So if you regard $x$ as a function of $w$, you just have $x(w)=we^w$. So what's plotted is just y=xe^x flipped.
Right.
What is decidedly less obvious is the follow. Both of those are still defined for $x<-1/e$, but they're complex there
However, due to how the branches are related, if you ask Mathematica to plot them you find that they've got equal real parts and opposite imaginary parts.
If in particular I ask for the real part for x<0 of both functions, I get this plot:
(Simplest way to do that is to start from $x=we^w$ and expand near $w=-1$ to get $x=-\frac{1}{e}+\frac{1}{2e}(w+1)^2+O((w+1)^3)$. From that we deduce $W(x)=-1\pm \sqrt{x+1/e}$ for the two real branches near $x=-1/e$)
Anyways, an observation: These two branches of Lambert-W have the same real parts and opposite imaginary parts for $x<-1/e$, and are real-valued for $-1/e<x<0$. So if I take the average of them, I'll get a function which is real valued for all $x<0$.
That gives this picture:
And now there's a surprise. The green function is continuous through the singularity!
So you can construct points on the the upper part of the green curve by drawing a horizontal line, intersecting that with $y=xe^x$, and constructing the midpoint.
@Astyx That'll give a plot like what I gave above. To flip it I used ParametricPlot
@AkivaWeinberger if I were to prove some conjecture in Z/1 and show that it being true in Z/n implied it were true in Z/{n+1} could I then conclude it is true for the integers?
@Semi when you're not busy do you think you could explain to me why Legendre transforms matter/are a thing (as a physicsy dude). We used them a bunch in a grad functional analysis class i took but they were super unmotivated
So the Legendre transform in this context tells you what a saddle point approximation of the given integral transform does to the exponent.
That's actually the context in which it arises for me right now. Not something I knew about until within the last year, or at least i'd not seen it presented as such.
(The reason I had all those minus signs is because when doing saddle-point analysis it's really more natural to think in terms of "where's the min" rather than "where's the max". As for why the Legendre transform is defiend to give a max... shrug)
For the latter, it's basically using the Legendre transform to give a mapping from the tangent bundle to the cotangent bundle. (I'm stealing from Wikipedia at this point.)
What I'm saying above is that I get $x^*=W_0(p)$ in the process of taking the Legendre transform $f^*(p)=px^*-f(x^*)$ of $f(x)=(x-1)e^x$. Or, at least, I can do so for $p>0$.
@Astyx well, there's Akiva's (proven) conjecture that for real $a$ and positive integer $n$, if $a^n$ and $(a+1)^n$ are both rational, then $a$ is rational
In the main MSE chat @AkivaWeinberger proposed the following conjecture they found from an old question.
If $a$ is a real number and $a^n$ and $(a+1)^n$ are both rational, then $a$ is a rational number.
I have tried several times to prove it and I just wish to know how to prove it.
Since I...
ahhhh this explains why ive heard people say it turns the Lagrangian formalism into the hamiltonian formalism (not that i understand those beyond simple definitions)
@EricSilva: At some point you might want to look at Spivak's book on mechanics. He spent his life learning differential geometry so that he could get to a more mathematical understanding of physics. His original plan was to write on relativity, but somehow I don't think he'll quite make it that far.
(Of course, there are still errors. And one can argue that we mathematicians don't always have the physicists' intuition or depth of understanding ... even if we're pedantic.)
The latter is more general (hence the name of the school), would probably allow me to earn more money, and wouldn't close any doors to research, but the curriculum being more general also means I would do less maths
Last but not least, the ENS is more prestigious for maths stuff
It's tougher in France, Astyx, because there are so few academic options. If you're good, you're funded by CRNS for years. But there aren't the zillions of varied-level teaching jobs the US has. On the other hand, the US has less and less commitment to tenured faculty.
@ERicSilva: I can no longer find Spivak's Publish or Perish website. I should email him and make sure he's ok.
if $\displaystyle \sin \alpha = p \bigg\lfloor \int^{1}_{0}\{\ln x\}dx\bigg\rfloor \;, \alpha \in (0,2\pi)$ .Then $p$ is ?
given $\lfloor x \rfloor $ is floor function of $x$ and $\{x\} = x-\lfloor x \rfloor$
using $\{\ln x \} = \ln (x) - \lfloor \ln x \rfloor $
and $0<x<1.$ So $-\infty<\ln(x)<
Oh right, this can still be arranged in the not-too-distant future, but do you have any opinion concerning what school I go to ? This has to be decided soon enough (like, by the end of the month)
i applied to like 4 or 5 but i didnt even think i would be able to go to college until like summer before my senior year of high school cause my parents were in a devastating financial situation
i ended up getting in everywhere i applied but i basically chose the places to apply based on food and if i could deal with their school colors for four years bc i didnt have a lot of time to "search around" as it were
i also didnt really have anyone to guide me at the time bc my parents knew nothing abt college, being from a different country and having never finished secondary school
I'm now butting in and giving advice to a first-year student at Berkeley (whom I know indirectly through one of my former students and one of his teachers in college while he was in high school).
i'm trying to think of how to form a topology on N with a basis of open sets iff sum of reciprocals of the members of the set converge, a intersection and finite union clearly are still open, i.e. the sum of convergent series is convergent, but i'm not sure how to apply this to the empty set or to the whole number line, any ideas on how to alter the definition
yeah, its the universe that always gives me problems. how do i treat the empty set and set N, no matter how i base a definition on convergence and divergence i cant quite get all the rules to work out
In the main MSE chat @AkivaWeinberger proposed the following conjecture they found from an old question.
If $a$ is a real number and $a^n$ and $(a+1)^n$ are both rational, then $a$ is a rational number.
I have tried several times to prove it and I just wish to know how to prove it.
Since I...
if $\displaystyle \sin \alpha = p \bigg\lfloor \int^{1}_{0}\{\ln x\}dx\bigg\rfloor \;, \alpha \in (0,2\pi)$ .Then $p$ is ?
given $\lfloor x \rfloor $ is floor function of $x$ and $\{x\} = x-\lfloor x \rfloor$
using $\{\ln x \} = \ln (x) - \lfloor \ln x \rfloor $
and $0<x<1.$ So $-\infty<\ln(x)<
