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From: Han de Bruijn on 2 Jun 2010 07:40
On Jun 2, 11:37 am, Torsten Hennig <Torsten.Hen...(a)umsicht.fhg.de>
wrote:
> > On Jun 2, 2:20 am, Torsten Hennig
> > <Torsten.Hen...(a)umsicht.fhg.de>
> > wrote:
> > > > Can someone help me with this.
> > > > I need to solve  an exponential equation for a
> > > > project of mine. It's
> > > > an initial value problem where I know the value
> > of y
> > > > at t=0 as well as
> > > > y at t=60s and 120s, but I don't know the value
> > of
> > > > y=n  at t=infinity.
> > > > The equation is y=C exp^(-kt) + n
> > > > I need to know if the value of n can be solved
> > given
> > > > the initial value
> > > > and one or two intermediate values of y .  At
> > > > infinity the exponential
> > > > term will disappear leaving only n, which is what
> > I
> > > > need.  I just need
> > > > to know if it can be done or is there
> > insufficient
> > > > data to determine
> > > > both the exponential term k and either C or n,
> > one
> > > > will lead to the
> > > > other.
> > > > Thanks
>
> > > If you have y at three different times, you can
> > > in principle solve for the three unknown
> > parameters:
> > > (1) y(t=0) = C+n
> > > (2) y(t=60) = C*exp(-k*60)+n
> > > (3) y(t=120) = C*exp(-k*120)+n.
>
> > > First solve (1) for n and insert in (2) and (3).
> > > You'll end up in a quadratic equation for
> > exp(-60*k)
> > > which can be solved analytically for k.
>
> > > But more measurements and a nonlinear regression
> > > to your function y=C*exp(-k*t)+n may give better
> > > results for the parameters C,k and n due to
> > > measurement errors at only three given times.
>
> > > Best wishes
> > > Torsten.
>
> > Problem is C and n are both unknown.
>
> I get
>
> k = 1/60 * log((y(t=60)-y(t=0))/(y(t=120)-y(t=0)))
> C = (y(t=60)-y(t=0))^2 / (y(t=120)-y(t=60))
> n = y(t=0) - (y(t=60)-y(t=0))^2 / ((y(t=120)-y(t=60))
>
> ln: natural logarithm

log: ??

> Best wishes
> Torsten.

Han de Bruijn
From: Torsten Hennig on 2 Jun 2010 03:59
> On Jun 2, 11:37 am, Torsten Hennig
> <Torsten.Hen...(a)umsicht.fhg.de>
> wrote:
> > > On Jun 2, 2:20 am, Torsten Hennig
> > > <Torsten.Hen...(a)umsicht.fhg.de>
> > > wrote:
> > > > > Can someone help me with this.
> > > > > I need to solve  an exponential equation for
> a
> > > > > project of mine. It's
> > > > > an initial value problem where I know the
> value
> > > of y
> > > > > at t=0 as well as
> > > > > y at t=60s and 120s, but I don't know the
> value
> > > of
> > > > > y=n  at t=infinity.
> > > > > The equation is y=C exp^(-kt) + n
> > > > > I need to know if the value of n can be
> solved
> > > given
> > > > > the initial value
> > > > > and one or two intermediate values of y .  At
> > > > > infinity the exponential
> > > > > term will disappear leaving only n, which is
> what
> > > I
> > > > > need.  I just need
> > > > > to know if it can be done or is there
> > > insufficient
> > > > > data to determine
> > > > > both the exponential term k and either C or
> n,
> > > one
> > > > > will lead to the
> > > > > other.
> > > > > Thanks
> >
> > > > If you have y at three different times, you can
> > > > in principle solve for the three unknown
> > > parameters:
> > > > (1) y(t=0) = C+n
> > > > (2) y(t=60) = C*exp(-k*60)+n
> > > > (3) y(t=120) = C*exp(-k*120)+n.
> >
> > > > First solve (1) for n and insert in (2) and
> (3).
> > > > You'll end up in a quadratic equation for
> > > exp(-60*k)
> > > > which can be solved analytically for k.
> >
> > > > But more measurements and a nonlinear
> regression
> > > > to your function y=C*exp(-k*t)+n may give
> better
> > > > results for the parameters C,k and n due to
> > > > measurement errors at only three given times.
> >
> > > > Best wishes
> > > > Torsten.
> >
> > > Problem is C and n are both unknown.
> >
> > I get
> >
> > k = 1/60 * log((y(t=60)-y(t=0))/(y(t=120)-y(t=0)))
> > C = (y(t=60)-y(t=0))^2 / (y(t=120)-y(t=60))
> > n = y(t=0) - (y(t=60)-y(t=0))^2 /
> ((y(t=120)-y(t=60))
> >
> > ln: natural logarithm
>
> log: ??
>

Should read
log : natural logarithm

> > Best wishes
> > Torsten.
>
> Han de Bruijn
From: Chip Eastham on 2 Jun 2010 09:53
On Jun 2, 4:20 am, Torsten Hennig <Torsten.Hen...(a)umsicht.fhg.de>
wrote:
> > Can someone help me with this.
> > I need to solve  an exponential equation for a
> > project of mine. It's
> > an initial value problem where I know the value of y
> > at t=0 as well as
> > y at t=60s and 120s, but I don't know the value of
> > y=n  at t=infinity.
> > The equation is y=C exp^(-kt) + n
> > I need to know if the value of n can be solved given
> > the initial value
> > and one or two intermediate values of y .  At
> > infinity the exponential
> > term will disappear leaving only n, which is what I
> > need.  I just need
> > to know if it can be done or is there insufficient
> > data to determine
> > both the exponential term k and either C or n, one
> > will lead to the
> > other.
> > Thanks
>
> If you have y at three different times, you can
> in principle solve for the three unknown parameters:
> (1) y(t=0) = C+n
> (2) y(t=60) = C*exp(-k*60)+n
> (3) y(t=120) = C*exp(-k*120)+n.
>
> First solve (1) for n and insert in (2) and (3).
> You'll end up in a quadratic equation for exp(-60*k)
> which can be solved analytically for k.
>
> But more measurements and a nonlinear regression
> to your function y=C*exp(-k*t)+n may give better
> results for the parameters C,k and n due to
> measurement errors at only three given times.
>
> Best wishes
> Torsten.

Taking resp. differences of (2) minus (1) and
(3) minus (2):

y(t=60) - y(t=0) = C*(exp(-60k) - 1)
y(t=120)-y(t=60) = C*(exp(-120k) - exp(-60k))

So taking the ratios of these:

exp(-60k) = [y(t=120)-y(t=60)]/[y(t=60)-y(t=0)]

allows us to solve for k:

k = 1/60 * ln([y(t=60)-y(t=0)]/[y(t=120)-y(t=60)])

and what remains is a linear system for C and n.
Note that this differs slightly in the denominator
from Torsten's solution:

> k = 1/60 * log((y(t=60)-y(t=0))/(y(t=120)-y(t=0)))

but this is presumably just a typo since elsewhere
he uses y(t=120)-y(t=60) as a denominator.

regards, chip
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