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Philip Thomas
Philip Thomas
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Arctangent formula (3) that can also be written as $$\tan^{-1}\left(x\right)=i\sum\limits_{n=1}^{\infty}{\frac{1}{2n-1}}\left(\frac{1}{\left(1+2i/x\right)^{2n-1}}-\frac{1}{\left(1-2i/x\right)^{2n-1}}\right).$$$$\tan^{-1}\left(x\right)=i\sum\limits_{n=1}^{\infty}{\frac{1}{2n-1}}\left(\frac{1}{\left(1+2i/x\right)^{2n-1}}-\frac{1}{\left(1-2i/x\right)^{2n-1}}\right)$$ Equation (3) isis more rapid in convergence than the equation (2). Therefore, it is more efficient in computation. Here is a link: How to compare convergence of two equations for the arctangent function?

Arctangent formula (3) can also be written as $$\tan^{-1}\left(x\right)=i\sum\limits_{n=1}^{\infty}{\frac{1}{2n-1}}\left(\frac{1}{\left(1+2i/x\right)^{2n-1}}-\frac{1}{\left(1-2i/x\right)^{2n-1}}\right).$$ Equation (3) is more rapid in convergence than equation (2). Therefore, it is more efficient in computation. Here is a link: How to compare convergence of two equations for the arctangent function?

Arctangent formula (3) that can also be written as $$\tan^{-1}\left(x\right)=i\sum\limits_{n=1}^{\infty}{\frac{1}{2n-1}}\left(\frac{1}{\left(1+2i/x\right)^{2n-1}}-\frac{1}{\left(1-2i/x\right)^{2n-1}}\right)$$ is more rapid in convergence than the equation (2). Therefore, it is more efficient in computation. Here is a link: How to compare convergence of two equations for the arctangent function?

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Philip Thomas
Philip Thomas
  • 43
  • 6

Arctangent formula (3) can also be written as $$\tan^{-1}\left(x\right)=i\sum\limits_{n=1}^{\infty}{\frac{1}{2n-1}}\left(\frac{1}{\left(1+2i/x\right)^{2n-1}}-\frac{1}{\left(1-2i/x\right)^{2n-1}}\right).$$ Equation (3) is more rapid in convergence than equation (2). Therefore, it is more efficient in computation. Here is a link: How to compare convergence of two equations for the arctangent function?