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conrad
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 Posted: Sun Jun 29, 2008 4:33 am Post subject: proof of chain rule |
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A flaw in the reasoning of:
du = g(x + dx) - g(x)
dy = f(u + du) - f(u)
dy/dx = lim_(dx->0) dy/dx
= lim_(dx->0) (dy/du) * (du/dx)
is that apparently du = 0 even when dx != 0
How can this be?
--
conrad |
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The World Wide Wade
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| Posted: Sun Jun 29, 2008 9:51 am Post subject: Re: proof of chain rule |
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In article
<31f191e3-a3c2-4b2e-92c0-08f16e402394@y38g2000hsy.googlegroups.com>,
conrad <conrad@lawyer.com> wrote:
| Quote: |
A flaw in the reasoning of:
du = g(x + dx) - g(x)
dy = f(u + du) - f(u)
dy/dx = lim_(dx->0) dy/dx
= lim_(dx->0) (dy/du) * (du/dx)
is that apparently du = 0 even when dx != 0
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You meant "apparently du can be = 0 even when dx != 0".
u = constant is one example of this. For nonconstant example, think
about u(x) = x^2*sin(1/x) for nonzero.
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William Elliot
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| Posted: Sun Jun 29, 2008 11:01 am Post subject: Re: proof of chain rule |
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Correction at **
On Sun, 29 Jun 2008, William Elliot wrote:
| Quote: |
On Sat, 28 Jun 2008, conrad wrote:
A flaw in the reasoning of:
du = g(x + dx) - g(x)
dy = f(u + du) - f(u)
dy/dx = lim_(dx->0) dy/dx
= lim_(dx->0) (dy/du) * (du/dx)
is that apparently du = 0 even when dx != 0
How can this be?
lim(h->0) (fg(x + h) - fg(x))/h
= lim(h->0) (fg(x + h) - fg(x))/(g(x + h) - g(x)) * (g(x + h) - g(x))/h
That was assuming a nhood N of x with for all y in N, g(y) /= g(x).
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** That was assuming a nhood N of x with for all y in N\x, g(y) /= g(x).
| Quote: |
What happens when there is no such nhood.
The case g is constant about x is no problem.
What is done for example, when g(x) = x^2.sin 1/x ?
= lim(h->0) (fg(x + h) - fg(x))/(g(x + h) - g(x))
* lim(h->0) (g(x + h) - g(x))/h
= g'(x) lim(h->0) (fg(x + h) - fg(x))/(g(x + h) - g(x))
= g'(x) lim(h->0) (f(g(x) + g(x + h) - g(x)) - fg(x))/(g(x + h) - g(x))
= g'(x) lim(k->0) (f(g(x) + k) - fg(x))/k
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William Elliot
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| Posted: Sun Jun 29, 2008 11:01 am Post subject: Re: proof of chain rule |
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On Sat, 28 Jun 2008, conrad wrote:
| Quote: |
A flaw in the reasoning of:
du = g(x + dx) - g(x)
dy = f(u + du) - f(u)
dy/dx = lim_(dx->0) dy/dx
= lim_(dx->0) (dy/du) * (du/dx)
is that apparently du = 0 even when dx != 0
How can this be?
lim(h->0) (fg(x + h) - fg(x))/h |
= lim(h->0) (fg(x + h) - fg(x))/(g(x + h) - g(x)) * (g(x + h) - g(x))/h
That was assuming a nhood N of x with for all y in N, g(y) /= g(x).
What happens when there is no such nhood.
The case g is constant about x is no problem.
What is done for example, when g(x) = x^2.sin 1/x ?
= lim(h->0) (fg(x + h) - fg(x))/(g(x + h) - g(x))
* lim(h->0) (g(x + h) - g(x))/h
= g'(x) lim(h->0) (fg(x + h) - fg(x))/(g(x + h) - g(x))
= g'(x) lim(h->0) (f(g(x) + g(x + h) - g(x)) - fg(x))/(g(x + h) - g(x))
= g'(x) lim(k->0) (f(g(x) + k) - fg(x))/k |
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Ted Hwa
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| Posted: Mon Jun 30, 2008 3:54 am Post subject: Re: proof of chain rule |
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conrad <conrad@lawyer.com> wrote:
: On Jun 29, 9:02am, David C. Ullrich <dullr...@sprynet.com> wrote:
: > Rephrase the definition of the derivative so
: > that there is no division involved: First, define
: > o(phi(h)) to mean "some function psi(h), with
: > the property that for every epsilon > 0 there
: > exists delta > 0 such that
: >
: > |psi(h)| < epsilon |phi(h)|
: >
: > whenever |phi(h)| < delta.
