
Is this an ok proof of $A \backslash (B \cap C) = (A \backslash B) \cup (A \backslash C)$? $$ \begin{aligned} \text{Suppose that } x \in A \backslash (B \cap C) & \iff x \in A \land \lnot ( x \in B \cap C ) \\ & \iff x \in A \land \lnot ( x \in B \land x \in \cap C ) \\ & \iff x \in A \land (x \not\in B \lor x \not\in \cap C ) \\ & \iff ( x \in A \land x \not\in B) \lor (x \in A \land x \not\in \cap C ) \\ & \iff (x \in A \backslash B) \lor (x \in A \backslash C) \\ & \iff x \in (A \backslash B) \cup (A \backslash C) \end{aligned} $$
Does this line of reasoning validate $A \backslash (B \cap C) \subseteq (A \backslash B) \cup (A \backslash C)$ and $A \backslash (B \cap C) \supseteq (A \backslash B) \cup (A \backslash C)$ thereby meaning that the two sets are equal?