Question

Solve Recursive Equations

Asked by: Telavian

Is there a general way to get a single equation for a recursive formula?

For instance, the equation X = 2x - 1 can be written as X = (2^N)x - (2^N-1)-1 to get a single equation. Is a general way to solve possible? I solved the above based on expansion, but would like a general way to solve.

Also, for the following loop system is a general solution possible?
x = y + z
z = y + x
y = 2 * x

The items can be written as follows to get them dependant on previous iteration only:
x = y + z
y = 2y + z
z = 2y + 2z

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Asked On: 2009-08-11 at 10:34:01ID24643935
Tags: Math
,; Equation
,; Puzzle
Topics: Math & Science
,; Puzzles & Riddles
,; Algorithms
Participating Experts: 2
Points: 500
Comments: 11

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Answers

by: InteractiveMindPosted on 2009-08-11 at 12:25:19ID: 25072404

First of all, I get a slightly different answer to you for your first example: X=(2^N)x - 2^(N-1).

With regards to a general solution: there are general approaches to recursive equations depending on their properties. Here's a list of solutions: http://en.wikipedia.org/wiki/Recurrence_relation#Solving

In the case of a system, let's first use the notation:

x_(n+1) = y_n + z_n
z_(n+1) = y_n + x_n
y_(n+1) = 2x_n

Then x_(n+1) = 2x_(n-1) + y_(n-1) + x_(n-1) = 4x_(n-1) + 2x_(n-2), which is equivalent to x_(n+3) = 3x_(n+1)+2x_n, which is a linear homogeneous recurrence relation (c.f. wiki)

by: InteractiveMindPosted on 2009-08-11 at 12:26:17ID: 25072418

(Then similarly for the y and z variables)

by: aburrPosted on 2009-08-11 at 15:48:49ID: 25074257

"Also, for the following loop system is a general solution possible?
x = y + z
z = y + x
y = 2 * x

Not only is no general solution possible but NO solution is possible. The equations are contradictory

by: TelavianPosted on 2009-08-11 at 16:16:14ID: 25074426

aburr, this is a system of recursive equations.

by: aburrPosted on 2009-08-11 at 19:29:47ID: 25075281

"aburr, this is a system of recursive equations."
you are right. How do you get the loop started?

by: TelavianPosted on 2009-08-11 at 20:52:23ID: 25075621

Each variable has some initial setting and then they loop from there:

int x = 1, y = 1, z = 1;
 
while (true) //Or some condition
{
  x = y + z;
  z = y + x;
  y = 2 * x;
}

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by: TelavianPosted on 2009-08-11 at 20:56:52ID: 25075642

InteractiveMind, just for completness can you post the solution for x, y, and z?

Thanks.

by: aburrPosted on 2009-08-11 at 21:06:06ID: 25075689

I am probably still confused by the loop but it seems to me that
the nth value of x, y, and z all = 2^n. If that is not so, do not bother to correct me as I probably do not have anything useful to contribute to this question.

by: InteractiveMindPosted on 2009-08-12 at 12:27:53ID: 25081960

>InteractiveMind, just for completness can you post the solution for x, y, and z?

y is found using the same method as x:

x_(n+3) - 3x_(n+1) - 2x_n = 0
y_(n+3) - 3y_(n+1) - 2y_n = 0

Finding z by rearrangement seems quite tedious though, but there's a slight trick: recall that z_(n+1)=x_n+y_n. So if we add the above two equations, we get:

[x_(n+3)+y_(n+3)] - 3[x_(n+1)+y_(n+1)] - 2[x_n+y_n] = 0

which is

z_(n+4) - 3z_(n+2) - 2z_(n+1) = 0

or

z_(n+3) - 3z_(n+1) - 2z_n = 0.

In other words, all three variables have the same solution. Which, as aburr correctly points out, has the solution x_n=y_n=z_n=2^n.

BTW, your code is not quite right.

int last_x = 1, last_y = 1, last_z = 1;
int x, y, z;
 
while (true) //Or some condition
{
  x = last_y + last_z;
  z = last_y + last_x;
  y = 2 * last_x;
  
  last_x = x;
  last_y = y;
  last_z = z;
}

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by: TelavianPosted on 2009-08-12 at 12:55:24ID: 25082224

Something is not right.
First off, my answer of X = (2^N)x - (2^N-1)-1 is correct.

My code is correct and that is what I am trying to find.
Maybe there is a misunderstanding.

