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/sci/ - Science & Math


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8967706 No.8967706 [Reply] [Original]

what's the hardest integral you've ever encountered, /sci/?

pic related, I just went through the derivation of the anomalous magnetic moment.

shit's so cash

>> No.8967708
File: 26 KB, 699x197, renormalized-phi-4.png [View same] [iqdb] [saucenao] [google]
8967708

most hectic i've done so far is renormalization of phi-4 theory

u jelly, non-physics fags?

>> No.8967710

>>8967706
>what's the hardest integral you've ever encountered, /sci/?

Other than pumping up your muscle (brain), that is the point of learning this? Does your job requires this knowledge...because you'll never use this in your life unless in that field.

>> No.8967719

>>8967710

>not understanding the intrinsic beauty of mathematics/physics

try again, brainlet

>> No.8967721

>>8967708
Are you using a cutoff?

>> No.8967729

>>8967719
>>not understanding the intrinsic beauty of mathematics/physics
>try again, brainlet

>"But the Solar System!" I protested.

"What the deuce is it to me?" he interrupted impatiently; "you say that we go round the sun. If we went round the moon it would not make a pennyworth of difference to me or to my work."

....so what's the point?

>> No.8967730

>>8967706
Probably one of those where only Cleo found a closed form when i was trying to get to them and prove it.

>> No.8967733

>>8967721
of course
pauli-villars regularization

>> No.8967737

>>8967729
>Using Sherlock Holmes as a reason not care about Maths and Physics.

God damn, you really are a brainlet.

>>8967733
Dimensional is the only way to go.

>> No.8967739

>>8967729
fine splitting in metals
that enough for ya, pal?
or do you need it to rub your balls as well?

>> No.8967740

>>8967706
Pro-tip: if an automatic solver like Matlab's or Mathematica's can do it, it's not a hard integral.

>> No.8967745

>>8967740
>being this retarded
if you think that Mathematica can give you a closed form solution to that integral you are brain dead.
also, why the fuck are you even comparing Matlab's solver to Mathematica's?
are you fucking 2?

>> No.8967750

>>8967729
jesus christ this is autistic

>> No.8967765

>>8967740
But sempai maple beats them all

>> No.8967766

>>8967706
I once did a triple integral. I'm planning to major in that.

>> No.8967770
File: 4 KB, 319x64, 3420423.png [View same] [iqdb] [saucenao] [google]
8967770

can you guys solve this?

>> No.8967787

>>8967770
is this one of the cleo ones

>> No.8967794

>>8967706
Why is physics notation so absolutely abhorrent?

>> No.8967802

>>8967745
>why the fuck are you even comparing Matlab's solver to Mathematica's?
I am?

>> No.8967807

>>8967770
4pi arccot sqrt(phi)

>> No.8967809

>>8967807
Prove it. In a different way to Ron Gordon.

>> No.8967829

>>8967770
The function is continuous so it exists. I shall call it [math] \vartheta [/math]

The answer is [math] \vartheta [/math].

>> No.8967880
File: 1016 KB, 800x2000, math maymay.jpg [View same] [iqdb] [saucenao] [google]
8967880

>>8967829
brilliant

>> No.8967884
File: 28 KB, 768x460, integral function.png [View same] [iqdb] [saucenao] [google]
8967884

the integral function
you can solve this for any abstract f(x)

>> No.8968005

>>8967770
Obviously the answer is

[eqn]2\frac{1}{x} \sqrt{\frac{1+x}{1-x}} \ln\left( \frac{2x^2+2x+1}{2x^2-2x+1} \right)[/eqn]

>> No.8968011
File: 247 KB, 576x440, norm.png [View same] [iqdb] [saucenao] [google]
8968011

>>8967880
Norm is a blessing and he should be writing the syllabus. Kids would be finding doing complex geometry before the end of primary school.

>> No.8968016

>>8967710
why must all endeavors serve a practical purpose?
why can't people partake in mathematics simply because they're interested? I mean, you started your shit reply with "Other than pumping up your brain", as if making yourself smarter is a pointless exercise.

let me guess, liberal arts major?

