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 realvalued 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
GRE 9367 42 and 64

 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 k1
(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.
(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 k1
(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.
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.
$$$
f(1)=f(0)+f'(0)+\frac{f''(0)}{2!}+\frac{f'''(x)}{3!}
$$$
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.
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.
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.
Re: GRE 9367 42 and 64
yea, I was about to add 0 is a perfectly reasonable eigenvalue