Next number in sequence problem/solving system of linear equations.

0 12 10 0 -12 -20

I calculate the differences, and find it's a 3rd degree polynomial.

An^3+Bn^2+Cn+D

So I know when n=1 the above equation=0, and when n=2, 12, and so on.

So I come up with these 4 equations

a + b+ c+d=0

8a+ 4b+2c+d=12

27a+ 9b+3c+d=10

64a+16b+4c+d=0

Now I have to solve this system of linear equations. I just keep trying and getting wrong answers. What's the best way to attack this problem?

Re: Next number in sequence problem/solving system of linear equations.

Use technology, continual substitution over a system of this order leaves too much room for error.

Re: Next number in sequence problem/solving system of linear equations.

Quote:

Originally Posted by

**StudentMCCS** a + b+ c+d=0 [1]

8a+ 4b+2c+d=12 [2]

27a+ 9b+3c+d=10 [3]

64a+16b+4c+d=0 [4]

Do subtractions: [2]-[1], [3]-[1] and [4]-[1]; will leave

7a + 3b + c = 12 [5]

26a + 8b + 2c = 10 [6]

63a + 15b + 3c = 0 [7]

Divide [6] by 2 and [7] by 3:

7a + 3b + c = 12 [5]

13a + 4b + c = 5 [6]

21a + 5b + c = 0 [7]

Get it? Take over...

Re: Next number in sequence problem/solving system of linear equations.

Since you have already calculated the differences, and seen that the "third differences" are all the same, I would suggest using "Newton's divided difference formula". The value and differences "at 0" are 0, 12, -14, 6 so the polynomial is

$\displaystyle 0+ 12x- (14/2!)x(x- 1)+ (6/3!)x(x-1)(x-2)= 12x- 7x(x- 1)+ x(x-1)(x-2)$$\displaystyle = 12x- 7x^2+ 7x+ x^3- 3x^2+ 2x= x^3- 10x^2+ 21x$

Check: $\displaystyle 0^3- 10(0^2)+ 21(0)= 0$, [itex]1^3- 10(1^2)+ 21(1)= 12[/tex], $\displaystyle 2^3- 10(2^2)+ 21(2)= 10[/itex], [itex]3^3- 10(3^2)+ 21(3)= 0$, $\displaystyle 4^3- 10(4^2)+ 21(4)= -12$, $\displaystyle 5^3- 10(5^2)+ 21(5)= -20$.

Re: Next number in sequence problem/solving system of linear equations.

using matrices will be helpful for solving such equations

Re: Next number in sequence problem/solving system of linear equations.