/sci/ - Science & Math » Thread #10854347

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Anonymous Thu 01 Aug 09:16:17 2019 No.10854347 View Reply Original Report

Quoted By: >>10854352 >>10854354 >>10854510 >>10856412 >>10859182

Let's go!

Anonymous

Anonymous Thu 01 Aug 2019 09:19:26 No.10854352 Report

Quoted By:

>>10854347
>pulls out laplace table

Anonymous

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Anonymous Thu 01 Aug 2019 09:20:28 No.10854354 Report

Quoted By: >>10854383 >>10856279 >>10861886

>>10854347

Anonymous

Anonymous Thu 01 Aug 2019 09:42:48 No.10854383 Report

Quoted By:

>>10854354
/thread

Anonymous

Anonymous Thu 01 Aug 2019 11:33:38 No.10854510 Report

Quoted By: >>10854542

>>10854347
Cannot be integrated, so no analytical form exist.

Anonymous

Anonymous Thu 01 Aug 2019 11:38:01 No.10854519 Report

Quoted By: >>10854540 >>10854547 >>10856562

how to do inverse laplace transforms and fourier transforms without tables? how do u solve the integral

Anonymous

Anonymous Thu 01 Aug 2019 11:52:48 No.10854540 Report

Quoted By:

>>10854519
By doing some analysis (basic stuff like integrating by part, subsitution etc... can help), but linearity, simplification and tables are really handy to calculate quickly.

Anonymous

Anonymous Thu 01 Aug 2019 11:54:10 No.10854542 Report

Quoted By: >>10860367

>>10854510
Nope

Anonymous

Anonymous Thu 01 Aug 2019 11:59:37 No.10854547 Report

Quoted By: >>10856232 >>10859203

>>10854519
Expanding the function via Taylor and using linearity of L

Anonymous

Anonymous Fri 02 Aug 2019 01:54:18 No.10856232 Report

Quoted By:

>>10854547
this
also QUESTION: could that be solved with imaginary numbers in its algebraic projection? e^omega*time*j and shit like that I studied in circuit analysis

Anonymous

Anonymous Fri 02 Aug 2019 02:14:08 No.10856279 Report

Quoted By:

>>10854354
based

Anonymous

Anonymous Fri 02 Aug 2019 02:48:53 No.10856347 Report

Quoted By:

These sort of integrals usually are done by puting the exponentials in a way that that you het something like 1/(1-x) or plus and put it as a geomtetric series.

Anonymous

Anonymous Fri 02 Aug 2019 02:53:47 No.10856357 Report

Quoted By:

Did not know there is a Leica function, the math behind the "L" in that red dot.

Anonymous

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Anonymous Fri 02 Aug 2019 02:59:57 No.10856367 Report

Quoted By:

$g(s) = \int_{0}^{\infty} \frac{1}{e^{-t+1}+1} e^{-st} dt [\math]. After some wizardry, g(s) is satisfies this equation: <script type="math/tex"> (1-s)g(s) = \frac{-1}{e+1} - \frac{1}{e^2 + e} +e^{-1} (s-1) g(1-s)$ .

Anonymous

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Anonymous Fri 02 Aug 2019 03:01:03 No.10856369 Report

Quoted By: >>10857516

$g(s) = \int_{0}^{\infty} \frac{1}{e^{-t+1}+1} e^{-st} dt$ .

After some wizardry, g(s) is satisfies this equation:
$(1-s)g(s) = \frac{-1}{e+1} - \frac{1}{e^2 + e} +e^{-1} (s-1) g(1-s)$ .

Anonymous

Anonymous Fri 02 Aug 2019 03:30:51 No.10856412 Report

Quoted By: >>10856431 >>10858932

>>10854347
sum{ ((-e)^n)/(s+n) }

Anonymous

Anonymous Fri 02 Aug 2019 03:41:36 No.10856431 Report

Quoted By: >>10856544

>>10856412
The sum $\sum\limits_{n=0}^{\infty} \frac{-e^n}{s+n}$ diverges....right?

Anonymous

Anonymous Fri 02 Aug 2019 05:11:29 No.10856544 Report

Quoted By: >>10856641 >>10858932 >>10861497

>>10856431
>diverges?
Yes.
e^(1-t) >1 for t <1 so doing a geometric expansion doesn't work over [0,inf).

If you split the integral into two parts, [0,1] and [1, inf), you can do a different geometric expansion on each part.

