Determine if S is a subset of P3

[vcb003104]

1. Homework Statement
Let S be the following subset of the vector space P3 of all real polynomials p of degree at most 3:

S = {p∈P3|p(1)=0,p'(1) = 0}

where p' is the derivative of p.
I need to first:

Determine whether S is a subspace of P3
Determine whether the polynomial q(x) = x - 2$x^{2}$ + $x^{3}$ is an element of S

2. Homework Equations

S = {p∈P3|p(1)=0,p'(1) = 0}

3. The Attempt at a Solution
How can I prove that it is a subspace?

I know that there are the axioms but how do I use it?

I.E. when I say
u is an element of V
v is an element of V
u+v is an element of V (How do I show it?)

 

[Fredrik]

One of the statements you want to prove is this: For all $u,v\in S$, we have $u+v\in S$. The first sentence in your proof should be this: Let u and v be arbitrary elements of S.

Then you just use the definition of S and the assumption that $u,v\in S$ to show that $u+v\in S$. If you understand the definition of S, you should know what it means to say that $u+v$ is an element of S.

 

[vcb003104]

One of the statements you want to prove is this: For all $u,v\in S$, we have $u+v\in S$. The first sentence in your proof should be this: Let u and v be arbitrary elements of S.

Then you just use the definition of S and the assumption that $u,v\in S$ to show that $u+v\in S$. If you understand the definition of S, you should know what it means to say that $u+v$ is an element of S.

Hi but how can I answer the first question? determine if S is a subspace of P3? They didn't give me any equations or anything. There is only p(1) = 0 and p'(1) = 0

 

[HallsofIvy]

Your first mistake was in the title! You do NOT want to determine if S is subset of P3, you are given that! Any member of P3, "polynomials of degree at most 3", can be written in the form $p(x)= ax^3+ bx^2+ cx+ d$.

NOW, what does "p(0)= 0 and p'(1)= 0" tell you about a, b, c, and d?

 

[vcb003104]

Your first mistake was in the title! You do NOT want to determine if S is subset of P3, you are given that! Any member of P3, "polynomials of degree at most 3", can be written in the form $p(x)= ax^3+ bx^2+ cx+ d$.

NOW, what does "p(0)= 0 and p'(1)= 0" tell you about a, b, c, and d?

Oops yep I meant subspace haha

So d = 0
3a + 2b + c = 0

So p(x) = a$x^{3}$+ b$x^{2}$+cx

 

[Fredrik]

Hi but how can I answer the first question? determine if S is a subspace of P3?

I thought you said that you knew the procedure. The fact that you mentioned "u+v is an element of S" seemed to support that.

S is a subspace of P3 if and only if the following statements are true.

$S\subseteq P3$.
For all $u,v\in S$, we have $u+v\in S$
For all $a\in\mathbb R$ and all $u\in S$, we have $au\in S$.
$0\in S$.​

You should try to prove these statements, one at a time. If one of them seems impossible to prove, this should make you suspect that you're not dealing with a subspace, and you should try to prove that one of the statements is false.

For the last of those statements, you need to think about what the symbol 0 represents.

To prove any of these statements, you have to make sure that you understand the definitions. In particular, can you explain what the statement $u+v\in S$ means?

 

[Fredrik]

Did you solve the problem?

I think it's easier to not use the thing you discussed with Halls, but it's perfectly fine to use it if you want to.

 

[vcb003104]

I thought you said that you knew the procedure. The fact that you mentioned "u+v is an element of S" seemed to support that.

S is a subspace of P3 if and only if the following statements are true.

$S\subseteq P3$.
For all $u,v\in S$, we have $u+v\in S$
For all $a\in\mathbb R$ and all $u\in S$, we have $au\in S$.
$0\in S$.​

You should try to prove these statements, one at a time. If one of them seems impossible to prove, this should make you suspect that you're not dealing with a subspace, and you should try to prove that one of the statements is false.

For the last of those statements, you need to think about what the symbol 0 represents.

