Ανάπτυγμα σε σειρές

Σειρές Taylor

Clear["Global`*"]
f[x_] := x^2 + 5 Sin[x]
Series[f[x], {x, Pi, 10}]
Series[f[x], {x, 0, 10}]

\pi^{2}+(-5 + 2*\pi)(x-\pi)+(x-\pi)^{2}+(5/6)(x-\pi)^{3}+(-1/24)(x-\pi)^{5}+(1/1008)(x-\pi)^{7}+(-1/72576)(x-\pi)^{9}+O((x-\pi)^{11})

5 x+x^{2}+(-5/6)x^{3}+(1/24)x^{5}+(-1/1008)x^{7}+(1/72576)x^{9}+O(x^{11})

Χρήση μέρους σειράς ως συνάρτηση.

Normal[Series[f[x], {x, 0, 10}]]
fs10 = Normal[Series[f[x], {x, 0, 10}]];
fs[x_] := Evaluate[fs10]
fs[2 x]
fs[3]

5 x+x^{2}-(\frac{5}{6}) (x^{3})+(\frac{1}{24}) (x^{5})-\frac{x^{7}}{1008}+\frac{x^{9}}{72576}

10 x+4 (x^{2})-(\frac{20}{3}) (x^{3})+(\frac{4}{3}) (x^{5})-(\frac{8}{63}) (x^{7})+(\frac{4}{567}) (x^{9})

\frac{1245}{128}

fsN[n_] := Normal[Series[f[x], {x, 0, n}]];
functList = Table[fsN[n], {n, 1, 4}]
Plot[{f[x], Evaluate[functList]}, {x, -6, 6}, PlotLegends -> functList]

{5 x,5 x+x^{2},5 x+x^{2}-(\frac{5}{6}) (x^{3}),5 x+x^{2}-(\frac{5}{6}) (x^{3})}

5 x

$5 x + x^2$

$5 x + x^2 - (5 x^3)/6$

Σειρές Fourier

Αυτόματη σειρά

Clear["Global`*"]
(*Πρέπει να 'χει συμμετρία περί του 0 το πεδίο ορισμού*)
f[x_] := Which[-3 < x <= 2, x^2, 2 < x < 3, 2 x + 2]
periodos = 6;
fH = FourierTrigSeries[f[x], x, 10, FourierParameters -> {1, 2 Pi/periodos}]
fHat[x_] := Evaluate[fH]
Plot[{f[x], fHat[x]}, {x, -4, 4}]

\frac{28}{9}-\frac{(9 (\sqrt{3})+27 \pi +(\sqrt{3}) ({\pi }^{2})) \cos((\frac{1}{3}) \pi x)}{{\pi }^{3}}+\frac{(9 (\sqrt{3})+42 \pi +4 (\sqrt{3}) ({\pi }^{2})) \cos((\frac{2}{3}) \pi x)}{8 ({\pi }^{3})}-\frac{2 \cos(\pi x)}{{\pi }^{2}}-\frac{(9 (\sqrt{3})-84 \pi +16 (\sqrt{3}) ({\pi }^{2})) \cos((\frac{4}{3}) \pi x)}{64 ({\pi }^{3})}+\frac{(9 (\sqrt{3})-135 \pi +25 (\sqrt{3}) ({\pi }^{2})) \cos((\frac{5}{3}) \pi x)}{125 ({\pi }^{3})}+\frac{5 \cos(2 \pi x)}{6 ({\pi }^{2})}-\frac{(9 (\sqrt{3})+189 \pi +49 (\sqrt{3}) ({\pi }^{2})) \cos((\frac{7}{3}) \pi x)}{343 ({\pi }^{3})}+\frac{(9 (\sqrt{3})+168 \pi +64 (\sqrt{3}) ({\pi }^{2})) \cos((\frac{8}{3}) \pi x)}{512 ({\pi }^{3})}-\frac{2 \cos(3 \pi x)}{9 ({\pi }^{2})}-\frac{(9 (\sqrt{3})-210 \pi +100 (\sqrt{3}) ({\pi }^{2})) \cos((\frac{10}{3}) \pi x)}{1000 ({\pi }^{3})}+\frac{(18+6 (\sqrt{3}) \pi -4 ({\pi }^{2})) \sin((\frac{1}{3}) \pi x)}{2 ({\pi }^{3})}-\frac{3 (9+2 (\sqrt{3}) \pi ) \sin((\frac{2}{3}) \pi x)}{8 ({\pi }^{3})}+\frac{(4+{\pi }^{2}) \sin(\pi x)}{3 ({\pi }^{3})}+\frac{3 (-9+4 (\sqrt{3}) \pi ) \sin((\frac{4}{3}) \pi x)}{64 ({\pi }^{3})}-\frac{(-9+15 (\sqrt{3}) \pi +50 ({\pi }^{2})) \sin((\frac{5}{3}) \pi x)}{125 ({\pi }^{3})}+\frac{\sin(2 \pi x)}{2 \pi }+\frac{(9+21 (\sqrt{3}) \pi -98 ({\pi }^{2})) \sin((\frac{7}{3}) \pi x)}{343 ({\pi }^{3})}-\frac{3 (9+8 (\sqrt{3}) \pi ) \sin((\frac{8}{3}) \pi x)}{512 ({\pi }^{3})}+\frac{(4+9 ({\pi }^{2})) \sin(3 \pi x)}{81 ({\pi }^{3})}+\frac{3 (-9+10 (\sqrt{3}) \pi ) \sin((\frac{10}{3}) \pi x)}{1000 ({\pi }^{3})}

