So therefore one can always get the dual basis from the original basis by using the cross product appropriately e.g. $\vec{b}_1=\vec{a}_2\times \vec{a}_3$.
@Semiclassic: Of course in the standard multivariable calc class (all in $\Bbb R^2$ and $\Bbb R^3$) I used cross product a lot. In my multivariable math class, I almost never did.
@ThomasRasberry okay the idea is to add and subract f(z_0) , so we get line integral of f (zeta) / ( zeta - z_0) dzeta = line integral f(z_0) / zeta -z_0 ) dzeta + line integal (f(zeta) - f(z_0) / (zeta - z_0 )
They're the basic objects just like manifolds are the basic objects in geometry/topology, @MikeM ... Unless you're going to go completely off the deep end.
@MikeM: Much to my shock (but not dismay), I actually ended up dealing with a non-reduced scheme and an embedded component in one of my papers (it was generalized Chern classes of the Whitney umbrella, basically).
@TedShifrin yeah I do some macros, can't keep my sanity without them. But it feels like getting interrupted on the fly. And the pathing ... three years and I still cannot figure out how to insert images. I have three (INSERT FIGURE HERE) typed into my dissertation currently. it is shameful
To see how any formula was written in any question or answer, including this one, right-click on the expression it and choose "Show Math As > TeX Commands". (When you do this, the '$' will not display. Make sure you add these. See the next point.)
For inline formulas, enclose the formula in $......
@TedShifrin In Kronheimer and Mrowka's beautiful book that gives the technical underpinning of their definition of Seiberg-Witten Floer homology, they give the Whitney Umbrella as a visualization of the sort of singularities that can occur when you try to make the natural compactification of the space of gradient flowlines a smooth manifold.
I will be heading out soon, @Thomas, but if you can't get any of your friends to help you at your university, send me a bit of your document and a graphics document in email and I'll see what I can do.
The structure of the moduli spaces they actually end up getting a handle on are something like stratified spaces; they tend to be awful but good enough that one can still make "counts of boundary points with orientation of a 1-dimensional thing = 0" work.
@MikeM: I'd post my article with that stuff in it, but it's done on an old-fashioned typewriter (back from the 80s) and you would all laugh me out of the business.
@TedShifrin ok, gotcha there, but where is Typeset? I know I'm supposed to know, but I've basically ben on autotype for about eleven months now since I last messed around with all this
@TedShifrin The kind of really bad singularity they get is special to the Seiberg-Witten case. In the Morse case, you get C^0 manifolds with corners, smooth on the interior. I seem to remember them saying that it could in principle be something like a Hawaiian earring with a whisker at the basepoint.
(I've uploaded a lot of snow-day and exam-day-at-work assignments from my phone over the past 4 years, doing this while working full time is rough lol)
Anyhow, @Thomas, you can find my email in my MSE profile. I'll try to help if you can't get it working. Just send me enough of a file (and an actual graphics file) to play with.
I have one little thing on a side-chapter of my dissertation left to prove. It seems so elementary in my head but I haven't seen a proof for it in literature (it is just assumed, which makes me think it is easy). I asked in Overflow and have had no positive responses yet
Yeah that's the closest term I can come up for what I'm doing @TedShifrin, since compressive sensing deals with the closest topic to my dissertation but of course \ell_1 doesnt function over finite fields
I invented a solver for min \|X\|_0 but it is terrible.
But it will get me my Ph.D in two months, at least.
A warning keeps popping up in my LaTeX editor, and it's really annoying. It says that the command \Bbb is obsolete and I should use \mathbb instead. I personally like the command \Bbb by the obvious reason that its shorter than the newer one. If the command \Bbb is obsolete then why they don't ta...
We know that $\aleph_0$ is smaller than $\mathfrak c$, the cardinality of the continuum. But are there some good upper-bounds?
For example, it is trivial that $\mathfrak c<2^{\mathfrak c}$, but I wonder if there are better bounds. Specifically, I have to wonder if there exists $\alpha\in\mathb...
Hi, a probabilistic question here. Consider zero mean random variables $X_1,\dots,X_N$ with unknown distributions. Actually they have random distributions such that the covariance between $X_i$ and $X_j$ is determined by some discrete random variable $Z_{ij}$. Then $E[X_iX_j]$ averages over $Z_{ij}$, while $E[X_iX_j|Z_{ij}]$ gives the "true" covariance in a sense that I use fixed distributions for $X_i$ and $X_j$ rather than random ones. Analogy to stationary zero mean times series: $E[X_tX_s]=r(|t-s|)$ for some known function $r$. The difference in my case is that this lag $|t-s|$ correspo…
There has to be something we can create specifically dealing with math that can add value to lots of people's lives not dealing with anything school related like tutoring, what is it?
So I've found some sort of contradiction in the definitions I know for Therefore, and Modus Ponens. In regards to therefore: I have it defined as; $A\therefore B \leftrightarrow (A\rightarrow B)\land A$ this definition simplifies to $A\therefore B\leftrightarrow AB$
In my Modus Ponens definition: I have $((A\rightarrow B)A)\rightarrow B$. This is the same as writing $(A\therefore B)\rightarrow B$
The problem is the whole thing simplifies down to $\lnot A \lor \lnot B \lor B$
Which does not imply that B must be true; which Modus Ponens is meant to do.
simplifying that expression give a tautology; but I cant see how the tautology tells us that B must be true. It seems that it would work regardless of the truth value of B
Besides tautalogies work regardless of the truth values of what is inside them by definition.
