Hi guys, I just want to ask something that might be very easy to all of you but it is hard for me to answer it. How can I answer question like this: "how much is exp(- pi/2)?" Yup, that is the question.
@Tunk-Fey It takes some work. You have to show that the $\Bbb Q$-extension of the exponentiation exists and is unique. If that's done then a natural $\Bbb R$-extension is induced from continuity and the $\Bbb C$-extension (which you don't need here) follows from analytically continuing it throughout the complex plane.
Fix some arbitrary base $z$. $f(x)$ be $z^x$. Then for integers $x$ you have $f(x + y) = f(x)f(y)$ so a $\Bbb Z \hookarrow \Bbb Q$ extension follows (e.g., $f(1) = f(1/2)^2$).
Since he is a physics prof, I doubt this question is asking about its numerical value or pure math. Perhaps he asks something that has a philosophical meaning.
But what you really need for reals it continuity. For some sequence ${a_n}$ converging to $a$, you need a unique extension of $f$ that satisfies $\lim_{a_n \to a} f(a_n) = a$
Well, just by contracting the intervals infinitely, we get a single point. (We don't get it for open intervals because it has no endpoints; it would just become nothing.) This is a continuous deformation, so they're homotopically equivalent. That's the intuitive way to think about it.
I'm not that well-versed at topology, though, so I'd be hard pressed to find a homotopy equivalence. I'll let the smarter people in the room handle that.
@Vrouvrou Didn't you receive a warning like this, when you were entering the question? i.sstatic.net/CSu6p.png
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@Chris'ssis yes .. but its a nice proof .. :D it is very similar to sos's proof as it evaluates the answer in the same form $\dfrac{1}{2}\zeta^2(2)+3\zeta(4)$ :-)
@r9m Depending on the work of someone, I might be appear along with someone else, but this is not sure yet. That means my proof you saw may be still simplified.
@r9m I also publish the other series you saw, the one involving the Riemann zeta function. I need to to that before a bigger project like publishing a book.
@r9m I need to publish a book, that would give me more confidence, strength in me to go on this way (especially for the fact that some may discourage you with things like "who cares this crazy mathematics?")
I have thousand of questions, I can make up a very beautiful book with crazy awesome questions & solutions.
@Chris'ssis I was going to say just the opposite ... you have no idea how many of us crazy (as you call it) mathematics fans are lurking around in the shadows .. all we need is an initiation :-) ... like that book of yours !! :D
@Chris'ssis well the best way to prove them wrong is to publish it and show them the success of the book :-) really that is the only way :) .. no one believes until you show them how it is done :)
@robjohn if i take $h,k\in F=\{h\in H:\langle f''(u)h,h\rangle<0\}$, $\langle f''(u)(\lambda h+\mu k),(\lambda h+\mu k)\langle$ but $(\lambda h+\mu k)\in H$ so $\lambda h+\mu k\in F$ so it is a subspace of $H$ no?
@Chris'ssis but how will I know when and where it is going to be published ?! .. its not like you are going to publish it with your pseudonym 'Chris'ssis' :P .. that one day I google and find a paper by Chris'ssis and xyz(the other mathematician) :P
@robjohn i do a mistake, i must show that $<f''(u)(\lambda h+\mu k),(\lambda h+\mu k)><0$, $<f''(u)(\lambda h+\mu k),(\lambda h+\mu k)>=<\lambda f''(u) h,\lambda h>+<\lambda f''(u) h,\mu k>+<\mu f''(u) k,\lambda h> +<\mu f''(u) k,\mu k>$
@robjohn i don't know if it is right
and also i don't know how to see that $<\mu f''(u)k,\lambda h>$ and $<\lambda f''(u) h,\mu k>$ are <0
@DanielFischer: I arrived at the part of my lecture notes about Sobolev-embeddings. Can you explain to me why compact embeddings are particularly interesting / powerful? I see that embeddings are useful, but what makes compact embeddings more useful?
@Huy Having a compact embedding $j\colon E \hookrightarrow F$ of Banach spaces means that each bounded sequence $(f_n)$ in $E$ has a subsequence such that $(j(f_n))$ is norm-convergent. That's pretty useful.
@r9m I'm noob too (but not that noob) ... Just saying, for this risk I protected all my information on this computer. Moreover, for the important stuff I use another computer.
@Huy Applications are not my friends, I don't remember examples. But iirc, a common story goes along the lines of having a family of functions that approximate a solution of an equation, optimisation problem or so, in the embedded space, a compact embedding guarantees (under some conditions on the problem) a weak solution in the larger space (with the coarser topology).
Well there is no way I can be cent percent secure .. certain figures of authority in our insti are capable of monitoring my activity on the internet if they want to :P lol .. but as long as I don't give them a reason to do that .. they have more important things to do :P
Princeton hired him at the rank of tenured full professor within only two years of finishing graduate school, which is considered a record in the Ivy League.
Hm. I tried a more general approach to what I'm looking for.
I want $a,b \in \mathbb{N}/2$ such that $\frac{a+b \pm \sqrt{a^2-ab+b^2}}{3}$ are "nice numbers", i.e. some simple fractions, ideally natural numbers or in $\mathbb{N}/2$.
Is there any non-brute-force way to accomplish this?
Hint: $p_1$ is one digit long, $p_2$ is two digits and $p_3$ is 3 digits where the first digit is $p_1$ and the third digit is the second digit of $p_2$. Hopefully that wasn't confusing.
There are only 3 numbers under $2\times 10^{13}$ that satisfy the condition that $(p-1)! +1$ is divisible by $p^2$. These are called Wilson Primes and it has been proved that there are infinitely many of them.
@skullpatrol "This is a difficult question. I don't think that everyone should become a mathematician, but I do believe that many students don't give mathematics a real chance. I did poorly in math for a couple of years in middle school; I was just not interested in thinking about it. I can see that without being excited mathematics can look pointless and cold. The beauty of mathematics only shows itself to more patient followers."
Then when I'm bored, I usually talk about all of life's problems. Like how difficult it must be for girls to wear tampons and why India can't just give Kashmir to Pakistan.
In 12th, if it's a not too smart teacher, he/she won't even give marks for the steps of large 6 mark questions if you get the final answer slightly off the answer key that was given to the teacher who corrected the paper.
I got 87/100 in my test. All my answers were right. Then why is my score low? Because my algebra was not simplified enough or not simplified right to match the answer key.
@robjohn: No, I just said that to get your attention. You're the kind of person high in the mathematician ranks that can make something like that happen.
@user2838619 construction of a bottle shouldn't be hard. the idea is to cobble up a bunch of linear abs-forms with two cubics near the edges to get a shape of a bottle.
