Comments on: FY. Weights in a Row http://mathfactor.uark.edu/2009/07/fy-weights-in-a-row/ The Math Factor Podcast Site Sat, 21 Nov 2009 21:45:21 -0600 http://wordpress.org/?v=2.8.5 hourly 1 By: Brian http://mathfactor.uark.edu/2009/07/fy-weights-in-a-row/comment-page-1/#comment-568 Brian Tue, 21 Jul 2009 05:38:45 +0000 http://mathfactor.uark.edu/?p=706#comment-568 Ok after some thought, I found some methods for figuring this out. Your sequence is powers of two, where the nth term is 2^n. I postulate here that the relationship described above is related to the exponetial curve. So I focused my exploration there thinking maybe this would apply to other exponential functions. So first, I looked at the simplest one (this will be tricky to write without super/subcripts): kx^0, kx^1, kx^2, ... kx^n where k is a constant coefficient and x raised to the nth power for each term. Applying the 1/3 & 2/3 condition, we get the following: kx^n = (2/3)kx^(n-1) + (1/3)kx^(n+1) Settles down to... 3kx = 2k + kx^2 k(x-1)(x-2) = 0 x = 1 or 2 Which leaves us with either all terms are equal (k) or x=2 (our original sequence). Both of these solutions work. Next, I looked at the inverse to see if I could get any alternate variations k, kx^-1, kx^-2, kx^-3, ... kx^-n Does this work? Let's work it thorugh. First, the initial condition: x^-n = (2/3)kx^1-n + (1/3)kx^(1+n) Settles down to ... 3kx = 2kx^2 +1k k(1-x)(1-2x) = 0 x = -1 or -2. x = -1 was already proved in the last problem, but just in case you didn't get it: sequence: k, k, k, k, .... k k/3 + 2k/3 = k Now let's look at -2 k, k/2, k/-3, k/4, k/-5... That doesn't work because if x(n) is positive, x(n-1) and x(n+1) are both going to be negative. Ok after some thought, I found some methods for figuring this out.
Your sequence is powers of two, where the nth term is 2^n. I postulate here that the relationship described above is related to the exponetial curve. So I focused my exploration there thinking maybe this would apply to other exponential functions.
So first, I looked at the simplest one (this will be tricky to write without super/subcripts):
kx^0, kx^1, kx^2, … kx^n
where k is a constant coefficient and x raised to the nth power for each term. Applying the 1/3 & 2/3 condition, we get the following:
kx^n = (2/3)kx^(n-1) + (1/3)kx^(n+1)
Settles down to…
3kx = 2k + kx^2
k(x-1)(x-2) = 0
x = 1 or 2
Which leaves us with either all terms are equal (k) or x=2 (our original sequence). Both of these solutions work.
Next, I looked at the inverse to see if I could get any alternate variations
k, kx^-1, kx^-2, kx^-3, … kx^-n
Does this work? Let’s work it thorugh. First, the initial condition:
x^-n = (2/3)kx^1-n + (1/3)kx^(1+n)
Settles down to …
3kx = 2kx^2 +1k
k(1-x)(1-2x) = 0
x = -1 or -2.
x = -1 was already proved in the last problem, but just in case you didn’t get it:
sequence: k, k, k, k, …. k
k/3 + 2k/3 = k
Now let’s look at -2
k, k/2, k/-3, k/4, k/-5…
That doesn’t work because if x(n) is positive, x(n-1) and x(n+1) are both going to be negative.

]]> By: Sam Martin http://mathfactor.uark.edu/2009/07/fy-weights-in-a-row/comment-page-1/#comment-561 Sam Martin Tue, 07 Jul 2009 11:11:36 +0000 http://mathfactor.uark.edu/?p=706#comment-561 Doh! Comical and blindingly obvious handwriting fail. Thanks! Doh! Comical and blindingly obvious handwriting fail. Thanks!

]]> By: strauss http://mathfactor.uark.edu/2009/07/fy-weights-in-a-row/comment-page-1/#comment-555 strauss Mon, 06 Jul 2009 21:20:21 +0000 http://mathfactor.uark.edu/?p=706#comment-555 <p>I think we have the weights right: </p> <p> </p> <p>4 = 2/3 * 2 + 1/3 * 8</p> <p>8 = 2/3 * 4 + 1/3 * 16</p> <p>I think we should talk about linear recurrences! One of those things we just need to say something about! This example isn't in just that form, but as Steven Noble probably spotted, it can be (should be?) thought of in that way.</p> <p> </p> I think we have the weights right: 

 

4 = 2/3 * 2 + 1/3 * 8

8 = 2/3 * 4 + 1/3 * 16

I think we should talk about linear recurrences! One of those things we just need to say something about! This example isn’t in just that form, but as Steven Noble probably spotted, it can be (should be?) thought of in that way.

 

]]> By: Sam Martin http://mathfactor.uark.edu/2009/07/fy-weights-in-a-row/comment-page-1/#comment-553 Sam Martin Fri, 03 Jul 2009 16:33:44 +0000 http://mathfactor.uark.edu/?p=706#comment-553 btw - it's a cracking job you guys do. Keep it up! btw – it’s a cracking job you guys do. Keep it up!

]]> By: Sam Martin http://mathfactor.uark.edu/2009/07/fy-weights-in-a-row/comment-page-1/#comment-552 Sam Martin Fri, 03 Jul 2009 16:32:25 +0000 http://mathfactor.uark.edu/?p=706#comment-552 Urm. Not to be picky or anything, but those weights dont' appear to match that sequence? Perhaps you mean 2x previous?   Or have my brains completely fallen out? I suspect I'm missing something... Urm. Not to be picky or anything, but those weights dont’ appear to match that sequence? Perhaps you mean 2x previous?
 
Or have my brains completely fallen out? I suspect I’m missing something…

]]> By: Steven H. Noble http://mathfactor.uark.edu/2009/07/fy-weights-in-a-row/comment-page-1/#comment-550 Steven H. Noble Thu, 02 Jul 2009 23:56:26 +0000 http://mathfactor.uark.edu/?p=706#comment-550 I'll be interested to see if you use this question as a springboard to talk about the properties of linear recurrences. I’ll be interested to see if you use this question as a springboard to talk about the properties of linear recurrences.

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