- #1
Unit
- 182
- 0
I understand how this works:
[tex]\cos x = \frac{1}{0!} - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!} + \ldots[/tex]
But what about this?
[tex]\frac{1}{\cos x} = \frac{1}{0!} + \frac{x^2}{2!} + \frac{5x^4}{4!} + \frac{61x^6}{6!} + \frac{1385x^8}{8!} + \frac{50521x^{10}}{10!} + \ldots[/tex]
Is there a way to take the reciprocal of an infinite series or is it necessary to take subsequent derivatives of secant and write the Taylor expansion that way?
Thanks,
Unit
[tex]\cos x = \frac{1}{0!} - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!} + \ldots[/tex]
But what about this?
[tex]\frac{1}{\cos x} = \frac{1}{0!} + \frac{x^2}{2!} + \frac{5x^4}{4!} + \frac{61x^6}{6!} + \frac{1385x^8}{8!} + \frac{50521x^{10}}{10!} + \ldots[/tex]
Is there a way to take the reciprocal of an infinite series or is it necessary to take subsequent derivatives of secant and write the Taylor expansion that way?
Thanks,
Unit