can some one tell me how to find values of si1(0) and si1(1) usng matlab?

[Image Analyst]

Which psi1? The top one or the bottom one? Did you try things like

c1 = sqrt(h - sqrt(h^2-1)) / sqrt(h^2-1);
c2 = sqrt(h + sqrt(h^2-1)) / sqrt(h^2-1);
A = sqrt(h^2-1) * (x + 1i * h * t);

and so on?

[Heya :)]

si1(0) which is written after where.

syms x t f;
h=1+f^2;
L=1i*h;
A=sqrt(h^2-1)*(x+1i*h*t);
C1=sqrt(h-sqrt(h^2-1))/sqrt(h^2-1);
C2=sqrt(h+sqrt(h^2-1))/sqrt(h^2-1);
X1=1i*(C1*exp(A)-C2*exp(-A))*exp(-1i*t/2);
Y1=(C2*exp(A)-C1*exp(-A))*exp(1i*t/2);
T1=taylor(X1,f);

I am using this code, but it says that the Taylor function for

T1 = taylor(X1,f);

does not exist.

[Image Analyst]

There is no such function unless you write it.

[Heya :)]

means we cannot exapnd it using taylor?

[Walter Roberson]

The article appears to be https://arxiv.org/pdf/1108.2867.pdf

Nonlinear Schrödinger Equation: Generalized Darboux Transformation and Rogue Wave Solutions, by Boling Guo, Liming Ling, and Q. P. Liu

[Rena Berman]

(Answers Dev) Restored Question.

Answer 1

[Walter Roberson]

taylor() can only be used for functions whose differentiation can be evaluated at the point of expansion.

Your expression has a denominator of sqrt(h^2-1) which is a problem if h is 1. You have h = 1 + f^2 so that is equal to 1 when f is 0. Therefore your expression has a division by 0 at f = 0. Such a function has no taylor expansion.

You can create a series() expansion of X1 around f = 0, but it will not look like what they show, and furthermore for the same reason the series does not really exist at 0. The expression involves csgn(), the Complex Sign operator, and a division by f; if you restrict to real valued f (we cannot read the image well enough to tell if that is justified) then you end up with signum(f)/f and then you have the question of what signum(0) is. You could potentially rewrite the signum(f)/f in terms of abs(), in which case it comes out as 1/abs(f), which gives you a leading term of

-(1/2)*exp(-(1/2*I)*t)*sqrt(2)*(-i+I)/abs(f)

Perhaps everything except for the abs(f) could be rewritten to look like what they wrote, but the abs(f) cannot be eliminated.

Your C1 has a limit of infinity near h = 1 and there really is no way around that: any expansion around that point is going to be infinite.

You could potentially expand around f = epsilon instead of around f = 0, and then take the limit as epsilon approaches 0... but the limit does not exist.

Either you have made a typing mistake or else they describe something that we viewers cannot read (the image quality is poor), or else what they describe is incorrect.

[Heya :)]

thanku walter for your ans inspite of ur variable health.actually we dont have to put f=0 in taylor series.let me give u an example

>> syms k c t y x;
>> y=(k*(x+4*k^2+6*c)*t);
>> taylor(sin(y),k)

ans =

(t*((t^4*(6*c + x)^4)/120 - (4*t^2*(6*c + x))/3)*(6*c + x) - (2*t^3*(6*c + x)^2)/3)*k^5 + (4*t - (t^3*(6*c + x)^3)/6)*k^3 + t*(6*c + x)*k

in this particular example i expand taylor series of sin with respect to k so si1[0]=t*(6*c + x) and si1[1]=(4*t - (t^3*(6*c + x)^3)/6)* and so on..... so in my pic ans i want to expand taylor series w.r.t f buh matlab is not solving f terms

>> syms x t f;
h=1+f^2;
L=1i*h;
A=sqrt(h^2-1)*(x+1i*h*t);
C1=sqrt(h-sqrt(h^2-1))/sqrt(h^2-1);
C2=sqrt(h+sqrt(h^2-1))/sqrt(h^2-1);
X1=1i*(C1*exp(A)-C2*exp(-A))*exp(-1i*t/2);
>> taylor(X1,2,f^2);
>> collect(ans,f)

ans =

(-(1/exp((t*i)/2))*(exp((t*(f^2*i + i) + f^2)*((f^2 + 1)^2 - 1)^(1/2))*(f^2 - ((f^2 + 1)^2 - 1)^(1/2) + 1)^(1/2)*i + ((((f^2 + 1)^2 - 1)^(1/2) + f^2 + 1)^(1/2)*i)/exp((t*(f^2*i + i) + f^2)*((f^2 + 1)^2 - 1)^(1/2))))*f^2 + (1/exp((t*i)/2))*((exp((t*(f^2*i + i) + f^2)*((f^2 + 1)^2 - 1)^(1/2))*(f^2 - ((f^2 + 1)^2 - 1)^(1/2) + 1)^(1/2)*i)/((f^2 + 1)^2 - 1)^(1/2) - ((((f^2 + 1)^2 - 1)^(1/2) + f^2 + 1)^(1/2)*i)/(exp((t*(f^2*i + i) + f^2)*((f^2 + 1)^2 - 1)^(1/2))*((f^2 + 1)^2 - 1)^(1/2))) + x*(1/exp((t*i)/2))*(exp((t*(f^2*i + i) + f^2)*((f^2 + 1)^2 - 1)^(1/2))*(f^2 - ((f^2 + 1)^2 - 1)^(1/2) + 1)^(1/2)*i + ((((f^2 + 1)^2 - 1)^(1/2) + f^2 + 1)^(1/2)*i)/exp((t*(f^2*i + i) + f^2)*((f^2 + 1)^2 - 1)^(1/2)))

