# Linear algebra interpolations and linear systems

1. Jul 6, 2010

### SpiffyEh

The problem statement, all variables and given/known data
Suppose you have a set S of three points in R2,
S = {(t1, p1), (t2, p2), (t3, p3)} ;
which you seek to interpolate with the quadratic polynomial p(t) = a0 + a1t + a2t2.

5.1. Interpolations and Linear Systems. Using S and p(t) define a linear system of equations in the a0; a1; a2 variables.

5.2. Existence and Uniqueness in Interpolation. Determine which of the following sets of points can be uniquely interpolated by p(t).

S1 = {(1, 12), (2, 15), (3, 16)}
S2 = {(1, 12), (1, 15), (3, 16)}
S3 = {(1, 12), (2, 15), (2, 15)}

2. Relevant equations

5.1 - Hint: This problem is meant to trick you. Clearly, p(t) is a nonlinear equation in the t-variable but once you have chosen a t-value then it is a
linear equation in the coecient variables. If you choose many t-values then you have many linear equations and now the tools of linear algebra apply.

5.2 - Your choice! We have two ways to approach this problem. First, you have a linear system and thus row-reduction and interpretation of pivot
structure. However, if you think about the geometry of the points and the possible graphs of quadratic polynomials you should be able to determine,
which of the points can be interpolated.

3. The attempt at a solution

I have no idea what to do or where to start. We never talked about interpolation. Can someone explain how to do this to me so that I can actually try it?

2. Jul 7, 2010

### Staff: Mentor

At the very least, please use t^2 to indicate that t is squared. The only way I could tell you meant t2 was that it's given that p(t) is a quadratic polynomial.
You have the formula for p(t) and three points on the graph of p: (t1, p1), (t2, p2), and (t3, p3). Surely you can think of something useful.
This one doesn't take a lot of math to do. Take a closer look at the three sets of points. One set can be discarded as a possible solution almost immediately, and another can be discarded not long after that.

3. Jul 7, 2010

### SpiffyEh

sorry about the t^2, I coppied it from a pdf and didn't realize it copied over as t2. I'm still unsure of what to do.... we never talked about interpolations or anything. Thats why i'm so lost

4. Jul 7, 2010

### Staff: Mentor

Don't worry about interpolation. Take a look at what I said in my previous post.

5. Jul 7, 2010

### SpiffyEh

I did try to read through and make undertanding of what you said but i'm still lost. I don't know why I don't get it but i really don't.

6. Jul 7, 2010

### hunt_mat

You have three points (x_1,y_1), (x_2,y_2), (x_3,y_3). Plug these into your quadratic formula, so for example, take (x_1,y_1), we get a_1+a_2*x_1+a_3*x_1^2=y_1. Each of the points will give a linear equation for a_1, a_2, a_3.

Three equations, three unknowns...

7. Jul 7, 2010

### SpiffyEh

ok so i end up with
p_1 = a_0 + a_1*t_1 + a_2*(T_1)^2
p_2 = a_0 + a_1*t_2 + a_2*(T_2)^2
p_3 = a_0 + a_1*t_3 + a_2*(T_3)^2

5.1 - Hint: This problem is meant to trick you. Clearly, p(t) is a nonlinear equation in the t-variable but once you have chosen a t-value then it is a
linear equation in the coecient variables. If you choose many t-values then you have many linear equations and now the tools of linear algebra apply.

This is still unclear to me, I probably sound dumb at this point but I'm really having a hard time with this. When it says once you have chosen a t-value then it is a linear equation what exactly does this mean? Because its not linear at this point with the t^2's. Is it really just solving these from here for 5.1?

8. Jul 7, 2010

u must be in my class lol. glad i'm not the only one that is completely...wondering what the hell.

9. Jul 7, 2010

### Staff: Mentor

This is what 5.1 is asking for, so you're done. If I were you, though, I would switch between lower and upper case on the t's. IOW you should write the system like this:
p_1 = a_0 + a_1*t_1 + a_2*(t_1)^2
p_2 = a_0 + a_1*t_2 + a_2*(t_2)^2
p_3 = a_0 + a_1*t_3 + a_2*(t_3)^2

For any given values of t_1, t_2, and t_3, the system above is linear in a_0, a_1, and a_2. For example, if t_1 = 1, the first equation is p_1 = a_0 + a_1 + a_2. By "linear" it is meant that each equation is a linear combination of the variables a_0, a_1, and a_2. Linear combination means a sum of constant multiples of the variables, and this is what you have for known values of t_1, t_2, and t_3.

10. Jul 7, 2010

### SpiffyEh

lol really? at CSM right?

11. Jul 7, 2010

### SpiffyEh

Oh! ok that makes sense. So for 5.2 do I basically just plug in the points for s1, s2 and s3 and see if they work?

12. Jul 7, 2010

### Staff: Mentor

Yeah, you could do that, or you could expend a little mental effort at thinking, instead. It's possible to see by inspection that two of the sets of points won't work.

13. Jul 7, 2010

### SpiffyEh

umm... so s3 doesn't seem like it would work because of the two same points and s2?

14. Jul 7, 2010

### Staff: Mentor

OK, S3 won't work because you are really given only two points. You need three points so that you can get three equations in three unknowns.

15. Jul 7, 2010

### SpiffyEh

I think s2 isn't because two of the points start with 1 but i'm not sure about that

16. Jul 7, 2010

### Staff: Mentor

You're getting warm. That is an astute observation.

17. Jul 7, 2010

### SpiffyEh

i'm really not sure what else to say about it

18. Jul 7, 2010

### Staff: Mentor

Can a function (such as p(t) = a0 + a1t + a2t2) have two different output values for a single input value?

Stated another way, can a vertical line intersect the graph of a function at more than one point?

19. Jul 7, 2010