In order to get your "obvious" answer -exp(-n(cos(phi)+sin(phi))) from the simplify command, it is pretty obvious to me that % must evaluate to -2*exp(-n(cos(phi)+sin(phi))), because it is pretty obvious to me that if

simplify(%+exp(-n(cos(phi)+sin(phi)))) = -exp(-n(cos(phi)+sin(phi)))

then

% = -2*exp(-n(cos(phi)+sin(phi)))

Which means that the humungous expression in n, (involving n⁸ and all sorts of intersting square root terms) must evaluate to -2 for all values of n. Possible, but at first sight a little unlikely - so it is not obvious to me

Perhaps if you posted the code in a usable format using the big green up-arrow (right-hand-end of the second row of toolbar icons), I might be able to check this interestingly complicated expression in n, which always evaluate to -2.

@zia9206314 

If you check my earlier post, you will see that the reason it has a solution for 0.2 is that (1+0.2)/(2*0.2)=4 which results in an integer power for eta. Whenever (1+n)/(2n)=p (an integer one gets a solution. Hence it has a solution whenever n=1/(2p-1) - so in fact there are an infinite number of values of n between 0 and 1, for which the original ODE has a power series solution, n=1/3, 1/5. 1/7 etc

And I missed all of them :-(

@Carl Love 

Should have made it clear in my earlier post that I can get series solutions for both equations for n=1. But for the reasons described in my earlier post, I suspect it works for no other integer n

I'm assuming that Mathematica can read MathML files - I don't have this Mathematica so I can't check, but I'd be surprised if it couldn't

Check the MathML:-Export  in maple help: this should(?) allow you to convert/export maple code to MathML which then (hopefully!) Mathematica can read.

I suspect the success/failure of this operation will be very dependent on the complexity of the code being transferred, so I'd be tempted to try it with a relatively simple example first!!!

@Rouben Rostamian  

I think another problem is the existence of eta^((n+1)/(2*n)) in your equation which means that powers of eta are indeterminate (cos n is unknown).

Setting n=1, so that eta^((n+1)/(2*n))=eta, I can get a power series solution

Setting n=any other (positive value) gives non-integer powers of eta, which I suspect means that no power series solution is possible.

The power series method starts with the hypothesis that f(eta) is a sum of non-negative powers of eta, so that the second differential is also a sum of non-negative powers of eta. However the remaining term in your ODE will be the product of such a sum with a rational (non-integer) power of eta: in other words it will be a sum of rational (non-integer) powers of eta.

Since the power series method for solving ODEs involves (more-or-less) equating like powers of the variable, this is doomed to fail, because the equation provides a set of distinct integer powers and a set of distinct rational (non-integer) powers. No equality will be possible:-(

@Carl Love 

That'll teach me to fix the obvious problem - and then stop looking!

@still0fthenight 

Just reading your code, and with no testing - you have defined CPeqz twice, so that the second one will overwrite the dirt one. I'm guessing that the second of these definitions should have been CPeqw??

Ignore my first attempt - parenthesis in the wrong place. I've added the necessary code to your worksheet - attache below

NMWO_mod.mw

1. No-one is going to retype your code in order to debug your problem, Don't post a picture of your code. Post the actual code (eq a .mw file) using the big green up-arrow - right hand end of the second row in the toolbar
2. This looks suspicioulsy like a "homework" problem - many here don't do "homework" problems on principle.

@Rouben Rostamian  

Like Rouben says lots of variables in this issue. Attached is a zip file comprising a maple worksheet and a text file. The text file was created from the Maple worksheet using the simple-minded process: left-click,swipe, control-C. It was dumped into the text editor (happens to be notepad++) using control-V.

try.zip

As you can see, for 1-D input it doesn't matter how one handles execution groups - they just "disappear". I do this sort of operation quite a lot because I develop stuff interactively in Maple, shift the code into a text file, prettify/comment the text file and save as xxxxx.mpl. Then insert an appropriate read statement in the Maple worksheet. Just happens to be the way I work, and everyone else when have the favourite work process, so I'm not particularly defending this one

I've never seen the issue of code collapsing into a single line, but

1. I only ever use 1-D math input in Maple: (I never use 2-D input or code regions, cos when I tried, I got too many hard-to-debug issues)
2. Not sure what you mean by IDE - I know that there is now a "Maple IDE" available, but I have never tried it, because everything I read says that it has "issues". I stick strictly to Maple 1-D input

If you posted a copy of the starting Maple worksheet, however you are developing it, 1-D input, 2-D input, code regions, IDE, whatever and the resulting text file then I might be able to help further

@Alejandro Jakubi 

All that the quoted thread demonstrates is that "theorem proving" - very roughly speaking "equivalence checking" is very difficult. This I have reluctantly come to accept. However such equivalence checking is a relatively common requirement for many Maple users, and I/they need to know what is the best pragmatic solution, given that perfection is unachievable.

My (admittedly unscientific conclusions) so far, are

1. In a programmatic situation with no foreknowledge of the expressions to be compared, and no opportunity to use a calibrated eyeball - then use 'is'
2. In an interactive situation then apply the calibrated eyeball and use verify/option, with the option depending on the nature of the expressions to be compared
3. I am open to any other pragmatic solutions

@Axel Vogt 

Many solution methods which can be proven to work on linear systems will also work on some (mildly?) non-linear systems. The problem is that they cannot be proven to work on all non-linear systems.

If your assertion is that the Gauss-Seidel method works on all non-linear systems, then I suggest that you put forward a proof. (also update the Wikipedia page http://en.wikipedia.org/wiki/Gauss%E2%80%93Seidel_method) whihc states (some emphasis added)

the Gauss–Seidel method, also known as the Liebmann method or the method of successive displacement, is an iterative method used to solve a linear system of equations.

and the Maple definition page (some emphasis added)

method of successive displacements
or Gauss-Seidel iteration, n. the variant of the Jacobi iteration, in which one immediately uses the new information. This usually out-performs the Jacobi method. Explicitly, this yields the iterative method of solving a linear matrix system
                            A x = b
 by repeatedly solving for xNEW from
"x[NEW(i)] = 1/(a[ii])[b[i] -(∑)a[ij]x[OLD(j)]-(∑)a[ij]x\

  [NEW(j)]]"
This assumes that
                           a[ii] <> 0
, and will not always converge.

In the paper you cite, the relevenat comment appears to be

An improvement, analogous to the Gauss-Seidel method for linear systems, of fixedpoint
iteration can be made.

Sorry, but a method using something analogous to the Gauss-Seidel Method, is not the Gauss-Seidel method

On the other hand, do I accept that methods recommended for linear problems will also (sometimes) solve non-linear problems  - of course I do. Would I recommend that you try this? - as a general rule no.

may be a language problem, but to me, this question does not make any sense. Please try rephrasing/clarifying

@mathieulibert 

Just because you want an analytic solution doesn't mean one exists:-(

If you just want to look at the sensitivity you could replace your solve command with

for f1 from 0.3 to 3 by 0.1 do
#
# need to change the below to store all solutions
#
    ABC:=fsolve({A, B, C}, {a, b, c});
od

which will provide solutions at all of the f1-values. With some rewriting of subsequent code, you could then use a pointplot command to plot a versus f, b versus f, c versus f

First thing I would do is fix the recursive assignment error - the fourth element in the section headed Calibration - see if that helps

@mathieulibert