fsN[n_] := FourierTrigSeries[f[x], x, n, FourierParameters -> {1, 2 Pi/periodos}];
functList = Table[fsN[n], {n, 1, 4}] // Simplify
Plot[{f[x], Evaluate[functList]}, {x, 0, periodos},  PlotLegends -> functList]

{\frac{28}{9}-\frac{(9 (\sqrt{3})+27 \pi +(\sqrt{3}) ({\pi }^{2})) \cos((\frac{1}{3}) \pi x)}{{\pi }^{3}}+\frac{(9+3 (\sqrt{3}) \pi -2 ({\pi }^{2})) \sin((\frac{1}{3}) \pi x)}{{\pi }^{3}},\frac{224 ({\pi }^{3})-72 (9 (\sqrt{3})+27 \pi +(\sqrt{3}) ({\pi }^{2})) \cos((\frac{1}{3}) \pi x)+9 (9 (\sqrt{3})+42 \pi +4 (\sqrt{3}) ({\pi }^{2})) \cos((\frac{2}{3}) \pi x)+72 (9+3 (\sqrt{3}) \pi -2 ({\pi }^{2})) \sin((\frac{1}{3}) \pi x)-27 (9+2 (\sqrt{3}) \pi ) \sin((\frac{2}{3}) \pi x)}{72 ({\pi }^{3})},\frac{224 ({\pi }^{3})-72 (9 (\sqrt{3})+27 \pi +(\sqrt{3}) ({\pi }^{2})) \cos((\frac{1}{3}) \pi x)+9 (9 (\sqrt{3})+42 \pi +4 (\sqrt{3}) ({\pi }^{2})) \cos((\frac{2}{3}) \pi x)-144 \pi \cos(\pi x)+72 (9+3 (\sqrt{3}) \pi -2 ({\pi }^{2})) \sin((\frac{1}{3}) \pi x)-27 (9+2 (\sqrt{3}) \pi ) \sin((\frac{2}{3}) \pi x)+24 (4+{\pi }^{2}) \sin(\pi x)}{72 ({\pi }^{3})},\frac{1792 ({\pi }^{3})-576 (9 (\sqrt{3})+27 \pi +(\sqrt{3}) ({\pi }^{2})) \cos((\frac{1}{3}) \pi x)+72 (9 (\sqrt{3})+42 \pi +4 (\sqrt{3}) ({\pi }^{2})) \cos((\frac{2}{3}) \pi x)-1152 \pi \cos(\pi x)-9 (9 (\sqrt{3})-84 \pi +16 (\sqrt{3}) ({\pi }^{2})) \cos((\frac{4}{3}) \pi x)+576 (9+3 (\sqrt{3}) \pi -2 ({\pi }^{2})) \sin((\frac{1}{3}) \pi x)-216 (9+2 (\sqrt{3}) \pi ) \sin((\frac{2}{3}) \pi x)+192 (4+{\pi }^{2}) \sin(\pi x)+27 (-9+4 (\sqrt{3}) \pi ) \sin((\frac{4}{3}) \pi x)}{576 ({\pi }^{3})}}

$28/9 - ((9 \sqrt{3} + 27 \pi + \sqrt{3} \pi^{2}) Cos[(\pi x)/3])/\pi^{3} + ((9 + 3 \sqrt{3} \pi - 2 \pi^{2}) Sin[(\pi x)/3])/\pi^{3}$

$(224 \pi^{3} - 72 (9 \sqrt{3} + 27 \pi + \sqrt{3} \pi^{2}) Cos[(\pi x)/3] + 9 (9 \sqrt{3} + 42 \pi + 4 \sqrt{3} \pi^{2}) Cos[(2 \pi x)/3] + 72 (9 + 3 \sqrt{3} \pi - 2 \pi^{2}) Sin[(\pi x)/3] - 27 (9 + 2 \sqrt{3} \pi ) Sin[(2 \pi x)/3])/(72 \pi^{3})$

