Given that the equation $x+\frac{1}{2} y^{2} +\frac{1}{2} z+\sin (z)=0$ can determine an implicit function $z(x,y)$ at {0, 0}, I now need to expand the implicit function $z(x,y)$ to a fourth-order Taylor series at {0, 0}. How can I do it?
x + 1/2 y^2 + 1/2 z[x, y] + Sin[z[x, y]] == 0
You can use AsymptoticSolve for this purpose:
AsymptoticSolve[x+1/2y^2+1/2z+Sin[z]==0,{z,0},{{x,y},{0,0},4}]
{{z -> -((2 x)/3) - (8 x^3)/243 - y^2/3 - (4 x^2 y^2)/81}}
mtaylor(solve(x + 1/2*y^2 + 1/2*z + sin(z) = 0, z), [x = 3, y = -2], 2) which results in $${\it RootOf} \left( 10+2\,\sin \left( {\it \_Z} \right) +{\it \_Z} \right) -2\,{\frac {x-3}{2\,\cos \left( {\it RootOf} \left( 10+2\, \sin \left( {\it \_Z} \right) +{\it \_Z} \right) \right) +1}}+4\,{ \frac {y+2}{2\,\cos \left( {\it RootOf} \left( 10+2\,\sin \left( {\it \_Z} \right) +{\it \_Z} \right) \right) +1}} $$ and after that evalf(allvalues(%)) produces $13.60567714- 3.091009870\,x+ 6.182019740\,y,...$ in Wolfram Language?
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Commented
Aug 9, 2020 at 13:22
s1 = z /. Solve[10 + 2 Sin[z] + z == 0, z, Reals] in Mathematica, you'll notice that you get three real solutions. The third one is the one you need to reproduce the Maple solution: z /. First @ AsymptoticSolve[x + y^2/2 + z/2 + Sin[z] == 0, {z, s1[[3]]}, {{x, y}, {3, -2}, 1}]. Of course, you can use the other two solutions in s1 to get two more possible series.
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Commented
Aug 9, 2020 at 14:05
N[%] should be added in the end in order to obtain a numeric result.
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Commented
Aug 9, 2020 at 14:10