I'm getting pi*(e^-s)/sin(pi*s) + sum{ ((-e)^(-n))/(n-s) ; n>=1}

Anonymous

Anonymous Fri 02 Aug 2019 05:25:44 No.10856562 Report

Quoted By:

>>10854519
for inverse laplace you can do so with contour integration over the Bromwich contour

Anonymous

Anonymous Fri 02 Aug 2019 06:20:08 No.10856641 Report

Quoted By:

>>10856544
This looks like it doesn't work when s is a positive integer but the singularities in the sum cancel with the singularities of the csc so the function converges when s>0.

Anonymous

Anonymous Fri 02 Aug 2019 06:22:54 No.10856645 Report

Quoted By: >>10856663

The thing I love about /sci/ is seeing threads that have problems I could never solve on my own, reading all of the much smarter anons solving it like it's nothing, and learning from that.
/sci/ is literally the best board on this shithole for that reason

Anonymous

Anonymous Fri 02 Aug 2019 06:34:19 No.10856663 Report

Quoted By: >>10856667

>>10856645
dude, a laplace transform is just checking a few conditions and solving the integral
$\int _{0} ^{\infty} f(x) e^{-st}$
for the desired function

I thought that these things were hard but you should learn this in your first year of math studies.

Anonymous

Anonymous Fri 02 Aug 2019 06:36:45 No.10856667 Report

Quoted By: >>10856674 >>10857516

>>10856663
I've done laplace transforms, I'm just drunk as shit so I can't even read the equations

Anonymous

Anonymous Fri 02 Aug 2019 06:38:43 No.10856674 Report

Quoted By:

>>10856667
cheers

Anonymous

Anonymous Fri 02 Aug 2019 15:18:54 No.10857516 Report

Quoted By: >>10858932

>>10856667
Just use a bunch of clever substations and integration by parts, you'll then arrive at the "recursion" identity: >>10856369

Anonymous

Anonymous Fri 02 Aug 2019 23:33:33 No.10858932 Report

Quoted By:

>>10857516
>"recursion" identity
It isn't recursive. It is a "reflection" identity that doesn't seem to admit a unique solution.

Splitting the integral, as >>10856544 says, gives a concrete result.
The pi/sin(pi*s) comes from
https://en.wikipedia.org/wiki/Trigonometric_functions#Power_series_expansion

Using >>10856544 you get a "recursion" identity:
g(s+1) = 1/(es) - g(s)/e
This (un)surprisingly agrees with >>10856412 where the "answer" was found by naively expanding the geometric series without worrying about convergence on [0, inf).

Anonymous

Anonymous Sat 03 Aug 2019 01:38:31 No.10859182 Report

Quoted By:

>>10854347
$ \mathcal{L} \left \{ \dfrac{1}{e^{-t+1}+1} \right \} $

Anonymous

Anonymous Sat 03 Aug 2019 01:49:12 No.10859203 Report

Quoted By: >>10859204 >>10860149

>>10854547

this only works if you prove beforehand that the taylor series exists

which engineerbrainolets in 100% of the cases don't do so it's invalid

Anonymous

Anonymous Sat 03 Aug 2019 01:50:25 No.10859204 Report

Quoted By:

>>10859203
>reinventing the wheel
hurr durr

Anonymous

Anonymous Sat 03 Aug 2019 11:10:14 No.10860149 Report

Quoted By:

>>10859203
>engineerbrainolets
no one but math autists do that

Anonymous

Anonymous Sat 03 Aug 2019 13:12:55 No.10860367 Report

Quoted By: >>10860453 >>10861497 >>10861508

>>10854542
No one in the thread as given out an analytical form thus far...

Anonymous

Anonymous Sat 03 Aug 2019 14:02:54 No.10860453 Report

Quoted By:

>>10860367
>proof by absence of counter proof

Anonymous

Anonymous Sat 03 Aug 2019 21:26:06 No.10861497 Report

Quoted By: >>10861508

>>10860367
See >>10856544
${\pi e^{-s} \over sin(\pi s)} + \sum\limits_{n=1}^{\infty} {e^{-n} \over n-s } (-1)^{n}$

Anonymous

Anonymous Sat 03 Aug 2019 21:34:14 No.10861508 Report

Quoted By:

>>10860367
>>10861497
If you want to give a name to the sum use the Lerch transcendent:
https://en.wikipedia.org/wiki/Lerch_zeta_function

Anonymous

Anonymous Sun 04 Aug 2019 01:01:38 No.10861886 Report

Quoted By:

You may laugh now, but >>10854354 has just put out a groundbreaking result in operator theory in the space of 241 by 225 pixels.

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