Simple curve - not so simple function?!!

by Miffymycat
Tags: curve, function, simple
 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!
 PF Gold P: 1,468 Try drawing it on a graph paper.
 P: 40 Thanks ... but how would that help, if I've already tried to fit it in Excel?
 PF Gold P: 1,468 Simple curve - not so simple function?!! You can try finding the nth term of the sequence(That's what I do).
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,363 So you are saying that $x_{n+2}= x_{n+1}+ (x_n+ x_{n+1})/2= (3/2)x_{n+1}+ (1/2)x_n$ That's a "second order difference equation" which has associated characteristic equation $s^2= (3/2)s+ 1/2$ or $s^2- (3/2)s- 1/2= 0$. The solutions to that equation are $s= \frac{3\pm \sqrt{17 }}{4}$. That means that the general solution to the difference equation is $8^{n/4}(C2^{\sqrt{17}n/4}+ D2^{-\sqrt{17}n/4})$
 Mentor P: 18,095 So you have a sequence defined by $$x_0 = 0,~x_1 = 8,~x_{n+1} = x_n + \frac{1}{2}(x_n - x_{n-1}) = \frac{3}{2}x_n - \frac{1}{2}x_{n-1}$$ Let's find a general term. We can express this sequence as follows: $$\left( \begin{array}{cc} x_{n+1}\\ x_n \end{array} \right) = \left( \begin{array}{cc} \frac{3}{2} & -\frac{1}{2}\\ 1 & 0 \end{array} \right) \left( \begin{array}{cc} x_{n}\\ x_{n-1} \end{array} \right)$$ So, we get $$\left( \begin{array}{cc} x_{n+1}\\ x_n \end{array} \right) = \left( \begin{array}{cc} \frac{3}{2} & -\frac{1}{2}\\ 1 & 0 \end{array} \right)^n \left( \begin{array}{cc} x_1\\ x_0 \end{array} \right)$$ By using the diagonalization theory of linear algebra, we can write this as $$\left( \begin{array}{cc} x_{n+1}\\ x_n \end{array} \right) = \left( \begin{array}{cc} \frac{1}{2} & 1\\ 1 & 1 \end{array} \right) \left( \begin{array}{cc} \frac{1}{2^n} & 0\\ 0 & 1 \end{array} \right) \left( \begin{array}{cc} -2 & 2\\ 2 & -1 \end{array} \right) \left( \begin{array}{cc} x_1\\ x_0 \end{array} \right)$$ and thus $$\left( \begin{array}{cc} x_{n+1}\\ x_n \end{array} \right) = \left( \begin{array}{cc} 2 - \frac{1}{2^n} & \frac{1}{2^n}- 1\\ 2 - \frac{1}{2^{n-1}} & \frac{1}{2^{n-1}}-1 \end{array} \right) \left( \begin{array}{cc} x_1\\ x_0 \end{array} \right)$$ Thus we get $$x_{n+1} = 2x_1 - \frac{x_1}{2^n} + \frac{x_0}{2^n} - x_0$$ Since ##x_1 = 8## and ##x_0 = 0##, we get $$x_{n+1} = 16 - \frac{8}{2^n}$$
 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?
 P: 354 You can manipulate Micromass's approach to get, for example: $$f(x) = 16 - 2^{4-x}$$ And, if f(1) = a, then: $$f(x) = a(2 - 2^{1-x}) = 2a(1 - \frac{1}{2^x})$$
 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?
 P: 40 Thanks Perok ... I was sending mine while yours was arriving - I was close! So does the function fit one of those categories?
 P: 354 I would say it's just a variation of a power function (powers of 1/2).
 P: 40 OK - a bit weird though, that it fits precisely to polynomial order 6! Coincidence?
P: 354
 Quote by Miffymycat 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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