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These are replies submitted by Abdoulaye

@ilyasweb  I don't understand what you wrote ??

@Kitonum Thanks  Kitonum 13190  !


Thanks @radaar 

I will check also the newton raphson method by approximating integral by middlesum.

@vv 5987

Yes it is true. An interpolation is needed to show the convergence.

@ vv 5982

I am thinking about Riemann Sum for the Integral term :   int ( F( x-z, 2*y*exp(-z) ), z=0..+infinity )


Thank you  vv 5982 

Let me check your compuation and see how we can compute F3(x,y). This is the main difficulty of the problem, since we cannot find S2(y) analytically. The ''fsolve or solve function'' will not work for S2(y).   This implies that we will not be able to compute U2(x,y) analytically, so F3(x,y) will not be analytic. 


This is the situation : 

F1(x,y)  [ is analytic ]      and     S1 (y) [ is analytic ]       

U1(x,y) [ is analytic ]   implies   F2(x,y)  [ is analytic ]     but     S2 (y) [ is not analytic ]   

U2(x,y) [ is not analytic ]   implies   F3(x,y)  [ is not analytic ]     and    S3 (y) [ is not analytic ]     

                                     . . . . . . . .

Un(x,y) [ is not analytic ]   implies   Fn+1(x,y)  [ is not analytic ]     and    Sn+1 (y) [ is not analytic ]     


N.B.  The approach for solving this problem is a Spline Interpolation solution and show convergence. 

@vv 5982. Thanks !

Your idea is interesting but I am not familiar with a Spline Interpolation. Here Fn at the data points (xi, yi) is of size M*N but can we get analytical expressions for all  Fn(x,y)  and Un(x,y) ?  Can you please Show F2(x,y), U1(x,y) and S2(y) with for example  the following  data ? 



@Rouben Rostamian  OK. Thank you !

@Rouben Rostamian  Do you have an idea to how to iterate numerically this process ? 

@Rouben Rostamian  Thank you. Yes, U1 depends on (x,y). I have attached the file






F1 := proc (x, y) options operator, arrow; (9/8)*y+9/8-(21/8)*y*exp(-2*x)-(21/8)*exp(-x) end proc:



S1 := proc (y) options operator, arrow; solve(F1(x, y), x)[1] end proc:




g := proc (U) options operator, arrow; U+S1(2*y*exp(-U)) end proc;

proc (U) options operator, arrow; U+S1(2*y*exp(-U)) end proc


U1 := proc (x) options operator, arrow; solve(g(U) = x, U)[1] end proc:




F2 := proc (x, y) options operator, arrow; (9/8)*y+9/8-(21/8)*y*exp(-2*x)-(21/8)*exp(-x)+(3/4)*(int(F1(x-z, 2*y*exp(-z)), z = 0 .. U1(x))) end proc;

proc (x, y) options operator, arrow; (9/8)*y+9/8-(21/8)*y*exp(-2*x)-(21/8)*exp(-x)+(3/4)*(int(F1(x-z, 2*y*exp(-z)), z = 0 .. U1(x))) end proc




F2(x, y);





Download question.mw

@Rouben Rostamian  

It is a typo. I have corrected it 

i) We define g( U ) = U + S1( 2*y*exp(-U) )

ii) We find the function U1(x) solution of g( U1(x) ) = x. 


Thanks !


@Carl Love Thank you Carl,


Can you please check the infnorm from the worksheet, I got   0.3096747114  ? See attached the file norm.mw


@Carl Love  That's true ! Thanks !


1)  Can we compute the error  ( x(t-2IPi) - x(t) )  without plotting  abs(  x( t-2*Pi ) - x(t) ) ?

     i.e. using norm 1, 2 or infinity  ?


2)  what is op( [ a, b, c] , F ),  where F is a piecewise function ?

@Carl Love OK I see... 

1) Do I need to add  1e-7  if I  alternate <= with < ?

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