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Deriving any Exponential Formula from a Pattern

Date: 01/18/2003 at 01:50:18
From: Lon
Subject: Deriving any exponential formula from a pattern

Dr. Math,

I write a lot of software, and often I identify growth/decay patterns 
that could easily be represented as exponetial formulas. But I always 
struggle with deriving them and end up turning to other people for 

I would like to know the standard process for deriving an exponential 

For example, I wrote a simple function which fades an image. The 
transparency of the image is a percent, where 100 is opaque. The 
simple formula I wrote runs in a loop, where:
  t = Target transparency (or alpha)
  c = Current transparency
  s = Speed (or rate) of fade
  F = final value

The formula is:
  F = c + ((t - c)/s)

I think that in the second iteration it would look like:
  F = (c + ((t - c)/s)) + ((t - (c + ((t - c)/s)))/s)

Please help!

Date: 01/18/2003 at 23:12:43
From: Doctor Peterson
Subject: Re: Deriving any exponential formula from a pattern

Hi, Lon.

As I understand it, your "F" means the next transparency, not really 
the "final" value; you could write the iteration as

  c[n+1] = c[n] + (t - c[n])/s

This can be simplified to

  c[n+1] = (1 - 1/s)c[n] + t/s

If we set

  A = 1 - 1/s  and  B = t/s

this becomes

  c[n+1] = Ac[n] + B

Repeating this, starting from a given the first value c[0] = f, we 

  c[0] = f
  c[1] = Af+B
  c[2] = A(Af+B)+B = A^2f + (A+1)B
  c[3] = A(A^2f + (A+1)B)+B = A^3f + (A^2+A+1)B
  c[4] = A(A^3f + (A^2+A+1)B)+B = A^4f + (A^3+A^2+A+1)B

Clearly (I won't bother to prove it)

  c[n] = A^nf + (1+A+A^2+...+A^(n-1))B

The geometric series can be summed using a familar formula, giving us

  c[n] = A^nf + B(A^n-1)/(A-1)

       = A^n(f + B/(A-1)) - B/(A-1)

Replacing A and B with their values, B/(A-1) = -t, and

  c[n] = (1 - 1/s)^n (f - t) + t

This clearly does what you want: it approaches t asymptotically (never 
actually getting there) in an exponential manner, starting at f. 
Your "speed" s, of course, is not really a speed of approach, but 
the reciprocal of the ratio by which the deviation from the target 
is decreased at each step. Setting s = 1/r,

  c[n] = t + (1-r)^n (f - t)

That should be the formula you want. And normally we would start with 
this, and form an iterative approach from it, rather than derive it 
from the iterative formula the way I have.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 

Date: 01/18/2003 at 23:27:06
From: Lon
Subject: Deriving any exponential formula from a pattern

Dr. Peterson,

Thank you so much for that awesome derivation! I see very clearly how 
you derived it, and I think it will solve my current dilemma. However, 
the general process is still not clear to me.

For example, how did you know to sub out the (1 -1/s)and the (t/s) ?

Date: 01/19/2003 at 22:48:50
From: Doctor Peterson
Subject: Re: Deriving any exponential formula from a pattern

Hi, Lon.

I defined temporary variables A and B merely to make the equation look 
simpler and allow me to (a) see more clearly what had to be done, and 
(b) avoid typing too many complicated expressions. The work could have 
been done without that substitution, but when you have done this sort 
of algebra enough you learn how to save work.

Also, my derivation could have been done more simply if I had chosen 
to use more insight into the problem (including your own insight that 
it is exponential) from the beginning, rather than just mechanically 
working out what successive iterations would do. Here's an 

Your iterative formula was:

  F = c + (t - c)/s

I write this iteratively as

  c[n+1] = c[n] + (t - c[n])/s

Your purpose here is to reduce the distance from t to c by the same 
ratio at each step, which is what makes this exponential. We can 
rewrite the equation to emphasize this:

  t - c[n+1] = t - c[n] - (t - c[n])/s

             = (t - c[n])(1 - 1/s)

So the difference t - c is multiplied by 1 - 1/s at each iteration.

Therefore if we temporarily define a new variable

  d[n] = t - c[n]

our iteration is

  d[n+1] = d[n] (1 - 1/s)

and we can easily see that

  d[n] = d[0] (1 - 1/s)^n

because we are multiplying by the same factor repeatedly.

Putting this back in terms of c,

  c[n] = t - d[n] = t - (t - c[0])(1 - 1/s)^n

That is the desired formula, equivalent to what I gave you last time.

- Doctor Peterson, The Math Forum 
Associated Topics:
High School Sequences, Series

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