Compute the first four nonzero terms in the power series for \(\tan z\) at \(z=0\) by formally inverting the power series for \[ \tan^{-1}z=\int\frac{dz}{1+z^2}. \]
Solution
The first four nonzero terms are \[ \boxed{ \tan z =z+\frac{z^3}{3}+\frac{2z^5}{15}+\frac{17z^7}{315}+O(z^9).} \]
1. The power series for the inverse tangent is\[ \tan^{-1}u =u-\frac{u^3}{3}+\frac{u^5}{5}-\frac{u^7}{7}+O(u^9). \]
For \(|u|<1\), \[ \frac1{1+u^2}=1-u^2+u^4-u^6+O(u^8). \] Integrating term-by-term from \(0\) to \(u\) gives the displayed series.
2. Write the inverse series in the form\[ u=\tan z=z+az^3+bz^5+cz^7+O(z^9). \]Then\[ \begin{aligned} u^3&=z^3+3az^5+(3b+3a^2)z^7+O(z^9),\\ u^5&=z^5+5az^7+O(z^9),\\ u^7&=z^7+O(z^9). \end{aligned} \]
These follow by formal multiplication, retaining only terms of degree at most \(7\).
3. Comparing the coefficient of \(z^3\) in\[ z=u-\frac{u^3}{3}+\frac{u^5}{5}-\frac{u^7}{7}+O(u^9) \]gives \(a=1/3\).
The \(z^3\) coefficient on the right is \(a-1/3\), so it must vanish.
4. Comparing the coefficient of \(z^5\) gives\[ b=a-\frac15=\frac{2}{15}. \]
The \(z^5\) coefficient is \[ b-a+\frac15, \] which must be zero. Substitute \(a=1/3\).
5. Comparing the coefficient of \(z^7\) gives\[ c=b+a^2-a+\frac17=\frac{17}{315}. \]
The \(z^7\) coefficient on the right is \[ c-(b+a^2)+a-\frac17. \] Setting it equal to zero and substituting \[ a=\frac13, \qquad b=\frac{2}{15} \] gives \[ c =\frac{2}{15}+\frac19-\frac13+\frac17 =\frac{17}{315}. \]
6. Therefore\[ \tan z =z+\frac{z^3}{3}+\frac{2z^5}{15}+\frac{17z^7}{315}+O(z^9). \]
Insert the coefficients from <1>3–<1>5 into the ansatz of 2.