To prove any of these statements, you have to make sure that you understand the definitions. In particular, can you explain what the statement $u+v\in S$ means?

Hi there,

I was thinking if I can prove the u+v is an element of S like this:

f(x) = $ax^3 + bx^2 + cx + d$ = 0
g(x)= $ex^3 + fx^2 + gx + h$ = 0

(f+g)(x) = $ax^3 + bx^2 + cx + d + ex^3 + fx^2 + gx + h$
= $(a+e)x^3 + (b+f)x^2 + (c+g)x + (d+h) = 0$


and to show that Ku is an element of S I can write something like:

kf(x)
= $k (ax^3 + bx^2 + cx + d)$
= $(ka)x^3 + (kb)x^2 + (kc)x + kd$
=0


The second part says to determine whether the polynomial q(x) = $x- 2x^2 +x^3$ is an element of S

Can I say something like:
q(x) = $x- 2x^2 + x^3$
q(1) = 1-2+1
=0

q'(x) = $1 - 4x + 3x^2$
q'(1) = 1 - 4 + 3
= 0

which corresponds with p (1) = 0 and p'(1) = 0
so it's an element

 

[Fredrik]

Hi there,

I was thinking if I can prove the u+v is an element of S like this:

f(x) = $ax^3 + bx^2 + cx + d$ = 0
g(x)= $ex^3 + fx^2 + gx + h$ = 0

No need to change the notation from u,v to f,g. Why =0? I also don't see where you're using the assumption that u and v (or f and g) are elements of S.

The second part says to determine whether the polynomial q(x) = $x- 2x^2 +x^3$ is an element of S

Can I say something like:
q(x) = $x- 2x^2 + x^3$
q(1) = 1-2+1
=0

q'(x) = $1 - 4x + 3x^2$
q'(1) = 1 - 4 + 3
= 0

which corresponds with p (1) = 0 and p'(1) = 0
so it's an element

This looks good. There is however no need to mention p, which is just a dummy variable in the definition of S. Note that by definition of S, the statement "q is an element of S" is equivalent to the statement "q(1)=0 and q'(1)=0".

 

[vcb003104]

No need to change the notation from u,v to f,g. Why =0? I also don't see where you're using the assumption that u and v (or f and g) are elements of S.


This looks good. There is however no need to mention p, which is just a dummy variable in the definition of S. Note that by definition of S, the statement "q is an element of S" is equivalent to the statement "q(1)=0 and q'(1)=0".

So my part b is alright?

In the first part I want to say something like f(x) is in S and g(x) is in S

To show that S is a subspace I need to show that (f+g)(x) is in S and kf(x) is in S right?

 

[HallsofIvy]

Oops yep I meant subspace haha

So d = 0
3a + 2b + c = 0

So p(x) = a$x^{3}$+ b$x^{2}$+cx

Yes, but, as you said, 3a+ 2b+ c= 0 so c= -3a- 2b. Every polynomial in the given set can be written $p(x)= ax^3+ bx^2- (3a+ 2b)x$

Hi there,

I was thinking if I can prove the u+v is an element of S like this:

f(x) = $ax^3 + bx^2 + cx + d$ = 0
g(x)= $ex^3 + fx^2 + gx + h$ = 0

What? Why equal to 0? Do you mean "for some value of x" or "for all x"?

(f+g)(x) = $ax^3 + bx^2 + cx + d + ex^3 + fx^2 + gx + h$
= $(a+e)x^3 + (b+f)x^2 + (c+g)x + (d+h) = 0$and to show that Ku is an element of S I can write something like:

kf(x)
= $k (ax^3 + bx^2 + cx + d)$
= $(ka)x^3 + (kb)x^2 + (kc)x + kd$
=0

Again, why "= 0"? What does that mean?

