How to Get function from data ?

[jemal.manjgaladze]

Hello .

I am doing some numerical experiment in Physics in university. So I obtained (x,y) pairs from computer calculations and i can draw graph of it using program. but i want to know is there any way or any website so that I input my (x,y) pair and it will give me what function is it?


for example I input (1,2) (2,4) (3,6) (4,8) ... and it gives me f(x)=2x.

 

[Deleted member 4993]

Hello .

I am doing some numerical experiment in Physics in university. So I obtained (x,y) pairs from computer calculations and i can draw graph of it using program. but i want to know is there any way or any website so that I input my (x,y) pair and it will give me what function is it?


for example I input (1,2) (2,4) (3,6) (4,8) ... and it gives me f(x)=2x.

Use a spreadsheet-software like Excel. In there, you will find a "function" called "trend-line". That will choose the "best-fit line" for those points and the functional description.

 

[LCKurtz]

You will need to have some idea of what kind of curve it should be. Linear? Quadratic? Polynomial? Exponential? Logarithmic? Periodic? If you know that, find a website that does least squares fit. Or maybe cubic splines. It all depends on what you are doing and what you expect to get out of it.

 

[topsquark]

Or you can work it out exactly. (Not so easy with this one.. I resorted to W|A.) I fit the points to a quartic. It looks a lot like a line, but if we go to higher degree equations there will be more curve in the plot.

-Dan

 

[LCKurtz]

I'm thinking those three points were just an example and the OP was wanting something that would work more generally as he gets output from his physics experiments.

 

[HallsofIvy]

Given any finite number of data points, there exist an infinite number of functions that give those points. In particular, given n points, there exist a unique polynomial of degree n-1. With your example, (1,2) (2,4) (3,6) and (4,8), if we didn't notice that they all lie on a straight line, we could try to find a, b, c, d, so that \(\displaystyle f(x)= ax^3+ bx^2+ cx+ d\) satisfied that.
We would have
f(1)= a+ b+ c+ d= 2
f(2)= 8a+ 4b+ 2c+ d= 4
f(3)= 27a+ 9b+ 3c+ d= 6
f(4)= 64a+ 16b+ 4c+d= 8,
four equations to solve for a, b, c, and d.

Subtracting the first equation from the second, we eliminate d and have
7a+ 3b+ c= 2.
Subtracting the third equation from the fourth, we also eliminate d and have
37a+ 7b+ c= 2
And subtracting the second equation from the third
19a+ 5b+ c= 2

Subtracting the first of those from the second, 30a+ 4b= 0.
Subtracting the third of those from the second, 18a+ 2b= 0

Subtracting twice the second of those from the first, -6a= 0 so a= 0.
Gosh who'd have guessed that!

Of course then 30a+ 4b= 4b= 0 so b= 0 also.

7a+ 3b+ c= c= 2.

And a+ b+ c+ d= 2+ d= 2 so d= 0.

\(\displaystyle f(x)= 0x^3+ 0x^2+ 2x+ 0= 2x\).

That was a lot of work for an obvious result!