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Simple curve - not so simple function?!!

by Miffymycat
Tags: curve, function, simple
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Miffymycat
#1
May25-14, 05:44 AM
P: 40
What is f(x) when eg

x = 0, 1, 2, 3, 4, 5, 6, etc
y = 0, 8, 12, 14, 15, 15.5, 15.75 etc (the y value at x =1 is arbitrary)

ie each successive y value adds half the difference of the preceding 2 values.

(It is effectively the inverse of a first order exponential decay where the y value halves at constant x intervals).

But it doesn’t fit an exponential ....! Could it be a type of hyperbola? or polynomial? or both? or something else entirely?

I have no decent curve fitting software and struggling to find this.
Thank you!
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adjacent
#2
May25-14, 06:03 AM
PF Gold
adjacent's Avatar
P: 1,496
Try drawing it on a graph paper.
Miffymycat
#3
May25-14, 06:12 AM
P: 40
Thanks ... but how would that help, if I've already tried to fit it in Excel?

adjacent
#4
May25-14, 06:28 AM
PF Gold
adjacent's Avatar
P: 1,496
Simple curve - not so simple function?!!

You can try finding the nth term of the sequence(That's what I do).
HallsofIvy
#5
May25-14, 08:22 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,553
So you are saying that [itex]x_{n+2}= x_{n+1}+ (x_n+ x_{n+1})/2= (3/2)x_{n+1}+ (1/2)x_n[/itex]

That's a "second order difference equation" which has associated characteristic equation [itex]s^2= (3/2)s+ 1/2[/itex] or [itex]s^2- (3/2)s- 1/2= 0[/itex]. The solutions to that equation are [itex]s= \frac{3\pm \sqrt{17
}}{4}[/itex]. That means that the general solution to the difference equation is [itex]8^{n/4}(C2^{\sqrt{17}n/4}+ D2^{-\sqrt{17}n/4})[/itex]
micromass
#6
May25-14, 08:30 AM
Mentor
micromass's Avatar
P: 18,299
So you have a sequence defined by

[tex]x_0 = 0,~x_1 = 8,~x_{n+1} = x_n + \frac{1}{2}(x_n - x_{n-1}) = \frac{3}{2}x_n - \frac{1}{2}x_{n-1}[/tex]

Let's find a general term. We can express this sequence as follows:

[tex]\left(
\begin{array}{cc}
x_{n+1}\\ x_n
\end{array}
\right)
=
\left(
\begin{array}{cc}
\frac{3}{2} & -\frac{1}{2}\\ 1 & 0
\end{array}
\right)
\left(
\begin{array}{cc}
x_{n}\\ x_{n-1}
\end{array}
\right)
[/tex]

So, we get

[tex]\left(
\begin{array}{cc}
x_{n+1}\\ x_n
\end{array}
\right)
=
\left(
\begin{array}{cc}
\frac{3}{2} & -\frac{1}{2}\\ 1 & 0
\end{array}
\right)^n
\left(
\begin{array}{cc}
x_1\\ x_0
\end{array}
\right)
[/tex]

By using the diagonalization theory of linear algebra, we can write this as

[tex]\left(
\begin{array}{cc}
x_{n+1}\\ x_n
\end{array}
\right)
=
\left(
\begin{array}{cc}
\frac{1}{2} & 1\\ 1 & 1
\end{array}
\right)
\left(
\begin{array}{cc}
\frac{1}{2^n} & 0\\ 0 & 1
\end{array}
\right)
\left(
\begin{array}{cc}
-2 & 2\\ 2 & -1
\end{array}
\right)
\left(
\begin{array}{cc}
x_1\\ x_0
\end{array}
\right)
[/tex]

and thus

[tex]\left(
\begin{array}{cc}
x_{n+1}\\ x_n
\end{array}
\right)
=
\left(
\begin{array}{cc}
2 - \frac{1}{2^n} & \frac{1}{2^n}- 1\\ 2 - \frac{1}{2^{n-1}} & \frac{1}{2^{n-1}}-1
\end{array}
\right)
\left(
\begin{array}{cc}
x_1\\ x_0
\end{array}
\right)
[/tex]

Thus we get

[tex]x_{n+1} = 2x_1 - \frac{x_1}{2^n} + \frac{x_0}{2^n} - x_0[/tex]

Since ##x_1 = 8## and ##x_0 = 0##, we get

[tex]x_{n+1} = 16 - \frac{8}{2^n}[/tex]
Miffymycat
#7
May25-14, 09:09 AM
P: 40
Thanks both! I'm not a mathematician ... just a chemist struggling to make sense of some data!

So Hallsofivy, is a "second order difference equation" the same thing as a second order polynomial? I tried to fit the data to a polynomial and it required 6th order before R2= 1 in Excel! (And not sure how you got from the series to the associated characteristic equation)

I see you guys have both written it as a series, but can it be expressed in terms of y=f(x)? And described as eg a power function or polynomial etc?
PeroK
#8
May25-14, 09:19 AM
P: 395
You can manipulate Micromass's approach to get, for example:

[tex]f(x) = 16 - 2^{4-x}[/tex]

And, if f(1) = a, then:

[tex]f(x) = a(2 - 2^{1-x}) = 2a(1 - \frac{1}{2^x})[/tex]
Miffymycat
#9
May25-14, 09:26 AM
P: 40
.... building on micromass's reply .... something like y = 2a -(a/2^x)? where a is an arbitrary constant. Is it a hyperbolic function??! Or a hyperbolic polynomial! does that exist?
Miffymycat
#10
May25-14, 09:54 AM
P: 40
Thanks Perok ... I was sending mine while yours was arriving - I was close! So does the function fit one of those categories?
PeroK
#11
May25-14, 10:04 AM
P: 395
I would say it's just a variation of a power function (powers of 1/2).
Miffymycat
#12
May25-14, 10:15 AM
P: 40
OK - a bit weird though, that it fits precisely to polynomial order 6! Coincidence?
PeroK
#13
May25-14, 10:23 AM
P: 395
Quote Quote by Miffymycat View Post
OK - a bit weird though, that it fits precisely to polynomial order 6! Coincidence?
Any set of n values will fit a polynomial of order n-1.


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