Taylor serie limitation??

Question

MartinM 2018-4-25

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https://ww2.mathworks.cn/matlabcentral/answers/397236-taylor-serie-limitation

编辑： John D'Errico 2018-4-25

Dear all I am blocked with a taylor serie. Sorry the code is not very nice, but to summarized it, all the parameter ngas, A, B ...depend of lam. Together they creat BETA, wich depends on lam too. I need to found the taylor serie of BETA, but I have this warning : "Warning: Cannot compute a Taylor expansion of : " I acn found the Taylor serie of ngas, or A, or k_0 etc but not BETA...et I need to finish with taylorBETA=taylor(taylorBETA,lam,'ExpansionPoint', 1030e-9); If someone have an idea...? Thanks

   clc,
  clf
  clear all
  close all
  format long
  n_g = 1.45  ; % clad index 
  c=300000000;
  n=1e-6;
  E=1e-9;
  %%%%%%%%%%%%%%%%%%TO BE CHANGED%%%%%%%%%%%%%%%%
  r=30;
  T=840 ;
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  a=2.405*2.405;  %coef pour les ordres de bessel
  t=T*E;
  rc=r*n;
  dc=2*rc;
  int=0.0001;
  L_start=800 ;
  L_stop= 1300 ;
  x1=L_start * E;
  x2=L_stop * E;
  xx=(L_stop-L_start)/int;
  lambda = (x1 : 5e-9 : x2)';
  %syms variable
  %subs (f,variable,ordre)
  pgas = 150;
  pgas=pgas *1e5;
  pgas= pgas/101325;
  syms lam
  syms lamm
  ngas= 1 + pgas.* (   (3.22869e-3 ./ (46.301-(lam.*1000).^-2) ) + (3.55393e-3 ./ (59.578-(lam.*1000).^-2) ) + (6.06764e-2 ./ (112.74-(lam.*1000).^-2)  )); %XENON
  k_0= 2*3.14./(lam);
  phi = k_0.*t.* sqrt(n_g.^2-ngas.^2) ;
  epsi= n_g.^2 ./ ngas.^2;
  A=a./( 2.*ngas .*(k_0.*rc).^2);
  B=a./ ( ngas(i,j).^2 .*(k_0(i,j).*rc).^3  );
  C = epsi;
  D = 0.5 .* (C+1) ./ sqrt(C-1);
  neff2= ngas - A - B.*D.*cot(phi);
 BETA= (ngas - A - B.*D.*cot(phi)).*k_0;
 taylorBETA=taylor(BETA,lam,'ExpansionPoint', 1030e-9);

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John D'Errico 2018-4-25

编辑：John D'Errico 2018-4-25

在 MATLAB Online 中打开

I don't think the problem is too much lam, as much as too highly nonlinear to ever have a chance off being convergent, no matter what.

We can find the first few coefficients of that Taylor series easily enough.

vpa(subs(BETA,lam,1.03e-6))
ans =
6096502.4190565957838395137805456
vpa(subs(diff(BETA,lam,1),lam,1.03e-6))
ans =
-5920135585248.8976060067604718056
vpa(subs(diff(BETA,lam,2),lam,1.03e-6))
ans =
11494640687453888043.683950132235
vpa(subs(diff(BETA,lam,3),lam,1.03e-6))
ans =
-33489337894938749515340200.941387

So those are the first 4 coefficients of (lam-1.03e-6)^n/factorial(n) in the Taylor series, starting from the zero'th order term.

At the same time, if we plot BETA in that vicinity, it seems to be huge, but at least smooth. Look carefully, because the y axis is multiplied by 1e10.

ezplot(BETA,[2e-7,2e-6])
grid on

So, around powers of (lam-1.03e-6)^n, here are the first few coefficients in that series:

vpa(subs(BETA,lam,1.03e-6))
ans =
6096502.4190565957838395137805456
for n = 1:5
  vpa(subs(diff(BETA,lam,n),lam,1.03e-6)/factorial(n))
end
ans =
-5920135585248.8976060067604718056
ans =
5747320343726944021.8419750661176
ans =
-5581556315823124919223366.8235646
ans =
5429269510660844571193986936400.4
ans =
-5.3413066400589435608804519200757e36

Can you say non-convergent? It does not seem so. My guess is that a Taylor series is a waste of time. I suppose if you stay within a very tiny radius of convergence, it might be convergent.

MartinM 2018-4-25

Thanks for that answer, I will read it carefully later. I will discuss it with my collegue. On on the point to take care, is that each coeff from the taylor serie will determine a physical parameter used in the second parto of the code. But this first solution seem nice. THANKS

Torsten 2018-4-25

在 MATLAB Online 中打开

Note that

lim (x->0+) cot(x) = Inf.

Best wishes

Torsten.

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Answer 1

John D'Errico 2018-4-25

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/397236-taylor-serie-limitation#answer_317118

编辑：John D'Errico 2018-4-25

在 MATLAB Online 中打开

To make this into an answer, I'll compute the first few terms of a series, expanded around lam = 1.03e-6. n=8 is as far as I bothered to go, and that took a few minutes to get an answer on my computer.

C(1) = vpa(subs(BETA,lam,1.03e-6));
for n = 1:8,
  C(n+1) = vpa(subs(diff(BETA,lam,n),lam,1.03e-6)/factorial(n));
end
C
C =
[ 6096502.4190565957838395137805456, -5920135585248.8976060067604718056, 5747320343726944021.8419750661176, -5581556315823124919223366.8235646, 5429269510660844571193986936400.4, -5.3413066400589435608804519200757e36, 5.6571437723662232853187257751051e42, -8.6696749577345104686166441304681e48, 2.981901322970591677439778972521e55]

So the Taylor series will be a polynomial in powers of (lam-1.03e-6)

C(1) + C(2)*(lam - 1.03e-6) + ...

with the general n'th term being:

C(n+1)*(lam - 1.03e-6)^(n)

MATLAB was bogging down for me to go past about the 6th terms or so, so I stopped it there.

As long as lam stays in the very near vicinity of 1.03e-6, this should be acceptably convergent. For example, if I plot that polynomial on top of BETA itself, we will see:

Cpol = flip(double(C));
ezplot(BETA,[2e-7,2e-6])
hold on
grid on
ezplot(@(x) polyval(Cpol,x - 1.03e-6),[2e-7,2e-6])

The green curve is the polynomial series, the blue is BETA. As you can see, even for small deviations from the expansion point, the series has insufficient terms to be convergent. But if you stay within a VERY small radius of convergence, the two curves do overlay reasonably well.

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