I am not going for the 'gro's. Erdos is rather my classical hero. =P
(Evidently the two most brilliant mathematical personalities from the past century are Grothendieck and Gromov, both having names starting with the same three letters)
^ apparently because both are ruski, well, grothendieck being part ruski.
I am trying to prove that $$(1)\qquad x\sum_{k=0}^\infty\frac{2^{-5k}(6k+1)((2k-1)!!)^3}{4(k!)^3} = 1 \implies x=\pi$$
Using W|A I got to a point where I have to prove that $$\color{red}{(2)\qquad \pi = -\frac{8 \left (\sqrt{16-3 K\left ({1\over4} (2-\sqrt{3})\right )^2\, _3 F_2\left ({3\over2},...
@Chris'ssis as far as I can remeber .. OL answered a question with a very nasty looking Lagrange Interpolation in the past too :D .. He 'sees' (Neo style :P) this stuff !!
@Semiclassical : Regarding http://math.stackexchange.com/questions/996985/proving-that-x-sum-k-0-infty-frac2-5k6k12k-134k3-1-i The reason for presenting it this way, is because I feel that it makes more sense with respect to lemma 2. However, I do understand that many see this as an odd way of presenting the problem. (Please note that I only do math recreationally, and my thought-process seem odd to most mathematicians.)
@r9m I think we can also use this result $$\frac{1}{a(a-b)(a-c)}+\frac{1}{b(b-a)(b-c)}+\frac{1}{c(c-a)(c-b)}=\frac{1}{a b c}$$ that we may then extend to an odd number of variables $(n\ge3)$.
I'm struggling with a Fourier series. I need to find the Fourier series of the following function.
That's the function under study:
$f(t)=\left[\sqrt{1-k^2\sin^2t}\,\right]$.
The function is even and $\pi$-periodic.
The Fourier series should be in this form:
$f(t)=\frac{a_0}2+\...
The @O.L.'s post makes me recollect the well-known identity
$$\frac{1}{a(a-b)(a-c)}+\frac{1}{b(b-a)(b-c)}+\frac{1}{c(c-a)(c-b)}=\frac{1}{a b c}$$
that can be generalized for any odd number of variable, $n\ge3$, and then if we set the variables
we have by $\displaystyle u_j=\cos \frac{x}{2^j}$ ...
O.L. explicitely stated that in his answer !! -_- .. $\begin{align}\sum_j\frac{u_j^{-1}}{\prod_{k\neq j}\left(u_j-u_k\right)}&=\frac{1}{\prod_{m}u_m} \sum_j \prod_{k\neq j}\frac{0-u_k}{u_j-u_k}=\frac{1}{\prod_{m}u_m}, \end{align}$ that is Lagrange Identity ! :O
I want to see OTHER approaches than this one. Make sure they are significantly different and not a direct restatement.
$$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}\tag{1}$$
$$\sum_{n=1}^x n=\frac{x(x+1)}{2} \; \forall x >0\tag{2}$$
$$(1)\stackrel{(2)}{\iff} \...
@r9m Let $X$ be a metric space with the property that every infinite subset of $X$ has an accumulation point in $X$. Let $\mathcal O$ be any open cover of $X$. Prove there exists $\varepsilon >0$ with the following property: for every $x\in X$; there is $V\in\mathcal O$ such that $B(x,\varepsilon)\subseteq V$.
@TedShifrin only what prof told me. that groups can be made into metric spaces by looking at the word metric on the cayley graph and that a notion of hyperbolicity is defined for groups.
the rest is given to me to work out and think about.
@Chris'ssis I refer to expressions of the sort $\displaystyle f(x) = \sum_{cyc} \dfrac{(x-a)(x-b)}{(a-b)(a-c)}$, which is a quadratic polynomial .. and $f(x) - 1$ has 3 roots ! $a,b,c$ .. so it has to be an identity .. :)
@Ted I found a book by Gallot and two other authors that does a Petersen-level introduction to Riemannian and so far does not make me want to shoot myself.
@TedShifrin Take a group $G$ and a finite generating set $S$. The cayley graph pf the pair can be made into a metric by defining the word metric $d$ on the nodes of the graph. in fact, that gives out a geodesic metric space structure over $G$. now to identify $\Gamma(G, S_1)$ and $\Gamma(G, S_2)$ for distinct generating sets S1 and S2 he defined quasi isometry to construct a whole category of geodesic metric spaces in which cayley graph of groups are unique.
nod. drive-by downvotes are annoying. though usually there it's because you put enough effort into your answers that a downvote would require some explanation
i'm fine with drive-by downvotes on answers people obviously put no effort into :P
How is continuity by $\varepsilon - \delta$ using the axiom of choice, and do we use the axiom of choice each time we define an injective function for range and domains of infinite cardinals?
Direct quote; Also, if you give important consequences I would add the fact that continuity by $\epsilon-\delta$ is equivalent to continuity by limits. Most people are not aware that this is a consequence of the axiom of choice.
@DanielFischer Ok, I understand. So, about this continuity question with the sup function. I am was just rewriting my answer into one that maps open sets to open sets.