: >
: > Now show that
: >
: > (*) f is differentiable at x,
: > with derivative d, if and only if
: >
: > f(x+h) = f(x) + dh + o(h).
: >
: I've never seen f(x+h) represented this way.
: Specifically, I do not understand how
: you came to the conclusion that it is equal to
: f(x) + dh + o(h)
: Could you elaborate on the connection
: that must be made to make them equivalent?
Just to be clear, the "dh" in the above formula is the
multiplication d*h (d is a number, the alleged derivative of f
at x), not a term like "dy" in your original post.
So let's see why this is the same as the usual definition
of derivative.
If f'(x) = d, then according to the usual definition,
d = lim (h->0) (f(x+h) - f(x))/h
In vague terms, this says that
(f(x+h) - f(x))/h is approximately d when h is close to 0
or rearranging the terms
f(x+h) is approximately f(x) + d*h when h is close to 0
This is what * is saying (but in more precise terms). The
o(h) makes precise what "approximately" means.
In precise terms, we can write:
f(x+h) = f(x) + d*h + psi(h)
where psi(h) is an error term. Rearranging this equation
(f(x+h) - f(x))/h = d + psi(h)/h
Taking the limit of both sides as h->0, the left side is
by definition f'(x) = d. Therefore lim (h->0) psi(h)/h = 0,
so psi(h) satisfies the definition o(h).
On the other hand, in the other direction,
If f(x+h) = f(x) + d*h + psi(h), where psi(h) satisfies the
definition of o(h), then rearranging,
(f(x+h) - f(x))/h = d + psi(h)/h
Taking the limit of both sides as h->0 gives f'(x)=d.
(The limit of psi(h)/h is 0 since psi(h) satisfies the definition
of o(h).)
: > Now assume that f is differentiable at x and
: > g is differentiable at f(x). Then two applications
: > of (*) show that
: >
: > gof(x+h) = g(f(x+h))
: >g(f(x) + h f'(x) + o(h))
: >g(f(x)) + (h f'(x) + o(h)) g'(f(x)) + o(h f'(x) + o(h))
: >gof(x) + h f'(x) g'(f(x)) + o(h) g'(f(x)) + o(h)
: >gof(x) + h f'(x) g'(f(x)) + o(h).
: >
: First, to make sure communication is clear.
: Two applications of * means differentiating
: f(x+h) twice?
No, it means apply * to f (in the first step), then to g (in the second step).
: Secondly,
: If f(x+h) = f(x) + dh + o(h)
: then g(f(x) + h f'(x) + o(h)) is obviously true.
: However, g(f(x)) + (h f'(x) + o(h)) g'(f(x)) + o(h f'(x) + o(h))
: is less obvious. Perhaps it has to do with my confusion related
: to the previously asked questions. What piece of information
: allowed you to conclude that you were to multiply g(f(x + h)) by
: its derivative?
Apply * to g, but with f(x) in place of x, and h f'(x) + o(h) in place of h.
Ted |
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The World Wide Wade
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| Posted: Mon Jun 30, 2008 7:34 am Post subject: Re: proof of chain rule |
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In article <3r4f64pfpfjqg0he4bl6er6599non2lqku@4ax.com>,
David C. Ullrich <dullrich@sprynet.com> wrote:
| Quote: |
On Sat, 28 Jun 2008 21:33:24 -0700 (PDT), conrad <conrad@lawyer.com
wrote:
A flaw in the reasoning of:
du = g(x + dx) - g(x)
dy = f(u + du) - f(u)
dy/dx = lim_(dx->0) dy/dx
= lim_(dx->0) (dy/du) * (du/dx)
is that apparently du = 0 even when dx != 0
Of course what's above is not the actual proof
of the chain rule - you can find a real proof in
any decent calculus book (by definition of
"decent calculus book").
