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- Thread starter Kaspelek
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- Feb 7, 2012

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The test for a subspace is that if it contains $x$ and $y$ then it also contains $x+y$ and $cx$ (where $c$ is any constant). Apply that test to each of those two subsets.Hi guys, back again....

Any help on this question would be appreciated.

Thanks in advance View attachment 803

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Hey, these are the laws closed under vector addition and multiplication?The test for a subspace is that if it contains $x$ and $y$ then it also contains $x+y$ and $cx$ (where $c$ is any constant). Apply that test to each of those two subsets.

However perhaps you can provide an example in this situation, not really sure how to go about it

- Jan 17, 2013

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First find all polynomials whose second derivative is 0 , do you know how to do that ?

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Doesn't it have to be a second degree polynomial?First find all polynomials whose second derivative is 0 , do you know how to do that ?

I know that for the one where derivative = 1 1/2(x^2) '' = 1

- Jan 17, 2013

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Start by the general form \(\displaystyle P(x)=ax^2+bx+c\) then try to find the constants , what do you get ?Doesn't it have to be a second degree polynomial?

I know that for the one where derivative = 1 1/2(x^2) '' = 1

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So if you double differentiate, 2a=0 for the second derivative to yield 0?Start by the general form \(\displaystyle P(x)=ax^2+bx+c\) then try to find the constants , what do you get ?

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So if you double differentiate, 2a=0 for the second derivative to yield 0?

There is no second degree polynomial that yields a second derivative =0?

- Jan 17, 2013

- 1,667

you can use integration

\(\displaystyle p''(x)=1\)

\(\displaystyle p'(x)= x +b\)

\(\displaystyle p(x)= \frac{x^2}{2} +bx+c\)

[HR][/HR]

\(\displaystyle p''(x)=0\)

\(\displaystyle p'(x)= a\)

\(\displaystyle p(x)= ax+c\)

Now test each one to see whether it forms a subspace .

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you can use integration

\(\displaystyle p''(x)=1\)

\(\displaystyle p'(x)= x +b\)

\(\displaystyle p(x)= \frac{x^2}{2} +bx+c\)

[HR][/HR]

\(\displaystyle p''(x)=0\)

\(\displaystyle p'(x)= a\)

\(\displaystyle p(x)= ax+c\)

Now test each one to see whether it forms a subspace .

I'm not trying to leech here, i'm just legitimately confused.

So i'm thinking of testing if its closed under multiplication first.

Am i correct in saying (assume a=alpha)

a* \(\displaystyle p(x)= \frac{x^2}{2} +bx+c\) does not equal a*p''(x) ?

How do i correctly show this proof?

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I'm not trying to leech here, i'm just legitimately confused.

So i'm thinking of testing if its closed under multiplication first.

Am i correct in saying (assume a=alpha)

a* \(\displaystyle p(x)= \frac{x^2}{2} +bx+c\) does not equal a*p''(x) ?

How do i correctly show this proof?

Does this correctly answer the question?

Test if subsets are closed under multiplication.

a*(\(\displaystyle \frac{x^2}{2} +bx+c\))

d^2/dx^2(a*(\(\displaystyle \frac{x^2}{2} +bx+c\)) = a

Therefore since a*p(x) does not equal p''(x) not closed. Hence not a subspace.

a*(\(\displaystyle ax+c\))

d^2/dx^2(a*(\(\displaystyle ax+c\))) = 0

Therefore since a*P(x)=P''(x) this subset is a subspace.

Thoughts?

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- Feb 7, 2012

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I think I may have misled you in my first comment by using bad notation. The situation is that if you have $P(x)$ and $Q(x)$, and $c$ is a constant, then you need to test whether $P(x)+Q(x)$ and $cP(x)$ satisfy the appropriate condition.

To avoid doing the exact same problem for you, suppose for example that you were given the set $U$ of all polynomials whose third derivative is 3: $U = \{P(x)\in\mathcal{P}_3|P'''(x)=3\}$. If $P(x)$ and $Q(x)$ satisfy that condition then $P'''(x) = Q'''(x)=3$, so $P'''(x)+Q'''(x) = 3+3=6$. So the third derivative of $P(x)+Q(x)$ is $6$. But $6\ne3$, so $P(x)+Q(x)$ does not satisfy the condition for being in $U$. Therefore $U$ is not a subspace. Incidentally, the third derivative of $cP(x)$ is $3c$, and since $3c\ne3$ (unless the constant $c$ happens to be equal to $1$), $U$ does not satisfy either of the conditions for being a subspace.

**Edit**. Your previous comment #10 is a mostly correct solution. It shows that the first set is not a subspace. But to show that the second set is a subspace you need to check that it satisfies both of the conditions for a subspace (vector addition as well as scalar multiplication). But you might also like to see the method I have used in this example above.

To avoid doing the exact same problem for you, suppose for example that you were given the set $U$ of all polynomials whose third derivative is 3: $U = \{P(x)\in\mathcal{P}_3|P'''(x)=3\}$. If $P(x)$ and $Q(x)$ satisfy that condition then $P'''(x) = Q'''(x)=3$, so $P'''(x)+Q'''(x) = 3+3=6$. So the third derivative of $P(x)+Q(x)$ is $6$. But $6\ne3$, so $P(x)+Q(x)$ does not satisfy the condition for being in $U$. Therefore $U$ is not a subspace. Incidentally, the third derivative of $cP(x)$ is $3c$, and since $3c\ne3$ (unless the constant $c$ happens to be equal to $1$), $U$ does not satisfy either of the conditions for being a subspace.

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- Jan 17, 2013

- 1,667

we have proved that all the polynomials of degree 2 that satisfy the conditions must be of the form

\(\displaystyle P(x) = \frac{x^2}{2}+bx+c\) were \(\displaystyle b\) and \(\displaystyle c\) are arbitrary numbers .

So the set contains only second degree polynomials with a leading coefficient \(\displaystyle =\frac{1}{2}\) .

Now , if we choose two polynomials in the set ,say, \(\displaystyle R(x) = \frac{x^2}{2}+x+1\) and \(\displaystyle Q(x) = \frac{x^2}{2}+x+2\)

We know by definition that a subspace must be closed under addition . Now, if we add the two polynomials , do we get a polynomial that is still in the set ?

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Oplagillustrated or you can do the following

we have proved that all the polynomials of degree 2 that satisfy the conditions must be of the form

\(\displaystyle P(x) = \frac{x^2}{2}+bx+c\) were \(\displaystyle b\) and \(\displaystyle c\) are arbitrary numbers .

So the set contains only second degree polynomials with a leading coefficient \(\displaystyle =\frac{1}{2}\) .

Now , if we choose two polynomials in the set ,say, \(\displaystyle R(x) = \frac{x^2}{2}+x+1\) and \(\displaystyle Q(x) = \frac{x^2}{2}+x+2\)

We know by definition that a subspace must be closed under addition . Now, if we add the two polynomials , do we get a polynomial that is still in the set ?

Both methods make sense to me, cheers for the help guys!