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/sci/ - Science & Math

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5487112 No.5487112 [Reply] [Original]

Hey /sci/,
I heard people here like a little math challenge. Can you guys solve the integral in pic related?
BTW, the delta function is inside the integral.

>> No.5487142

so you can take the sigma out of the integrand correct? assuming sigma is a constant of some sort.
then double integral from 1 to -1 of (x^2+y^2-Y)dxdy
split it up.

(1)it would be (x^3)/3 from x=1 to -1, 1/3+1/3=2/3
integral of 2/3dy is 2/3y evaluated from y=1 to -1 which is 4/3

(2) integral of y^2dx is xy^2, where x=1 to -1
y^2+y^2
the integral of 2y^2 dy
you get 2[(y^3)/3] evaluated at 1 to -1
you get 4/3

(3) integral of Ydx is Yx, evaluate to get 2Y.
integral of 2Ydy is, 2Yy evaluate to get 4Y.

so you get f(Y)=sigma(2*(4/3)+4Y)

please help me >>>5487070

>> No.5487147

>>5487142
That is incorrect. The delta is Dirac's delta function.

>> No.5487149

Go go go, engineers.

>> No.5487153

>>5487149
I always find it funny that mathfags can't do integrals.

>> No.5487156

>implying I want to do basic substitution

This is what computers are for.

>> No.5487162

>>5487156
>I can't add without calculator
This is why today's kids can't reason or solve any real life problems.

>> No.5487165

>>5487147
hahah oh wow, thanks for reminding me why i wanted to go into math. sincerely happy that you pointed that out, looked it up on wikipedia and nostalgia'd hard thinking about my old science fair project in high school on fourier'rs transform. i left this place years ago, but i might come back, now that litfag trolling is gone.

>> No.5487166

>>5487153

I'd say something sarcastic, but you got me there...

>> No.5487174

>>5487162
Not the guy you addressed, but that's not why. Try a lack of genuine interest.

>> No.5487176

>>5487174
Accomplishment breeds interest. When your only accomplishment is to punch a calculator you are not going to get interested in anything.

>> No.5487181

>>5487176
>Accomplishment breeds interest
Nah, other way around m8. I can accomplish something and not be remotely interested in it.

>> No.5487183

>>5487162
Ok, add up the first 100 primes.

You should be able to solve this.

>> No.5487190

f(y) = infinity (on circle about origin of radius Y)
f(y) = zero (everywhere else)

>> No.5487193

>>5487183
24,133

>> No.5487195

>>5487190
It's not infinity inside the circle but it is zero outside a certain circle.

>> No.5487196

>>5487183
Pn denotes primes up to a certain n,
Pn<pi(x)
lim as x->infinity of pi(x)/[x/log(x)]=1
P100<100/log100
approximately how many numbers you have to test, before sum.

>> No.5487197

>>5487190
done goofed, i mean f(x,y)

f(Y) = infintity for Y=0
= 0 everywhere else

>> No.5487198

/g/res/31149652

>> No.5487212

>>5487198
The guys on /g/ are getting closer.
Seems like /sci/ is just for discussing religion vs atheism.

>> No.5487220

>>5487212

Not possible, what are our engineers doing?

>> No.5487224

>>5487220
What about the physicists?
This integral appeared in a statistical mechanics problem.

>> No.5487227

>>5487224

I wouldn't know that, but in that case: what are our engineers/physicists doing?

>> No.5487231
File: 124 KB, 539x465, 1359484004641.png [View same] [iqdb] [saucenao] [google]
5487231

Leonard Susskind does a decent job at explaining the properties of the Delta function.
https://www.youtube.com/watch?v=oWe9brUwO0Q#t=474s

>> No.5487236

>>5487227
The problem can be cast as one of probability theory.
Why are the mathematicians slacking off? Are they reading Bourbaki or something?

>> No.5487239

>>5487236

I'm grading hws, but other than that I haven't solved an integral in 5 years.

>> No.5487242

>>5487236
Probability theory isn't math.

>> No.5487245

This is easy
<span class="math">f(Y) = 0[/spoiler] for <span class="math">Y < 0[/spoiler]
<span class="math">f(Y) = 2 \pi Y [/spoiler] for <span class="math">0 \leq Y \leq 1[/spoiler]
<span class="math">f(Y) = 2 \pi - 8 \arccos \left({1 \over Y} \right)[/spoiler] for <span class="math">1 < Y \leq \sqrt{2}[/spoiler]
<span class="math">f(Y) = 0[/spoiler] for <span class="math">\sqrt{2} < Y[/spoiler]

Learn to conversion in polar coordinates.

>> No.5487247

/g/ is very close now.

>> No.5487251

>>5487245
close but no cigar

>> No.5487255

>>5487247

Someone go spy on them.

>> No.5487260

>>5487255
>>>31149652

>> No.5487261

>>5487255
/sci/ is closer now: >>5487245

>> No.5487265

>>5487242
Why not?

>> No.5487263

>>5487260
i cannot into cross board linking, sorry

>> No.5487268

>>5487260
>>>/g/31149652

>> No.5487269
File: 100 KB, 547x511, 1339900578786.jpg [View same] [iqdb] [saucenao] [google]
5487269

>>5487261

Woooo, go /sci/.

>> No.5487288

>>5487265
Bourbaki didn't consider probability theory rigorous enough to be considered math. They even heckled speakers at conferences telling them that.

