Question

How do you use the Maclaurin series sinx to find the maclaurin series for `xsin(x/2)`?

Answer 1

`xsin(x/2) = x^2/2-x^4/(2^3 3!)+x^6/(2^5 5!)-x^8/(2^7 7!) + x^9/(2^9 9!) - ...`

Or in sigma notation:

`xsin(x/2) = sum_(k=0)^oo (-1)^k x^(2k+2)/(2^(2k+1)(2k+1)!)`

Explanation:

Starting with the Maclaurin Series for `sinx`,

`sinx = x-x^3/(3!)+x^5/(5!)-x^7/(7!) + x^9/(9!) -... \ \ x in RR`

Then the series for `sin(x/2)` is:

`sin(x/2) = (x/2)-(x/2)^3/(3!)+(x/2)^5/(5!)-(x/2)^7/(7!) + (x/2)^9/(9!) - ...`
`" " = x/2-x^3/(2^3 3!)+x^5/(2^5 5!)-x^7/(2^7 7!) + x^9/(2^9 9!) - ...`

And so:

`xsin(x/2) = x{x/2-x^3/(2^3 3!)+x^5/(2^5 5!)-x^7/(2^7 7!) + x^9/(2^9 9!) - ...}`
`" " = x^2/2-x^4/(2^3 3!)+x^6/(2^5 5!)-x^8/(2^7 7!) + x^9/(2^9 9!) - ... \ x in RR`

Or in sigma notation:

`xsin(x/2) = sum_(k=0)^oo (-1)^k x^(2k+2)/(2^(2k+1)(2k+1)!)`