You have already shown that any member of this set is of the form "$p(x)= ax^3+ bx^2- (3a+ 2b)x$. What you want to do is show that if $p(x)= ax^3+ bx^2- (3a+ 2b)x$ and $q(x)= ex^3+ fx^2- (3e+ 2f)x$ and k is a number then
$p(x)+ q(x)= ax^3+ bx^3- (3a+ 3b)x+ ex^3+ fx^2- (3e+ 2f)x= (a+ e)x^3+ (b+ f)x^3- (3(a+ e)+ 2(b+ f))x$ and $kp(x)= k(ax^3+ bx^2-(3a+ 3b)x)= (ak)x^3+ (bk)x^2+ (3(ak)+ 2(bk))x$ are of the same form and so also in the set.

The second part says to determine whether the polynomial q(x) = $x- 2x^2 +x^3$ is an element of S

Can I say something like:
q(x) = $x- 2x^2 + x^3$
q(1) = 1-2+1
=0

?? Why are you finding q(1)? The condition was that q(0)= 0 wasn't it?

q'(x) = $1 - 4x + 3x^2$
q'(1) = 1 - 4 + 3
= 0

which corresponds with p (1) = 0 and p'(1) = 0
so it's an element

By the way, one thing you should be looking at soon is "dimension" and "basis". We have already determined that any polynomial in this set must be of the form "$p(x)= ax^3+ bx^3- (3a+ 2b)x$ can be written as $p(x)= a(x^3- 3x)+ b(x^2- 2)$ showing that any polynomial in this set can be written, in a unique way, as a linear combination of $x^3- 3x$ and $x^2- 2x$, showing that $\{x^3- 3x, x^2- 2\}$ is a "basis" for this subspace and so it has dimension 2.

 

[pasmith]

So my part b is alright?

In the first part I want to say something like f(x) is in S and g(x) is in S

To show that S is a subspace I need to show that (f+g)(x) is in S and kf(x) is in S right?

Yes.

You can if you wish exploit the fact that you're dealing with a space of polynomials to conclude that if $p \in S$ then either $p$ is zero or it has a factor of $(x - 1)^2$.

 

[pasmith]

?? Why are you finding q(1)? The condition was that q(0)= 0 wasn't it?

The condition stated in the problem is indeed $p(1) = p'(1) = 0$.

 

[Fredrik]

So my part b is alright?

Yes.

In the first part I want to say something like f(x) is in S and g(x) is in S

Good, but I would recommend that you be a bit more careful with statements like that. What you should be saying is that f and g are in S, not that f(x) and g(x) are in S. If f is a function, then f(x) is an element of the range of that function. So if $f\in S$, then $f(x)\in\mathbb R$ for all $x\in\mathbb R$.

You can start the proof with the statement "Let f and g be two arbitrary elements of S". Now you need to prove that f+g is in S. To do that, you just need to know what it means to say that f+g is in S. Look at the definition of S.

To show that S is a subspace I need to show that (f+g)(x) is in S and kf(x) is in S right?

You need to show that f+g, kf and the zero vector of P3 are in S. Think carefully about which element of P3 is the zero vector. About a year ago I tried to help three different people in two days with problems similar to this, and not only did they all get it wrong, they all got it wrong in the same way. (Hint: The zero vector is NOT the number 0. It's not a number at all, because the elements of P3 aren't numbers).

 

[vcb003104]

Yes.


Good, but I would recommend that you be a bit more careful with statements like that. What you should be saying is that f and g are in S, not that f(x) and g(x) are in S. If f is a function, then f(x) is an element of the range of that function. So if $f\in S$, then $f(x)\in\mathbb R$ for all $x\in\mathbb R$.

You can start the proof with the statement "Let f and g be two arbitrary elements of S". Now you need to prove that f+g is in S. To do that, you just need to know what it means to say that f+g is in S. Look at the definition of S.


You need to show that f+g, kf and the zero vector of P3 are in S. Think carefully about which element of P3 is the zero vector. About a year ago I tried to help three different people in two days with problems similar to this, and not only did they all get it wrong, they all got it wrong in the same way. (Hint: The zero vector is NOT the number 0. It's not a number at all, because the elements of P3 aren't numbers).

