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Curve Fitting


Date: 03/08/97 at 20:23:39
From: Aaron Hardinger
Subject: Calculating the function for a curved line

Hello,
     
I am using the equation ax^3+bx^2+cx+d and also its derivative 
3ax^2+2bx+x.
     
Asuming that f(0) = 0 and f'(0) = 0, I need to get an equation that 
will give me 'a' and another one to give me 'b'.  By this I mean an 
equation that I can plug c and d into to get either a or b out.  
Then the second equation will give me the value of the remaining 
variable if I plug in c, d and either a or b.
     
I have tried this several times. I keep getting a different equation 
and worse, the results, a and b, are different with each set with a 
given value of slopes and ending x.

As you can see, I am trying to write a computer algorithm that will 
produce a curved line with a starting point of (0,0) and an ending 
point of some x and some y.  This line also starts with slope1 and 
ends with slope2.

Given this equation, ax^3+bx^2+cx+d, are there any limits to the 
slopes?  It seems that in some cases the curve looks good while others 
are way out there.  I found that if the starting slope is zero, then 
the ending slope can be anything and looks great!  But the moment that 
the initial slope changes to something not equal to 0, the curve looks 
very bad!  
     
If you could please help me with this I would be very thankful. I have 
been stuggling for a long time trying to get this to work.  If the 
equation I mentioned earlier is not be that versatile, could you 
please give me some other equations and explain how to derive the end 
equation from it?

Thank you very much for your time,
Aaron Hardinger


Date: 03/10/97 at 08:45:11
From: Doctor Jerry
Subject: Re: Calculating the function for a curved line

Hi Aaron,

From what you say to begin with, it seems you are are trying to fit a 
cubic to certain data.  For example, you want the graph to pass 
through the origin (f(0) = 0) and you want the slope to be 0 there 
(f'(0) = 0).  This requires:

  f(0) = d = 0  and  f'(0) = c = 0

(The derivative of ax^3+bx^2+cx+d is 3ax^2+2bx+c, not 3ax^2+2bx+x, as 
you wrote.)

So, the cubic must be f(x) = ax^3+bx^2.  Any choice of a and b  will 
give a cubic polynomial f for which f(0) = 0 and f'(0) = 0.

But from what you say at the end of your message, maybe you mean that
f'(0) = 1.  In this case:

  f(0) = d= 0  and  f'(0) = c = 1

So, the cubic is f(x) = ax^3+bx^2+x.

To choose a and b you need to require more than f(0) = 0 and 
f'(0) = 1.

The slope function for the above result would be 3ax^2+2bx+1. 
Depending on a and b, yes, the slope can be very large away from the 
origin. At the origin, we have forced it to be 1.

It is known that fitting data with polynomials often does not give 
very satisfactory results.  Some use Bezier curves, others use 
splines. Depending on your data, you might want to fit using some kind 
of exponential model.

You are working with a subject with a long history and quite a few 
possible pitfalls. 

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Equations, Graphs, Translations

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