>> pretty(ans)

/              /                                                                                          2     2     1/2    2     1/2        \ \ 
|       1      |          2           2     2     2     1/2    2      2     2     1/2     1/2         (((f  + 1)  - 1)    + f  + 1)    i      | |  2 
| - ---------- | exp((t (f  i + i) + f ) ((f  + 1)  - 1)   ) (f  - ((f  + 1)  - 1)    + 1)    i + ------------------------------------------- | | f  + 
|      / t i \ |                                                                                           2           2     2     2     1/2  | | 
|   exp| --- | \                                                                                  exp((t (f  i + i) + f ) ((f  + 1)  - 1)   ) / | 
\      \  2  /                                                                                                                                  / 

                /          2           2     2     2     1/2    2      2     2     1/2     1/2                       2     2     1/2    2     1/2                 \ 
         1      | exp((t (f  i + i) + f ) ((f  + 1)  - 1)   ) (f  - ((f  + 1)  - 1)    + 1)    i                 (((f  + 1)  - 1)    + f  + 1)    i               | 
     ---------- | ------------------------------------------------------------------------------ - -------------------------------------------------------------- | + 
        / t i \ |                                  2     2     1/2                                          2           2     2     2     1/2     2     2     1/2 | 
     exp| --- | \                               ((f  + 1)  - 1)                                    exp((t (f  i + i) + f ) ((f  + 1)  - 1)   ) ((f  + 1)  - 1)    / 
        \  2  / 

                  /                                                                                          2     2     1/2    2     1/2        \ 
           1      |          2           2     2     2     1/2    2      2     2     1/2     1/2         (((f  + 1)  - 1)    + f  + 1)    i      | 
     x ---------- | exp((t (f  i + i) + f ) ((f  + 1)  - 1)   ) (f  - ((f  + 1)  - 1)    + 1)    i + ------------------------------------------- | 
          / t i \ |                                                                                           2           2     2     2     1/2  | 
       exp| --- | \                                                                                  exp((t (f  i + i) + f ) ((f  + 1)  - 1)   ) / 
          \  2  /

i.e its no solving (f^2)^1/2 terms

[Walter Roberson]

My system rejects doing the taylor at all there. Which MATLAB version are you using?

[Heya :)]

i am using R2011a

[Heya :)]

thanku for your rply....

[Heya :)]

Please help me coz I have to finish my thesis :( nd I am pretty much bad at MATLAB :(

[Walter Roberson]

The reason you are seeing a taylor expansion is that your X1 has multiple variables and you are not telling taylor which variable to use, so it is expanding around a different variable than your are expecting. You should be using

expression = X1;
variable = f;
order = 2;
expansion_location = 0;
taylor(expression, variable, order, expansion_location)

which will tell you

??? Error using ==> mupadmex
Error in MuPAD command: exp(-(t*I)/2)*((exp((x + t*(f^2*I + I))*((f^2 + 1)^2 - 1)^(1/2))*(f^2 - ((f^2 + 1)^2 - 1)^(1/2) + 1)^(1/2)*I)/((f^2 + 1)^2 - 1)^(1/2) -
((((f^2 + 1)^2 - 1)^(1/2) + f^2 + 1)^(1/2)*I)/(exp((x + t*(f^2*I + I))*((f^2 + 1)^2 - 1)^(1/2))*((f^2 + 1)^2 - 1)^(1/2))) does not have a Taylor series
expansion... [taylor]

The X1 you are giving does not have a taylor or series expansion near f = 0, for the reasons I explained above.

(A) You might have made a typing mistake. Or (B) in the part of the image that is not readable to us, there is some information or expression that would allow us to proceed. Or else (C) what they describe is incorrect.

I cannot assist you with (A) or (B) without seeing a clearer image. What you posted is dark and does not include much of the text, and my eyesight is not good today.

[Heya :)]

thanku again for ur rply walter. sorry for ot providing u with the clear picture earlier..

[Walter Roberson]

I have some things to do so it will be a few hours before I can look at it

[Walter Roberson]

I do not understand yet how they constructed ψ[1] in (14), so I do not understand where C1, C2 and A come from. It is true that individually C1 and C2 are singular at f = 0. However, within ψ[1] of (14), the limit as f approaches 0 is well defined, and equal to the values shown in (15) .

It appears to me that (15) is not the result of a taylor expansion or series expansion: that instead it is writing out the generalized Darboux transformation from just before "Proposition 1" or from Theorem 2 or from (8), and that the values are being read off in (15) by a matching process.

Anyhow, do not taylor() X1 from (14), just take the limit as f approaches 0; although C1 and C2 are individually singular at 0, they are only used as multipliers of exp() values and the limit is defined.

[Heya :)]

Okay I will look into it.btw thanku soooooo much for all ur time and effort.stay blessed.

[Heya :)]

walter you are genius :) look got my expression :P.i am so happy :)

[Heya :)]

>> syms x t f;
h=1+f^2;
L=1i*h;
A=sqrt(h^2-1)*(x+1i*h*t);
C1=sqrt(h-sqrt(h^2-1))/sqrt(h^2-1);
C2=sqrt(h+sqrt(h^2-1))/sqrt(h^2-1);
X1=1i*(C1*exp(A)-C2*exp(-A))*exp(-1i*t/2);
Y1=(C2*exp(A)-C1*exp(-A))*exp(1i*t/2);
limit(X1,f,0)
limit(Y1,f,0)

ans =

-(1/exp((t*i)/2))*(2*t - 2*x*i + i)

ans =

exp((t*i)/2)*(2*x + 1 + 2*t*i)

[Heya :)]

but i am not getting si1(1)....