@mick Perhaps it was poorly phrased. But $\ln(z-1)$ is not even defined on the whole complex plane. The only functions that have always-defined inverses are the linear ones. That's the point he was making.
isnt " include a math term ??(not) including addition , substraction etc , isnt that used to define that general quintics cannot be solved by radicals ? @BalarkaSen
If you have a differential equation that you can solve by multiplying through by an integrating factor, say $e^{\int \frac{1}{t} \text{ d}t} = e^{\log|t|} = |t|$, do we have to just go on ahead and assume $t$ is positive?
Problem
Suppose we some function $F\left(\sum\limits_{k=1}^n x_k\right)$ over the positive orthant $[0,\infty)^n$. Show that this this is proportional to the integral $\int\limits_0^\infty s^{n-1}F(s)\,ds$. What is the constant of proportionality? If this is a well-known result, a referen...
@BalarkaSen it says no solutions in radicals .... ( from wiki ) and then the link for radicals gives me nth roots ... not so formal ?? or is wiki wrong ?
@BalarkaSen: I have a different impression. A lot of things on wikipedia are by no means BS. Maybe some aren't 100% rigorous, but some things are. And most other things are detailled enough, imo.
@BalarkaSen But common , i do not want to ask about Galois theory ! I want to ask about contour integral .... I shouldnt be an expert on Galois theory to ask about contour integral ??
then the identity becomes $$V = \frac{1}{\alpha} \int_0^\alpha d\tau \, e^{\tau H} V e^{-\tau H}$$ since $e^{\alpha H} e^{-\alpha H}=1$ ($H$ commutes with itself, after all)
if I ask for a nonpolynomial function that does not contain a sqrt , i do not want to see sqrt(x) or sqrt(x^2+1) or sqrt(ln(x)) ... ln(sqrt(x)) is arguable because it can be simplified ...
@Huy I'd really appreciate if you can put your tone down a few notches. I have expressed my personal views on wiki and just because I can't provide examples RIGHT AT THIS MOMENT doesn't mean that you're going to reply in SUCH an attacking tone.
@BalarkaSen I understand how you feel , but maybe you feel better as seeing it as a defense for me / wiki rather than an attack on you ... which is probably closer to the truth ...
@BalarkaSen: I'm not attacking. I'm just saying I would never state something such as "this [...] website is complete BS" if I didn't have a single example in my mind. Because if anyone asked and I couldn't provide one, I'd be a bit embarassed.
If you have a differential equation that you can solve by multiplying through by an integrating factor, say $e^{\int \frac{1}{t} \text{ d}t} = e^{\log|t|} = |t|$, do we have to just go on ahead and assume $t$ is positive?
@Chris'ssis If you name something after yourself, no one will ever call it that. If someone else who writes a book or well-renowned paper names it after you, then it might catch on.
That's how I've seen it done in loads of examples, both online and in lecture. It's not just $t$ either. I've seen my lecturer let $e^{\int \frac{1}{x} \text{ d}x} = e^{|x|} = x$, @Huy.
(I know I know, but the constant cancels out anyway and I'm comfortable with that.)
@mick For example, this works well even with your restrictions : $g(x) = \int_{\Bbb R} \lim_{h\to 0} \frac{f(f^{-2}(f(x+h)-f(x+h-1)))-f(f^{-2}(f(x)-f(x-1)))}{h} dx$
@mick people who need to get honor by building monuments to themselves rarely deserve the honor. It needs to be built by someone else to be taken seriously.
Let's say we're going to multiplying our DE by the integrating factor $e^{\int 2x \text{ d}x} = e^{x^2 + \mathcal{C}} = Ae^{x^2}$. The constant $A$ will cancel out anyway, @Huy.
having something named after you isn't a sign that you've accomplished something, but that you've done something which the community wants to be able to reference in the future
@mick Sure, whoever discovers something deserves to get the credit, but the monument effect is what sways the masses, and masses are what makes a name stick.
I've got two special matrices I'm trying to diagonalize :
The Z matrix :$$\begin{bmatrix}
1&1&\cdots&1&1\
\\&&&1
\\&&\diagup
\\&1
\\1&1&\cdots&1&1
\end{bmatrix}$$ (all the other members are $0$)
And the X matrix :
$$\begin{bmatrix}
a&0&\cdots&0&b\
\\0&\ddots&&\diagup&0
\\\vdots&&&&\vdots
\\0&...
I am interested in how blind people learn mathematics at any level, but particularly before college. Math is often taught using a lot of visualization; how does this work with blind people?
My interest in this is a little round-about. I have nonverbal learning disabilities (NLD) and am writing ...
Multiplying through by that gives: $$ A|t| \dfrac{\text{d}x}{\text{d}t} + \dfrac{A|t|}{t} x = 0$$ The $A$s clearly cancel out as they aren't equal to 0, but what of the $|t|$?
@BalarkaSen Because Perelman is sick and tired by so many ungrateful, unthankful people in this community, and I also believe he terribly hates the thieves.
@robjohn: By any chance, do you know how I can show this? $$\frac{d}{d\lambda} e^{\alpha H} = \int_0^\alpha d\tau \, e^{\tau H} V e^{-\tau H} e^{\alpha H}$$ for $H = H_0 + \lambda V$.
@Khallil The integral diverges at $0$ so there can't be a continuous solution across $0$. The integral $\log|t|+C$ seems to indicate that that is okay, but it is not.
@Alizter: Yau didn't really critisise him, but he interpreted Perelman's work as not a complete proof and worked out the "missing bits" himself, thinking he (Yau) completed the proof of the conjecture.
Problem
Suppose we some function $F\left(\sum\limits_{k=1}^n x_k\right)$ over the positive orthant $[0,\infty)^n$. Show that this this is proportional to the integral $\int\limits_0^\infty s^{n-1}F(s)\,ds$. What is the constant of proportionality? If this is a well-known result, a referen...
the solution is $\left\{\left\{y\to -\frac{x W\left(-\frac{\log (x)}{x}\right)}{\log
(x)}\right\}\right\}$
I've got two special matrices I'm trying to diagonalize :
The Z matrix :$$\begin{bmatrix}
1&1&\cdots&1&1\
\\&&&1
\\&&\diagup
\\&1
\\1&1&\cdots&1&1
\end{bmatrix}$$ (all the other members are $0$)
And the X matrix :
$$\begin{bmatrix}
a&0&\cdots&0&b\
\\0&\ddots&&\diagup&0
\\\vdots&&&&\vdots
\\0&...