$(224 \pi^{3} - 72 (9 \sqrt{3} + 27 \pi + \sqrt{3} \pi^{2}) Cos[(\pi x)/3] + 9 (9 \sqrt{3} + 42 \pi + 4 \sqrt{3} \pi^{2}) Cos[(2 \pi x)/3] - 144 \pi Cos[\pi x] + 72 (9 + 3 \sqrt{3} \pi - 2 \pi^{2}) Sin[(\pi x)/3] - 27 (9 + 2 \sqrt{3} \pi ) Sin[(2 \pi x)/3] + 24 (4 + \pi^{2}) Sin[\pi x])/(72 \pi^{3})$

$(1792 \pi^{3} - 576 (9 \sqrt{3} + 27 \pi + \sqrt{3} \pi^{2}) Cos[(\pi x)/3] + 72 (9 \sqrt{3} + 42 \pi + 4 \sqrt{3} \pi^{2}) Cos[(2 \pi x)/3] - 1152 \pi Cos[\pi x] - 9 (9 \sqrt{3} - 84 \pi + 16 \sqrt{3} \pi^{2}) Cos[(4 \pi x)/3] + 576 (9 + 3 \sqrt{3} \pi - 2 \pi^{2}) Sin[(\pi x)/3] - 216 (9 + 2 \sqrt{3} \pi ) Sin[(2 \pi x)/3] + 192 (4 + \pi^{2}) Sin[\pi x] + 27 (-9 + 4 \sqrt{3} \pi ) Sin[(4 \pi x)/3])/(576 \pi^{3})$

Αυτόματοι συντελεστές

Η εντολή FourierCoefficient αναφέρεται στην εκθετική μορφή της σειράς Fourier:

$F(x)=\sum_{n=-\infty}^{+\infty}(c_n e^{\frac{2n\pi x i}{L}})$

Οπότε θα χρησιμοποιήσουμε τους μετασχηματισμούς:

$a_0 = c_0$
$a_n = c_n+c_{-n}$ για $n>0$
$b_n =(c_n-c_{-n})i$ για $n>0$

ώστε να συμπληρώσουμε τους όρους του τύπου:

$F(x)=a_ 0+\sum_{n=1}^{+\infty}(a_n \cos {\dfrac{2n\pi x}{L}}+b_n \sin {\dfrac{2n\pi x}{L}} )$.

ypologismosAn[synart_, periodos_, n_] := 
 Which[n > 0, 
  Simplify[
   FourierCoefficient[synart, x, n, FourierParameters -> {1, 2*Pi/periodos}] + 
   FourierCoefficient[synart, x, -n, FourierParameters -> {1, 2*Pi/periodos}]], n == 0, 
  1/2 Simplify[
    FourierCoefficient[synart, x, n, FourierParameters -> {1, 2*Pi/periodos}] + 
     FourierCoefficient[synart, x, -n, FourierParameters -> {1, 2*Pi/periodos}]]]
ypologismosBn[synart_, periodos_, n_] := 
 Simplify[
  I*FourierCoefficient[synart, x, n, FourierParameters -> {1, 2*Pi/periodos}] - 
   I*FourierCoefficient[synart, x, -n, FourierParameters -> {1, 2*Pi/periodos}]]
ypologismosAn[f[x], 6, 0]
ypologismosAn[f[x], 6, 1]
ypologismosBn[f[x], 6, 1]

\frac{28}{9}

-\frac{9 (\sqrt{3})+27 \pi +(\sqrt{3}) ({\pi }^{2})}{{\pi }^{3}}

\frac{9+3 (\sqrt{3}) \pi -2 ({\pi }^{2})}{{\pi }^{3}}

Συντελεστές στο χέρι

syntelestes[synart_, periodos_, k0_] := 
 Module[{f = synart, L = periodos, k = k0, a, b}, 
  a[n_] := 
   Which[n == 0, (1/L)*Integrate[f, {x, -L/2, L/2}], n > 0, (2/L)*Integrate[f*Cos[(2 n*Pi*x)/L], {x, -L/2, L/2}]];
  b[n_] := (2/L)*Integrate[f*Sin[(2 n*Pi*x)/L], {x, -L/2, L/2}];
  {Simplify[a[k]], Simplify[b[k]]}]
syntelestes[f[x], 6, 0]
syntelestes[f[x], 6, 1]
syntelestes[f[x], 6, 2]
syntelestes[f[x], 6, 3]

{\frac{28}{9},0}

{-\frac{9 (\sqrt{3})+27 \pi +(\sqrt{3}) ({\pi }^{2})}{{\pi }^{3}},\frac{9+3 (\sqrt{3}) \pi -2 ({\pi }^{2})}{{\pi }^{3}}}

{\frac{9 (\sqrt{3})+42 \pi +4 (\sqrt{3}) ({\pi }^{2})}{8 ({\pi }^{3})},-\frac{3 (9+2 (\sqrt{3}) \pi )}{8 ({\pi }^{3})}}

{-\frac{2}{{\pi }^{2}},\frac{4+{\pi }^{2}}{3 ({\pi }^{3})}}