Finding a Basis

1. Jan 21, 2014

Nexttime35

1. The problem statement, all variables and given/known data
The problem asks, find a basis for the P2 subspace that consists of polynomials, p(x) such that p'(5)=0.

3. The attempt at a solution

I know that a set of vectors is a basis if it's linear independent and spans the vector space.

So I let p(x) = ax2 + bx +c .

Then p'(x) = 2ax + b.
p'(2) = 4a + b

I set that equal to 0. so 4a+b = 0. then 4a=-b, and a= =1/4*b.

Now I am stuck as to where to go from here when finding a basis (prof hasn't really taught this yet)

I was wondering what the next step would be?

Thank you.

2. Jan 21, 2014

tiny-tim

Hi Nexttime35!

(do you mean p'(2) = 0 ?)

Hints:

i] is 1 in the subspace??

ii] how many elements do you think the basis has?

3. Jan 21, 2014

Nexttime35

I think the basis has two elements, a and b.

However, I am unsure how to apply your first hint to this question... Do you mean, can 4a+b=1?

4. Jan 21, 2014

tiny-tim

Hi Nexttime35!
No, I mean is p = 1 an element of P2 such that p'(2) = 0 ?

5. Jan 21, 2014

Staff: Mentor

What tiny-tim is asking about is the inconsistency between what you're given, p'(5) = 0, and what your work shows, p'(2) = 0.

6. Jan 21, 2014

Nexttime35

Ahh, ok.

Then p=1 is not an element of p2 such that p'(2) = 0.

7. Jan 21, 2014

Nexttime35

So a basis for this problem must consist of a function with only a degree 1, such as p(x) = x, or p(x) = 2x?

8. Jan 21, 2014

Nexttime35

Oh sorry, I meant p'(2) = 0 in the problem header. Sorry about that.

9. Jan 21, 2014

tiny-tim

i don't mind whether it's 2 or 5, it makes little difference, i was just checking
yes it is!!

p = 1 is a polynomial, and its derivative is 0 everywhere!

10. Jan 21, 2014

Nexttime35

Duh! My mistake! For some reason I understood that differently. So p(x) = 1 for some x, and then obviously p'(x)=0 . I am confused about how that helps me find a basis, though. Thoughts?

11. Jan 21, 2014

vela

Staff Emeritus
An element of P2 is a polynomial. We can think of these polynomials as vectors because with the appropriate definitions of vector addition and scalar multiplication, P2 has all the properties required of a vector space.

A basis is a set of linearly independent vectors that span the vector space. So in this problem, the vectors are polynomials, so you're looking for a set of polynomials that are linearly independent and that span the given subspace. Clearly $a$ and $b$ (the way you used them) aren't elements of the subspace, so they can't be elements of the basis.

12. Jan 21, 2014

Nexttime35

So vela,

there isn't just one basis. There are essentially many, many bases that can be formed? So I could make a set of vectors {v1,v2,v3} and that would be a basis if each of those vectors (or polynomials in this case) satisfy p'(2)=0, and that are linearly independent?

13. Jan 21, 2014

tiny-tim

well, p = 1 is in the subspace, so is p = 2, p = √2, etc

so does that help with one of the elements of the basis?

14. Jan 21, 2014

vela

Staff Emeritus
Yes, there are many possible bases for a vector space.

You have a subset S of P2 such that any element in S satisfies p'(2) = 0. A basis for S should be a set of linearly independent vectors that you can express any element of S as a linear combination of the basis vectors. This is close to what you said but not exactly. You need to make sure you can express every element of S in terms of the basis vectors.

15. Jan 21, 2014

Nexttime35

Yes!

So one element of the basis could be a constant, since that derivative of a constant is equal to 0. One element of the basis could be the function p(x) = x , because that satisfies the p'(2) = 0 condition. So, could {1, x} be a basis for the subspace of P2 in this case?

16. Jan 21, 2014

tiny-tim

Hi Nexttime35!
Yes!!

(obviously, any constant will do, but equally obviously most people would choose 1 !)
erm nooo!

you had that part correct at the beginning …

do you need some sleep? :zzz:​

17. Jan 21, 2014

vela

Staff Emeritus
If p(x)=x, then p'(x)=1. That doesn't satisfy p'(2)=0.

Suppose for the moment that p(x)=x did satisfy the condition. If {1, x} was a basis, that means any polynomial satisfying p'(2)=0 can be written as a linear combination of 1 and x. You need to prove that is the case for whatever prospective basis you come up with.

18. Jan 21, 2014

Nexttime35

Oh gosh. I think I do need some sleep...

So one element of the basis will be a constant. Say p(x) = 4x^2. Then plugging in x=2, we get p(2) = 16. p'(2) = 0. OK I understand. I had a brain fart... Say p(x) = 4x. Then plugging in x=2, we get p(2) = 8. p'(2) = 0. So, could a basis be {1,x,x^2}? Or am I just overthinking this?

19. Jan 21, 2014

tiny-tim

Sorry, I'm calling a time-out

GET SOME SLEEP! :zzz:

NOW!!

20. Jan 21, 2014

Nexttime35

I sincerely apologize for my lapse in mathematical judgement.

The only derivative that can result in 0 is the derivative of a constant. Thus, is the basis for this subspace of P2 a one-element vector, a constant?