GRE 9367 42 and 64

Forum for the GRE subject test in mathematics.
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GRE 9367 42 and 64

Postby toughluck » Fri Sep 25, 2009 9:30 am

Need some help figuring out how to solve the following problems

42. What is the greatest value of b for which any real value function f that satisfies the following properties must also satisfy f(1)<5?

(i) f is infinitely differentiable on the real numbers
(ii) f(0)=1, f'(0)=1, and f''(0)=2; and
(iii) |f'''(x)|<b for all x in [0,1]

a. 1
b. 2
c. 6
d. 12
e. 24

64. Let V be the real vector space of all real-valued functions defined on the real numbers and having derivatives of all orders. If D is the mapping from V into V that maps every function in V to its derivative, what are all the eigenvectors of D?

a. All nonzero functions in V
b. All nonzero constant functions in V
c. All nonzero functions of the form ke^(lk), where k and l are real numbers
d. All nonzero functions of the form sum from i=0 to k of c sub i times x^i where where k>0 and csub i's are real numbers
e. There are no eigenvectors of D

the answers are d and c any help is greatly appreciated

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Re: GRE 9367 42 and 64

Postby contrapositive » Fri Sep 25, 2009 1:04 pm

For #64 i think the reasoning goes like this (I guess its useful to recall that an eigenvector f of the linear map D must satisfy Df=lf for some nonzero l):

(By the way, in your statement of option (c), i think you meant to say "All nonzero functions of the form ke^{lx}, where k and l are real numbers")

(a) False, take f(x) = x^2.
(b) False, if f(x)=c then Dc=0=0.c, but by definition an eigenvalue must be nontrivial (nonzero).
(d) False, since applying D to a polynomial of degree k gives us a polynomial of degree k-1
(c) Functions of this type are indeed eigenvectors of D (with eigenvalue l) since
but there could be others as well. However, upon viewing (e), we deduce that these must be all the eigenvectors of D, so (c) must be the answer.

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Re: GRE 9367 42 and 64

Postby lunarmono » Sat Sep 26, 2009 1:08 pm

Problem 42 is hinting at the truncated Taylor series:
$<br />f(1)=f(0)+f'(0)+\frac{f''(0)}{2!}+\frac{f'''(x)}{3!}<br />$
$<br />0 \leq x \leq 1<br />$
Since we are given that:
we have:
$<br />f(1)=1+1+1+\frac{f'''(x)}{3!}<5<br />$
so for:
$\frac{|f'''(x)|}{3!}<2$ or $|f'''(x)|<12$
this is clearly satisfied.

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Re: GRE 9367 42 and 64

Postby conrado.math » Tue Sep 29, 2009 2:26 pm

Actually, in question 64 there are no other eigenvectors. We can easily see this from the fact that f is an eigenvector of D if and only if it satisfies the differential equation f' = lf for some real number l. It is known that the solutions to this equation all have the form f(x) = ke^{lx}, where k is a real number satisfying f(0) = k.

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Re: GRE 9367 42 and 64

Postby Legendre » Fri Aug 17, 2012 11:51 am

Actually, the reasoning in this thread is a little off.

The functions in (B) are in fact eigenvectors, with eigenvalue 0. Eigenvalues are allowed to 0. (eigenvectors are not).

However, (C) is the answer because (B) is actually a subset of (C) if you take lambda to be 0.

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Re: GRE 9367 42 and 64

Postby PNT » Tue Aug 21, 2012 10:19 pm

yea, I was about to add 0 is a perfectly reasonable eigenvalue

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