I just tripple checked everything and it appears my definitions are right; its just the understanding of why Modus Ponens tells us that the consequent is true is off.
Suppose $B$ is false. Then the only way for that implication to be true is if either $A\implies B$ is false or $A$ is false. But if $A$ is true then $A\implies B$ is certainly false, so either way the statement is true.
Which is to say: It seems to indeed work just fine if $B$ is false.
Or, putting it better: If $A$ is false, then the implication is true regardless of $B$. But if $A$ is true, then you've just got $B\implies B$ which is trivial. So it doesn't constrain $B$ at all.
Anyway; was that question in chat explained well enough to be posted on SE? If I just wrote it the same way with better formating; would other people be able to understand what I'm asking?
Combined with the deduction you gave earlier that "$A$ therefore $B$" is equivalent to "$A$ and $B$", modus ponens would amount to "If A and B then B."
Well, careful. If $A$ is true, then "A therefore B" would seem to mean just "$B$" in which case you've just got "$B$ implies $B$". That's true regardless of $B$.
It is in fsct the case: I can give two reasons why:
first of which is the definition of therefore: A∴B↔(A→B)∧A. Which is read as "If A, then B, And A." Which is basicly saying that "If A then B" is true, and we know A to be true, then we can say A therefore B.
Another example, is the conjuction of sets when we use therefore, in our working: When we write functions and their definitions in our working, then use the therefore symbol; we always take the conjunction of the sets in the next line. This is becuase we are "ANDing" the two together.
well the first example is enough to prove the eqivalence between the two anyway. So we can just go with that
I think the best thing to do at the moment is not to solve the problem but figure out the best way to ask it and what we are wanting to ask.
That way I can put this up on SE and more people can see it
I think I should drop including therefore in the question; since it can be asked without it and it seems to be confusing people.
As it stands the question will just be: Modus Ponens is defined as ((A→B)∧A)→B "reference". How is it that this expression allows us to write things like "If today is Tuesday, then John will go to work. Today is Tuesday. Therefore, John will go to work"
And if this is the case why does the english statement seem to tell us that "it is true that John will go to work"; while the logical expression seems to be true independent of the value of the consequent?
@DHMO you can solve for the coefficients using (input,output) pairs and linear algebra, in which case if all integer inputs yield integer outputs then the coefficients must be rational
that is, if $f\in\Bbb Q[x]$ is a polynomial such that $f(\Bbb Z)\subseteq\Bbb Z$, then $f(x)=\sum_k a_k\binom{x}{k}$ for some integers $a_k\in\Bbb Z$ (all but finitely many $0$ of course). essentially you can induct on degree, or apply forward difference operators and "integrate" back
the forward difference operator is $\Delta f(x):=f(x+1)-f(x)$. notice this reduces the degree of $f$, and in particular $\Delta \binom{x}{k}=\binom{x}{k-1}$. thus, we can apply $\Delta$ enough times to $f$ to get $\Delta^n f(x)=c$ for some constant $c$, then write $\Delta^{n-1}f(x)=c\binom{x}{1}+d$, then $\Delta^{n-2}f(x)=c\binom{x}{2}+d\binom{x}{1}+e$, etc.
I've done that stuff before, just hadn't seen this application of it.
Here's a question for you, in this context: One way of solving difference equations is via generating functions. What's the differential equations analog of that?
(I have an answer, I'm just curious what you think)
sure. multiplying both sides of the recurrence by x^n and then summing over n is the discrete analogue of multiplying both sides of a differential equation by a^-x and integrating over x
You have an integral rather than a sum, but in both scenarios one converts the problem into an algebra form, solve for the transform, then 'invert' the transform to get the asnwer.
@user400188: I don't think you got a very clear answer. Your issue is that you're conflating the formal system and the meta-system. In propositional logic, given any propositions A,B the string "A implies B" is also a proposition. There are many formal systems for propositional logic. Of them, Hilbert-style systems have a single inference rule called modus ponens. A rule is not a proposition, nor a string, unlike an axiom, which is a proposition.
@user400188: Modus ponens (MP) exists as an object in the meta-system, which is part of the formal system being studied, in this case a Hilbert-style system, which is also an object in the meta-system. MP can be represented as a relation such that MP(x,y,z) is true iff there are propositions A,B such that x="A" and y="A implies B" as z="B".
(I'm being slightly sloppy here. Technically one should distinguish clearly between variables and quoted symbols. Precisely, I should say: MP can be represented as a relation such that MP(x,y,z) is true iff there are propositions A,B such that x=A and y="("A+"->"+B+")" and z=B... But for ease of reading we shall just use our brain to figure out which symbols are quotations.)
@user400188: Now in a Hilbert-system, the collection of theorems is simply the smallest collection of propositions that includes the axioms and is closed under MP. You can think of MP as a generator that generates more strings from previously generated strings. Taking to the 'limit' gives the resulting collection of theorems.
So remember, MP is not an axiom. Given any propositions A,B, "(A implies B) and A implies B" is a tautology, but is not at all MP. It can be said to be an internal reflection of MP, though.
Sorry for me to say "closed under" I should have defined MP as a function such that MP(x,y)=z iff there are propositions A,B such that x="A" and y="A implies B" and z="B".
@user21820 Sorry I can't amend this message and the next one.
We know that $\aleph_0$ is smaller than $\mathfrak c$, the cardinality of the continuum. But are there some good upper-bounds?
For example, it is trivial that $\mathfrak c<2^{\mathfrak c}$, but I wonder if there are better bounds. Specifically, I have to wonder if there exists $\alpha\in\mathb...
