## GRE 9367 42 and 64

Forum for the GRE subject test in mathematics.
toughluck
Posts: 9
Joined: Thu Jun 25, 2009 10:57 am

### GRE 9367 42 and 64

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

contrapositive
Posts: 7
Joined: Mon Sep 21, 2009 7:31 am

### Re: GRE 9367 42 and 64

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
$$D(ke^{lx})=lke^{lx},$$
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.

lunarmono
Posts: 2
Joined: Fri Feb 13, 2009 10:43 am

### Re: GRE 9367 42 and 64

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

Posts: 9
Joined: Tue Sep 29, 2009 1:07 pm

### Re: GRE 9367 42 and 64

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.

Legendre
Posts: 217
Joined: Wed Jun 03, 2009 1:05 am

### Re: GRE 9367 42 and 64

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.

PNT
Posts: 37
Joined: Fri Mar 11, 2011 9:01 pm

### Re: GRE 9367 42 and 64

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