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Q: advanced calculus -- proofs and/or calculations ( Answered 5 out of 5 stars,   0 Comments )
Question  
Subject: advanced calculus -- proofs and/or calculations
Category: Science > Math
Asked by: aoyama-ga
List Price: $50.00
Posted: 21 Oct 2003 17:13 PDT
Expires: 20 Nov 2003 16:13 PST
Question ID: 268406
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.)

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.

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.)
Answer  
Subject: Re: advanced calculus -- proofs and/or calculations
Answered By: mathtalk-ga on 22 Oct 2003 11:45 PDT
Rated:5 out of 5 stars
 
Hi, aoyama-ga:

For clarity I've recopied your original questions below and
interspersed my answers between them:

* * * * * * * * * * * * * * * * * * * * * * * * * * * * *

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.) 

* * * * * * * * * * * * * * * * * * * * * * * * * * * * *

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)).


* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
 
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. 

* * * * * * * * * * * * * * * * * * * * * * * * * * * * *

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.


* * * * * * * * * * * * * * * * * * * * * * * * * * * * *
 
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.)

* * * * * * * * * * * * * * * * * * * * * * * * * * * * *

(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
aoyama-ga rated this answer:5 out of 5 stars

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