f

Re: Re: Please help: How to use Mathematica to get Parametric solution for a transcendental equation?

Sorry. I just copied from Mathematica and pasted it. I have no idea why it

The function is just one:

A*(1+G)+a*[(1+G)^t]=T, where A,a,t,T are all positive real numbers, and t is
no larger than 1.

Thanks.

Leon

On 9/2/06, Jens-Peer Kuska <kuska@informatik.uni-leipzig.de> wrote:
>
> Hi,
>
> can you try to use correct Mathematica syntax ??
> This means is D7(1+G) an algebraic bracket or a function
> call and is D7[(1+G)^t] an algebraic bracket or a function
> call and what does "====" mean together with "=" and
> have you a single equation or several ones ??
> Regards
>   Jens
>
>
> > Leonxf wrote:
> > Dear All,
> >
> > I am new to the group. Nice to have all of you here.
> >
> > I have an urgent question: I have a transcendental equation, which is
> >
> > A=D7(1 + G) + a=D7[(1 + G)^t] ==== T, where A,a,T,t are unknown, but
> A,a,T,=
> > t
> > are all positive real numbers, and t is no greater than 1. (I use ^ sign
> > here to indicate "to the power of").
> >
> > What I hope to get is a parametric expression for G, say, G==f(A,a,t,T)
> for
> > some fucntion f().
> >
> > Is there anyway to do that? I noticed someone asked a similar question
> > before, and an answer is to get Taylor Series, if that is the only way,
> how
> > to do that?
> >
> > Thank you in advance.
> >
> > Leon
> >
>
> 0 9/5/2006 9:54:50 AM comp.soft-sys.math.mathematica  28821 articles. 0 followers. 2 Replies 443 Views Similar Articles

[PageSpeed] 20

Hi,

and you still don't like the Mathematica syntax ??
You mean
A*(1 + G) + a*((1 + G)^t) == T

because the "[", "]" symbols indicate a function
call and

*not* a algebraic grouping. And you can't find a
algebraic solution

to a transcendental equation. You can try to use a
Taylor series like

Solve[Normal[Series[A*(1 + G) + a*((1 + G)^t), {G,
1, 4}]] == T, G]

and look how wide is the convergence radius.

Regards

Jens

"Leonxf" <leonxf@gmail.com> schrieb im Newsbeitrag
news:edjhha\$lvs\$1@smc.vnet.net...
| Sorry. I just copied from Mathematica and pasted
it. I have no idea why it
| becomes something unreadable.
|
| The function is just one:
|
| A*(1+G)+a*[(1+G)^t]=T, where A,a,t,T are all
positive real numbers, and t is
| no larger than 1.
|
| Thanks.
|
| Leon
|
|
| On 9/2/06, Jens-Peer Kuska
<kuska@informatik.uni-leipzig.de> wrote:
| >
| > Hi,
| >
| > can you try to use correct Mathematica syntax
??
| > This means is D7(1+G) an algebraic bracket or
a function
| > call and is D7[(1+G)^t] an algebraic bracket
or a function
| > call and what does "====" mean together with
"=" and
| > have you a single equation or several ones ??
| > Regards
| >   Jens
| >
| >
| > > Leonxf wrote:
| > > Dear All,
| > >
| > > I am new to the group. Nice to have all of
you here.
| > >
| > > I have an urgent question: I have a
transcendental equation, which is
| > >
| > > A=D7(1 + G) + a=D7[(1 + G)^t] ==== T, where
A,a,T,t are unknown, but
| > A,a,T,=
| > > t
| > > are all positive real numbers, and t is no
greater than 1. (I use ^ sign
| > > here to indicate "to the power of").
| > >
| > > What I hope to get is a parametric
expression for G, say, G==f(A,a,t,T)
| > for
| > > some fucntion f().
| > >
| > > Is there anyway to do that? I noticed
someone asked a similar question
| > > before, and an answer is to get Taylor
Series, if that is the only way,
| > how
| > > to do that?
| > >
| > > Thank you in advance.
| > >
| > > Leon
| > >
| >
| >
| 0 9/6/2006 8:35:51 AM
In article <edjhha\$lvs\$1@smc.vnet.net>, Leonxf <leonxf@gmail.com>
wrote:

> The function is just one:
>
> A*(1+G)+a*[(1+G)^t]=T, where A,a,t,T are all positive real numbers, and t is
> no larger than 1.

Consider the equivalent transcendental equation

a g + b g^t == c

where a > 0, b > 0, c > 0, 0 < t <= 1, and you want to find g.

Using series reversion:

http://mathworld.wolfram.com/SeriesReversion.html

one obtains a series expansion of the solution (about g == 1 to 2nd
order):

ser = InverseSeries[ Series[ a g + b g^t, {g, 1, 2}], c ]

2              2
(-a - b + c)   (b t - b t ) (-a - b + c)                 3
1 + ------------ + -------------------------- + O[-a - b + c]
a + b t                        3
2 (a + b t)

One can choose a different expansion point and compute to higher order.

For, say, a -> 3, b -> 1, c -> 2.5 and t -> 0.7 one obtains

Normal[ser]/. {a -> 3, b -> 1, c -> 2.5, t -> 0.7}

0.5992586816180681

Using FindRoot, one obtains

FindRoot[a g+b g^t == c /. {a -> 3, b -> 1, c -> 2.5, t -> 0.7}, {g, 1}]

{g -> 0.600165573071457}

Cheers,
Paul

_______________________________________________________________________
Paul Abbott                                      Phone:  61 8 6488 2734
School of Physics, M013                            Fax: +61 8 6488 1014
The University of Western Australia         (CRICOS Provider No 00126G)
AUSTRALIA                               http://physics.uwa.edu.au/~paul 0 9/7/2006 8:47:04 AM