Hi, aoyama-ga:
For clarity I've recopied your original questions below and
interspersed my answers between them:
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1. Suppose f(0) = 0 and the second derivative of f(x) is greater than
0, x >= 0. (a) Can you prove that f(x) >0, for any x > 0? (Prove or
find a counterexample.) (b) Can you prove any x* such that if x > x*
then f(x) > 0? (Prove or find a counterexample.)
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Without additional assumptions it cannot be proven that f(x) > 0 for
all x > 0, or even that there exists x* s.t. f(x) > 0 for x > x*.
Consider the counterexample f(x) = exp(-x) - 1. Although f(0) = 0
and the second derivative of f(x) is exp(-x) > 0, it's actually the
case that f(x) < 0 for all x > 0.
Some extra conditions that might be introduced are:
- If f'(0), the first derivative of f(x) at x = 0, were nonnegative
then f(x) > 0 for all x > 0.
- If f(x) were a polynomial function, then there exists x* such that
f(x) > 0 for all x > x* (but x* depends on the polynomial f(x)).
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2. For b is the element of real numbers, find lim( x - (square root
of ((x^2)-2bx)) as x approaches to the infinity. Prove your answer is
correct.
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For sufficiently large x, say x > 2|b|, we have the identities:
[x - sqrt(x^2 - 2bx)]*[x + sqrt(x^2 - 2bx)] = 2bx
x - sqrt(x^2 - 2bx) = 2bx / [x + sqrt(x^2 - 2bx)]
= 2b / [1 + sqrt(1 - 2b/x)]
Therefore as x approaches infinity (x -> +oo), the limit:
LIMIT (x - sqrt(x^2 - 2bx)) AS x -> +oo
is equal to 2b divided by:
LIMIT (1 + sqrt(1 - 2b/x)) AS x -> +oo
This last limit is seen "by inspection" to be 2 (since 2b/x
becomes arbitrarily close to zero). Thus the original limit
is 2b/2 = b.
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3. Suppose that for every interval (x,y) there exists z is the element
of (x,y) such that |f(z)| < y-x. (a) Can you prove that f(x) = 0 for
all x? ( Prove or find a counterexample.) (b) Now, suppose in
addition that f is continuous. Can you prove that f(x) = 0 for all x?
( Prove or find a counterexample.)
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(a) If function f is allowed to be a discontinuous function, we can
define f(x) = 0 if x is rational, 1 if x is irrational. Since the
rational numbers are "dense" among the real numbers, one has for
every (nonempty) interval (x,y) a rational number z in (x,y) s.t.
|f(z)| = 0 < y - x
Yet f(x) is clearly not identically 0, e.g. f(sqrt(2)) = 1.
(b) With the extra condition that f is continuous, the claim f(x) = 0
for all x can be proven. Without continuity the conditions require
only that for any nonempty open interval (x,y) and any epsilon > 0,
there exists z in (x,y) such that |f(z)| < epsilon. With continuity
as an additional assumption, this forces f(x) = 0 for every x, since
the value f(x) must be in every neighborhood of zero.
Perhaps the most concise way to present a rigorous proof is to argue
by contradiction. Suppose all the conditions given in the problem,
ie. that within every open interval (x,y) there lies z such that:
|f(z)| < y - x
and also assume that function f is continuous at a some point w at
which f(w) is nonzero. By the definition of continuity there must
be an interval (w-delta,w+delta) for some delta > 0 such that:
|f(x) - f(w)| < |f(w)|/2
for all x in (w-delta,w+delta). By the triangle inequality, then:
|f(w)| <= |f(x)| + |f(w) - f(x)| < |f(x)| + |f(w)|/2
for all x in (w-delta,w+delta), whence:
|f(x)| > |f(w)|/2
throughout this delta-neighborhood of w. Now simply apply the
given condition to find z in the narrower of these intervals:
(w - delta,w + delta) INTERSECT (w - |f(w)|/4, w + |f(w)|/4)
so that |f(z)| < |f(w)|/2, the width of the latter interval.
This is a contradiction, since we showed |f(z)| > |f(w)|/2
for every point in this interval. So it must be that f is
zero at every point where f is continuous.
regards, mathtalk-ga |
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