/sci/ - Science & Math » Thread #13257049

798KiB, 1200x960, kurisu.png

View Same Google iqdb SauceNAO

Does this have no analytical (closed form)) solutions?

Anonymous Wed 09 Jun 15:41:54 2021 No.13257049 View Reply Original Report

Quoted By: >>13257147 >>13257188 >>13257247 >>13258718 >>13261314 >>13262109 >>13262280

Seriously - I have like 5 pages of A4 paper full of attempts that led me nowhere. Help, mathbros?
>inb4 homework
This is not homework.

OP Wed 09 Jun 2021 15:42:25 No.13257052 Report

Quoted By:

I saw this: https://math.stackexchange.com/questions/8213/logistic-function-passing-through-two-points
and thought - what if the unknown coefficients were the other way around - the amplitude and growth factor instead of growth factor and center.

Anonymous

Anonymous Wed 09 Jun 2021 16:07:33 No.13257147 Report

Quoted By: >>13257186

>>13257049
>anaytical
>2021
ngmi

OP Wed 09 Jun 2021 16:17:22 No.13257186 Report

Quoted By:

>>13257147
analytical is required here in order to save computation time

Anonymous

Anonymous Wed 09 Jun 2021 16:17:33 No.13257188 Report

Quoted By: >>13257214

>>13257049
Let
z = e^(-y)
Take the quotient of those equations to get
(1 + z^(c-1))/(1 + z^c) = a/b
(1 + z^(c-1))(b/a) = 1 + z^c
z^c - (b/a) z^(c-1) + 1 - (b/a) = 0
Now if c is either 1,2,3 or 4 you can simply use the linear, quadratic, cubic or quartic formula.
After that you get x and y back
x = a (1 + z^c)
y = -log(z)

OP Wed 09 Jun 2021 16:27:08 No.13257214 Report

Quoted By: >>13257221

>>13257188
>if c is either 1,2,3 etc
c is a positive real number - can be positive or negaive, usualy a fraction or can be irrational. Usually between -1 and 1, but not neceserally.
I never considered this solution in particular, because c is not an integer. No real way of getting past:
>z^c - (b/a) z^(c-1) + 1 - (b/a) = 0
because c can be something like 0.12 or 0.1333(3)

OP Wed 09 Jun 2021 16:28:17 No.13257221 Report

Quoted By:

>>13257214
I meant c is just real number - can be positive or negative, as written in pic related

Anonymous

Anonymous Wed 09 Jun 2021 16:30:33 No.13257231 Report

Quoted By:

bumping for actual math

Anonymous

Anonymous Wed 09 Jun 2021 16:36:24 No.13257247 Report

Quoted By: >>13257272 >>13257300 >>13257338

>>13257049
$ \frac{1+e^{y(1-c)}}{1+e^{-yc}}=\frac{e^y-1}{e^{cy}+1}+1 =\frac{a}{b}\\ \frac{t-1}{t^c+1}=\frac{a}{b}-1\\ t-1 = \frac{a}{b}t^c+\frac{a}{b}-t+1\\ t^c (1+\frac{a}{b})-2t+2=0\\ $
I don't think so

Anonymous

Anonymous Wed 09 Jun 2021 16:42:17 No.13257272 Report

Quoted By:

>>13257247
Neat, this is what i also did

OP Wed 09 Jun 2021 16:49:40 No.13257300 Report

Quoted By: >>13257333

>>13257247
I have gotten to this point once before, but could not figure out a way forward. Is seems simple initially - 2 equations and 2 unknowns, but turns out it isnt so simple after all. Also, a similar problem (my first comment here) is solvable very easily

Anonymous

Anonymous Wed 09 Jun 2021 16:56:30 No.13257333 Report

Quoted By: >>13257366

>>13257300
>but could not figure out a way forward
I don't think there's a general way.
For c>4 there is no analytical solution at least unless I'm missing something.

Anonymous

Anonymous Wed 09 Jun 2021 16:57:06 No.13257338 Report

Quoted By: >>13257344

>>13257247
Algebra error

Anonymous

Anonymous Wed 09 Jun 2021 16:58:15 No.13257340 Report

Quoted By: >>13257360

Probably not, dumbass. The fact you spent so much time when it should be obvious there isn't is really embarrassing, desu

Anonymous

Anonymous Wed 09 Jun 2021 16:58:57 No.13257344 Report

Quoted By: >>13257376 >>13257381

>>13257338
Where?

OP Wed 09 Jun 2021 17:02:20 No.13257360 Report

Quoted By:

>>13257340
dont underestimte me, I got it to work half-way: I came up with formulas for x and y so that one of the equations was satisfied, but the other one failed. I thought someone here could maybe find a better way

OP Wed 09 Jun 2021 17:03:30 No.13257366 Report

Quoted By:

>>13257333
yes, but in my scenario c is some small number, usually in the range [-2;2]

Anonymous

Anonymous Wed 09 Jun 2021 17:05:05 No.13257376 Report

Quoted By:

>>13257344
Third line.