Can I show f+g by saying that:

(f+g)(x)
= $ax^3 + bx^2 + cx + d + ex^3 + fx^2 + gx + h$
= $(a + e)x^3 + (b + f)x^2 + (c + g)x + (d + h)$
=f(x) + g(x)

(Do I need to do the same for f'(x) and g'(x)


kf(x)
= $k(ax^3 + bx^2 + cx d)$
=$akx^3 +bkx^2 + ckx + d$
=(kf)(x)

(Do I need to do the same for f'(x) again?)

and to say that S is non empty I just write p(1) = p'(1) = 0 so S is non empty?

 

[Fredrik]

Can I show f+g by saying that:

(f+g)(x)
= $ax^3 + bx^2 + cx + d + ex^3 + fx^2 + gx + h$
= $(a + e)x^3 + (b + f)x^2 + (c + g)x + (d + h)$
=f(x) + g(x)
[/itex]

This calculation shows that if f and g are in P3, then f+g is in P3. You want to show that if f and g are in S, then f+g is in S. You should start the proof with the statement "Let f and g be two arbitrary elements of S". Then you need to use that f and g are in S somewhere in your calculation.

(Do I need to do the same for f'(x) and g'(x)

Not sure what you mean by "the same". If you mean prove that if f' and g' are in S, then f'+g' is in S, then no, you don't have to prove that.

kf(x)
= $k(ax^3 + bx^2 + cx d)$
=$akx^3 +bkx^2 + ckx + d$
=(kf)(x)

This shows that if k is a real number and f is in P3, then kf is in P3. You want to show that if k is in S, then kf is in S.

and to say that S is non empty I just write p(1) = p'(1) = 0 so S is non empty?

I don't understand what you're doing here. What is p? OK, if you take some specific element of P3 (you have to say which one) and call it p, and then you show that p(1)=p'(1)=0, then you will have shown that p is in S, and therefore also that S is non-empty.

One more comment about showing your work: Let's say that you do a calculation in two steps. First you find that A=B, and then you find that B=C. Then you should always present this result as A=B=C, not as A=C=B. The latter statement is equally true, but your readers will find it more difficult to understand. For example, you shouldn't write (f+g)(x) = something = f(x)+g(x) when the calculation went (f+g)(x) = f(x)+g(x) and then f(x)+g(x) = something.

 

[HallsofIvy]

Can I show f+g by saying that:

(f+g)(x)
= $ax^3 + bx^2 + cx + d + ex^3 + fx^2 + gx + h$
= $(a + e)x^3 + (b + f)x^2 + (c + g)x + (d + h)$
=f(x) + g(x)

The fact that (f+ g)(x)= f(x)+ g(x) is pretty much the definition of "f+ g". That is NOT what you want to prove. You want to prove that "if $f\in S$ and $g\in S$ then $f+ g\in S$" which mean you want to prove that "if f is a cubic polynomial such that f(1)= 0 and f'(1)= 0 and g is a cubic polynomial such that g(1)= 0 and g'(1)= 0 then f+ g is a cubic polynomial such that (f+g)(1)= 0 and (f+ g)'(1)= 0.

(Do I need to do the same for f'(x) and g'(x)kf(x)
= $k(ax^3 + bx^2 + cx d)$
=$akx^3 +bkx^2 + ckx + d$
=(kf)(x)

(Do I need to do the same for f'(x) again?)

and to say that S is non empty I just write p(1) = p'(1) = 0 so S is non empty?

 

[vcb003104]

This calculation shows that if f and g are in P3, then f+g is in P3. You want to show that if f and g are in S, then f+g is in S. You should start the proof with the statement "Let f and g be two arbitrary elements of S". Then you need to use that f and g are in S somewhere in your calculation.


Not sure what you mean by "the same". If you mean prove that if f' and g' are in S, then f'+g' is in S, then no, you don't have to prove that.


This shows that if k is a real number and f is in P3, then kf is in P3. You want to show that if k is in S, then kf is in S.


I don't understand what you're doing here. What is p? OK, if you take some specific element of P3 (you have to say which one) and call it p, and then you show that p(1)=p'(1)=0, then you will have shown that p is in S, and therefore also that S is non-empty.

