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Q: Formula of a curve, from a set of xy values ( No Answer,   6 Comments )
Subject: Formula of a curve, from a set of xy values
Category: Science > Math
Asked by: jimboston-ga
List Price: $10.00
Posted: 25 May 2006 12:25 PDT
Expires: 24 Jun 2006 12:25 PDT
Question ID: 732367
How can I find a formula for a curve if I have a set of X and Y
values? I'd like to find Y for any given X up to about x = 70 The
values I have are:

when x=3 y=1300
when x=6 y=700
when x=12 y=350
when x=24 y=155
when x=40 y=75
There is no answer at this time.

Subject: Re: Formula of a curve, from a set of xy values
From: frankcorrao-ga on 25 May 2006 16:20 PDT
There are no guarantees though.
Subject: Re: Formula of a curve, from a set of xy values
From: falcoboy7-ga on 25 May 2006 18:31 PDT
Using a TI-83/83+/84, plug the values into a stat->edit
Then go to Stat->Calc->(regression model-for example QuadReg)
Then put in the values of the table columns so you have...
QuarReg L1,L2,Y1
This plugs the formula into Y1
You may need to experiment a bit with this.
Subject: Re: Formula of a curve, from a set of xy values
From: jimboston-ga on 27 May 2006 14:07 PDT
I'm able to use interpolation functions within excel to get answers,
but I'd like to know how to derive a single formula for expressing
this type of curve from a set of values.
Subject: Re: Formula of a curve, from a set of xy values
From: brix24-ga on 27 May 2006 19:31 PDT
Hello jimboston,

I suspect there are several possible answers, depending on the context
of your question.

Based only on the information given, I would infer the type of curve
and the equation to use by examining a plot of the data. To me, a plot
in Excel suggests the data approximates a simple hyperbola confined to
the first quadrant. The fit won?t be exact for such a simple hyperbola
(xy=c) since the product of the xy values is not truly constant, so a
"best fit" determination would be needed.  (In a real world situation,
these deviations from a constant value would probably reflect
experimental variation.)

 x	 y	  xy
 3	1300	 3900
 6	 700	 4200
12	 350	 4200
24	 155	 3720
40	  75	 3000

(My apologies if the above table doesn?t appear correctly. I haven?t
figured out how to get data to stay aligned correctly after a comment
is posted.)

If a hyperbola of the simple type, xy=<constant>, is chosen as the
curve, you?ll have to do a best fit to get the constant.

In most situations, the type of curve that you would fit the data to
would be based on the type of xy relationship expected from whatever
physical, chemical, or biological process was being modeled. Since
that information is lacking here, I chose a simple curve suggested by
graphing the data.

Note: there are other solutions, including at least two curves that
exactly match the five data points given.

Exact curve 1: If you absolutely had to have a curve that fit the data
exactly, it?s always possible to derive a fourth degree polynomial in
x that fits 5 xy values exactly. You can get a good approximation to
this equation by entering your values at 

(You?ll get y = 0.0157103006367711 x^4 - 1.34519104188221 x^3  +
39.0410079196841 x^2 - 472.984707396469 x + 2402.63267469149).
However, this is not a solution most of us would prefer. For instance,
if you solved this equation for y using x=35, you?ll get y = -426;
because the curve is more complex than a simple ?one-bend? hyperbola,
the interpolated y value from the polynomial is not what one would
expect by looking the trends in the original data.

Exact curve 2: You can get an exactly matching hyperbola in this case
by using the general formula for a hyperbola.

The xy=c formula for a hyperbola is one that I remember from high
school; this hyperbola does not cross either the x axis or the y axis.
A hyperbola can, of course, be positioned elsewhere. A google search

hyperbola equation

leads to other equations for a hyperbola, such as, x^2/a^2 ? y^2/b^2 =
1 (which is also a special case) and to a general formula for a
hyperbola at

?Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 
such that B2 > 4AC, where all of the coefficients are real, and where
more than one solution, defining a pair of points (x, y) on the
hyperbola, exists.?