There are at least two ways to fix the problem;
what I think is the best one is not the one you
find in a lot of books.
Rephrase the definition of the derivative so
that there is no division involved: First, define
o(phi(h)) to mean "some function psi(h), with
the property that for every epsilon > 0 there
exists delta > 0 such that
|psi(h)| < epsilon |phi(h)|
whenever |phi(h)| < delta.
Now show that
(*) f is differentiable at x,
with derivative d, if and only if
f(x+h) = f(x) + dh + o(h).
Now assume that f is differentiable at x and
g is differentiable at f(x). Then two applications
of (*) show that
gof(x+h) = g(f(x+h))
= g(f(x) + h f'(x) + o(h))
= g(f(x)) + (h f'(x) + o(h)) g'(f(x)) + o(h f'(x) + o(h))
= gof(x) + h f'(x) g'(f(x)) + o(h) g'(f(x)) + o(h)
= gof(x) + h f'(x) g'(f(x)) + o(h).
Another application of (*) now shows that gof is
differentiable at x, with derivative f'(x) g'(f(x)).
How can this be?
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
|
I respectfully disagree that this is the best fix for someone at the
OP's level. Note that the pseudo proof works in the case where f'(x)
is nonzero. In the case f'(x) = 0, we can use the fact that for y near
f(x), there exists a constant C such that |g(y) - g(f(x))| <= C*|y -
f(x)|. This is immediate from the definition of g'(f(x)). It follows
that for small nonzero h,
|[g(f(x+h)) - g(f(x))]/h| <= C*|f(x+h) - f(x)|/|h|.
Because f'(x) = 0, the last expression -> 0 as h -> 0, which is what we
want in this case. |
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Michael Press
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| Posted: Fri Jul 04, 2008 9:01 am Post subject: Re: proof of chain rule |
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In article <3r4f64pfpfjqg0he4bl6er6599non2lqku@4ax.com>,
David C. Ullrich <dullrich@sprynet.com> wrote:
| Quote: |
On Sat, 28 Jun 2008 21:33:24 -0700 (PDT), conrad <conrad@lawyer.com
wrote:
A flaw in the reasoning of:
du = g(x + dx) - g(x)
dy = f(u + du) - f(u)
dy/dx = lim_(dx->0) dy/dx
= lim_(dx->0) (dy/du) * (du/dx)
is that apparently du = 0 even when dx != 0
Of course what's above is not the actual proof
of the chain rule - you can find a real proof in
any decent calculus book (by definition of
"decent calculus book").
There are at least two ways to fix the problem;
what I think is the best one is not the one you
find in a lot of books.
Rephrase the definition of the derivative so
that there is no division involved: First, define
o(phi(h)) to mean "some function psi(h), with
the property that for every epsilon > 0 there
exists delta > 0 such that
|psi(h)| < epsilon |phi(h)|
whenever |phi(h)| < delta.
Now show that
(*) f is differentiable at x,
with derivative d, if and only if
f(x+h) = f(x) + dh + o(h).
Now assume that f is differentiable at x and
g is differentiable at f(x). Then two applications
of (*) show that
gof(x+h) = g(f(x+h))
= g(f(x) + h f'(x) + o(h))
= g(f(x)) + (h f'(x) + o(h)) g'(f(x)) + o(h f'(x) + o(h))
= gof(x) + h f'(x) g'(f(x)) + o(h) g'(f(x)) + o(h)
= gof(x) + h f'(x) g'(f(x)) + o(h).
Another application of (*) now shows that gof is
differentiable at x, with derivative f'(x) g'(f(x)).
|
Thanks. This is neat.
The proof of (*) is a one-liner, which surprised me.
f is differentiable at x, with derivative d
iff lim_{h -> 0} {f(x+h) - f(x)}/h = d
iff for each epsilon > 0 there exists delta > 0 such that h < delta
==> |{f(x+h) - f(x)}/h - d| < epsilon
iff for each epsilon > 0 there exists delta > 0 such that h < delta
==> |f(x+h) - f(x) - dh| < epsilon.|h|
iff f(x+h) - f(x) - dh = o(h).
--
Michael Press |
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