Re: making a e-funtion of two coördinates

• To: mathgroup at smc.vnet.net
• Subject: [mg31630] Re: making a e-funtion of two coördinates
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Mon, 19 Nov 2001 03:11:11 -0500 (EST)
• References: <9t86cv\$r6l\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Roderik,

Make the equations derived from the data

e = y == U*E^(a*t) /. {{t -> t1, y -> y1}, {t -> t2, y -> y2}}

{y1 == E^(a*t1)*U, y2 == E^(a*t2)*U}

We need to help Solve by taking Logs

e2=Map[Log, eqns,{2}]

{Log[y1] == Log[E^(a*t1)*U], Log[y2] == Log[E^(a*t2)*U]}

The problem is that, by default,  Mathematic works with complex numbers
but we can FullSimplify assuming that a,t1,t2 are real

e3=FullSimplify[eqns2, {a,t1,t2}\[Element]Reals]

{Log[y1] == a*t1 + Log[U], Log[y2] == a*t2 + Log[U]}

Now we can solve

s=Solve[e3, {a,U}]

{{a -> (-Log[y1] + Log[y2])/(-t1 + t2),
U -> E^(((-t2)*Log[y1] + t1*Log[y2])/(t1 - t2))}}

Check that this is a solution

e/.s

{{y1 == E^((t1*(-Log[y1] + Log[y2]))/(-t1 + t2) +
((-t2)*Log[y1] + t1*Log[y2])/(t1 - t2)),
y2 == E^((t2*(-Log[y1] + Log[y2]))/(-t1 + t2) +
((-t2)*Log[y1] + t1*Log[y2])/(t1 - t2))}}

Simplify[%]

{{True, True}}

So it is a solution in with no assumptions needed.
Of course if we allow complex numbers, as Mathematica does, then  there will
be other solutions.

--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"TheSquaredBun" <veezdREMOVETHIS at hotmail.com> wrote in message
news:9t86cv\$r6l\$1 at smc.vnet.net...
>
> I have got two coordinates out of graph. The line through the
> coordinates could be described as:
>
> y = U * e^(a * t)
>
> (e is Exponetial e (approx. 2.37))
>
> t is the x-value, I have give two points (t1,y1) and (t2,y2)
>
> I have tried several things (Solve and Fit for example), but I can't
> get the function.
> Does anyone know what should be the input to let Mathematica solve the
> a and U in the formula.
>