Like this:
Let $O \subseteq \mathbb R$ be open and $x_0 \in f^{-1}O$. Since $O$ is open there exists $\varepsilon > 0$ such that $(f(x_0)-\varepsilon, f(x_0)+\varepsilon)\subseteq O$. Now let $\delta = {\varepsilon \over 2}$. Then (same argument as in the $\varepsilon$-$\delta$-definition-argument) for $x$ such that $d(x,x_0) < \delta$ and for all $a \in A$:
$$ d(x_0, a) \le d(x_0,x) + d(x,a)$$
hence
$$d(x_0, a)- d(x,a) \le d(x_0,x) < \delta$$
Then take the supremum on the left hand side twice to get $\displaystyle d(x_0, a)- \sup_{a \in A} d(x,a) \le \delta$ and also $\displaystyle \sup_{a \in A}d(x_0, a)- \sup_{a \in A} d(x,a) \le \delta$.
Hence $f(x_0) - f(x) \le \delta < \varepsilon$. Now repeat the same argument with $x$ and $x_0$ swapped so as to get $f(x) - f(x_0) < \varepsilon$. Putting both together yields $|f(x) - f(x_0)|<\varepsilon$.
But why should $B(x_0, \delta)$ be contained in $f^{-1}O$?
@DanielFischer Good point. I'll let you in on a secret: I am so lazy that I hardly ever use paper. But it's good to be reminded that paper is a good idea. I should have it handy at all times.
@DanielFischer No, I do most things in my head. Once I have a solution I either type it up in latex if I have to, or content myself with a solution in my head.
Except new stuff.
When I do stuff that I'm not so familiar with (so that it cannot comfortably fit into my working memory) then I always use paper.
@DanielFischer The odd thing is that in this case paper might not have helped: I already had it all in front of me in latex and was still confused.
Sometimes coffee helps. I'm never too lazy to make coffee : D
I do this no-paper thing too. If I answer something that's pretty simple, I'll solve and $\LaTeX$ at the same time. But if it's more than 3 lines or so and it requires algebra/calc I'll have to write a briefing in about 30 seconds or so on paper.
So what I gather: If we use $d_h(f)$ to denote the directional derivative of $f$ in direction $h$ then the guy is asking how to show $d_{h + k}(f) = d_h(f) + d_k (f)$. So far so good. He wants to use the limit definition of the directional derivative to do so. You answer: This is not possible. The directional derivative is only linear if the Fréchet derivative of the function exists at the point where you want to take the directional derivative.
@TedShifrin Now the missing piece in my understanding is: I understand the first sentence in the question to mean that he assumes that the Fréchet derivative exists.
Hi, I'd like to check some easy exercise, show that the arithmetic sum in $\mathbb{R}$ is continuous in the topological sense. In what follows I assume that $\mathbb{R}^2$ has the taxicab metric to simplify the calculation. Let $c\in\mathbb{R}$, and $\epsilon>0$. Let $P=\langle p_1,p_2\rangle \in (+)^{-1}\big (B_{\mathbb{R}}(c,\epsilon)\big)$ , i.e., $|c-(p_1+p_2)|<\epsilon$, let $\delta =\epsilon-|c-(p_1+p_2)|$
@MattN. He was misunderstanding the definition of "differentiable" to mean "Has all directional derivatives". He then wanted to use this latter definition to prove that $v \mapsto D_vf$ is linear.
I should have said 'a' sensible interpretation. I read the first sentence as shorthand for $f(x+tv)-f(x)$ as a function of $v$. But Ted's interpretation seems more likely.
"Charitable" is not a word I want to use to refer to a colleague...
@TedShifrin "So you must use linearity as part of the definition;" this means "use the linearity of the Fréchet derivative to show the linearity of the directional derivative", right?
My professor said "don't worry about induction on the exam" so everyone assumed he wasn't going to put induction on the exam, but I guess he meant "induction is on the exam, but it's easy so don't worry about it" LOL
I can't do integration of stuff where it does not tell you the method. I always pick the wrong one and then end up writing a long dull computation just to end up with the integral I started with.
@MattN. For the first one, you have to be a teeny bit clever. But this exact problem was a previous homework problem, and the particular cleverness was mentioned repeatedly in class.
Hi, I'd appreciate if someone could say me if the following is correct: Show that $(+): \mathbb{R}^2\to \mathbb{R}$ is continuous in the topological sense. We may assume that $\mathbb{R}^2$ has the taxicab metric to simplify the calculation.
Let $c\in\mathbb{R}$, and $\epsilon>0$. Let $P=\langle p_1,p_2\rangle \in (+)^{-1}\big (B_{\mathbb{R}}(c,\epsilon)\big)$, i.e., $|c-(p_1+p_2)|<\epsilon$, let $\delta =\epsilon-|c-(p_1+p_2)|$ we shall show that
I'm struggling with a Fourier series. I need to find the Fourier series of the following function.
That's the function under study:
$f(t)=\left[\sqrt{1-k^2\sin^2t}\,\right]$.
The function is even and $\pi$-periodic.
The Fourier series should be in this form:
$f(t)=\frac{a_0}2+\...
The @O.L.'s post makes me recollect the well-known identity
$$\frac{1}{a(a-b)(a-c)}+\frac{1}{b(b-a)(b-c)}+\frac{1}{c(c-a)(c-b)}=\frac{1}{a b c}$$
that can be generalized for any odd number of variable, $n\ge3$, and then if we set the variables
we have by $\displaystyle u_j=\cos \frac{x}{2^j}$ ...
