« first day (1731 days earlier)      last day (3586 days later) » 

19:09
I have to find the forces at the points A and B. Is it as followed?


A: F_1*a+F_2*(I-b)=F_B*I
B: F_1*(I-a)+F_2*b=F_A*I

@robjohn
Hej! @kahen
$F_A=F_B \cdot I-F_1 \cdot a-F_2 \cdot (I-b)=4.5 \cdot F_B-35 \cdot 0.8-21 \cdot 3$
$F_B=F_A \cdot I -F_1 \cdot (I-a)-F_2 \cdot b=\cdot 4.5 F_A-3.7 \cdot 35-1.5 \cdot 21$

Is this correct?? @robjohn
@robjohn Hey! I have posted the question on M.S.E. if you're interested :)
19:24
@MaryStar HINT: The total torque at A is the sum of the four torques, and same for B.
So isn't it

$F_A=F_B \cdot I-F_1 \cdot a-F_2 \cdot (I-b)=4.5 \cdot F_B-35 \cdot 0.8-21 \cdot 3$
$F_B=F_A \cdot I -F_1 \cdot (I-a)-F_2 \cdot b=\cdot 4.5 F_A-3.7 \cdot 35-1.5 \cdot 21$

?? @DonLarynx
@robjohn Another question, why do people post stuff on the chat in LaTeX if it doesn't display correctly, or is it an issue with my browser ?
the link I provided tells you how to render latex in chat
19:39
@Gamamal That's news to me. Thanks a lot :)
No problem.
 
1 hour later…
20:59
Does anyone remember the name of the identity for $\idotsint_0^1 f\left(x_1,\cdots,x_n\right)\:dx_1\:\cdots\:dx_n$?
I've been looking for it for the past half an hour, I just can't remember what it's called!
21:12
@jm324354 what sort of "identity"? without information on $f$ I don't see any way to simplify it
Anybody could answer my bounty question? math.stackexchange.com/questions/1258191/…
@JMoravitz Unfortunately not. The identity came out something like $\idotsint_0^1f\left(x_1,\cdots,x_n\right)\:dx_1\:\cdots\:dx_n=\int_0^1 \ftrac{\left(x-t\right)^n}{n}$\left(\right)\:dt$ and I can't remember what went in the parenthesis, but it was something like that...
woops $\int_0^1\frac{\left(x-t\right)^n}{n}\left(\right)\:dt$
or something like that
Then again, maybe it was n-integrals over one variable.
22:01
back
@TedShifrin are you here?
Maybe some of you wanna answer it:
1
Q: A possible dilogarithm identity?

Chris's sisI'm curious to find out if the sum can be expressed in some known constants. What do you think about that? Is it feasible? Have you met it before? $$2 \left(\text{Li}_2\left(2-\sqrt{2}\right)+\text{Li}_2\left(\frac{1}{2+\sqrt{2}}\right)\right)+\text{Li}_2\left(3-2 \sqrt{2}\right)$$

@Chris'ssis I was wondering do you normally keep solution set to the problems you solve?
in like a folder or something or do you remove it ?
@KarimMansour I have both problems and solutions saved partially in some folders on a safe computer.
@MaryStar sorry I am back was away
I see yeah I should do that
@KarimMansour I am looking again at the exercise http://physics.stackexchange.com/questions/179540/how-can-we-use-the-distances
Are the $F_Z, s$ correct?? I am still confused how to find $F_H$...
22:17
well you need to find $F_H$ to make this system in equilibrium
that is since it doesn't accelerate you need to find $F_H$ to make it stationary
I would use newton second law and newton 3rd law
Ok... Are the formulas of $F_Z$ and $s$ correct?? @KarimMansour
1 moment let me see
how did you derive it ?
and what is s here?
$s$ is how long the rope will be moved, so that the hook will get 10cm upwards.

Isn't that the general formula?? @KarimMansour
oh the stretch
that is given by hooke's law
and is the number of ropes equal to 5 ?? @KarimMansour
22:23
yeah I don't know though I thought you would solve this problem using tension and newtons laws
I didn't know about hooke's law
I thought to use these formulas because they are in my notes... @KarimMansour
I see I don't know I don't see how here hooke's law will help in this problem honestly
So, is this wrong to use these formulas?? @KarimMansour
no I don't think hooke's law is useful here
Ok...
Let the forces about the hook be $F_1$ and $F_2$. Do we use the distance as followed??

