Finding a Basis for P2 Subspace with p'(5)=0

[Nexttime35]

Homework Statement

The problem asks, find a basis for the P2 subspace that consists of polynomials, p(x) such that p'(5)=0.

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.

 

[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? ​

 

[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?

 

[tiny-tim]

Hi Nexttime35!

Do you mean, can 4a+b=1?

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

 

[Mark44]

Homework Statement

The problem asks, find a basis for the P2 subspace that consists of polynomials, p(x) such that p'(5)=0.

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.

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.

 

[Nexttime35]

Ahh, ok.

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

 

[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?

 

[Nexttime35]

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

 

[tiny-tim]

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.

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

The problem asks, find a basis for the P2 subspace that consists of polynomials, p(x) such that p'(5)=0.

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

yes it is!

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

 

[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?

 

[vela]

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?

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.

 

[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?

 

[tiny-tim]

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?

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? ​

 

[vela]

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.

 

[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?

 

[tiny-tim]

Hi Nexttime35!

Yes!

So one element of the basis could be a constant, since that derivative of a constant is equal to 0.

Yes!

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

One element of the basis could be the function p(x) = x , because that satisfies the p'(2) = 0 condition.

erm … nooo!

you had that part correct at the beginning …

do you need some sleep? :zzz:​

 

[vela]

One element of the basis could be the function p(x) = x , because that satisfies the p'(2) = 0 condition.

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

So, could {1, x} be a basis for the subspace of P2 in this case?

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.

 

[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?

 

[tiny-tim]

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.

Sorry, I'm calling a time-out …

GET SOME SLEEP! :zzz:

NOW!​

 

[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?

 

[tiny-tim]

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?

SERIOUSLY!

go and have a meal, have a walk, have a sleep, do anything but this

and come back to this tomorrow! ​