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How to derive a function from an irregular table of values? 
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#1
Mar1214, 10:22 PM

P: 5

It's easy to get a function from a table that follows a clear pattern, but what if the values for y (or x) don't follow a pattern? For example;
x = 1 , y = 13 x = 2 , y = 9 x = 3 , y = 3 x = 4 , y = 8 x = 5 , y = 1 x = 6 , y = 5 x = 7 , y = 12 If you look at these points on a graph, they don't make any normal line. 


#2
Mar1214, 10:53 PM

P: 561

You should be able to fit a 6th degree polynomial that goes through all those points. You can do this using the LINEST function in Excel.



#3
Mar1214, 11:21 PM

P: 5

I put the values into excel and did =LINEST(A2:A8,B2:B8,,FALSE) and it gave me 0.464285714 as the slope and 9.142857 as the yintercept but that can't be right, so I must have done something wrong... I'm not great with excel.



#4
Mar1214, 11:57 PM

P: 561

How to derive a function from an irregular table of values?
Yes, you want a 6th degree polynomial not a one degree polynomial:
ax^6 + bx^5 + cx^4 + dx^3 + ex^2 + fx + g You will probably need to research to figure out how to get Excel to do this for you. 


#5
Mar1314, 03:43 AM

P: 439

http://spreadsheetpage.com/index.php...line_formulas/
Excel can do up to 16th power. After that you're on your own :) 


#6
Mar1314, 05:55 AM

P: 996

...unless you have reason to believe that the underlying data is not random and actually came from a sixth degree polynomial in the first place. 


#7
Mar1414, 06:44 AM

P: 5




#8
Mar1414, 07:05 AM

P: 439

So what you're looking for is this right?
$$ \frac{53}{240}x^6+\frac{1277}{240}x^5\frac{2419}{48}x^4+\frac{3801}{16}x^3\frac{34733}{60}x^2+\frac{40477}{60}x275 $$ Have you tried Lagrange Interpolation? http://mathworld.wolfram.com/Lagrang...olynomial.html 


#9
Mar1414, 10:45 PM

P: 5




#10
Mar1514, 06:49 AM

P: 996

Call the functions f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6} and f_{7}. You want to put together a combination k_{1}f_{1} + k_{2}f_{2} + k_{3}f_{3} + k_{4}f_{4} + k_{5}f_{5} + k_{6}f_{6} + k_{7}f_{7} such that the combination has the right values at each of your seven data points. You can find the right combination by solving a set of seven simultaneous equations in seven unknowns. Say your data points are (x_{1},y_{1}) through (x_{7},y_{7}) Then the equations are: y_{1} = k_{1}f_{1}(x_{1}) + k_{2}f_{2}(x_{1}) + ... + k_{7}f_{7}(x_{1}) y_{2} = k_{1}f_{1}(x_{2}) + k_{2}f_{2}(x_{2}) + ... + k_{7}f_{7}(x_{2}) [...] y_{7} = k_{1}f_{1}(x_{7}) + k_{2}f_{2}(x_{7}) + ... + k_{7}f_{7}(x_{7}) You know all the x's and y's. Those are your data points. You know the values for each of the functions at each of the x's. That's just a matter of evaluating each function at each point. The k values are the unknowns. Solve for for k_{1} through k_{7} Lagrange Interpolation is what you get if you take f_{1} through f_{7} to be: f_{1}(x) = 1 f_{2}(x) = x f_{3}(x) = x^{2} ... f_{n}(x) = x^{n1} [Edit, fixed error on the final exponent above] Obviously this will work for any number of data points, not just seven. 


#11
Mar1514, 02:30 PM

P: 561

That is what the excel function is doing except it does a least squares approach for the case where the number of data points exceeds the degree of the polynomial by one or more.



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