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I have some points and want to find a function from them. For example, $(-2,4),(0,0),(1,1),(5,25)$. I don't have a function that generates the points.

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Look up Interpolation[] or FindSequenceFunction[]. –  J. M. Apr 26 '13 at 16:03
    
Or, if you have some idea of the form of the function, look at FindFit, LinearModelFit, and NonlinearModelFit. –  rcollyer Apr 26 '13 at 16:06
    
Do you just want to fit those points or do you want a function that is good for interpolaton in between? The appropriate fit can be quite different. –  george2079 Apr 26 '13 at 18:32
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closed as too localized by Ajasja, Oleksandr R., Artes, Yves Klett, m_goldberg Apr 26 '13 at 20:30

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2 Answers

Try Interpolation; the second form should do exactly what you want. This will return an InterpolatingFunction object, which you can apply to numerical arguments just like any other function.

if = Interpolation[{{-2, 4}, {0, 0}, {1, 1}, {5, 25}}]

I have if[3.2] returning 10.24.

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sorry ,i don't understand.if[3.2] , 10.24 what are them mean? –  amir Apr 26 '13 at 18:54
    
evaluating the interpolating function if at 3.2 returns 10.24. –  s0rce Apr 26 '13 at 19:09
    
how can i obtain this evaluating? sorry because i know a little programming –  amir Apr 26 '13 at 19:23
    
@amir, just type in if[3.2] and hit Shift + Enter. –  Pillsy Apr 26 '13 at 19:35
1  
Look up Plot in the documentation. –  s0rce Apr 26 '13 at 20:29
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If you can hazard a guess as to the form of the function, then you can fit almost anything. For example, here is your data:

points = {{-2, 4}, {0, 0}, {1, 1}, {5, 25}}

Let's see how well a second order polynomial fits:

poly = NonlinearModelFit[points, a x^2 + b x + c, {a, b, c}, x]

The answer is: it fits really well! You can also pick other functional forms, to see how well they fit. You can read about this kind of function-finding in the help file for NonlinearModelFit section of the help.

Here you can see how well the function fits the points by plotting:

Show[ListPlot[points], Plot[poly[x], {x, -2, 5}]]

enter image description here

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Since this is a linear model, one can even use LinearModelFit... –  Oleksandr R. Apr 26 '13 at 16:39
    
my point is {0.652121,0.00259997},{0.653734,0.0398411},{0.655347,0.0770823},{0.665245,0.1161‌​71},{0.672903,0.1683},{0.680445,0.229739},{0.686049,0.280009},{0.695993,0.315373}‌​,{0.713827,0.384244},{0.727623,0.443811},{0.747697,0.499642},{0.76994,0.548021},{‌​0.792253,0.590813},{0.822595,0.655937},{0.846962,0.700588},{0.89182,0.76755},{0.9‌​40832,0.834504},{0.994044,0.897727},{1.03286,0.949802},{1.07592,0.99442} and i guess poly = NonlinearModelFit[points, a x^5 + b x^4 + c x^3+d x^2+ex+f, {a, b, c,d,e,f}, x] but i had this error: Syntax::sntxf: "points = " cannot be followed by .what is my fualt? –  amir Apr 26 '13 at 18:51
    
You are missing a curly parenthesis - look carefully at the way I defined points -- it is a list of pairs. –  bill s Apr 26 '13 at 19:16
    
@amir points = {{0.652121,0.00259997},{0.653734,0.0398411},{0.655347,0.0770823},{0.665245,0.116‌​171},{0.672903,0.1683},{0.680445,0.229739},{0.686049,0.280009},{0.695993,0.315373‌​},{0.713827,0.384244}, {0.727623, 0.443811},{0.747697,0.499642},{0.76994,0.548021}, {0.792253,0.590813},{0.822595,0.655937},{0.846962,0.700588},{0.89182,0.76755},{0‌​.940832, 0.834504},{0.994044,0.897727}, {1.03286, 0.949802},{1.07592,0.99442}}; together with nlm = NonlinearModelFit[points,a+b x+c x^2+d x^3+e x^4+f x^5,{a,b,c,d,e,f},x] and Show[ListPlot@points,Plot[nlm[x], {x,0.6,1.1}]] should do the job. –  chromate Apr 26 '13 at 19:28
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You need to look more carefully. There are many different kinds of parenthesis and brackets and they all mean different things. You have [{... it should be {{... –  bill s Apr 26 '13 at 20:26
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