>> No.5487293

http://mathpages.com/home/kmath663/kmath663.htm

>> No.5487295

>>5487288
How the fuck can measure theory restricted to the unit interval not be rigorous?

>> No.5487298

>>5487295
Something about the region of integration not being locally compact.

>> No.5487300

>>5487298
>Something about the region of integration not being locally compact.
>>5487293
There's ways to make it work.

>> No.5487319

>>5487300
Maybe Bourbaki was just a bully.
Anyway, that's why they didn't include Probability Theory in their books.

>> No.5487369

>>5487298
>[0,1] not locally compact
u wot m8

>> No.5487371

>>5487112
I'm the guy from /g/ who answered PI and now I just realised that this is a a function in terms of gamma and that gamma does not represent the EulerGamma constant. If it would've been the case my answer would be right.

Back to the papers.

>> No.5487382

>>5487369
[0,1] with what topology, m8?
Integrating with what measure?

>> No.5487391

>>5487382
Subset topology inherited from <span class="math">\mathbb{R}[/spoiler], lebesgue measure of course.

>> No.5487407

>>5487391
But m8, Bourbaki didn't use Lebesgue work in their book on integration but something based on locally compact Radon measures.

Many problems in probability theory use integration in sets that are not locally compact and the Bourbakis heckled people because of that.

>> No.5487413

>>5487407
fair enough. fuck probblebility then.

>> No.5487415

>>5487382
>>5487407
>m8
stop that you underage tool

>> No.5487421

>>5487371
Alright I think I got it this time :

(-infinity, 0] = 0
(0, 1] = PI
(1, 2) = goes from PI to 0 in a somewhat linear way
[2, infinity) = 0

>> No.5487422

>>5487413
>fuck probability
And that's exactly what Bourbaki said, m8. And set probability theory back for decades in France.

>> No.5487425

>>5487421
>(1, 2) = goes from PI to 0 in a somewhat linear way
Not linear, you only get partial credit.

>> No.5487436

>>5487425
Dear god, I'm the only one who came up with a decent answer and I'm an undergraduate Computer Science student, give me some slack

>> No.5487437

I got
f(Y) = pi for 0<Y<1
and
f(Y) = pi - 4 arccos(1/sqrt(Y)) for 1<Y<2

>> No.5487431
File: 18 KB, 452x339, 135792056026.jpg [View same] [iqdb] [saucenao] [google]
5487431

>>5487422

>> No.5487440

>>5487436

So what? Is this supposed to be some grad-level problem?
But good job, anon.

>> No.5487449

What the fuck is Y?
Way to define your problem correctly.

>> No.5487454

>>5487440
While the integral itself it is not grad level, it came up in a stat mech book that's been used for grad level courses.

>> No.5487458

>>5487449
Y is a variable > 0.

>> No.5487464

>>5487437
We have a winner!
(minus some points for not answering for Y>2)

>> No.5487477

>>5487464

Victory!

>> No.5487486

>>5487477
The /g/ losers couldn't even get close.

>> No.5487513

>>5487437
Would you mine posting how you got the solution?

>> No.5487548

>>5487513
Substitute z = x^2 + y^2

<div class="math">\int_{-1}^1 \int_{-1}^1 \delta(x^2+y^2 - Y)\; dx dy </div>
<div class="math">=4 \int_{0}^1 \int_{0}^1 \delta(x^2+y^2 - Y)\; dx dy </div>
<div class="math">=2 \int_{0}^1 \int_{y^2}^{1+y^2} \frac{1}{\sqrt{z-y^2}}\delta(z - Y)\; dz dy </div>
<div class="math">=2 \int_{0}^1 \int_{y^2}^{1} \frac{1}{\sqrt{z-y^2}}\delta(z - Y)\; dz dy +2 \int_{0}^1 \int_{1}^{1+y^2} \frac{1}{\sqrt{z-y^2}}\delta(z - Y)\; dz dy</div>

Note that the first integral is always 0 unless 0<Y<1 and that the second integral is 0 unless 1<Y<2.

For 0<Y<1 :
<div class="math">2 \int_{0}^1 \int_{y^2}^{1} \frac{1}{\sqrt{z-y^2}}\delta(z - Y)\; dz dy</div>
<div class="math">=2 \int_{0}^{\sqrt{Y}} \int_{y^2}^{1} \frac{1}{\sqrt{z-y^2}}\delta(z - Y)\; dz dy</div>
<div class="math">=2 \int_{0}^{\sqrt{Y}} \frac{1}{\sqrt{Y-y^2}} \; dy</div>
<div class="math">=\pi</div>

For 1<Y<2 :
<div class="math">2 \int_{0}^1 \int_{1}^{1+y^2} \frac{1}{\sqrt{z-y^2}}\delta(z - Y)\; dz dy</div>
<div class="math">=2 \int_{\sqrt{Y-1}}^1 \int_{1}^{1+y^2} \frac{1}{\sqrt{z-y^2}}\delta(z - Y)\; dz dy</div>
<div class="math">=2 \int_{\sqrt{Y-1}}^1 \frac{1}{\sqrt{Y-y^2}} \; dy</div>
<div class="math">=\pi - 4 \arccos\left({1 \over \sqrt{Y}} \right)</div>

>> No.5487596

Excuse my ignorance but what branch of math is this called?

No, I don't belong here.

>> No.5487601

>>5487596
Integral calculus

>> No.5487607

>>5487596
It is called Calculus.