$F_1=F_2-\frac{3}{5}F_L$
$F_2=F_1-\frac{2}{5}F_L$

@KarimMansour
22:32
which hook the above one?
Let $H_1,H_2,H_3$ be the hooke's of the system with 1 is the highest one etc
so the forces on $H_1$ are just
two forces
3 forces
the tension due to the ropes from left and right and the normal force upward
heya @Karim. No, I'm not here.
I meant above the the forces about the object. @KarimMansour
if we go to $H_2$ then what are the forces can you tell me ?
:D @TedShifrin
Hooke's law is not relevant in pulley problems :P
yeah
well @TedShifrin it could be relevant if you want to add more details to the system.
it depends on how precise you want your system to be
22:38
I didn't see any springs, mr Karim :P
rope could also be a spring but I guess the k will be very small
only if it's very elastic, Karim :P
@Karim: In general physics, no.
Ropes are generally not elastic.
heya @DonL
Hi @Ted! School finished today!
22:41
Hey @DonLarynx
Well, I have finals on Tuesday, and then I'm done (after grading) forever :P
Then you're gonna live in a beach off a sunny island in China wat?
how is the average of differential geometry ? @TedShifrin
We'll find out next Tuesday/Wednesday, @Karim.
yeah, right, @Don.
22:42
oh I see
I would go live in Hawaii @TedShifrin
I'm moving where there are good friends, @Karim ...
oh I see
I don't have friends @TedShifrin :D
I know that's false.
Unless you're now Jasper.
well, I had friends in uni but you know they keep like doing weird stuff like talking behind my back etc.
there's nothing good to say behind math geeks' backs, anyhow
22:45
so I decided not to have friends better to have my gf,math,science and that is it :D
well, a gf counts as a f.
yeah I guess so
So does a bf.
no, like I had one friend in uni that I helped sometimes in math then when I ask him a question he never replies and tries to make me get confused and this stuff.
if he never replies, how can he make you confused?
anyhow, did you figure out the lcm stuff?
22:47
Hi folks.
h @Jeff
I figured it out :D
If you're sure you're correct, I won't ask, @Karim :P
so here is how I did it
can you create a link to a specific answer?
22:51
proof: case 1:
Suppose $gcd(a,b) \neq 1$
let $gcd(a,b) = d$

$ab = dq_1b = dbq_1 = ddq_1q_2$
Claim $lcm(a,b) = bq_1 = ddq_1q_2$.

