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Simple curve  not so simple function?!! 
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#1
May2514, 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! 


#3
May2514, 06:12 AM

P: 40

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



#4
May2514, 06:28 AM

PF Gold
P: 1,516

Simple curve  not so simple function?!!
You can try finding the nth term of the sequence(That's what I do).



#5
May2514, 08:22 AM

Math
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Thanks
PF Gold
P: 39,682

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] 


#6
May2514, 08:30 AM

Mentor
P: 18,346

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_{n1}) = \frac{3}{2}x_n  \frac{1}{2}x_{n1}[/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_{n1} \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^{n1}} & \frac{1}{2^{n1}}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] 


#7
May2514, 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? 


#8
May2514, 09:19 AM

P: 446

You can manipulate Micromass's approach to get, for example:
[tex]f(x) = 16  2^{4x}[/tex] And, if f(1) = a, then: [tex]f(x) = a(2  2^{1x}) = 2a(1  \frac{1}{2^x})[/tex] 


#9
May2514, 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?



#10
May2514, 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?



#11
May2514, 10:04 AM

P: 446

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



#12
May2514, 10:15 AM

P: 40

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



#13
May2514, 10:23 AM

P: 446




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