The initial system is a loop and so should be "executed" top to bottom then again.
It can be rewritten as follows to get it dependent on previous: (Is this wrong?)

x_n = y_(n-1) + z_(n-1)
z_n = y_(n-1) + x_(n-1)
y_n = 2x_(n-1)

If all three are the same then running the loop would yield the same answers which it doesn't.
My goal is to remove the loop and use a single equation for x, a single for y, and so on for z.

if x = y = z = 1 then I get

1:
x: 2
y: 4
z: 3

2:
x: 7
y: 14
z: 11

3:
x: 25
y: 50
z: 39

4:
x: 89
y: 178
z: 139

5:
x: 317
y: 634
z: 495

6:
x: 1129
y: 2258
z: 1763

7:
x: 4021
y: 8042
z: 6279

8:
x: 14321
y: 28642
z: 22363

9:
x: 51005
y: 102010
z: 79647

10:
x: 181657
y: 363314
z: 283667

This is the exact code to calculate the loop.

      int x = 1;
      int y = 1;
      int z = 1;
 
      for (int c = 0; c < 10; c++)
      {
        x = y + z;
        z = y + x;
        y = 2 * x;
 
        Console.WriteLine("{0}:\r\nx: {1}\r\ny: {2}\r\nz: {3}\r\n", c+1, x, y, z);
      }
 
      Console.ReadLine();

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by: InteractiveMindPosted on 2009-08-13 at 06:36:46ID: 25088022

Sorry, I had misunderstood you (or rather, made incorrect assumptions). Your code is not the same as,

x_n = y_(n-1) + z_(n-1)
z_n = y_(n-1) + x_(n-1)
y_n = 2x_(n-1),

instead, it is equivalent to:

x_n = y_(n-1) + z_(n-1)
z_n = y_(n-1) + x_n
y_n = 2x_n,

which can be solved in a similar way (below). Let me know if you have any other questions.

===Derivation===

First, rearrange to get a recursive equation for x:

x_(n+2) - 3x_(n+1) - 2x_n = 0

as described in the wiki article, we solve this by trying the ansatz x_n=k^n:

k^(n+2) - 3k^(n+1) - 2k^n = 0

which gives the quadratic

k^2 - 3k - 2 = 0

hence, k=[3 +/- sqrt(17)]/2. So the solution for x_n is:

x_n = A*([3+sqrt(17)]/2)^n + B*([3-sqrt(17)]/2)^n,

where A and B are constants which account for our initial conditions. Before we find A and B though, you must note that for our system we require that y_n=2x_n, and so x_0=y_0=1 is not possible (unless we take n=1 to be the first element in our series, but this makes it a lot trickier to solve A and B below), so let's take x_0=2, y_0=4, z_0=3 (which is your 3-tuple for n=1). Then:

x_0 = 2 = A + B, and
x_1 = y_0 + z_0 = 7 = A*([3+sqrt(17)]/2) + B*([3-sqrt(17)]/2).

Solving simultaneously for A and B we find:

A = 1 + 8/(2*sqrt(17)),
B = 1 - 8/(2*sqrt(17)).

Now that we know x_n, we find y_n and z_n as functions of x_n.

===Final Solution===

So putting it all together:

x_n = {1 + 8/(2*sqrt(17))}*([3+sqrt(17)]/2)^n + {1 - 8/(2*sqrt(17))}*([3-sqrt(17)]/2)^n,
y_n = 2*x_n, and z_n = 2*x_(n-1) + x_n,
z_n = 2x_(n-1) + x_n.

Important: Remember that the {1+/-8/(2*sqrt(17)} terms need to be changed if you change the initial variables!

===Test===

This solution may look complicated, but it's a fairly typical solution for this type of problem, moreover, it works!

Recall that because I chose x_0 to be your x_1 (and similarly for y_0 as y_1 et cetera), we'd expect x_n from the above equation to give x_(n+1) in the numbers you computed above.

So let's take n=6 (you get x_7=4021):

x_6 = {1 + 8/(2*sqrt(17))}*([3+sqrt(17)]/2)^6 + {1 - 8/(2*sqrt(17))}*([3-sqrt(17)]/2)^6 = 4021,

as expected. Then y_n and z_n are easily found as per above.

I've attached a Java implementation if it helps (I don't have C# installed on this computer).

class Test
{
	public static final double SQRT17 = Math.sqrt(17);
 
	public Test(int n)
	{
		System.out.println("n\tvariable\t(observed)\t(predicted)");
		
		int x = 2, y = 4, z = 3;
 
		System.out.println(0+"\tx\t\t"+x+"\t\t"+x(0));
		System.out.println("\ty\t\t"+y+"\t\t"+y(0));
		System.out.println("\tz\t\t"+z+"\t\t"+z(0));
 
		for (int c = 1; c < n; c++)
		{
			x = y + z;
			z = y + x;
			y = 2 * x;
 
			System.out.println(c+"\tx\t\t"+x+"\t\t"+x(c));
			System.out.println("\ty\t\t"+y+"\t\t"+y(c));
			System.out.println("\tz\t\t"+z+"\t\t"+z(c));
		}
 
	}
 
	public int x(int n)
	{
		return (int)((1.0 + 8.0 / (2.0 * SQRT17)) * Math.pow((3.0 + SQRT17) / 2.0, n)
				+ (1.0 - 8.0 / (2.0 * SQRT17))
				* Math.pow((3.0 - SQRT17) / 2.0, n));
	}
 
	public int y(int n)
	{
		return 2 * x(n);
	}
 
	public int z(int n)
	{
		return y(n - 1) + x(n);
	}
 
	public static void main(String[] a)
	{
		new Test(Integer.parseInt(a[0]));
	}
 
}

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Solve Recursive Equations

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