/sci/ - Science & Math » Thread #13804809

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Math question

Anonymous Sat 30 Oct 21:45:35 2021 No.13804809 View Reply Original Report

Quoted By: >>13804995 >>13805078 >>13806017

If I have a process that changes with time as f(x), how do I find a function g(x) so that if I apply g(x) to f(x) (multiply?) the result is a linear function y=kx? Should g(x) be the inverse of f(x)? For example f(x) is 1-e^-x. Pic related.

Anonymous

Anonymous Sat 30 Oct 2021 21:50:48 No.13804835 Report

Quoted By: >>13804960

More specifically, it is a heating process, and if the applied power is constant, the temperature increases according to 1-e^-x. But I need to make sure it changes linearly. So I have to ramp up power according to a certain function. This is the alternative to using a PID since the temperature change with time is quite predictable and consistent.

Anonymous

Anonymous Sat 30 Oct 2021 22:34:55 No.13804960 Report

Quoted By: >>13805065

>>13804835
Yes, you can always just set p(x) - the power - to be the inverse. So that your heat value is the same as time.

More interestingly, if you want it to be a constant×time, you can use the chain rule twice and solve a second order differential equation since you're setting the second derivative to 0 everywhere.

I.e. (f(p(x)))' ' = (f'(p(x))×p'(x))'
= p' '(x)×f'(p(x)) + p'(x)^2×f' '(p(x)) = 0.

Fuck off, not typing mathjax on phone

Anonymous

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Anonymous Sat 30 Oct 2021 22:51:27 No.13804995 Report

Quoted By: >>13805065

>>13804809
If i understand the problem correctly you are solving the equation f(g(x)) = kx with f(x) = 1-e^(-x).
The solution is pretty simple (see pic).

Anonymous

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Anonymous Sat 30 Oct 2021 23:30:02 No.13805065 Report

Quoted By: >>13805973

>>13804995
>>13804960
Yeah thanks anons, using the inverse sounds intuitive but thanks for writing this down for me. Dumb me thought of multiplying f(x)g(x) instead of f(g(x)), it all makes sense now.
I think you have a typo
>f(x)=1-e^x
it is actually 1-e^-x
but the result is correct: -ln(1-kx)

>More interestingly, if you want it to be a constant×time, you can
i don't follow, could you explain this?

Anonymous

Anonymous Sat 30 Oct 2021 23:39:34 No.13805078 Report

Quoted By: >>13805101

>>13804809
g(x) = f(x)/x

Anonymous

Anonymous Sat 30 Oct 2021 23:45:29 No.13805101 Report

Quoted By: >>13805345

>>13805078
you mean x/f(x)? but i don't really need f(x)*g(x) what i actually need is f(g(x))
f(x) is the temp(time)
g(x) is the power(time)

Anonymous

Anonymous Sat 30 Oct 2021 23:51:31 No.13805117 Report

Quoted By:

fg = 1 - e^(-g) means g = -log(1 - kx)

Anonymous

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Anonymous Sat 30 Oct 2021 23:53:03 No.13805120 Report

Quoted By:

mathsh

Anonymous

Anonymous Sun 31 Oct 2021 01:46:17 No.13805345 Report

Quoted By:

>>13805101
You just need g(x)=f^-1(x). (Not f(x)^-1). So if you have y=f(x) just solve for x

Anonymous

Anonymous Sun 31 Oct 2021 09:43:33 No.13805973 Report

Quoted By:

>>13805065
I can't i was being retarded.

Anonymous

Anonymous Sun 31 Oct 2021 10:13:35 No.13806017 Report

Quoted By:

>>13804809
yes, it should just be the inverse, as their product is always a linear function

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