Expansion of cosine function by shifted chebyshev polynomials

Question

$f(t)=f_1(t)+f_2(t) $

$w_1=1$, $w_2=6$

$f_1(t)=A \cos(w_1 t)$

$f_2(t)=B \cos(w_2 t)$

$A$ and $B$ are constant $2 \times 2$ matrix

After normalizing the above equation by $t=T*y$ where $T$ is the principal period $T=\frac{2\pi}{w_1}$, the equation will be in the following form ,

$f_1(y)= A \cos(T w_1 y)$

$f_2(y)= B \cos(T w_2 y) $ here $y$ varies from $0$ to $1$

now if I expand the $f_1(y)$ and $f_2(y)$ in shifted chebyshev polynomials it should give me convergent series.It gives me correct result for $f_1(y)$ but when I expand $f_2(y)$ I am not able to find convergence and in addition to this the solution oscillates with large amplitude after $y=0.5$ (approximate value)

the formula that I used is following, https://i.stack.imgur.com/k7V9v.jpg

$36$ terms in chebyshev expansion https://i.stack.imgur.com/k1jiE.jpg

here's my code, I did coding in Mathematica software

w1 = 1; w2 = 6;

T = (2 Pi/w1)
x = T w2;
\[Alpha] = 0;
\[Beta] = 1;
\[Gamma]1 = (\[Beta] + \[Alpha])/2;
\[Gamma]2 = (\[Beta] - \[Alpha])/2;
\[Xi] = (2*y - \[Alpha] - \[Beta])/(\[Beta] - \[Alpha]);
x1 = Cos[x y]

n1 = 18;
x2 = Simplify[
  Cos[\[Gamma]1*x]*(Sum[
      2*(((-1)^n)*BesselJ[2*n, \[Gamma]2*x]*P[2*n]), {n, 1, 
       n1 - 1}]) + Cos[\[Gamma]1*x]*BesselJ[0, \[Gamma]2*x]*P[0] - 
   Sin[\[Gamma]1*
      x]*2*(Sum[(((-1)^n)*BesselJ[2*n + 1, \[Gamma]2*x]*
        P[2*n + 1]), {n, 0, n1 - 1}])]

ptable = Table[P[i], {i, 0, 2 (n1 - 1) + 1}];
tst = Expand[Table[ChebyshevT[i, \[Xi]], {i, 0, 2 (n1 - 1) + 1}]];
sol = Solve[ptable - tst == 0, ptable];

x3 = N[Expand[x2 /. sol], 15]
Plot[{Expand[x3], x1}, {y, \[Alpha], \[Beta]}, PlotRange -> All]

I am not able to figure out the correct way to do this.Can anyone help me in this? Thanks

Answer 1

I played a little with your code - I think your code is ok. Change the value of $w_2$ to something smaller to verify everything works correct.

Your problem is saturation, look at the coefficients, e.g. $$-1.42339308536533*10^{17} y^{18} + 3.68617004545264*10^{17} y^{19}$$ .

Solution, normalize using $$T=\frac{2\pi}{max\{w_1,w_2\}}$$ However, you will always have a problem when $|w_1-w_2|$ is big.

Answer 2

Thanks michael. I figured out a way.divide the interval into subintervals and then normalize it with length of subinterval so then our interval will be between 0 and 1. And everything works fine. Thanks