Determining Formula for a graph

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• Jun 10th 2011, 10:44 PM
IBstudent
Determining Formula for a graph
Can anyone please show me how to determine the formula of the plot dots in the graph, the dots I think is divided into three parts, I put a line between the parts to avoid confusion....

A looks something like a x^3 graph
B looks like a sine curve
C looks weird.....

Thanks in advance, the files are attached
• Jun 10th 2011, 11:14 PM
Prove It
Count the number of points in each graph. The polynomial that fits exactly will be of order one less than the number of points. Then you can set up a series of equations to solve simultaneously.
• Jun 10th 2011, 11:30 PM
IBstudent
Quote:

Originally Posted by Prove It
Count the number of points in each graph. The polynomial that fits exactly will be of order one less than the number of points. Then you can set up a series of equations to solve simultaneously.

Thanks,
sorry but I have a question, if you don't mind....
so one less than the number of points... the division A in the graph has 9 points, 9-1=8, so what now?
• Jun 11th 2011, 04:36 AM
HallsofIvy
There are 9 points so you can fit a unique 8th degree polynomial through those nine points. One way to do that is what Prove It suggested: $y= ax^8+ bx^7+ cx^6+ dx^5+ ex^4+ fx^3+ gx^2+ hx+ i$. Putting the x and y values of the 9 points into that gives 9 equations to solve for the 9 numbers a, b, c, d, e, f, g, h, and i.
• Jun 11th 2011, 06:37 AM
IBstudent
Quote:

Originally Posted by HallsofIvy
There are 9 points so you can fit a unique 8th degree polynomial through those nine points. One way to do that is what Prove It suggested: $y= ax^8+ bx^7+ cx^6+ dx^5+ ex^4+ fx^3+ gx^2+ hx+ i$. Putting the x and y values of the 9 points into that gives 9 equations to solve for the 9 numbers a, b, c, d, e, f, g, h, and i.

o_0 this is gonna take a long long time.... I'll keep you guys updated if it worked or not :-)
• Jun 11th 2011, 06:38 AM
IBstudent
Quote:

Originally Posted by HallsofIvy
There are 9 points so you can fit a unique 8th degree polynomial through those nine points. One way to do that is what Prove It suggested: $y= ax^8+ bx^7+ cx^6+ dx^5+ ex^4+ fx^3+ gx^2+ hx+ i$. Putting the x and y values of the 9 points into that gives 9 equations to solve for the 9 numbers a, b, c, d, e, f, g, h, and i.

o_0 this is gonna take a long long time.... I'll keep you guys updated if it worked or not :-)
• Jun 11th 2011, 06:59 AM
IBstudent
Okay so now I got 9 very long equations, I would use simultaneous equations... but I don't know where to start! any suggestions? thanks
• Jun 11th 2011, 07:35 AM
Ackbeet
• Jun 11th 2011, 08:11 AM
IBstudent
Quote:

Originally Posted by Ackbeet

I reached the limit before even finishing the second formula :-(
• Jun 11th 2011, 08:13 AM
Ackbeet
Hmm. What software is available to you? Mathematica? MATLAB? Maple? Excel? I think all of these packages could solve a 9 x 9 system of linear equations for you. I think even my TI-85 calculator could do that, though I grant you that it would be exceptionally laborious to do by hand.
• Jun 11th 2011, 08:26 AM
IBstudent
Solving polynomial simultaneous equations
I have a very many (nine to be precise) simultaneous equations, I want to solve them to get a,b,c,d,e,f,g,h and i. I tried everything from websites to trying myself, but couldn't :-(

it is attached if you want to see it, please if anyone has a method of finding these values let me know....

• Jun 11th 2011, 08:38 AM
Prove It
Otherwise, write down your augmented matrix and upper triangularise the system using row operations yourself.
• Jun 11th 2011, 08:55 AM
IBstudent
Quote:

Originally Posted by Prove It
Otherwise, write down your augmented matrix and upper triangularise the system using row operations yourself.

You say it like its so simple, but I don't get a simple word!.... lol
but thanks for the throught :-)
• Jun 11th 2011, 08:56 AM
Prove It
Quote:

Originally Posted by IBstudent
You say it like its so simple, but I don't get a simple word!.... lol
but thanks for the throught :-)

Then look them up and consider it a learning experience :)
• Jun 11th 2011, 12:14 PM
Ackbeet
The following commands in Mathematica solved the system:

f[x_]:= Aye x^8 + Bee x^7 + Cee x^6 + Dee x^5 + Eee x^4 + Eff x^3 + Gee x^2 + H x + Eye;
Solve[{f[2] == 426.8, f[3] == 470.2, f[4] == 503.4, f[5] == 557.3, f[6] == 564.7, f[7] == 575.4, f[8] == 579.8, f[9] == 624.7, f[10] == 669.9}, {Aye, Bee, Cee, Dee, Eee, Eff, Gee, H, Eye}]

It did claim that the coefficient matrix was badly conditioned. You might be better off plugging in different values for x, unless you don't have any control over where you sample your function.
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