Maclaurin series for function f(x) = (x - cos(x))/x^2 for x != 0, 1/5 for x = 0

[paytheprice]

9. Obtain the MacLaurin series for the function:

. . . . .\(\displaystyle f(x)\, =\, \begin{cases}\dfrac{x\, -\, \cos(x)}{x^2}&\mbox{for }\, x\, \neq\, 0 \\ \dfrac{1}{5}&\mbox{for }\, x\, =\, 0\end{cases}\)

i have no idea how to approach this question, do i still have to differentiate the equation when x is not =0? is it considered even a Maclaurin series?

thanks guys.

 

[Ishuda]

9. Obtain the MacLaurin series for the function:

. . . . .\(\displaystyle f(x)\, =\, \begin{cases}\dfrac{x\, -\, \cos(x)}{x^2}&\mbox{for }\, x\, \neq\, 0 \\ \dfrac{1}{5}&\mbox{for }\, x\, =\, 0\end{cases}\)

i have no idea how to approach this question, do i still have to differentiate the equation when x is not =0? is it considered even a Maclaurin series?

thanks guys.

Strictly speaking, there is no Maclaurin series, see
http://mathworld.wolfram.com/MaclaurinSeries.html
however there is a Laurent series
http://mathworld.wolfram.com/LaurentSeries.html
where you would need to choose your cut.

 

[HallsofIvy]

Do you know that the MacLaurin series for cos(x) is \(\displaystyle 1- x^2/2+ x^4/4!- \cdot\cdot\cdot+ (-1)^n x^{2n}/(2n)!+ \cdot\cdot\cdot\)? If so then the MacLaurin series for \(\displaystyle x- cos(x)\) is \(\displaystyle -1+ x+ x^2/2- x^4/4!+ \cdot\cdot\cdot- (-1)^n x^{2n}/(2n)!\). And, from that, the series for \(\displaystyle \frac{x- cos(x)}{x^2}\) is \(\displaystyle -x^{-2}+ x^{-1}+ 1/2- x^2/4!+ \cdot\cdot\cdot- (-1)^nx^{2n-2}/(2n)!\). As Ishuda said, that isn't really a "MacLaurin series", it is a "Laurent series" because it has negative powers of x.

 

[paytheprice]

Yea, i was told there was no maclaurin series as when x=0 f(x)=1/5 only. was just wondering how do i explain it in words.

thank you guys!!!