a | $bq_1$ since $a = dq_1$ | $dq_1q_2$ and b | $bq_1$ trivially.
now
@KarimMansour There is the following picture is my notes:
suppose c = lcm(a,b)
@Karim: There are typos in that.
What's your definition of gcd and lcm, btw?
and then for the second graph it stands:
$F_Z=\frac{F_G}{4}, s=4h$
and for the third one it stands:
$F_Z=\frac{F_G}{8}, s=8h$ @KarimMansour
22:53
definition of gcd is the largest divisor of two numbers and lcm is the smallest number m such that a | m and b | m. @TedShifrin
but I proved earlier thats same thing as I)a | t and b | t II) If a | c and b | c, then t | c
ok, @Karim ... make sure you actually use those definitions in your proof.
ok
I still recommend you do the gcd=1 case first.
Do we not have the same situation at the exercise as in the picture above?? @KarimMansour
alright I will try that again first I got the general case I proved that lcm(a,b) = $bq_1$ by proving that a divides it and b divides it and I supposed that c = lcm(a,b) then I showed that $bq_1$ $\leq$ c.
the other direction is easy by definition of lcm(a,b) being c.
but I still don't see how to do it for gcd(a,b) = 1 @TedShifrin?
22:56
@TedShifrin, let's say I want to have a little box around my $\implies$ symbol, for when I'm doing the directions in an if and only if proof. Do you know how I'd go about it?
LOL ... well, that's pretty ridiculous, since that's the easy case :P
don't even say hi, @Kaj? ...
haha, hey @TedShifrin
Hello @Ted :) how did your final go?
@Kaj, the grad students raided my office and took 70% of my books today.
Oh wow
22:57
Only 8 students took it, @AlexW. 23+5 more on Tuesday.
how many books were there @TedShifrin ?
Wow. Why such a long gap?
Hundreds, @Karim.
Well, not quite 70%, but what's left is mostly calculus and undergraduate stuff.
@Kaj, you should take my TeX books :P
We have exams over 5 days, @AlexW.
btw @TedShifrin I bought a cat :D
you bought a cat?
22:58
yeah
shout @Ted I guess they're using them to study. Will probably return them. :D
Perhaps I should. I'm always in here asking about TeX lol
no, @Jeff, I sent an email telling them to come get stuff. Thousands of dollars of books ... but now they have good homes.
In LaTeX there are box commands, @Kaj.
Oh, do you mean you're giving another final on Tuesday, or have I misunderstood again?
I tried something like fbox or some such earlier, but it wasn't centering well
23:00
@KarimMansour wrong group of friends.
@Ted Oh. You shoulda told me. Do you have any teacher's editions? I would have taken those
right, @AlexW, the diff geo is on Tuesday, and 5 of the multivariable students are taking it then.
No teachers' editions.
fbox is framebox; that should work.
AMSTeX has its own Boxed command.
Gotcha, that makes more sense. Impressive to see students take multivariable and diff geo at the same time. :)
I suppose in some sense, there's no better time to take it, though!
sometimes you have to tweak spacing in LaTeX, @Kaj. That's what they paid me the big bucks to do when I typeset the books :P
no, no, @AlexW ...
yeah @DonLarynx definitely
23:01
some of the multivariable students are taking their final late ... silly.
that is why I have you guys as my friends :D
Damn, I'm off today.
However, in the past, I've had 2 or 3 students who've taken both simultaneously from me. They had big psychiatry bills that semester :P
We're enemies, @Karim.
@Ted: I can't resist... "LaTeX: A Geometric Approach" in the works? :))
OK, can someone explain what is meant by "the maximality of $T$" in the one answer to this: math.stackexchange.com/questions/1253452/…
23:02
why @TedShifrin :S I thought we were friend :S
ROFL, @AlexW ... with that, you should come to my retirement roast :P
LOL @AlexWertheim
glares @Kaj I see how this is going.
Don't worry @TedShifrin. I plan on writing Geometry: An Algebraic Approach
23:03
Actually, I've written part of that.
Can a polynomial of the form $x^2+\sqrt{a}x+b$, where $a$ is not a square , have roots in $\mathbb{Q}$ ?
Read chapter 8 :P
cool question, @evinda
@TedShifrin Do you have an idea? :)
@evinda Probably not. The roots would have to be two numbers that multiplied to $b$ and added up to $\sqrt a$. Two numbers that add up to $\sqrt a$ cannot be rational.
23:05
Could you take a look at my question : physics.stackexchange.com/questions/179540/… ?? @robjohn
@MaryStar: You really have to stop pinging everyone to do your homework.
Technically, @Jeff, they'd add up to $-\sqrt a$. :)
@evinda: You also need to specify that $a,b\in\Bbb Q$?
@TedShifrin Yes, it holds that $a,b\in\Bbb Q$.
I don't want that you do my homework. I want to do them on my own. But I got stuck on some points. That's why I'm asking... @TedShifrin
@TedShifrin Good point.
But, seriously, @MaryStar, you are always in here asking many of us to help with each and every problem.
@Karim: I'm going to do dinner, but you should focus on proving that if $\gcd(a,b)=1$, then $ab$ satisfies the definition of lcm. And don't ask @Kaj for help.
23:09
@Jeff @TedShifrin Ok, I see.. Thank you :)
okay @TedShifrin asking @KajHansen will be prohibated
The temptation...
23:25
@MaryStar It looks like $s=10$ cm and $F_Z=100$ N, to me. That is if the supports stay horizontal. If not, it would seem they'd just flop.
The first pulley system reduces by half the amount of distance moved by $s$ and the second pulley system doubles that.
i still am in disbelief this semester is over. time to party!!!!!!!
Hey @DanielFischer
Could I ask you something about the complexity of an algorithm?
http://pastebin.com/Eu4s3ht1
Could you help me to find the complexity of the  algorithm?
How can we find the time that is required for the for-loop:

For each list x in X:
            Add [v, x] to the set Y.

? :confused:
@robjohn Now I found the solution... It is the following:
How have we find these results?? @robjohn
@evinda order n complexity
where $n$ is the number of lists
@DonLarynx Why ? are there n possible topological sortings?
23:37
@evinda Each list adds [v, x] to the set $Y$ so that's one operation. Now there's n lists.
It's linear.
Evening @Mike. How'd your talk go?
@DonLarynx How could we implement the for-loop:
For each list x in X:
            Add [v, x] to the set Y.

 in C?
@evinda for(int i = 0; i < numberOfLists; ++i){list[i] = "v, x";} where numberOfLists is of type int and list is of type char**
@DonLarynx How can we initialize a char** list?
@evinda char** list[sizeKnownAtCompileTime_ActuallyANumber]; or char** list = new char*[sizeKnownAtRuntime_canBeAVariable]; /*do stuff*/ for(int i = 0; i < sizeKnownAtRuntime_canBeAVariable; ++i){delete list;} delete[] list;
23:51
@MaryStar I forgot to account for the movement of the free pulleys in one of the calculations.
statically declared arrays (the former) are a lot less complicated and less bug-prone than the latter
What is [sizeKnownAtCompileTime_ActuallyANumber / sizeKnownAtRuntime_canBeAVariable? @DonLarynx
@MaryStar Yes, taking that into account, I get the same result.
It went well, @AlexW. They want me to give the parts of it where I don't prove anything to the math alumni assoc in a couple weeks.
That's great! Yet more incentive for that blog post. :)
23:55
@evinda The former has to be a number e.g. 5, 10, 12, 4 and the latter can be a number and/or an already-defined variable e.g. 5, 6, myNumber, thisLimit
So, are the formulas of $F_Z$ and $s$ correct as I have written them in the post?? @robjohn
one day...

« first day (1731 days earlier)      last day (3586 days later) »