Anonymous

Anonymous Wed 09 Jun 2021 17:06:03 No.13257381 Report

Quoted By: >>13258342

>>13257344
$\frac{t-1}{t^c - 1} = \frac{a}{b} - 1 \rightarrow t-1 = t^c\left( \frac{a}{b}-1\right) - \frac{a}{b} + 1 \rightarrow t^c\left(\frac{a}{b} - 1\right) -t+\frac{a}{b}$

Anonymous

Anonymous Wed 09 Jun 2021 20:30:49 No.13258209 Report

Quoted By:

bump

Anonymous

View Same Google iqdb SauceNAO hä.jpg, 13KiB, 167x182

Anonymous Wed 09 Jun 2021 20:44:45 No.13258282 Report

Quoted By: >>13260167

If a, b and c are known, why don't I know them?

Anonymous

Anonymous Wed 09 Jun 2021 20:57:49 No.13258342 Report

Quoted By:

>>13257381
It's t^c+1 not -1

Anonymous

View Same Google iqdb SauceNAO 1594037666498.jpg, 457KiB, 715x1000

Anonymous Wed 09 Jun 2021 22:34:10 No.13258718 Report

Quoted By: >>13260171

>>13257049
I don't know
I don't care
Post more Kurisu.

OP Thu 10 Jun 2021 06:03:25 No.13260167 Report

Quoted By:

>>13258282
I meant "known" as in "they are not variables to be determined, just numbers with defined properties"

OP Thu 10 Jun 2021 06:04:50 No.13260171 Report

Quoted By:

>>13258718
well, his thread devolved very quickly

OP Thu 10 Jun 2021 06:06:01 No.13260175 Report

Quoted By: >>13261137

I dont suppose anyone else has some new insight into solving this problem?

Anonymous

Anonymous Thu 10 Jun 2021 10:29:55 No.13260819 Report

Quoted By:

bump

Anonymous

Anonymous Thu 10 Jun 2021 12:46:03 No.13261137 Report

Quoted By: >>13261310 >>13261312

>>13260175
I tried earlier without posting about it. Mathematica wasn't able to solve it which is a bad sign

OP Thu 10 Jun 2021 14:00:13 No.13261310 Report

Quoted By:

>>13261137
I too have tried online solvers, but sometimes they fail because not enough assumptions are put in or some simplifications arent made

OP Thu 10 Jun 2021 14:01:29 No.13261312 Report

Quoted By: >>13261362

>>13261137
shoulkd be possible though - 2 equations, 2 variables

Anonymous

Anonymous Thu 10 Jun 2021 14:01:51 No.13261314 Report

Quoted By: >>13261480

>>13257049
It doesn't

Anonymous

Anonymous Thu 10 Jun 2021 14:20:46 No.13261362 Report

Quoted By: >>13261488

>>13261312
That doesn't imply an analytical solution

OP Thu 10 Jun 2021 14:52:04 No.13261480 Report

Quoted By: >>13261530

>>13261314
how do you know this?

OP Thu 10 Jun 2021 14:53:08 No.13261488 Report

Quoted By: >>13261530

>>13261362
but it does imply the existence of a solution. Im just hoping it is analytical

Anonymous

Anonymous Thu 10 Jun 2021 15:03:39 No.13261530 Report

Quoted By: >>13262052

>>13261480
I just know.

>>13261488
It doesn't imply anything of that sort. Here are 2 equations in 2 variables which have no (real) solutions: xy = 2, x^2 + y^2 = 1.

OP Thu 10 Jun 2021 17:29:09 No.13262052 Report

Quoted By: >>13262119

>>13261530
you dont just "know" anything, it either can be proven or its just a hunch

It does, imply, you just said it yourself
>which have no (real) solutions
so then it has complex solutions

Anonymous

View Same Google iqdb SauceNAO 1591692693915.png, 96KiB, 250x250

Anonymous Thu 10 Jun 2021 17:42:10 No.13262109 Report

Quoted By: >>13262258

>>13257049
If they are known, then why did you use letters instead?

Anonymous

Anonymous Thu 10 Jun 2021 17:45:40 No.13262119 Report

Quoted By: >>13262262

>>13262052
e^z = 0 has no solution in C either. 1 equation, 1 variable.

Anonymous

Anonymous Thu 10 Jun 2021 17:46:01 No.13262121 Report

Quoted By:

throw it at python

OP Thu 10 Jun 2021 18:22:40 No.13262258 Report

Quoted By:

>>13262109
because they dont have to be defined to specific values, just known

OP Thu 10 Jun 2021 18:23:43 No.13262262 Report

Quoted By:

>>13262119
ok, you win

Anonymous

Anonymous Thu 10 Jun 2021 18:26:39 No.13262280 Report

Quoted By: >>13263243 >>13264760

>>13257049
${ x \over { 1 + \mathrm{ e }^{ - yc }}} = a \Leftrightarrow \mathrm{ e }^{ - yc } = \frac{1}{a} x - 1$ The same for the second equation
The quotient of the second and first equation yields
$\mathrm{ e }^y = { \frac{1}{b} x - 1 \over \frac{1}{a} x- 1 }$
And then I do not know how to simplify the second term

Anonymous

Anonymous Thu 10 Jun 2021 22:48:24 No.13263243 Report

Quoted By: >>13264762

>>13262280
The last equation gives:
$y=\ln a(x-b)-\ln b(x-a)$
so it looks like _c_ is arbitrary.

OP Fri 11 Jun 2021 06:20:59 No.13264760 Report

Quoted By:

>>13262280
been here done that. it only led to further failure

OP Fri 11 Jun 2021 06:22:02 No.13264762 Report

Quoted By:

>>13263243
you are right, but then x becomes inexpressible when I put this expression for Y into the original equations

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches