[superprismatic]

The numbers 1, Y, and Y^2, Y a real

number, are used as the basis for a

sequence in which every value after the

first three is the average of the

previous three numbers. This is

continued indefinitely and the sequence

is found to converge to the value 321.

Find all possible values for Y.

SUPERPRISMATIC NOTE: This is much more

of a challenge for the programming

solvers out there than Penney Puzzle

#7 from yesterday was!

[Guest]

How is this one different from the earlier question? The only difference I noticed was Y being addressed as "a real number" rather than "a positive integer" in the previous case. If the answer to the previous case was 25, why would it be any different now?

I'm not a programmer so may be I am missing something in the question!!

[superprismatic]

How is this one different from the earlier question? The only difference I noticed was Y being addressed as "a real number" rather than "a positive integer" in the previous case. If the answer to the previous case was 25, why would it be any different now?

I'm not a programmer so may be I am missing something in the question!!

There is only one answer in the positive integers. But there may be more in the reals.

In fact, there is at least one other.

[Guest]

There is only one answer in the positive integers. But there may be more in the reals.

In fact, there is at least one other.

Ah, silly me! Of course there could (read would) be value < 0 which would satisy this condition.

I have a hunch that the negative value should be close to the positive value, the only question is how close!

But the dificult part is that as easy as my previous method for part 1 was, it won't come in handy if the second solution is not an integer!

Edited August 6, 2009 by DeeGee

[Guest]

Just to add a little spice, may be a part 3 for complex numbers would be in order!

[superprismatic]

Just to add a little spice, may be a part 3 for complex numbers would be in order!

By that remark, I guess you're intrigued by this little problem. I never thought about it, but I guess if the convergent value were something like 321+87i, there may be some fruitful puzzling there! What next? Quaternions?

[HoustonHokie]

Y could be -25.666666....

[Guest]

By that remark, I guess you're intrigued by this little problem. I never thought about it, but I guess if the convergent value were something like 321+87i, there may be some fruitful puzzling there! What next? Quaternions?

Y could be -25.666666....

How did you solve? Was it through a program or mathematical analysis?

[Guest]

How did you solve? Was it through a program or mathematical analysis?

I got as far as seeing:

nth term = (ax^2 + bx + c)/d

a = x^2 coeef: 1 4 16 37 121

b = x coeef: 1 4 7 28 85

c = addend: 1 1 4 16 37

d = divisor: 3 9 27 81 243

divisor = 3^n

coeeficients: nth coeeficient = 9 * (n-3) coefficent + 3*(n-2) coeefcient + (n-1) coeeficent.

Then I threw my hands up in the air,

and went back to work

[Guest]

I got as far as seeing:

nth term = (ax^2 + bx + c)/d

a = x^2 coeef: 1 4 16 37 121

b = x coeef: 1 4 7 28 85

c = addend: 1 1 4 16 37

d = divisor: 3 9 27 81 243

divisor = 3^n

coeeficients: nth coeeficient = 9 * (n-3) coefficent + 3*(n-2) coeefcient + (n-1) coeeficent.

Then I threw my hands up in the air,

and went back to work

Same here! I also couldn't find any general expression for the coefficients of nth term (after the first three).

Another observation was that a+b+c of nth term = 3ⁿ but that was all

Edited August 6, 2009 by DeeGee

[Guest]

Same here! I also couldn't find any general expression for the coefficients of nth term (after the first three).

Another observation was that a+b+c of nth term = 3ⁿ but that was all

One other aha moment....isn't this a form of a Cauchy sequence????

(I wouldn't know how that could help...and it's just sitting at the back of my brain)

(Edit: Add parenthetical)

Edited August 6, 2009 by tpaxatb

[HoustonHokie]

How did you solve? Was it through a program or mathematical analysis?

spreadsheet I'm working on a mathematical analysis to make it more rigorous. From what I've seen (and I can tell others have seen it too), the general form of the converging function is f(n) = [a(n) + b(n)Y + c(n)Y²]/d(n), where a(n), b(n), c(n), and d(n) are distinct numerical sequences in their own right. The patterns become pretty regular for n>5, so I was thinking that if I could take limits on a(n), b(n), c(n), and d(n) as n approached infinity, I might be able to set that limit equal to 321 and directly solve for Y. Then you would obtain the solution from last time and the one I just posted, and possibly any complex solutions that might exist. But I've yet to define each of these sequences in a simple manner - just haven't spent the time on it yet. Obviously, d(n) is rather simple: d(n) = 3^n-3. And a(n) = c(n-1), so there's some simplification there. Just need to work it out...

[HoustonHokie]

Well, here's something interesting:

a(n), b(n), c(n), and d(n) can be approximated by exponential functions of the form f(n) = Ae^Bx. So you can approximate that:

a(n) = 0.006187 e ^1.0985n

b(n) = 0.01226 e ^1.0989n

c(n) = 0.01858 e ^1.0985n

d(n) = e ^1.0986n

I did this using regression analysis after taking the natural log of each of the numbers in the various series up to n = 35. The R² value was 0.999999 for a, b, & c, and 1.0 for d. Noticing that the exponents were all very close to 1.0986n, I decided to check what happens if I removed some of the lower values of n from the equation. As I suspected, the exponents eventually all became 1.0986n (1.0986 being the natural log of 3). So, we have 4 parallel functions when they are all graphed on a semi-logarithmic scale. I'm thinking the answer somehow has to do with the intercepts of the various equations, because the slopes are all the same. But I am not quite there yet.

[HoustonHokie]

A correction to my earlier post, which I can no longer edit...

a(n), b(n), c(n), and d(n) can be approximated by exponential functions of the form f(n) = Ae^Bx. So you can approximate that:

a(n) = 0.16666 e ^1.0986n

b(n) = 0.33333 e ^1.0986n

c(n) = 0.5 e ^1.0986n

d(n) = e ^1.0986n

I realized that I had messed up one of the calculations, and it looks much prettier now. I still think the answer lies with the intercepts of these functions.

And it does...

If you set up an equation:

.167 + .333Y + .5Y² = 321, then Y = 25 or -25.666 will solve.

Edited August 6, 2009 by HoustonHokie

[Quantum.Mechanic]

This is trivial to program, even easier if you have a symbolic solver, or even a numeric one such as (gulp) MS Excel "goal seek". (Granted, you have to know there's a negative solution for some of these to find it, but still...)

The only problem after that is machine resolution, and determining the exact answer.

[Guest]

right but i dont know y

so at first glance it look like (when i say y= i mean the coeff)

y²= summation( i 1->inf) 1/3ⁱ=1/2

y=(summation( i 1->inf)1/3ⁱ)-1/6=1/3 (its 1/3 starts off being considered only twice to y² 3)

integer=(summation i 1->inf)1/3ⁱ-1/3=1/6 (it starts off being considered only once to y²3)

now on second glance there are a couple of logic jumps but then i thought of it this way and it works

now at start y^2 is consider 3 times for ever 2 y and 1 int. So after averageing this number with other averages of this number y would be 2/3 as big and int would be 1/3 as big.

the y²=1/2 I cant really prove but it logically makes sense(at least to me)

i tried recursion but my brain hurts currently, so this is the best i got now

[Quantum.Mechanic]

Sorry, I didn't do the table formatting right, I'll have to work on that.

Using my noggin' and a spreadsheet, I figured it out. Here's the generic sequence:

S0 = c

S1 = x

S2 = x^2

S3 = (x^2 + x + 1)/3

S4 = ( (x^2 + x + 1)/3 + x^2 + x )/3

= ( (4x^2 + 4x + 1)/3 )/3

If we rewrite this as

S4 = ( Aix^2 + Bix^2 + C ) / D

then watch the pattern of A,B,C,D, we get a table like this:

S Ai Bi Ci Di

0 1

1 1

2 1

3 1 1 1 3

4 4 4 1 9

5 16 7 4 27

6 37 28 16 81

7 121 85 37 243

Note that each Ai = A[i-1] + 3A[i-2] + 9A[i-3], same for Bi and Ci.

Now compute Bi/Ai, Ci/Ai, and Di/Ai:

S Ai Bi Ci Di b/a c/a d/a

0 1

1 1

2 1

3 1 1 1 3 1 1 3

4 4 4 1 9 1 0.25 2.25

5 16 7 4 27 0.438 0.250 1.688

6 37 28 16 81 0.757 0.432 2.189

7 121 85 37 243 0.702 0.306 2.008

where the ratios converge pretty quickly to this:

2/3 1/3 2

With T = target,

T = (x^2 + 2x/3 + 1/3)/2

and rearranging:

3y^2 + 2y - (6*T-1) = 0

For T = 321, this yields y = (25,-25-2/3).

Edited August 6, 2009 by Quantum.Mechanic

[superprismatic]

Sorry, I didn't do the table formatting right, I'll have to work on that.

Using my noggin' and a spreadsheet, I figured it out. Here's the generic sequence:

S0 = c

S1 = x

S2 = x^2

S3 = (x^2 + x + 1)/3

S4 = ( (x^2 + x + 1)/3 + x^2 + x )/3

= ( (4x^2 + 4x + 1)/3 )/3

If we rewrite this as

S4 = ( Aix^2 + Bix^2 + C ) / D

then watch the pattern of A,B,C,D, we get a table like this:

S Ai Bi Ci Di

0 1

1 1

2 1

3 1 1 1 3

4 4 4 1 9

5 16 7 4 27

6 37 28 16 81

7 121 85 37 243

Note that each Ai = A[i-1] + 3A[i-2] + 9A[i-3], same for Bi and Ci.

Now compute Bi/Ai, Ci/Ai, and Di/Ai:

S Ai Bi Ci Di b/a c/a d/a

0 1

1 1

2 1

3 1 1 1 3 1 1 3

4 4 4 1 9 1 0.25 2.25

5 16 7 4 27 0.438 0.250 1.688

6 37 28 16 81 0.757 0.432 2.189

7 121 85 37 243 0.702 0.306 2.008

where the ratios converge pretty quickly to this:

2/3 1/3 2

With T = target,

T = (x^2 + 2x/3 + 1/3)/2

and rearranging:

3y^2 + 2y - (6*T-1) = 0

For T = 321, this yields y = (25,-25-2/3).

Well, actually nothing ever actually converges until way, way out there at infinity. So, you are actually relying on a guess that this thing converges to (1+2*x+3*x^2)/6. If it differed by a small amount from this, the solutions may have been 25.00001, -25.6666678 say.

Edited August 6, 2009 by superprismatic

[Guest]

Analytic solution in word document

What a Sequence.doc

[bushindo]

Analytic solution in word document

What a Sequence.doc

Marvelous work. Well done.

[Guest]

Marvelous work. Well done.

Because of my layout of coefficients, i didn't see the fact that c_n = a_n-1