Question

Maclaurin series for xcosx

Answer 1

Note that Maclaurin Series for cos(x) is:\[\bf \cos(x)=1-\frac{ x^2 }{ 2! }+\frac{ x^4 }{ 4! }+....+(-1)^n \frac{x^n}{n!}\]Multiplying both sides of the Maclaurin series for cos(x) yields the Maclaurin series for xcos(x):\[\bf xcos(x)=x-\frac{x^3}{2!}+\frac{x^5}{4!}+....(-1)^n \frac{x^{n+1}}{n!}\]

Answer 2

@lasorsat You should also take note that you should, by heart, know the Maclaurin series of 1/x, cos(x), sin(x) and other common ones. If you would like to find the Maclaurin series yourself, then you can assume that, in this case, cos(x) equals to some polynomial of finite degree. Let's assume it's of degree 4. Then we get the following approximation:\[\bf \cos(x)=ax^4+bx^3+cx^2+dx+e\]Now plug in x = 0; the left side equals 1 and the right side is just 'e', hence e = 1. Now differentiate both sides of the equation to get:\[\bf -\sin(x)=4ax^3+3bx^2+2cx+d\]Now once again set x = 0; everything cancels out except 'd' hence d = 0. Keep differentiating until you solve for each coefficient. After doing so, you will start seeing the pattern in the Taylor polynomial which will allow you to find the term number for the Maclaurin Series. This will allow you to figure out the series yourself.

Now at this point, you probably figured that the process is labour intensive which is why it's useful to remember some common series. Hope this helps!