Expand power series of 1/(cosx) analytic at x=0. Find the first 4 terms of the series.
I know the taylor series for cosx, but isn't this just the same thing but with all the fractions flipped?
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Expand power series of 1/(cosx) analytic at x=0. Find the first 4 terms of the series.
I know the taylor series for cosx, but isn't this just the same thing but with all the fractions flipped?
Lets suppose that the McLaurin expansion of $\displaystyle \frac{1}{\cos x}$ exists and write...
$\displaystyle \frac{1}{\cos x} = \sum_{n=0}^{\infty} a_{n}\cdot x^{n}$ (1)
In this case we can derive the $\displaystyle a_{n}$ from the McLaurin expansion of $\displaystyle \cos x$...
$\displaystyle \cos x = \sum_{k=0}^{\infty} b_{k}\cdot x^{k}$ (2)
... where $\displaystyle b_{k} = \frac{(-1)^{\frac{k}{2}}}{k!}$ for $\displaystyle k$ even and $\displaystyle b_{k}=0$ for $\displaystyle k$ odd. If we write the product of the series (1) and (2) we obtain...
$\displaystyle \sum_{n=0}^{\infty} a_{n}\cdot x^{n} \cdot \sum_{k=0}^{\infty} b_{k}\cdot x^{k} = \sum_{n=0}^{\infty} x^{n}\cdot \sum_{k=0}^{n} a_{k}\cdot b_{n-k} = 1$ (3)
... and from (3) we obtain...
$\displaystyle a_{0}\cdot b_{0}= 1 \rightarrow a_{0} = 1$
$\displaystyle a_{0}\cdot b_{1} + a_{1}\cdot b_{0} = 0 \rightarrow a_{1} = 0$
$\displaystyle a_{0}\cdot b_{2} + a_{1}\cdot b_{1} + a_{2}\cdot b_{0} = 0 \rightarrow a_{2}= \frac{1}{2}$
$\displaystyle a_{0}\cdot b_{3} + a_{1}\cdot b_{2} + a_{2}\cdot b_{1} + a_{3}\cdot b_{0} = 0 \rightarrow a_{3} = 0 $
$\displaystyle a_{0}\cdot b_{4} + a_{1}\cdot b_{3} + a_{2}\cdot b_{2} + a_{3}\cdot b_{1} + a_{4}\cdot b_{0} = 0 \rightarrow a_{4} = - \frac{5}{24}$
The McLaurin expansion of $\displaystyle \frac{1}{\cos x}$ with maximum exponent 4 is then...
$\displaystyle \frac{1}{\cos x} = 1 + \frac{1}{2}\cdot x^{2} - \frac{5}{24}\cdot x^{4} + \dots$ (4)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