One more comment about showing your work: Let's say that you do a calculation in two steps. First you find that A=B, and then you find that B=C. Then you should always present this result as A=B=C, not as A=C=B. The latter statement is equally true, but your readers will find it more difficult to understand. For example, you shouldn't write (f+g)(x) = something = f(x)+g(x) when the calculation went (f+g)(x) = f(x)+g(x) and then f(x)+g(x) = something.

Hi there, but how can I say that f and g is part of s? p(1) and p'(1) are both in S right? so aren't they something like $ax^3 + bx^2 + cx + d = 0$ ?

because S = {p∈P3|p(1)=0,p'(1) = 0} so since p(1) = 0 does it mean that it is non empty?

 

[vcb003104]

The fact that (f+ g)(x)= f(x)+ g(x) is pretty much the definition of "f+ g". That is NOT what you want to prove. You want to prove that "if $f\in S$ and $g\in S$ then $f+ g\in S$" which mean you want to prove that "if f is a cubic polynomial such that f(1)= 0 and f'(1)= 0 and g is a cubic polynomial such that g(1)= 0 and g'(1)= 0 then f+ g is a cubic polynomial such that (f+g)(1)= 0 and (f+ g)'(1)= 0.

So if f(1) = 0 that's
a + b + c + d = 0

f'(x) = 0 that's 3a + 2b + c = 0 right

and g(1) = 0 that's
e + f + g + h=0

g'(1) = 0 is
3e + 2f + g = 0


wouldn't (f + g)(1) just be a + b + c + d + e + f +g + h = 0?
and (f + g)'(1) be 3(a + e) + 2(b + f) + (c + g) = 0?

 

[Fredrik]

Hi there, but how can I say that f and g is part of s?

You just use the definition of S. I'm not sure I can tell you much more about that. We don't give away complete solutions in the homework forums. We can only give you hints about what you should do, and point out mistakes in the attempts you show us.

What can you say about two functions f,g that are elements of S? Answer that first, and then think about whether you can say something similar about f+g?

p(1) and p'(1) are both in S right?

p(1) and p'(1) are numbers, and none of the elements in S are numbers. If you meant to ask if p and p' are in S, then I would have to ask you what p you're talking about. Some functions are in S and some are not. So the answer to the question of whether p and p' are in S depends on what function the symbol p represents.

so aren't they something like $ax^3 + bx^2 + cx + d = 0$ ?

Yes, but so are all the elements of P3 that are not in S.

because S = {p∈P3|p(1)=0,p'(1) = 0} so since p(1) = 0 does it mean that it is non empty?

I don't follow you here. What function does the symbol "p" represent when you say "since p(1)=0"? Is it an arbitrary element of S? The definition of S ensures that the statement "for all p in S, p(1)=0" is true, but this doesn't imply that S is non-empty, because that statement is certainly true when S is empty. (If S is empty, there's no p in S such that p(1)≠0. This means that the statement "for all p in S, p(1)=0" is not false, and if it's not false, it has to be true).

 

[Fredrik]

So if f(1) = 0 that's
a + b + c + d = 0

f'(x) = 0 that's 3a + 2b + c = 0 right

and g(1) = 0 that's
e + f + g + h=0

g'(1) = 0 is
3e + 2f + g = 0


wouldn't (f + g)(1) just be a + b + c + d + e + f +g + h = 0?
and (f + g)'(1) be 3(a + e) + 2(b + f) + (c + g) = 0?

OK, here it looks like you're on the right track. This is what HallsofIvy has been trying to get you to do. (But you could perhaps present the result in a way that makes it easier to see what you're doing). I think my approach is simpler, but this works too. So let's finish it this way first, and then discuss another approach.

 

[vcb003104]

OK, here it looks like you're on the right track. This is what HallsofIvy has been trying to get you to do. (But you could perhaps present the result in a way that makes it easier to see what you're doing). I think my approach is simpler, but this works too. So let's finish it this way first, and then discuss another approach.

