# Simple curve - not so simple function?

1. May 25, 2014

### Miffymycat

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!

2. May 25, 2014

Try drawing it on a graph paper.

3. May 25, 2014

### Miffymycat

Thanks ... but how would that help, if I've already tried to fit it in Excel?

4. May 25, 2014

You can try finding the nth term of the sequence(That's what I do).

5. May 25, 2014

### HallsofIvy

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 $8^{n/4}(C2^{\sqrt{17}n/4}+ D2^{-\sqrt{17}n/4})$

Last edited by a moderator: May 25, 2014
6. May 25, 2014

### micromass

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

$$x_{n+1} = 16 - \frac{8}{2^n}$$

7. May 25, 2014

### Miffymycat

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?

8. May 25, 2014

### PeroK

You can manipulate Micromass's approach to get, for example:

$$f(x) = 16 - 2^{4-x}$$

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

$$f(x) = a(2 - 2^{1-x}) = 2a(1 - \frac{1}{2^x})$$

Last edited: May 25, 2014
9. May 25, 2014

### Miffymycat

.... 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?

10. May 25, 2014

### Miffymycat

Thanks Perok ... I was sending mine while yours was arriving - I was close! So does the function fit one of those categories?

11. May 25, 2014

### PeroK

I would say it's just a variation of a power function (powers of 1/2).

12. May 25, 2014

### Miffymycat

OK - a bit weird though, that it fits precisely to polynomial order 6! Coincidence?

13. May 25, 2014

### PeroK

Any set of n values will fit a polynomial of order n-1.