0568 #22

Forum for the GRE subject test in mathematics.
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Joined: Thu Sep 18, 2008 5:58 pm

0568 #22

Postby gaucho85 » Thu Oct 16, 2008 2:48 pm

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.

Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Postby Nameless » Thu Oct 16, 2008 4:22 pm

Basically, your solution is absolutely right.

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

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