Right, so like:

p(1) = a + b + c + d = 0
p'(1) = 3a + 2b + c = 0

→c = -(3a + 2b)
∴ p(x) = $ax^3 + bx^2 -(3a + 2b)x + d$

let f and g be an element in S

where f(x) = $ax^3 + bx^2 -(3a + 2b)x + d$
and g(X) = $ex^3 + fx^2 -(3e + 2f)x + h$

f(x) + g(x)
= $ax^3 + bx^2 -(3a + 2b)x + d + ex^3 + fx^2 -(3e + 2f)x + h$
= $(a + e)x^3 + (b + f)x^2 - [3(a + e) + 2(b + f)]x + (d + h)$
(which is the same form so that it is closed under addition?)

kf(x)
=$k(ax^3 + bx^2 - (3a + 2b)x + d)$
= $(ak)x^3 + (bk)x^2 - [3(ak) + 2(bk)]x + (dk)$
which is the same form again?

 

[Fredrik]

You could also have eliminated d and h, so that the formula for f(x) contains a and b, but not c or d, and the formula for g(x) contains e and f, but not g and h.

Showing that f+g and kf are "of the same form" as f and g is the right way to prove that f+g and kf are in S. So you definitely have the right idea about how to do this.

Note however that when you find the "form" of f, using the assumption that f is in S, all you have proved is that if f is in S, then f is of that form. You should also prove (or at least make a comment about it) that every function of that form is in S. Then you will have proved that a 3rd-degree polynomial is an element of S if and only if it's of that specific form.

 

[vcb003104]

You could also have eliminated d and h, so that the formula for f(x) contains a and b, but not c or d, and the formula for g(x) contains e and f, but not g and h.

Showing that f+g and kf are "of the same form" as f and g is the right way to prove that f+g and kf are in S. So you definitely have the right idea about how to do this.

Note however that when you find the "form" of f, using the assumption that f is in S, all you have proved is that if f is in S, then f is of that form. You should also prove (or at least make a comment about it) that every function of that form is in S. Then you will have proved that a 3rd-degree polynomial is an element of S if and only if it's of that specific form.

Yep , so I found that
since a + b + c + d= 0

3a + 2b + c = 0
c = -(3a + 2b)

d = -(a + b - 3a - 2b)
= 2a + b

P(x) = [itex] ax^3 + bx^2 - (3a + 2b)x + (2a + b)

like this?

but may I ask why do we need to eliminate c and d and g and h?

Is so that it looks 'neater' or actually helps to prove it?

 

[vela]

but may I ask why do we need to eliminate c and d and g and h?

Is so that it looks 'neater' or actually helps to prove it?

Earlier, using the fact that if $p \in S$ then p'(1)=0, you found that $p(x) = ax^3+bx^2-(3a+2b)x+d$. This will guarantee that p'(1)=0 for any a, b, and d, but you're not guaranteed that p(1)=0. That shouldn't be surprising because you never used the fact that p(1)=0 in deriving the form for p. For example, if a=b=d=1, then $p(x)=x^3+x^2-5x+1$. You get $p'(x)=3x^2+2x-5$, which indeed satisfied p'(1) = 0, but you get p(1)=-2. So to say p has the form $p(x) = ax^3+bx^2-(3a+2b)x+d$ isn't enough to conclude that $p \in S$.

However, when you eliminated d, using the fact that p(x)=0 when $p \in S$, you found that p has the form $p(x) = ax^3 + bx^2 - (3a + 2b)x + (2a + b)$. This form is guaranteed to satisfy p(1)=0 and p'(1)=0 for any a and b, so if you can show a polynomial q has this form, you can conclude that q is an element of S.

 

[Fredrik]

I would like to discuss the simpler solution, but we're still not completely done with this one. You also have to show that S is non-empty. Do you see how to do that?

I'll add a hint for the simple solution as well: I said earlier that when you want to prove the statement "if f and g are in S, then f+g is in S", you should think about what the statement "f+g is in S" means. What does the definition of S tell you that it means?