Thread: Finding A Polynomial Model

Hi everybody!

I'm provided a table of values:

x: 0 1 2 3 4 5 6
y: 0 5 14 33 68 125 210

I used the Polynomial difference theorum to find that by subtracting consecutive y values to find the difference 3 times, they equalled out to 6's. So, I knew that it was degree 6, or ax^3 + bx+2 + cx + d.

So far so good.

Next, I'm instructed to use Matrices to find the coefficients. This is what's giving me trouble.

I set up the system:

T(4): 33 = 64a + 16b + 4c + d
T(3): 14 = 27a + 9b + 3c + d
T(2): 5 = 8a + 4b + 2c + d
T(1): 0 = 0a + 0b + 0c + 0d


Puting that into matrices, I have:

A: (4 by 4)

64, 16, 4, 1
27, 9, 3, 1
8, 4, 2, 1
0, 0, 0, 0

---

B: (4 by 1):

33,
14,
5,
0


When I multiply the inverse of matrix A times matrix B, I get an error message: "ERROR 03 SINGULAR MAT"


I know I can find the coefficients by using STAT, typing in the x values and the y values and using cubic regression, but I'm supposed to solve this using matrices. Any idea why this isn't working?

Thanks,
Ethan


Edit:
Sorry about the formatting, I had the data values spaced out nicely, but they clumped back together after I posted!

T(1): 0 = 0a + 0b + 0c + 0d

This is incorrect it should be

T(1): 0 = 0a + 0b + 0c + d

or d =0

Thanks,

That was my bad.

If I replace that 0 with a 1 in the matrix, and I multiply the inverse of A times B, I get:

a = .70833
b = -1.375
c = 2.416
d = 0

So, that's telling me my cubic regression should be:
y = .70833x^3 - 1.375x^2 + 2.416x

However, when I put the x values and the y values into Stat under L1 and L2, and I choose cubic regression, I get much NICER values:

a= 1
b=-1
c=5
d=0

So, according to STAT, my regression equation should be:

y = 1x^3 - 1x^2 + 5x

If they are both degree 3, why am I getting different results? Again, I think I am doing something wrong when I solve by Matrices...

Thanks for the help,
Ethan

T(4): 33 = 64a + 16b + 4c + d
T(3): 14 = 27a + 9b + 3c + d
T(2): 5 = 8a + 4b + 2c + d
T(1): 0 = 0a + 0b + 0c + 0d


The Equations should be ...

T(3): 33 = 27a + 9b + 3c + d
T(2): 14 = 8a + 4b + 2c + d
T(1): 5 = a + b + c + d
T(0): 0 = 0a + 0b + 0c + 1

Thank you very much!