Solving an nonlinear equation

[amirhossein20n]

hello everybody

can anybody help me

i want to solve this nonlinear equation in form of closed solution , i mean not by usual nonlinears solution such newton method and other estimation methods ... . the form of solution that by changing the defind parameter , the solution can be observed.

sincerely



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reminding that a,b & c are arbitrary numbers but cant be eliminated or combined. thanks

 

[c_mitra]

1. j is square root of -1?
2. tanhx is the hyperbolic function?
3. you want to find x: can you use approximations?

Case 1: x is small;

expand the exponential and the hyperbolic function and keep upto first order and solve...

 

[amirhossein20n]

hi c_mitra
yes, 1 to 3 assumption are correct ,
and i tried for expanding the expo fuction,but i didnt reach to the solution
thanks

 

[amirhossein20n]

though by multiplying the left side of equa, in exp(x) we have :

 

[pancho_hideboo]

Second order approximation.

 

[albbg]

We know that

tanh(x) = [exp(2*x)-1]/[exp(2*x)+1] from which

[a+b*exp(-2*x)]/[a-b*exp(-2*x)] = c/(jx)*[exp(2*x)-1]/[exp(2*x)+1]

if jx <> 0 we can write:

jx*[a+b*exp(-2*x)]*[exp(2*x)+1]=c*[a-b*exp(-2*x)]*[exp(2*x)-1].......(eq. 1)

we can see the left side is only imaginary while the right is only real. The form is:

0+j*x*A=B+j*0

so we have to solve the system of two equations j*x*A = 0 and B = 0.

that means, exapanding eq. 1:

a*exp(2*x)+b*exp(-2*x)+(a+b) = 0
a*exp(2*x)+b*exp(-2*x)-(a+b) = 0

can be noticed that the parameter "c" plays no role.
From the system can be seen that must be (a+b) = 0 ==> a = -b otherwise there is no solution. But if a = -b we can write:

exp(2*x)-exp(-2*x)=0

calling y=exp(2*x) we have

y-1/y=0 with y <> 0

y^2-1 = 0 from which y=+/-1

but x = ln/2 so the only possible solution is y=1 that lead to

x=ln(1)/2 = 0

but we see that must be jx <> 0

so seems that your equation has no solution.

 

[c_mitra]

so seems that your equation has no solution.

That can be easily verified:

You are using matlab, right (It is so expensive, I just use octave)? Plot the left hand side as a function of x; also plot the right hand side as a function of x. The point where they intersect is the solution.

You need to plot several times and expand to see that the intersection is real. That will give you a rough idea.

 

[albbg]

That can be easily verified:

You are using matlab, right (It is so expensive, I just use octave)? Plot the left hand side as a function of x; also plot the right hand side as a function of x. The point where they intersect is the solution.

You need to plot several times and expand to see that the intersection is real. That will give you a rough idea.

Not completely true. By numerical methods you can verify for some a, b and c value not for ALL a, b, c values.

However we can also see this way:

[a+b*exp(-2*x)]/[a-b*exp(-2*x)] = c/(jx)*tanh(x) can be written as:

[a+b*exp(-2*x)]/[a-b*exp(-2*x)]*1/tanh(x) = c/(jx)

as did in my previous post we have to equate imaginary and real parts to 0, that is:

[a+b*exp(-2*x)]/[a-b*exp(-2*x)]*1/tanh(x)=0
c/x=0

We can see that the last one can be verified only for x --> +/- inf
that is no finite solutions are possible.

 

[amirhossein20n]

hi pancho_hideboo

excuse me i cant understand what does it do ??
can i ask what the software is ?
thanks

 

[pancho_hideboo]

excuse me i cant understand what does it do ??
can i ask what the software is ?

Wolfram Mathematica.
http://en.wikipedia.org/wiki/Wolfram_Mathematica

Mathematica and Maple are major as symbolic mathematical analysis tool.
http://en.wikipedia.org/wiki/Maple_(software)
http://en.wikipedia.org/wiki/List_of_computer_algebra_systems

Maxima is a freeware for same purpose.
http://en.wikipedia.org/wiki/Maxima_(software)
http://maxima.sourceforge.net/

 

[amirhossein20n]

dear friend

excuse me ,i think my explanation was a little ambiguous .

i wanna illustrate clearly that

a,b are real and are given by question a = 50 , b=-1 ;

now we wanna write(modify) x as a function of c : x=X(c);

so that c is input parameter and x is output ;

thanks so much

amir

 

[amirhossein20n]

hi albbg

you supposed that a = -b but in my problem a=50 , b =-1 see that theyre not equal .
i think this imagination lead you to mistakes and causes no solution

 

[c_mitra]

Using albbg idea:

assume a, b, c are real: then left hand side is real but the right hand side is imaginary (contains j)

if you want a real solution x must be real;

a+b*(exp(-2*x)=0;
exp(-2*x)=-a/b;
-2*x=ln(-a/b);

Note that we do not use the right hand side at all.

If a, b, c are all complex and the solution is also, x (we use z in such cases) is also complex;

then the albbg idea does not work.

 

[pancho_hideboo]

Using albbg idea:
assume a, b, c are real: then left hand side is real but the right hand side is imaginary (contains j)
if you want a real solution x must be real;

Real value solution for x can not exist.
x must be complex number.

a,b are real and are given by question a = 50 , b=-1 ;
now we wanna write(modify) x as a function of c : x=X(c);

Such closed form solution can not exist for your nonlinear equation.

However we can solve this equation numerically while varying c value.

 

[albbg]

then the albbg idea does not work.

Why not ?

You cannot ignore the imaginary part coming from rhs

amirhossein20n: I just proof that if a solution exist it was only in case a=-b. But we se nevet that case has solution

I presumed x as real number. In case x can be in the form x=a+jb the solution can exists.

 

[c_mitra]

Real value solution for x can not exist.
x must be complex number..

As the OP has supplied the both a and b, you can see the equation(s) has a solution;

In fact this is a pair of equation (because of j) and the LHS is equal to RHS equal to ZERO

You can plot the left hand side as a graph and see where it intersects the zero line.

You can also superimpose the plot of RHS (for several values of c) on the same graph.

 

[pancho_hideboo]

As the OP has supplied the both a and b, you can see the equation(s) has a solution;

In fact this is a pair of equation (because of j) and the LHS is equal to RHS equal to ZERO

You can plot the left hand side as a graph and see where it intersects the zero line.

You can also superimpose the plot of RHS (for several values of c) on the same graph.

Do you really think so ?
RHS is -j*c*tanh(x)/x.
-j*c*tanh(0)=0.
-j*c*tanh(0)/0=-j*c.

tanh(x)/x can not be zero for real number x.

Real number solution for x can never exist for this nonlinear equation as far as a, b, and c are all real number.

 

[amirhossein20n]

yes a,b,c are real

and x is complex in form of Q+jk , for exapmle.

 

[albbg]

I'm sorry but for x real and c <> 0 a soltution does NOT exist.

 

[c_mitra]

Do you really think so ?
RHS is -j*c*tanh(x)/x.
-j*c*tanh(0)=0.
-j*c*tanh(0)/0=-j*c.
.

The next step you missed: c must be zero.

from wolframalpha.com, I see that tanh(x)/x is zero only at +/- inf.

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I'm sorry but for x real and c <> 0 a soltution does NOT exist.

a+ib=c+id means a=c and b=d. (if a,b,c,d are real)

the complex equation is in reality two real equations. The two equations (a=c and b=d) may not have identical solutions (they will have two solutions)

In the present case, the right hand side does not have a solution in x except the trivial solution of c=0

The solution of the LHS is not in dispute.