>> No.8968019
File: 40 KB, 400x402, 1456769940970.jpg [View same] [iqdb] [saucenao] [google]
8968019

>>8967729
you raging autist

>> No.8968051

>>8967706
When I took Calculus 2 I had to solve both of these:

[math]\int x \tan x \cdot dx[/math]
[math]\int \tan^{-1} x \cdot dx[/math]

>> No.8968064

>>8967884
No you can't. f has to be a locally integrable function.

>> No.8968082

>>8967706
wait till you do renormalization in qcd, those calculations are fucking hard as hell

>> No.8968092

>>8967733
dimensional regularization is only regularization that doesnt shit all over lorentz symmetry

>> No.8968286

>>8968005
good job man

>> No.8968382

>>8968092
Isn't zeta function and dimensional regularisation equivilent?

>> No.8968886

>>8968064
yeah well come back after uve finished primary school mathematics

>> No.8968915

>>8967719
>manipulating equations is intrinsic math/physics beauty
i want undergrads to leave

>> No.8968923

>>8968092
How though? How can you have lorentz symmetry without an integer number of dimensions

>> No.8968929

>>8968016
i found funny that this question usually not being asked for sports, games or whatever useless in general hobby

>> No.8968937

>>8967770
Why is the dx written right after the integral and not after whatever the fuck is being integrated?

>> No.8968938

>>8968937
standard practice for physicists

>> No.8968939

>computing integrals
>hard

wtf is wrong with you guys?

>> No.8968967

>>8968937
In physics it is common functions of multiple variables and with many parameters being integrated in only some of them, so it makes sense to know right away what you are integrating.

>> No.8968990

>>8967710
It's okay, the world needs HR reps and sales guys too.

>> No.8969022

>>8967710
Why the fuck are you even here?

>> No.8969359

>>8968937
because physicists like cocks in their ass and this is how they signal it to each other

>> No.8969579

>>8967884
>any
Tried a non-measurable function, brainlet?
Or a nowhere continuous function, you fucking brainlet?

>> No.8969582

>>8968939
Are you a brainlet?

Most (Riemann)integrals have no analytic solution and besides numerically solving them there is nothing you can do.

>> No.8970022
File: 75 KB, 960x960, 1486982716408.jpg [View same] [iqdb] [saucenao] [google]
8970022

>>8969579
If your notion of integral depends on measure theory, then it isn't the most general notion of integral.

>> No.8970152

>>8967708
no, i am the opposite of jelly

there is nothing more retarded than people who think big integrals=more smart

>> No.8970226

Proving that if an operator which maps functions to non-negative functions is weak (1,1) and weak (infinity, infinity), then it is strong (p,p) for p>1. I'll skip the set up and just post the integral computation:
[math]
\|Mf\|_p^p = \int Mf^p\,d\mu = \int_0^\infty \mu(\{x: Mf(x)^p > y)\}dy = \int_0^\infty pt^{p-1}\mu(\{x: Mf(x) > t\})\,dt \leq \int_0^\infty pt^{p-1} \int_{x: |f(x)|>t/2} 2c|f(x)|/t \,dx\,dt
[/math]
[math]
= 2cp \int_0^\infty \int_{x: |f(x)|>t/2} |f(x)|t^{p-2}\,dx\,dt = 2cp \int_{\mathbb{R}} \int_0^{2|f(x)|} |f(x)|t^{p-2}\,dt\,dx = \int_{\mathbb{R}} |f(x)| \frac{2^{p-1}|f(x)|^{p-1}}{p-1}\,dx = \frac{2^p cp}{p-1} \int_{\mathbb{R}} |f(x)|^p \,dx = \frac{2^p cp}{p-1}\|f\|_p^p.
[/math]
So taking the p-th root of both sides gives the strong (p,p) bound.

>> No.8971537

>>8967710
>why do something if your employer doesn't require it