recursive math funcction

**whatman** · 07-29-2003

hi, i am writing a function to return the value of a function A(n), where:
A(n) = 2A(n-1) - 3A(n-2) for n>1

when n = 0, A = a and when n = 1, A = 2.

I've came up with the function below, but I am getting different results to when I am work out the answers by hand.

Code:

int A(int n)
{	

	if(n==0||n==1)
	{
		n +=1;
	}
	
	else
	{
		n = (2*A(n-1)) - (3*A(n-2));
	}

	return n;
}

I am not sure where I am going wrong and would appreciate a push in the right direction.
The answers I am getting from the program are -4, -11, -10 and 13 for n = 3 to 6 and my answers from working them out are-4, -29, -192 and -419. I think it has something to do with multiplying the function, and its probably really simple, but I've spent a good six hours trying and testing code and I've ran out of ideas.
Thanks.

**money?** · 07-29-2003

>when n = 0, A = a
What do you mean A = a? what's "a" ?

**whatman** · 07-29-2003

Sorry, its meant to be n = 0, A = 1.
Thanks for the help Salem, I shall try it. I did try it when I started, but the program kept crashing, but I will try again and let you know.

**money?** · 07-29-2003

Here's how i would have made it:

Code:

int A( int n )
{
	if( n==0 || n==1 )
		return n;
	else
		return( 2*A(n-1) - 3*A(n-2) );
}

**Dave_Sinkula** · 07-29-2003

Originally posted by whatman
The answers I am getting from the program are -4, -11, -10 and 13 for n = 3 to 6 and my answers from working them out are-4, -29, -192 and -419.

Are you sure that the answers you are working out are correct?

Code:

A(0) = 1, A(1) = 2, A(n) = 2A(n-1) - 3A(n-2) for n > 1 :
A(2) = 2 * A(1) - 3 * A(0) = 2 * (  2) - 3 * (  1) =   4 -   3 =   1
A(3) = 2 * A(2) - 3 * A(1) = 2 * (  1) - 3 * (  2) =   2 -   6 =  -4
A(4) = 2 * A(3) - 3 * A(2) = 2 * ( -4) - 3 * (  1) =  -8 -   3 = -11
A(5) = 2 * A(4) - 3 * A(3) = 2 * (-11) - 3 * ( -4) = -22 - -12 = -10
A(6) = 2 * A(5) - 3 * A(4) = 2 * (-10) - 3 * (-11) = -20 - -33 =  13

**whatman** · 07-29-2003

Thanks guys, it turns out it was my maths. I was substituting the equation as opposed to the value of the equation, which got complicated. My bad, but thanks again.

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