
Subject: Advanced Calculus
Hi, my name is Kay. Please helpthese problems are driving me crazzzzy!!!! Your help would be greatly appreciated!
 Let a,b be contained in R, a<b, and let f,g be continuous realvalued functions on [a,b] that are differentiable on (a,b). Prove that there exists a number c in (a,b) such that f'(c)[g(b)g(a)] = g'(c)[f(b)f(a)].
 Show that f is a continuous realvalued function on the interval [a,b] and f(x)>=0 for all x in [a,b], then
lim(int a>b (f(x))^{n} dx)^{1/n} = max {f(x):x in [a,b]}.
 Prove that the equation cosxx1/2=0 has a unique real solution. Show that the fixedpoint theorem is applicable to the function F(x)=cosx1/2 and the interval [0,(pi)/4] and thereby find this solution to three decimal places.
If you can helpThank You!!!
Hi Kay,
 When g(b) is different from g(a), you can rewrite this as
which is the generalized
form of the mean value theorem.
Mean Value Theorem
If a < b, and h is continuous on [a,b] and differentiable on (a,b),
then there exists c in (a,b) such that
For your problem, it is tempting to apply the mean value theorem
to both f and g:
And then divide the two equations, cancelling out the terms (ba),
but unfortunately this does not work, because the c value you
get for f may not be the same as the c value you get for g.
However, there is another way to use the mean value theorem
to solve your problem: You apply the mean value theorem to a particular
function, namely
Can you see how it works?
P.S.
This "generalized mean value theorem" is the key to
proving that De l'Hospital's rule works: What happens
if f(a) = g(a) = 0, but the limit of f'(c)/g'(c) as c approaches a
exists and you are trying to evaluate the limit of
f(b)/g(b) as b approaches a?
 Let M = max {f(x):x in [a,b]}. (And suppose that M is greater than 0,
otherwise the result is obvious.
Then (int a>b (f(x))^{n} dx)^{1/n}
<= (int a>b M^{n} dx)^{1/n}
= (M^{n} (ba))^{1/n} = M (ba)^{1/n}
for all n.
Therefore lim(int a>b (f(x))^{n} dx)^{1/n}
<= lim M (ba)^{1/n} = M.
For the other inequality, you use continuity:
Take B such that 0 <= B < M. Since f is continuous and
reaches its maximum M at some point c, there exists
an interval [u,v] containing c such that f(x) > B for all
x in [u,v].
Therefore
(int a>b (f(x))^{n} dx)^{1/n}
>= (int u>v (f(x))^{n} dx)^{1/n}
>= (int u>v B^{n} dx)^{1/n}
= B (vu)^{1/n}
for all n, whence
lim (int a>b (f(x))^{n} dx)^{1/n}
>= lim B (vu)^{1/n} = B
And since this holds for ALL B smaller than M, we have
lim (int a>b (f(x))^{n} dx)^{1/n} >= M.
P.S. For all n, f_n = (int a>b (f(x))^{n} dx)^{1/n}
are norms on the vector space of continuous functions
on [a,b], and so is
f_infinity = max {f(x):x in [a,b]}.
This exercise should help to clarify the name f_infinity.
 The last part of the problem is the fun part.
The theoretic limbo above just says that anyone can
approximate the solution of cosxx1/2=0 to a high precision:
Make sure your calculator is in RAD mode, and keep pushing the buttons
(cos) () (.) (5) (=)
(cos) () (.) (5) (=)
(cos) () (.) (5) (=) ...
Eventually, the numbers will start repeating.
On mine, I get x = 0.41508289, and this is an approximate
solution to the equation cosxx1/2=0.
Now the mathematical question is "why does this work?"
cosxx1/2=0 means cosx  1/2 = x, that is x is a fixed
point of the function F(x) = cosx  1/2.
By pushing the calculators buttons as above,
you are just finding consecutive terms
in the sequence 0, F(0), F(F(0)), F(F(F(0))), ...
The (Banach) fixed point theorem says that
such a sequence converges to a unique fixed point
if F is a function from a complete metric space to itself
which satisfies the Lipschitz condition:
F(b)  F(a) <= lambda * b  a ,
with lambda < 1.
Now F([0, pi/4]) is contained in [0,pi/4] (this is important),
and we can verify the Lipschitz condition using the mean
value theorem: F'(x) = sinx, so that for x in [0,pi/4],
we have 0 <= F'(x) <= 1/2^{1/2}.
Now for a, b in [0,pi/4], there exists c between a and b
such that
F(b)  F(a) = F'(c) * (b  a),
whence
F(b)  F(a) = F'(c) * b  a <= 1/2^{1/2} * b  a,
and F satisfies the Lipschitz condition with lambda = 1/2^{1/2}.
To find out just how much iterations you have to do before
being sure that the three first decimals are correct,
you need to look into the proof of the fixed point theorem.
Cheers,
Claude
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