Mathematica question

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  • Thread starter radou
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  • #1
radou
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Main Question or Discussion Point

OK, here's a problem I'm trying to solve for my work.

Given a set of n points {(x1, y1), (x2, y2), ... (xn, yn)} I need to obtain a linear function f from the broken line which connects all these points, so that I can the value of f at any point.

I need to do this for two sets of points, i.e. obtain functions f and g, because I have an expression which involves both the values of f and g at specific points.

Thanks in advance for any help.
 

Answers and Replies

  • #2
1,796
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Use the functions Fit and FindFit or other related functions. See help on Fit function:

mydata = {{1, 1}, {2, 4}, {3, 8.5}}

myfunction = Fit[mydata, {1, x, x^2}, x]
N[myfunction /. x -> 2.1]
Plot[myfunction, {x, 0, 10}]
 
  • #3
radou
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jackmell, thanks for the reply.

The fitting functons are not what I need, I need to obtain a linear function which directly connects these given points, and not an interpolation for this set of points. Any ideas how to do this?
 
  • #4
1,796
53
jackmell, thanks for the reply.

The fitting functons are not what I need, I need to obtain a linear function which directly connects these given points, and not an interpolation for this set of points. Any ideas how to do this?
Directly as in "exactly"? Well I know you can fit n points exactly to an n-degree polynomial. Gotta' solve a system of n equations in n unknowns for the coefficients I think. Be interesting to set that up in Mathematica. I don't think it has a built-in function to do this however but not sure.
 
  • #5
Hepth
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39
jackmell, thanks for the reply.

The fitting functons are not what I need, I need to obtain a linear function which directly connects these given points, and not an interpolation for this set of points. Any ideas how to do this?
So like a bunch of lines together? One "linear" function cannot connect all of them unless they all lie on one line.

You can construct a total function that includes those, connecting with heaviside theta functions.

ASSUMING THEY'RE ORDERED in increasing X values:
Code:
data = {{1, 1}, {2, 3}, {3, 7}, {6, 10}};
slopes = Table[(data[[i + 1]][[2]] - data[[i]][[2]])/(data[[i + 1]][[1]] - data[[i]][[1]]), {i, 1, Length[data] - 1}];
yint = Table[(data[[i]][[2]] - slopes[[i]] (data[[i]][[1]])), {i, 1, Length[slopes]}];
functions = Table[slopes[[i]] x + yint[[i]], {i, 1, Length[slopes]}];
totalfunction[x_] =  Sum[HeavisideTheta[data[[i + 1]][[1]] - x] HeavisideTheta[
    x - data[[i]][[1]]] functions[[i]], {i, 1, Length[slopes]}];
Plot[totalfunction[x], {x, 0, 4}]
is something i just whipped up, you can obviously turn everything into just one equation

Code:
fullfunction[x_] = 
 Sum[HeavisideTheta[data[[i + 1]][[1]] - x] HeavisideTheta[x - data[[i]][[1]]] ((data[[i + 1]][[2]] -  data[[i]][[2]])/(data[[i + 1]][[1]] - data[[i]][[1]]) x + (data[[i]][[2]] - slopes[[i]] (data[[i]][[1]])))  , {i, 1, Length[data] - 1}]
 
  • #6
radou
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Hepth, thanks.

When evaluating, I get error messages of type:

"Plot::plnr: totalfunction[x] is not a machine-size real number at x = \
1.6666666666666665`*^-7."

Btw, how can I get the value of this function at a point? For example, the line

"N[totalfunction[2], 2]"

returns

"6.0 HeavisideTheta[0] HeavisideTheta[
1.0] + 6.0 HeavisideTheta[-1.0] HeavisideTheta[4.0]"
 
  • #7
Hepth
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Ah, its because heavisidetheta isnt defined at those points. Try piecewise instead:
replace the totalfunction in the above code with this
Code:
totalfunction[x_] = 
 Piecewise[Table[{functions[[i]], data[[i]][[1]] <= x <= data[[i + 1]][[1]]}, {i, 1, Length[slopes]}]]
 
  • #8
radou
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Hm, for some reason, I still get the same error message when trying to plot the functions.

Btw:

"In[18]:=
totalfunction[3]

Out[18]=
Piecewise[{{5,False},{7,True},{7,True}}]"

Is there a way Mathematica can recognize that 3 is in the domain of the second function, and give its value at this function only?

This function will be a parameter in another function, so I only need the value of one of the piecewise it consists of, depending in which domain the value is.
 
  • #9
Hepth
Gold Member
448
39
try resetting mathematica by
evaluation> quit kernel local (or whatever)

The FALSE, TRUE stuff means you have defined "x" somewhere. If you need to in all my code rename "x" to "X" or something.

or Clear[x]

From a clear start I have
Code:
data = {{1, 1}, {2, 3}, {3, 7}, {6, 10}};
slopes = Table[(data[[i + 1]][[2]] - 
      data[[i]][[2]])/(data[[i + 1]][[1]] - data[[i]][[1]]), {i, 1, 
    Length[data] - 1}];
yint = Table[(data[[i]][[2]] - slopes[[i]] (data[[i]][[1]])), {i, 1, 
    Length[slopes]}];
functions = Table[slopes[[i]] x + yint[[i]], {i, 1, Length[slopes]}];
totalfunction[x_] = 
 Piecewise[
  Table[{functions[[i]], 
    data[[i]][[1]] <= x <= data[[i + 1]][[1]]}, {i, 1, 
    Length[slopes]}]]
Plot[totalfunction[x], {x, 0, 4}]
totalfunction[3]
which gives
Out->
[tex]
\begin{cases}
2 x-1 & 1\leq x\leq 2 \\
4 x-5 & 2\leq x\leq 3 \\
x+4 & 3\leq x\leq 6
\end{cases}
[/tex]
and the plot
then
7
 
Last edited:
  • #10
Hepth
Gold Member
448
39
Also, 3 is in the domain of both functions, I guess you can just change one of your less than equal to signs to less than, that way it wont be inclusive of both functions (though they're the same value), I just did it so it included the end points in each function as well, though you can prescribe it to only one if needed.:

Code:
data = {{1, 1}, {2, 3}, {3, 7}, {6, 10}};
slopes = Table[(data[[i + 1]][[2]] - 
      data[[i]][[2]])/(data[[i + 1]][[1]] - data[[i]][[1]]), {i, 1, 
    Length[data] - 1}];
yint = Table[(data[[i]][[2]] - slopes[[i]] (data[[i]][[1]])), {i, 1, 
    Length[slopes]}];
functions = Table[slopes[[i]] x + yint[[i]], {i, 1, Length[slopes]}];
totalfunction[x_] = 
 Piecewise[
  Table[{functions[[i]], 
    If[i == Length[slopes], data[[i]][[1]] <= x <= data[[i + 1]][[1]],
      data[[i]][[1]] <= x < data[[i + 1]][[1]]]}, {i, 1, 
    Length[slopes]}]]
Plot[totalfunction[x], {x, 0, 4}]
totalfunction[3]
That way the last function includes both the preceeding point as well as the final one, though every other segment only includes the leading point.
 
  • #11
radou
Homework Helper
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I'll try this out tomorrow.

Thanks a lot for your help. :)
 

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