While tidying up my old University Maths notes the other day I came across this problem. (I didn't get it right first time round!)
It's quite interesting and I thought I'd also use the opportunity to setup nice looking maths on the blog. So, the question:
Find the first two terms of the series for tanx in powers of x.
Now,
$$tanx=\frac{sinx}{cosx}$$
and the power series expansions of sinx and cosx begin as:
$$sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$$
$$cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...$$
Put this together:
$$tanx=\frac{x-\frac{x^3}{6}+...}{1-\frac{x^2}{2}+...}$$
Multiply by $\frac{1+\frac{x^2}{2}}{1+\frac{x^2}{2}}$ (This is just multiplying by 1 and chosen so we can deal with the $x^2$ term in the denominator), so:
$$tanx=\frac{(1+\frac{x^2}{2})(x-\frac{x^3}{6}+...)}{(1+\frac{x^2}{2})(1-\frac{x^2}{2}+...)}$$
If we now just multiply out the first term we get
$$tanx=\frac{x+\frac{x^3}{2}-\frac{x^3}{6}+...}{1+\frac{x^2}{2}-\frac{x^2}{2}+...}$$
$$tanx=\frac{x+\frac{x^3}{3}+...}{1+...}$$
The other terms in $x^4$ and $x^5$ can be ignored, giving:
$$tanx=x+\frac{x^3}{3}+...$$
To display the formulae I've added MathJax to the blog.
Go to Theme then from the Customise dropdown select Edit HTML. Add the following just below the <head> element:
Finally, here's a useful link: MathJax basic tutorial and quick reference - Mathematics Meta Stack Exchange
Oops, missed the first x when multiplying out the numerator in the first version of this post. Now updated!
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