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12:00 AM
"wow, maybe it actually is difficult to understand a series of equations, or right hand sides of equations, with random stuff happening all over the place"
 
 
2 hours later…
2:24 AM
What were you interviewing for?
 
2:52 AM
i was interviewing for internships or full time jobs at a law firm. ted may have been interviewing for applicants to the school where he went for undergrad. dunno.
 
 
4 hours later…
dtn
6:23 AM
@leslietownes Quite right, but after a little digging, I realized that this is tantamount to finding a solution to any polynomial, with arbitrary coefficients ... Do you understand? I was hoping to just find a solution to a problem that hasn't been resolved in years. :(((
There are already a lot of problems with the 3rd order system of equatio.
 
 
3 hours later…
9:30 AM
Hello Friends.
How can i derive the following implication out of this theorem:

Theorem (proven): is $f:[a,b] \rightarrow \mathbb{R} $ riemann integrable and $f$ in the point $x_o \in [a,b]$ Continious, then is $F(x)= \int_a^x f(t)dt $ in that point diffrentiable and $F'(x_o)=f(x_o)$
Implication: for $f:[a,b]\times[c,d] \rightarrow \mathbb{R}$ Continious and all the partial derivates exist and are continious, we define
$F(y) = \int_a^b f(x,y)dx$
We want to show that $F'(y)= \int_a^b \frac{\partial f}{\partial y}(x,y) dx$
I tried the following thought process:

We have that $ [\int_a^b \frac{\partial f}{\partial y}(x,y)dx] $ is continious on that interval $[y,y+h]$ such that for each point $\tilde{y}$ in that interval we have according to the theorem that with the definition $F(y+h)-F(y) := \phi(y)$ that $\phi'(\tilde{y})= [\int_a^b \frac{\partial f}{\partial \tilde{y}}(x,\tilde{y})dx$
say then the left side is $lim_{k\rightarrow 0}\frac{(F(\tilde{y}+k+h) - F(\tilde{y}))-F(y+h)-F(y)}{k}$
now for k going to zero we recieve ?
Basically i am having trouble with the rigorous writing of the LHS of the equation :D
 
 
2 hours later…
11:55 AM
is the domain of x^2+y^2+z^2+1-x-0.5y-(1/3)z a tetrahedron ?
I think it is just the 1st octant
I think it is inverse function problem
I wish I had paper and pen with me ;_;
 
 
2 hours later…
jay
2:13 PM
for a probability density $p$ with finite second moments does anyone know of any inequalities of the form $\int_{\mathbb{R}^d}\frac{\|p(x)\|^2}{p(x)}- \|x\|^2p(x) dx \geq 0$ ?
 
 
2 hours later…
4:14 PM
@MadSpaces The expression for $F(y+h)-F(y)$ looks wrong.
 
4:35 PM
14
A: An approximation related to Euler's constant and the Harmonic number

robjohnApplying Riemann-Stieltjes Integrals: $$ \begin{align} \sum_{k=1}^n\frac1k &=\int_{1^-}^{n^+}\frac1x\,\mathrm{d}\lfloor x\rfloor\tag1\\ &=\int_1^n\frac1x\,\mathrm{d}x-\int_{1^-}^{n^+}\frac1x\,\mathrm{d}\!\left(\{x\}-\tfrac12\right)\tag2\\ &=\log(n)+\frac1{2n}+\frac12-\int_{1^-}^{n^+}\frac{\{x\}-\...

How to go from 6) to 7)? @robjohn
 
5:05 PM
Ah, so that assumes that $H_n-\log n\to \gamma$.
I think I got it now. :)
 
 
2 hours later…
7:15 PM
Does anyone have an opinion on if this is positive, negative, or symmetrically skewed?
I was going to say negative, but the mode = median = mean, which would indicate that it is symmetric.
Well, the mode and mean are exactly 21 and the median is 21.3
Nevermind, this is symmetric
I created a better histogram that more clearly shows the distributions around the median.
 
7:38 PM
It looks like you have 20 datapoints. Hardly enough to draw any serious conclusions, if you ask me.
In any event, you might be better off asking a statistician.
E.g. at stats.se
 
It's just a hw assignment, I know it doesn't really mean too much.
I didn't know there was a stats.se though
Stats is low-key boring
 
8:30 PM
Is there a way to reduce the number of notifications per bounty question I get?
 
i need a hint on how to prove that $\frac{2ab}{a+b}<\sqrt{\frac{a^2+b^2}{2}}$
for all a,b>0
thank you
 
 
2 hours later…
10:32 PM
@Vrouvrou you probably also need an assumption that $a\neq b$. otherwise the inequality isn't strct
as far as proving it, it may be easier to prove something a bit stronger: $2ab/(a+b)\leq \sqrt{ab}\leq \sqrt{(a^2+b^2)/2}$, with equality only if $a=b$
(a direct proof is possible, though)
 
 
1 hour later…
11:56 PM
@Vrouvrou: the left inequality is HM < GM and the right inequality is AMGM
 

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