We can derive an exact match from this equation because we have 5
pairs of xy values that give 5 equations, which, in turn, can be
solved for the five coefficients, A ? E. (F can be chosen almost
arbitrarily; all that matters is that F is not arbitrarily chosen as
zero. I don?t know what a mathematician would call F, but I?ll call it
a scalable Factor. It is equivalent to the right-hand term in these
equations, which all represent the same straight line: 2x + 3y = 6, x
+ 1.5y = 3, 10x + 15y = 30.)

I set F to 1 initially and solved the five simultaneous linear
equations. At the end, I multiplied all terms by 100000 to get
coefficients that were easier for me to compare. Here is the final
-6.25969943795458 x^2 + 18.0792113361007 xy -0.00592726833278236 y^2 +
1344.03199473387 x + 27.3326158325765 y = 100000

(A necessary check: B^2 is indeed > 4AC).

The key point is to know what type of curve / relationship would be
expected from the properties of the system you are observing or,
barring that, what type of curve you deduce from looking at a plot of
the data. Then it?s a matter of finding the equation (general if need
be) that describes that curve and either solving exactly (possible in
some situations) or finding the "best fit" for the chosen curve type
to the data.

Note that the hyperbola equation that exactly matches the data is not
in the form, y=f(x), making interpolation more complicated. On the
other hand, a best fit of the data to xy=c makes calculation of
interpolated y values simpler since y is easily calculated as y=c/x.
(Ideally, I should have done some interpolation checks for the "exact
match" hyperbola, but I'm "mathed out" at the moment.)

If your data described observations of a real world system, I would
probably have chosen fitting the ?xy=<a constant>? equation to the
data rather than finding a hyperbola that matched exactly.

Interpolation and extrapolation: Interpolation of y values for x
between ~3 and ~40 is relatively safe, but extrapolation much outside
that range might not be reliable in a real world situation. In
general, I wouldn?t place much confidence in extrapolation at an x
value of 70 without further evidence that the equation truly
represents the data in the ?known? region.
Subject: Re: Formula of a curve, from a set of xy values
From: brix24-ga on 27 May 2006 19:40 PDT
Sorry, the last sentence should have ended:

"...without further evidence that the equation truly
represents the data _outside_ the ?known? region.
Subject: Re: Formula of a curve, from a set of xy values
From: activealexaoki-ga on 11 Jun 2006 13:47 PDT
As for myself, I hate to rely on instruments to solve math, like so
many crowds of people. The sole reason is that you won't know why, in
which case people can neglect the reasoning and just steal the result.
There is a method in Numerical Analysis. Someone mentioned before -

My method of interpolation, I could not confirm the validity of the
equation online. Many sites are just introducing the principle or
Lagrange interpolation, which seems very simple but not
comprehensible. My method is not going to guarantee the result but it
should be mathematically correct.

Step 1. Recognize polynomial
By recognizing the points, you can guess how far your polynomial
should consider to ensure that your points are all interpolated. Since
one point cannot draw any non-trivial line, two points are needed to
draw straight line. Two points cannot guess unique quadric equation,
you need three points. So and so. The highest degree of polynomial is
always one less than the number of known points - at least. In your
case, since you have 5 known points, you may only take fourth power
polynomial. (Of course, by plotting points, if you see the graph
exponential, you can write the equation in terms of exponential
function. You may figure that out by yourself, or let me know.)
Step 2. Reasoning of method
From Step 1, I decided to take polynomial such that
Let x's and y's are vectors, the underline reason of interpolation is
that the vector sum satisfies. By simplifying the equation the vector
product of the interpolation will look like:
As you can see, you only need to solve the augmented matrix:
[x^4,x^3,x^2,x^1,x^0 | y]
Since you have five points,
|x1^0 x1^1 x1^2 x1^3 x1^4 || y1|
|x2^0 x2^1 x2^2 x2^3 x2^4 || y2|
|x3^0 x3^1 x3^2 x3^3 x3^4 || y3|
|x4^0 x4^1 x4^2 x4^3 x4^4 || y4|
|x5^0 x5^1 x5^2 x5^3 x5^4 || y5|

Just solve the augmented matrix above and you can find the
interpolated function manually. If you read more about Lagrange
Interpolation, you find it simpler. If you find explanation for
Incremental Newton Interpolation, you find it very accessible.

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