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From: Gerry on 19 Jul 2010 19:33
On Jul 20, 1:40 am, cbb <stefan.lhach...(a)gmail.com> wrote:
> On Jul 19, 4:59 pm, "Greg Neill" <gneil...(a)MOVEsympatico.ca> wrote:
>
>
>
>
>
> > Stefan.D...(a)gmail.com wrote:
> > > Hi,
> > > there is one thing that puzzles me in the derivation of exponential
> > > growth.
> > > Let's  start with
> > > dx/dt = kx
>
> > > where t is time and the rate.
> > > Then
> > > dx/x= k dt
> > > and then
> > > Int[dx/x]=Int[k dt]
> > > ln x = kt + constant
>
> > > Can somebody explain or direct me to a source explain why Int[dx/x]=ln
> > > x ?
> > > Thanks and best,
> > > stefan
>
> > Are you aware that d(e^x)/dx = e^x  ?
>
> > Consider the relationship y = e^x.  If we differentiate
> > both sides:
>
> > dy/dx = e^x = y
>
> > so we can write:
>
> > dy/y = dx
>
> > Integrating both sides:
>
> > INT(dy/y) = x + c
>
> > but from above, y = e^x, so taking the natural log of each
> > side, ln(y) = x.  Thus
>
> > INT(dy/y) = ln(y) + c
>
> thanks for the quick response. So if I get it right, this only holds
> when I define Y=exp(x) (given some scale parameter for the growth
> rate). It wouldn't hold for any other function, say, y=x^2?

Well, you said you start with dx/dt = k x
(with k, presumably, a constant).
If x = t^2, you don't get dx/dt = k x, do you?
--
GM
From: Noone on 20 Jul 2010 04:16
On Mon, 19 Jul 2010 08:40:46 -0700 (PDT), cbb
<stefan.lhachimi(a)gmail.com> wrote:

>On Jul 19, 4:59 pm, "Greg Neill" <gneil...(a)MOVEsympatico.ca> wrote:
>> Stefan.D...(a)gmail.com wrote:
>> > Hi,
>> > there is one thing that puzzles me in the derivation of exponential
>> > growth.
>> > Let's  start with
>> > dx/dt = kx
>>
>> > where t is time and the rate.
>> > Then
>> > dx/x= k dt
>> > and then
>> > Int[dx/x]=Int[k dt]
>> > ln x = kt + constant
>>
>> > Can somebody explain or direct me to a source explain why Int[dx/x]=ln
>> > x ?
>> > Thanks and best,
>> > stefan
>>
>> Are you aware that d(e^x)/dx = e^x  ?
>>
>> Consider the relationship y = e^x.  If we differentiate
>> both sides:
>>
>> dy/dx = e^x = y
>>
>> so we can write:
>>
>> dy/y = dx
>>
>> Integrating both sides:
>>
>> INT(dy/y) = x + c
>>
>> but from above, y = e^x, so taking the natural log of each
>> side, ln(y) = x.  Thus
>>
>> INT(dy/y) = ln(y) + c
>
>thanks for the quick response. So if I get it right, this only holds
>when I define Y=exp(x) (given some scale parameter for the growth
>rate). It wouldn't hold for any other function, say, y=x^2?

In the following, a and c represent constants.

If a function y = f(x) satisfies a differential equation of the form
y' = a y, then y has the form y = f(x) = c e^(a x). Conversely,
every function of the form c e^(a x) satisfies the equation
y' = a y.

From: Daniel Royer on 20 Jul 2010 04:31
On 19.07.2010 16:41, Stefan.Duke(a)gmail.com wrote:
> Hi,
> there is one thing that puzzles me in the derivation of exponential
> growth.
> Let's start with
> dx/dt = kx
>
> where t is time and the rate.
> Then
> dx/x= k dt
> and then
> Int[dx/x]=Int[k dt]
> ln x = kt + constant
>
> Can somebody explain or direct me to a source explain why Int[dx/x]=ln
> x ?
> Thanks and best,
> stefan
Because the derivative of ln(x) is 1/x.

--



---------------------------
Daniel Royer
University of Geneva
daniel at royer dot ch
From: mjc on 20 Jul 2010 10:53
Another way to look at ln is to start with its functional equation:
f(xy) = f(x)+f(y). (This also works for exp, atanh, and others.)

Assume the function is differentiable. Then
f(x+h)-f(x) = f(1+h/x) = f(1+h/x)-f(1) (since f(1)=0)
so f'(x) = lim_(h->0) (f(x+h)-f(x))/h
= lim_(h->0) (f(1+h/x)-f(1))/h
= (lim_(h->0) (f(1+h/x)-f(1))/(h/x) )/x
= f'(1)/x.

Thus f(x) = integral from 1 to x of f'(1)/t dt.

The natural log is distinguished by f'(1) = 1.
From: Stefan.Duke on 20 Jul 2010 12:00
On Jul 20, 4:53 pm, mjc <mjco...(a)acm.org> wrote:
> Another way to look at ln is to start with its functional equation:
> f(xy) = f(x)+f(y). (This also works for exp, atanh, and others.)
>
> Assume the function is differentiable. Then
> f(x+h)-f(x) = f(1+h/x) = f(1+h/x)-f(1) (since f(1)=0)
> so f'(x) = lim_(h->0) (f(x+h)-f(x))/h
>          = lim_(h->0) (f(1+h/x)-f(1))/h
>          = (lim_(h->0) (f(1+h/x)-f(1))/(h/x) )/x
>          = f'(1)/x.
>
> Thus f(x) = integral from 1 to x of f'(1)/t dt.
>
> The natural log is distinguished by f'(1) = 1.

Thank you all so much!
Things are clearer now.
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