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5:04 PM
@satan29 The second derivative doesn't converge?
 
Hello
So I just read about quotient topology
and my mind is blown
Is there a nice way to get more intuition of it?
 
@ComFreek quite false!
 
only true for $h\to0^+$
 
Bob
@copper.hat I got two responses to my original post. One was yours and I believe I am doing it the way you suggested. The other approach was over my head. I did not understand it.
 
@satan29 i was referring to the original integral, not the one @robjohn mentioned recently.
 
5:16 PM
@satan29 the integral is $\frac{s}{s^2+t^2}$ by two IBPs
 
@robjohn yes
@copper.hat wait, you said f'(t) doesnt converge
 
$|{x \over x^2+4} \cos (tx)|$.
 
@robjohn No as in, my point was, when we used feynmans trick we ended up with $f"(t)=4f(t) + (divergent)$, What I was asking was what exactly is the problem in the original integral f(t) which renders feynmans trick unusable? something along the lines of non differentiability or something..
@copper.hat ok, this is less than x/x^2+4
@copper.hat sin(tx) btw
 
same issue.
 
ok, so this is less than $\int _{0}^{\infty} xdx/x^2+4 $ i presume, and this one diverges?
 
5:25 PM
@satan29 As I said, the integral of the second derivative of the function either doesn't converge or converges ins some regularized fashion.
 
for $x\ge 2$ you have ${x \over x^2+4} \ge {1 \over 2x}$ but that is not directly relevant.
 
What notation is used more often when introducing a new function, or what would be considered more "natural"?: (1) Define f: D\to R as f(x)=... or (2) Define f: D\to R as x\mapsto ...
 
@satan29 the $e^{-st}$ was introduced to regularize the integral
 
I see
@copper.hat hmm yes
 
@satan29 As Robjohn mentioned, you need to either deal with the improper integral or regularise in some form.
 
5:30 PM
I see.
 
Consider, as Robjohn suggested the Laplace transform of ${1 \over x^2+4}$.
 
yeah
he used feynman's trick too, but somehow he got around the problem and was able to determine $f''(t)$..
If anyone wants to watch, the actual discussionn of the problem starts from 11:48
 
@robjohn $\cdot \dots \cdot$ is rendered wrong on the site, right? (Sorry for the ping)
 
5:56 PM
@satan29 you can't just do the first differentiation under the integral sign without justifying it, which the presenter did not do.
 
is a vessel a thing that holds something inside?
 
i'm just guessing here, but do you have access to the internet?
 
6:12 PM
ah indeed it holds something inside, but what is the difference then between a bottle and a box, don't they both hold soomething inside
a box can be closed too just like the bottle
 
6:24 PM
@TedShifrin i stand corrected :-)
 
 
2 hours later…
8:28 PM
@Wolgwang looks right. What do you think is wrong?
 
English is weird about that. i would never call a box a vessel although the reasoning somewhat holds up. in the context of containers (and not ships) it is usually used for things that hold liquid. i may be wrong about that.
 
I hate to agree…
 
a friend of a friend emailed me this morning asking for a recommendation on a multivariable calculus textbook. i expect royalties.
you did remember to make people pay for your books, right? right??
i need this i have a power bill to pay
 
9:11 PM
@robjohn Glad to see that at least one moderator, namely you, hasn't abandoned interactions with users. That's a complement! And @Ted, I pretty much place you as a moderator too, so kudos to you, as well, for your dedication here!
@leslietownes What book is that? I know Ted has written a bunch of things, but I haven't been here often enough to know of them all.
 
his "Multivariable Mathematics" book............that is the title of it...not trying to be facetious or anything
 
yeah, that one.
i'm waiting for my royalty check. for book purchases and also product promotion.
as long as i'm asking i also want a pony for my daughter.
 
@dc3rd Makes sense as a title!
 
I will be pedantic and say that a compliment is often not complementary!
@leslie Maybe royalties for the free diff geo pdf.
 
I have a question about a gradient question I did Ted in comparison to your solution.
 
9:22 PM
Are we grading on a gradient ?
 
gradually
 
Probably downhill ….
 
no downhill over here....I'm moving up hill :)

it was asking the vector tangent to the curve of steepest ascent on the hill at $(\mathbf{a},c)$. You gave the gradient as $Df(a) = (3,-4)$, so the gradient being normal to my curve means that a taagent vector would be a 90 degree rotation. I used $(4,3)$....in your solution you actually stated it in 3 dimensions: $(3,-4,25)$ wondering if this is a typo? Well part of it seems to be, but where did the 25 come from?
 
We’re climbing the mountain. What slope?
No typo whatsoever.
 
slope would be $\|(3,-4)\|$ i.e the norm of the gradient
 
9:28 PM
Grrrr.
OK.
So what vector in 3D has that slope with $(x,y)$ component $(3,-4)$?
 
I have the picture in my head already, i'm just trying to phrase it in the proper form,
 
10:03 PM
wow...I went through a myriad of looking at things in overly complex forms to just return to the simple notion of "what "rise" value do I need to satisfy the equation slope = rise/run............smh.....getting better at it, but need to chill more on jumping straight to sophisticated things..
 
is the slope $5$?
 
also I realize why my tangent vector is wrong....you asked for the tangent vector to the curve of steepest ascent.....not the tangent vector to the level set.........couple sips of yerba mate helped me see the way.
yes
robjohn
 
Sounds like progress.
 
back to the book I go....
 
10:22 PM
Is this a problem from your diff geo book, @Ted?
 
Sec 3.4 - Exercise 4 - Multivariable Mathematics @robjohn
 
No, the multivariable math course (and WebWork problems I wrote for it).
 
Ch. 3
 
@dc3rd ah, thanks.
@TedShifrin okay. Just curious what I was watching here 8-)
 
Mere discombobulation?
 
10:31 PM
Well, without knowing the exact problem, it is a bit of puzzle working.
 
can't you just post the ultimate answer?
 
11:11 PM
42?
 
Can't be as perfect as 28.
 
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