@amWhy I think you can reason it as such: The probability of trying three keys and only the third working should be the same as him picking 3 keys, putting the first two back in his pocket without trying them, and then the third one working
The first two choices can in that sense be ignored.
if $$AB=AC=EF,AE=BE,\angle BAC=120^{\circ},\angle ADB=\angle BEF=90^{\circ}$$
find $\angle ABE$
Following is one methods
Let $\angle ABE=a,\angle AFE=\angle DBE=\dfrac{\pi}{6}-a,\angle EAF=\dfrac{\pi}{3}-a$ use sine thereom we have
$$\dfrac{\sin{a}}{\sin{(\pi-2a)}}=\dfrac{AE}{AB}=\dfrac{AE}{EF...
Finally back home and able to write things with a real keyboard which is much much better.
(ah, I didn't have a conference - some asked me if I attended some conferences - but I met some important (or if you don't like this adjective, you can replace it by known) mathematicians today imho)
i take that back, it seems fine. I created a Geogebra version of it (or at least an approximation thereof) and the angle does seem to be 6 degrees
(which i guess leads to a question on my part: how would one give a construction of the desired configuration? I had to adjust things to make the 90-degree angle correct
While on that subject, why are there periodic continuous functions that always have some trigonometric functions associated with it? And not otherwise?
There are periodic functions which are analytic but aren't trigonometric. Elliptic functions and theta functions are the big examples if memory serves.
Point 1: Elliptic functions are not holomorphic everywhere in the complex plane. They can't be, since they're also doubly periodic and therefore would be bounded entire functions.
Holomorphic function by definition is a function complex-differentiable on a subset of $\Bbb C$. The term for holomorphic everywhere in $\Bbb C$ is "entire".
Anyways. Theta functions have product representations, and since elliptic functions can be written as quotients of theta functions you also get product representations for these.
I have this figure. AB = CB, AC = DC. And I don't know how to prove that angle BAE and angle BCE are equal. Any hint? I have tried a long time and I don't find congruent triangles...
@Agawa001 It was a great year for my research so far, although there were also tons of negative points, I could have started (some serious) work on a second book ...
i hate when one's fate fell as a prey in other careless people's hands, i relied on people once and they made me suffer 3 years, because they simply didnt care about my diploma, they delayed , and delayed, and delayed more i insisted it was fruitless
@Agawa001 I know your point. I like when I go out and talk to my dogs, I feel they understand me better than anyone. They are really smart! They staring at me as if every single word would be understood in the deepest details. :-)
@Agawa001 Take a pet and you'll feel much better, no matter the experiences you mentioned above.
last night I dreamt about a weird metric where there is some topology, and the distance function is defined by the trace of some positive trace matrix A, where each pair of points in the topology has a unique A assigned to it
@Agawa001 Usually when talking about dogs some expect me to tell them I have some famous breed, but no, I love these ones, no need for famous breed, and I wouldn't give them for anything in the world.
By the way, in the meantime I've just created a marvellous question that I bet could be published everywhere without any difficulty.
Again, it's related to the Catalan constant. Not even mentioning that I also derived in a different way one of the identities Ramanujan derived, again related to Catalan constant.
It's cool to have a pack with such questions, you don't see them everywhere.
@TobiasKildetoft how many problems did you create so far? How many ideas did you promote in the mathematical community that could revolutionize your math area? No one? Hope I'm wrong.
@0celo7 I got you don't like hard integrals. I understood that. Now you can do something for your math better than continuing this nonsense conversation.
My questions probably started out well-intentioned. But given that I never get any answers, it is true that I am starting to suspect that they do not have any good answers
@TobiasKildetoft You starred Juan when he said that what I do is trivial, that his student are able to finish all I do in 2, 3 days (or approved him in some way - I remember that).
@TobiasKildetoft This is my problem, don't you think? I approach my math the way I want, and yeah, I could ask people on the street to calculate some integrals.
I don't recall interacting with Juan at all. For what it is worth, I do not agree at all that your problems are easy. I just do not find them interesting (and you seem unwilling to give any motivation for them).
@TobiasKildetoft The anwer is in The Man Who Knew Infinity.
@TobiasKildetoft Hardy to Ramanujan: What they (referring to Euler and Jacobi) had in common, what I see in you, is a love of form. It's all through your notebooks. Let me ask you something. Why do you do it, any of this? Ramanujan to Hardy: Because I have to. I see it. Hardy to Ramanujan: Like Euler. Form for its own sake. An art unto itself.
@DanielFischer getting in skirmish sometimes is a daily custom to fuel this chatoom as a natural reaction for boedom when nothing is curent here apart the closed circle of : topology -> abstact algebra -> topology ...
I have to admit that @DanielFischer is right. Some might be sick and tired by all these discussions. You can ignore me since you usually open these discussions, and all will be fine for all.
@Agawa001 Also there are less inequalities on site, it's a weak point. Just as I do integrals and series, there are lots of mathematicians that are specialized on doing inequalities. Actually they create and solve inequalities all day long. You can find them on internet.
@Agawa001 lol, and I just started to write a proof that I suspect it will take me an hour or so to nicely arrange, and it's a bit later. Maybe a coffee might help?
if $$AB=AC=EF,AE=BE,\angle BAC=120^{\circ},\angle ADB=\angle BEF=90^{\circ}$$
find $\angle ABE$
Following is one methods
Let $\angle ABE=a,\angle AFE=\angle DBE=\dfrac{\pi}{6}-a,\angle EAF=\dfrac{\pi}{3}-a$ use sine thereom we have
$$\dfrac{\sin{a}}{\sin{(\pi-2a)}}=\dfrac{AE}{AB}=\dfrac{AE}{EF...
