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It needs to be understand that solution $\epsilon$ for $(q+\epsilon)\sin(\epsilon) + \cos(\epsilon)=0$ represents a power series in $\frac{1}{q}$. It is therefore convenient to introduce $r:=\frac{1}{q}$ and solve equation: $$(\frac{1}{r} + \epsilon)\sin(\epsilon) + \cos(\epsilon) = 0.$$ To do so, we rewrite the equation as $$r = - \frac{\sin(\epsilon)}{ \epsilon\sin(\epsilon) + \cos(\epsilon)}$$ and perform series reversion of the right hand side.

Running the following code

eps = QQ[['ϵ']].0
a = - sin(eps) / (cos(eps) + eps*sin(eps))
b = a.reverse()
print(b)

gives power series -ϵ - 2/3*ϵ^3 - 13/15*ϵ^5 - 146/105*ϵ^7 - ..., where the variable needs to be understand as $r$, that is the solution is $$\epsilon = -r - \frac23 r^3 - \frac{13}{15} r^5 - \frac{146}{105} r^7 - \dots$$

To verify it, we can run

print( (1/eps + b)*sin(b) + cos(b) )

which gives O(ϵ^19) as expected.

It needs to be understand understood that solution $\epsilon$ for to $(q+\epsilon)\sin(\epsilon) + \cos(\epsilon)=0$ represents a power series in $\frac{1}{q}$. $\frac{1}{q}$ and not $q$. It is therefore convenient to introduce $r:=\frac{1}{q}$ and solve the equation: $$(\frac{1}{r} + \epsilon)\sin(\epsilon) + \cos(\epsilon) = 0.$$ To do so, we rewrite the equation as $$r = - \frac{\sin(\epsilon)}{ \epsilon\sin(\epsilon) + \cos(\epsilon)}$$ and perform series reversion of the right hand side.

Running the following code

eps = QQ[['ϵ']].0
a = - sin(eps) / (cos(eps) + eps*sin(eps))
b = a.reverse()
print(b)

gives power series -ϵ - 2/3*ϵ^3 - 13/15*ϵ^5 - 146/105*ϵ^7 - ..., where the variable needs to be understand as $r$, that is the solution is $$\epsilon = -r - \frac23 r^3 - \frac{13}{15} r^5 - \frac{146}{105} r^7 - \dots$$

To verify it, we can run

print( (1/eps + b)*sin(b) + cos(b) )

which gives O(ϵ^19) as expected.

It needs to be understood that solution $\epsilon$ to $(q+\epsilon)\sin(\epsilon) + \cos(\epsilon)=0$ represents a power series in $\frac{1}{q}$ and not $q$. It is therefore convenient to introduce $r:=\frac{1}{q}$ and solve the equation: $$(\frac{1}{r} + \epsilon)\sin(\epsilon) + \cos(\epsilon) = 0.$$ To do so, we rewrite the equation as $$r = - \frac{\sin(\epsilon)}{ \epsilon\sin(\epsilon) + \cos(\epsilon)}$$ and perform series reversion of the right hand side.

Running the following code

eps = QQ[['ϵ']].0
a = - sin(eps) / (cos(eps) + eps*sin(eps))
b = a.reverse()
print(b)

gives power series -ϵ - 2/3*ϵ^3 - 13/15*ϵ^5 - 146/105*ϵ^7 - ..., where the variable needs viewed as $r$ (I did not find easy way to be understand as $r$, replace variable in a power series), that is the solution is $$\epsilon = -r - \frac23 r^3 - \frac{13}{15} r^5 - \frac{146}{105} r^7 - \dots$$

To verify it, we can run

print( (1/eps + b)*sin(b) + cos(b) )

which gives O(ϵ^19) as expected.

It needs to be understood that solution $\epsilon$ to $(q+\epsilon)\sin(\epsilon) + \cos(\epsilon)=0$ represents a power series in $\frac{1}{q}$ and not $q$. It is therefore convenient to introduce $r:=\frac{1}{q}$ and solve the equation: $$(\frac{1}{r} + \epsilon)\sin(\epsilon) + \cos(\epsilon) = 0.$$ To do so, we rewrite the equation as $$r = - \frac{\sin(\epsilon)}{ \epsilon\sin(\epsilon) + \cos(\epsilon)}$$ and perform series reversion of the right hand side.

Running the following code

eps = QQ[['ϵ']].0
r = QQ[['r']].0
a = - sin(eps) / (cos(eps) + eps*sin(eps))
b = a.reverse()
a.reverse()(r)
print(b)

gives power series -ϵ -r - 2/3*ϵ^3 2/3*r^3 - 13/15*ϵ^5 13/15*r^5 - 146/105*ϵ^7 146/105*r^7 - ..., where the variable needs viewed as $r$ (I did not find easy way to replace variable in a power series), that is the solution is $$\epsilon = -r - \frac23 r^3 - \frac{13}{15} r^5 - \frac{146}{105} r^7 - \dots$$

To verify it, we can run

print( (1/eps (1/r + b)*sin(b) + cos(b) )

which gives O(ϵ^19)O(r^19) as expected.

It needs to be understood that solution $\epsilon$ to $(q+\epsilon)\sin(\epsilon) + \cos(\epsilon)=0$ represents a power series in $\frac{1}{q}$ and not $q$. It is therefore convenient to introduce $r:=\frac{1}{q}$ and solve the equation: $$(\frac{1}{r} + \epsilon)\sin(\epsilon) + \cos(\epsilon) = 0.$$ To do so, we rewrite the equation as $$r = - \frac{\sin(\epsilon)}{ \epsilon\sin(\epsilon) + \cos(\epsilon)}$$ and perform series reversion of the right hand side.

Running the following code

eps = QQ[['ϵ']].0
r = QQ[['r']].0
a = - sin(eps) / (cos(eps) + eps*sin(eps))
b = a.reverse()(r)
print(b)

gives power series -r - 2/3*r^3 - 13/15*r^5 - 146/105*r^7 - ..., that is the solution is $$\epsilon = -r - \frac23 r^3 - \frac{13}{15} r^5 - \frac{146}{105} r^7 - \dots$$

To verify it, we can run

print( (1/r + b)*sin(b) + cos(b) )

which gives O(r^19) as expected.