Thread Subject:

Numerical Iteration

Hello,

I want to solve this equation using numerical iteration:

x= ln ( (A*B - C*x)/C) * D

Thanks in advance for anyone that helps.

"Abdalla Mohamed" <mangawy_21@hotmail.com> wrote in message <gj6nh1$b0r$1@fred.mathworks.com>...
> Hello,
>
> I want to solve this equation using numerical iteration:
>
> x= ln ( (A*B - C*x)/C) * D
>
> Thanks in advance for anyone that helps.

  One way is to use the Newton-Raphson method. Define f(x) as

 f(x) = log((A*B-C*x)/C)*D-x

Use your knowledge of calculus (or the Symbolic Toolbox) to compute the derivative, f'(x). Then make an initial guess, x(0), and start the iteration

 x(n+1) = x(n) - f(x(n))/f'(x(n))

Keep it going until x(n) converges to some limiting value, and that value will be your solution, since f(x(n)) will necessarily have approached zero.

  Hint: use a 'while' loop to accomplish this iteration using as its exiting condition whatever your criterion is for convergence.

  Have you despaired of using the 'lambertw' function? It works pretty well. I've tried it.

Roger Stafford

"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <gj6qgl$6v4$1@fred.mathworks.com>...
> "Abdalla Mohamed" <mangawy_21@hotmail.com> wrote in message <gj6nh1$b0r$1@fred.mathworks.com>...
> > Hello,
> >
> > I want to solve this equation using numerical iteration:
> >
> > x= ln ( (A*B - C*x)/C) * D
> >
> > Thanks in advance for anyone that helps.
>
> One way is to use the Newton-Raphson method. Define f(x) as
>
> f(x) = log((A*B-C*x)/C)*D-x
>
> Use your knowledge of calculus (or the Symbolic Toolbox) to compute the derivative, f'(x). Then make an initial guess, x(0), and start the iteration
>
> x(n+1) = x(n) - f(x(n))/f'(x(n))
>
> Keep it going until x(n) converges to some limiting value, and that value will be your solution, since f(x(n)) will necessarily have approached zero.
>
> Hint: use a 'while' loop to accomplish this iteration using as its exiting condition whatever your criterion is for convergence.
>
> Have you despaired of using the 'lambertw' function? It works pretty well. I've tried it.
>
> Roger Stafford



Hey Rogers,

I wrote this code following ur instructions, given that i am a new user,and i am getting an error:

A=3.9387;
B=0.099077;
C=2.5776;

f(x)=log((A-x)/B)*C-x; %% Undefined function or variable 'x'.
f1(x)=(-C/(A-x))-1;

x(0)= 21.7;
n=0:1:10;
x(n+1) = x(n) - y(x(n))/y1(x(n));
 
Please help me debug this error, and also, is this going to function properly ?

Thanks for your time

Abdalla Mohamed wrote:
> "Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <gj6qgl$6v4$1@fred.mathworks.com>...
>> "Abdalla Mohamed" <mangawy_21@hotmail.com> wrote in message <gj6nh1$b0r$1@fred.mathworks.com>...
>>> Hello,
>>>
>>> I want to solve this equation using numerical iteration:
>>>
>>> x= ln ( (A*B - C*x)/C) * D
>>>
>>> Thanks in advance for anyone that helps.
>> One way is to use the Newton-Raphson method. Define f(x) as
>>
>> f(x) = log((A*B-C*x)/C)*D-x
>>
>> Use your knowledge of calculus (or the Symbolic Toolbox) to compute the derivative, f'(x). Then make an initial guess, x(0), and start the iteration
>>
>> x(n+1) = x(n) - f(x(n))/f'(x(n))
>>
>> Keep it going until x(n) converges to some limiting value, and that value will be your solution, since f(x(n)) will necessarily have approached zero.
>>
>> Hint: use a 'while' loop to accomplish this iteration using as its exiting condition whatever your criterion is for convergence.
>>
>> Have you despaired of using the 'lambertw' function? It works pretty well. I've tried it.
>>
>> Roger Stafford
>
>
>
> Hey Rogers,
>
> I wrote this code following ur instructions, given that i am a new user,and i am getting an error:
>
> A=3.9387;
> B=0.099077;
> C=2.5776;
>
> f(x)=log((A-x)/B)*C-x; %% Undefined function or variable 'x'.
> f1(x)=(-C/(A-x))-1;
 >
> x(0)= 21.7;

How did you choose this initial value? Did you plot the function f(x)?
Or is this just a typo and you meant 2.17?

With the above values for A, B, and C, assuming you are looking for real
values, the function f(x) is 2.5776*log(39.75392878-10.09315987*x)-x,
and it is defined only when 39.75392878 - 10.09315987 * x < 0.

> n=0:1:10;
> x(n+1) = x(n) - y(x(n))/y1(x(n));

Looks like some typos here: y and y1 should be f and f1, shouldn't they?

Also, note that Roger spoke about a *while* loop and you did not specify
a convergence test (stopping criterion) either.

> Please help me debug this error, and also, is this going to function properly ?
>
> Thanks for your time

Regards,
-- Jean-Marc