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tan_forward

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Tangent and Hyperbolic Tangent Forward Taylor Polynomial Theory

Derivatives

\begin{eqnarray} \tan^{(1)} ( u ) & = & [ \cos (u)^2 + \sin (u)^2 ] / \cos (u)^2 \\ & = & 1 + \tan (u)^2 \\ \tanh^{(1)} ( u ) & = & [ \cosh (u)^2 - \sinh (u)^2 ] / \cosh (u)^2 \\ & = & 1 - \tanh (u)^2 \end{eqnarray}

If \(F(u)\) is \(\tan (u)\) or \(\tanh (u)\) the corresponding derivative is given by

\[F^{(1)} (u) = 1 \pm F(u)^2\]

Given \(X(t)\), we define the function \(Z(t) = F[ X(t) ]\). It follows that

\[Z^{(1)} (t) = F^{(1)} [ X(t) ] X^{(1)} (t) = [ 1 \pm Y(t) ] X^{(1)} (t)\]

where we define the function \(Y(t) = Z(t)^2\).

Taylor Coefficients Recursion

Suppose that we are given the Taylor coefficients up to order \(j\) for the function \(X(t)\) and up to order \(j-1\) for the functions \(Y(t)\) and \(Z(t)\). We need a formula that computes the coefficient of order \(j\) for \(Y(t)\) and \(Z(t)\). Using the equation above for \(Z^{(1)} (t)\), we have

\begin{eqnarray} \sum_{k=1}^j k z^{(k)} t^{k-1} & = & \sum_{k=1}^j k x^{(k)} t^{k-1} \pm \left[ \sum_{k=0}^{j-1} y^{(k)} t^k \right] \left[ \sum_{k=1}^j k x^{(k)} t^{k-1} \right] + o( t^{j-1} ) \end{eqnarray}

Setting the coefficients of \(t^{j-1}\) equal, we have

\begin{eqnarray} j z^{(j)} = j x^{(j)} \pm \sum_{k=1}^j k x^{(k)} y^{(j-k)} \\ z^{(j)} = x^{(j)} \pm \frac{1}{j} \sum_{k=1}^j k x^{(k)} y^{(j-k)} \end{eqnarray}

Once we have computed \(z^{(j)}\), we can compute \(y^{(j)}\) as follows:

\[y^{(j)} = \sum_{k=0}^j z^{(k)} z^{(j-k)}\]