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Arctangent Function Forward Taylor Polynomial Theory
If  F(x) = \arctan(x)   \[
     (1 + x * x ) * F^{(1)} (x) - 0 * F (x)  = 1
\] 
and in the standard math function differential equation ,  A(x) = 0 ,  B(x) = 1 + x * x  , and  D(x) = 1 . We use  a ,  b ,  d and  z to denote the Taylor coefficients for  A [ X (t) ]  ,  B [ X (t) ] ,  D [ X (t) ]  , and  F [ X(t) ]  respectively. It now follows from the general Taylor coefficients recursion formula that for  j = 0 , 1, \ldots ,  \[
\begin{array}{rcl}
z^{(0)} & = & \arctan ( x^{(0)} )
\\
b^{(j)}
& = &  \left\{ \begin{array}{ll}
     1 + x^{(0)} * x^{(0)}          & {\rm if} \; j = 0 \\
     \sum_{k=0}^j x^{(k)} x^{(j-k)} & {\rm otherwise}
\end{array} \right.
\\
e^{(j)} 
& = & d^{(j)} + \sum_{k=0}^{j} a^{(j-k)} * z^{(k)}
\\
& = & \left\{ \begin{array}{ll}
     1 & {\rm if} \; j = 0 \\
     0 & {\rm otherwise}
\end{array} \right.
\\
z^{(j+1)} & = & \frac{1}{j+1} \frac{1}{ b^{(0)} } 
\left(
     \sum_{k=0}^j e^{(k)} (j+1-k) x^{(j+1-k)} 
     - \sum_{k=1}^j b^{(k)} (j+1-k) z^{(j+1-k)}  
\right)
\\
z^{(j+1)} & = & \frac{1}{j+1} \frac{1}{ b^{(0)} } 
\left(
     (j+1) x^{(j+1)}
     - \sum_{k=1}^j k z^{(k)}  b^{(j+1-k)} 
\right)
\end{array}
\] 

Input File: omh/atan_forward.omh