Simple curve - not so simple function?

[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!

 

[adjacent]

Try drawing it on a graph paper.

 

[Miffymycat]

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

 

[adjacent]

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

 

[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<br /> }}{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})

 

[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(<br /> \begin{array}{cc}<br /> x_{n+1}\\ x_n<br /> \end{array}<br /> \right)<br /> =<br /> \left(<br /> \begin{array}{cc}<br /> \frac{3}{2} &amp; -\frac{1}{2}\\ 1 &amp; 0<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{cc}<br /> x_{n}\\ x_{n-1}<br /> \end{array}<br /> \right)<br />

So, we get

\left(<br /> \begin{array}{cc}<br /> x_{n+1}\\ x_n<br /> \end{array}<br /> \right)<br /> =<br /> \left(<br /> \begin{array}{cc}<br /> \frac{3}{2} &amp; -\frac{1}{2}\\ 1 &amp; 0<br /> \end{array}<br /> \right)^n<br /> \left(<br /> \begin{array}{cc}<br /> x_1\\ x_0<br /> \end{array}<br /> \right)<br />

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

\left(<br /> \begin{array}{cc}<br /> x_{n+1}\\ x_n<br /> \end{array}<br /> \right)<br /> =<br /> \left(<br /> \begin{array}{cc}<br /> \frac{1}{2} &amp; 1\\ 1 &amp; 1<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{cc}<br /> \frac{1}{2^n} &amp; 0\\ 0 &amp; 1<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{cc}<br /> -2 &amp; 2\\ 2 &amp; -1<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{cc}<br /> x_1\\ x_0<br /> \end{array}<br /> \right)<br />

and thus

\left(<br /> \begin{array}{cc}<br /> x_{n+1}\\ x_n<br /> \end{array}<br /> \right)<br /> =<br /> \left(<br /> \begin{array}{cc}<br /> 2 - \frac{1}{2^n} &amp; \frac{1}{2^n}- 1\\ 2 - \frac{1}{2^{n-1}} &amp; \frac{1}{2^{n-1}}-1<br /> \end{array}<br /> \right)<br /> \left(<br /> \begin{array}{cc}<br /> x_1\\ x_0<br /> \end{array}<br /> \right)<br />

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}

 

[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?

 

[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})

 

[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?

 

[Miffymycat]

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

 

[PeroK]

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

 

[Miffymycat]

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

 

[PeroK]

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.