# Determining polynomials

1. Jul 28, 2004

### Mathman23

Hi

I got a Linear Algebra question.

I'm suppose to find two polynomials p1 and p2 both of highest degree 3, and which satisfies the following:

p1(-1) = 1

p1'(-1) = 0

p2(1) = 3

p2'(1) = 0

p1(0) = p2(0)

p1'(0) = p2'(0)

I hope that there is somebody out there who can explain to me how I do that ?

Fred

2. Jul 28, 2004

### arildno

Hint:
If your polynomials are of highest degree 3, they may be written as:
$$p_{1}=a_{1}x^{3}+b_{1}x^{2}+c_{1}x+d_{1}$$
$$p_{2}=a_{2}x^{3}+b_{2}x^{2}+c_{2}x+d_{2}$$

The letters are the constants you must determine with the aid of the given relations (you will gain a matrix system you''ll need to solve)

3. Jul 28, 2004

### Mathman23

If I understand You correctly.

I take these two polynomials p1 and p1'.

I then insert the terms into the polynomial equations? So I get the following

p1 = a1*(-1)^3 + b2*(-1)^2 + c1*(-1) +d = 1

p1' = 3*a1*(-1)^2 + 2*b1*(-1) +c1 = 0

I use a matrix system to solve these two equations so they satisfy the given conditions?

And then find the variables for p2 ?

Sincerely
Fred

4. Jul 28, 2004

### matt grime

you have far more than just those two equations. for instance the last two tell you that d_1=d_2 and c_1=c_2

5. Jul 28, 2004

### Mathman23

You mean I have these equations:

p1 = a1*(-1)^3 + b1*(-1)^2 + c1*(-1) +d1 = 1

p1' = 3*a2*(-1)^2 + 2*b2*(-1) +c2 = 0

p2 = a1*(1) ^3 +b1 * (1) ^2 + c1*(1) + d1 = 3

p2' = 3*a2*(1)^2 + 2*b2*(-1) +c2 = 0

a1*(0)^3 + b1*(0)^2 + c1*(0) +d1 = a1*(0) ^3 +b1 * (0) ^2 + c1*(0) + d1

3*a2*(0)^2 + 2*b2*(0) +c2 = 3*a2*(0)^2 + 2*b2*(0) +c2

and solve these as one system of equations ?

/Fred

6. Jul 28, 2004

### matt grime

not quite since you appear to have several mistakes in there, let us use arildno's notation properly

1=-a_1+b_1-c_1+d_1
from (p_1(-1)=1)

0=3a_1-2b_1+c_1
from (p_1'(-1)=0)

3=a_2+b_2+c_2+d_2
from (p_2(1)=3)

0=3a_2+2b_2+c_2
from (p_2'(1)=0)

d_1=d_2
from (p_1(0)=p_2(0))

c_1=d_1
from (p_1'(0)=p_2'(0))

note you have 8 unkowns and 6 equations

7. Jul 28, 2004

### Mathman23

okay Thank You.

Fred

8. Jul 28, 2004

### Mathman23

I get the following solution values:

a1, a2 = -1/2

b1, b2 = 3/2

c1,c2 = 0

d1, d2 = 2

Can anybody tell me if they are correct ?

Thanks again.

Sincerely

Fred

9. Jul 29, 2004

### matt grime

But you've all these identities that allow you to check your answer by yourself, you don't need us to verify it.