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0568 #22

Posted: Thu Oct 16, 2008 2:48 pm
by gaucho85
This is the one where you have to decide which of the following sets are subspaces of C(R) with scalar mult and pointwise addition.

I. f''(x) - 2f'(x) +3f(x) = 0
II. g''(x) = 3g'(x)
III. h''(x) = h(x) + 1

If you turn these into polynomials, where f''(x) is x^3, f'(x) = x^2, f(x)= x, etc, you get
x^3 - 2x^2 + 3x = 0
x^3 - 3x^2 = 0
x^3 - x = 1

and then you can see that III is not closed under pointwise addition.

Is this legit? It seems to get the right answer here, but does it make any sense, or is it just coincidence? Anyone have a better way to solve this?

Thanks again.

Posted: Thu Oct 16, 2008 4:22 pm
by Nameless
Basically, your solution is absolutely right.

III is the answer since you pick up two functions
h1, h2 satisfy
h1''=h1+1
h2''=h2+1
then (h1+h2)''=h1''+h2''= h1+1+h2+1 =(h1+h2)+2 != (h1+h2)+1