/sci/ - Science & Math » Thread #11204881

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Anonymous Fri 06 Dec 18:39:54 2019 No.11204881 View Reply Original Report

Quoted By: >>11204911 >>11206135 >>11206144 >>11206205 >>11206230 >>11206244 >>11206950 >>11207398

How would I find a function where the derivative is the inverse of the function or is it not possible

Anonymous

Anonymous Fri 06 Dec 2019 18:54:22 No.11204910 Report

Quoted By: >>11204911

So you want some function $x(t)$ such that
$x\Big(\frac{\text{d}x}{\text{d}t}\Big=1$
I dunno. That's an interesting question

Anonymous

Anonymous Fri 06 Dec 2019 18:55:25 No.11204911 Report

Quoted By: >>11206144

>>11204910
>>11204881
I too am interested in this, does anyone know

Anonymous

Anonymous Fri 06 Dec 2019 18:55:35 No.11204913 Report

Quoted By:

So you want some function $x(t)$ such that
$x\Big(\frac{\text{d}x}{\text{d}t}\Big=1$
I dunno. That's an interesting question

Anonymous

Anonymous Fri 06 Dec 2019 18:57:10 No.11204917 Report

Quoted By:

So you want some function $x(t)$ such that
$x\Big(\frac{\text{d}x}{\text{d}t}\Big)=1$
I dunno. That's an interesting question

FUCK latex

Anonymous

Anonymous Fri 06 Dec 2019 18:58:58 No.11204923 Report

Quoted By:

my initial guess is that this is hard as fuck

Anonymous

Anonymous Fri 06 Dec 2019 19:30:36 No.11204972 Report

Quoted By: >>11205254 >>11207400

$y' \;=\; y^{-1}$
$y'' \;=\; \left( y^{-1} \right)' \;=\; \frac1{y'}$
Let $z \;=\; y'$ , solve by separation of variables.

Anonymous

Anonymous Fri 06 Dec 2019 21:54:16 No.11205254 Report

Quoted By: >>11206151

>>11204972
That's not the inverse

Anonymous

Anonymous Sat 07 Dec 2019 05:36:30 No.11206135 Report

Quoted By: >>11206154

>>11204881
try y(x) = k*x^a
y'(x) = ka*x^(a-1)
inverse_y(x) = (x/k)^(1/a)
first solve for a by letting a-1 = 1/a
a = phi (the golden ratio) or 1-phi

next solve for k by letting ka = k^(1/a)
k = a^(1/(1/a - 1)) = a^(a/(1-a)) = a^(a/(-1/a)) = a^(-a^2)

Anonymous

Anonymous Sat 07 Dec 2019 05:39:51 No.11206144 Report

Quoted By:

>>11204881
>>11204911
I know the answer

Anonymous

Anonymous Sat 07 Dec 2019 05:42:51 No.11206151 Report

Quoted By: >>11206157

>>11205254
Than what is ut

Anonymous

Anonymous Sat 07 Dec 2019 05:43:33 No.11206154 Report

Quoted By: >>11206164

>>11206135
my bad
ka= k^(-1/a)
k = a^(1/(-1/a - 1)) = a^(-a/(1+a)) = a^(-a/(a^2)) = a^(-1/a) = a^(1-a)

Anonymous

Anonymous Sat 07 Dec 2019 05:44:14 No.11206157 Report

Quoted By:

>>11206151
the reciprocal

Anonymous

Anonymous Sat 07 Dec 2019 05:47:43 No.11206164 Report

Quoted By: >>11206314 >>11206604

>>11206154
This gives y(x) = phi*(x/phi)^phi where phi is the golden ratio or 1 - (the golden ratio)

Anonymous

Anonymous Sat 07 Dec 2019 06:11:03 No.11206205 Report

Quoted By: >>11206216

>>11204881
Zero function

Anonymous

Anonymous Sat 07 Dec 2019 06:16:31 No.11206216 Report

Quoted By:

>>11206205
no inverse

Anonymous

Anonymous Sat 07 Dec 2019 06:21:45 No.11206230 Report

Quoted By:

>>11204881
$\integral f(x)$

Anonymous

Anonymous Sat 07 Dec 2019 06:28:39 No.11206244 Report

Quoted By: >>11206313 >>11206898

>>11204881
sqrt(2x)

Anonymous

Anonymous Sat 07 Dec 2019 06:53:56 No.11206278 Report

Quoted By:

my guess would be some clusterfuck of sinusoids so they carry through the derivative

Anonymous

Anonymous Sat 07 Dec 2019 07:18:34 No.11206313 Report

Quoted By: >>11206314 >>11206898

>>11206244
/thread

Anonymous

Anonymous Sat 07 Dec 2019 07:19:29 No.11206314 Report

Quoted By: >>11206898

>>11206313
lol
>>11206164
/thread

Anonymous

Anonymous Sat 07 Dec 2019 10:43:47 No.11206604 Report

Quoted By:

>>11206164
Nicely done.

Anonymous

Anonymous Sat 07 Dec 2019 13:46:34 No.11206898 Report

Quoted By:

>>11206244
>>11206313
>>11206314
He said inverse not reciprocal brainlets. Do you even understand half the shit you spew out?

Anonymous

Anonymous Sat 07 Dec 2019 14:19:05 No.11206947 Report

Quoted By:

this is what an inverse means

f(g(x)) = x

Anonymous

Anonymous Sat 07 Dec 2019 14:20:45 No.11206950 Report

Quoted By:

>>11204881
I think this thread serves as a powerful illustration that /sci/ posters are braindead retards.

Anonymous

Anonymous Sat 07 Dec 2019 16:09:30 No.11207134 Report

Quoted By:

x?/???1

Anonymous

Anonymous Sat 07 Dec 2019 18:21:25 No.11207398 Report

Quoted By:

>>11204881
OP, I'd post it in /mg/ if you want any chance at getting a non-retarded answer. I'm curious to know if it's possible though. Or if there are functions that oscillate between inverse and original with each derivative (each degree derivative just takes the inverse again).

Anonymous

Anonymous Sat 07 Dec 2019 18:21:43 No.11207400 Report

Quoted By:

>>11204972
brainlet

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