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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 advanceView attachment 803
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.
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 ?
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
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 ?
So if you double differentiate, 2a=0 for the second derivative to yield 0?
you can use integration
\(\displaystyle p''(x)=1\)
\(\displaystyle p'(x)= x +b\)
\(\displaystyle p(x)= \frac{x^2}{2} +bx+c\)
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\(\displaystyle p''(x)=0\)
\(\displaystyle p'(x)= a\)
\(\displaystyle p(x)= ax+c\)
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?
You can use the method Oplag illustrated 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 ?