Re: mean value like and differentiable function

On Mon, 22 May 2006 13:14:47 EDT, eugene <jane1806@xxxxxxx> wrote:

On Mon, 22 May 2006 06:05:42 EDT, eugene
<jane1806@xxxxxxx> wrote:

Let f be twice continuously differentiable fucntion
on [a,b] sich that f(a) = f(b) = 0 and x_0 any fixed
real from (a,b).

How can i prove that there exists c from (a,b) which
satisfies

f(x_0) = (x_0-a)(x_0-b)*f''(c)/2.

A hint is that

[(f(x_0)/(x_0-a) - f(x_0)/(x_0-b)]/(b-a)

can be written as the integral of f'' times a certain
weight.

(Start with f(x_0) = f(x_0) - f(a) = int_a_{x_0} f',
and
make a change of variables so the integral goes from
0 to 1.
Do the same with f(b) - f(x_0). Subtract, then write
f'(here) - f'(there) as the integral of f''.)

Now if w(c) is positive and

I = int_0^1 w(c) f''(c) dc

then there must exist c such that f''(c) int_0^1 w =
I,
because if m and M are the minimum and maximum of f''
then
m int_0^1 w <= I ,+ M int_0^1 w.

David C. Ullrich


Thanks a lot, David, i've got it.

I was trying to use Taylor formula but failed.

When I wrote that up it seemed like it should
be just Taylor's formula, but I didn't see
right away how that would go. I saw later:

Consider Taylor's formula, centered at x_0,
and evaluated at a and at b:

0 = f(a) = f(x_0) + (a-x_0) f'(x_0) + E_a
0 = f(b) = f(x_0) + (b-x_0) f'(x_0) + E_b.

Divide the first equation by one thing and the
second by something else, subtract and you get

(*) f(x_0)(1/(x_0-a) - 1/(x_0-b)) = E_a/(x_0-a) - E_b/(x_0-b).

Now use the integral form for the remainder in Taylor's
theorem, and the right side of (*) becomes an expression
of the form int_a^b w(c) f''(c) dc, where w(c) =
whatever it comes out to be...
Thanks


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

David C. Ullrich
.

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