Hi guys could you show me how to do the expansion of the Taylor series of $\cot x $ at the point $x=0$. My idea was to use $\dfrac{\cos x}{\sin x} $ and I want to expand it to the second term because I have to find the limit of $\dfrac{x\cot x-1}{x^2}$ but when I do the expansion I get $\dfrac{1}{x} - \dfrac{x}{2}$ instead of $\dfrac{1}{x} - \dfrac{x}{3}$. Any idea why?
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$\begingroup$You're right we use the equality $$\cot x=\frac{\cos x}{\sin x}$$ and since $$\frac{1}{\sin x}=\frac{1}{x-\frac{x^3}{6}+o(x^3)}=\frac{1}{x}\left(1+\frac{x^2}{6}+o(x^3)\right)$$ hence we expand and we find $$\cot x=\left(1-\frac{x^2}{2}+o(x^3)\right)\frac{1}{x}\left(1+\frac{x^2}{6}+o(x^3)\right)=\frac{1}{x}-\frac{x}{3}+o(x)$$
$\endgroup$ 5 $\begingroup$You should consider the Taylor expansion series for both $\cos{x}$ and $\sin{x}$ at $x=0$, separately. Then, divide term by term to obtain the Taylor series for $\cot{x}$.
Cheers!
Edit:
By using division term by term, I meant this:
$$ (1-x^2/2 + x^4/24 + O(x^6) : (x-x^3/6 + x^5/120 + O(x^6)) \approx 1/x-x/3,$$
being the rest of the polynomial long division:
$$r = -x^4/45 + O(x^6)$$
I hope this is useful.
$\endgroup$ 0 $\begingroup$Try this instead...$$\frac{(xcot x-1)}{x^2} = \frac{(x-\tan x)}{(x^2×\tan x)}$$$$= \frac{(x-(x + (x^3)/3 + ....)}{(x^2)×\tan x} =\frac{-x^3×(1+2x^2/5+.....)}{(3×x^2×tan x)}$$$$= \frac{-1}{(3×(\frac{tan x}{x})}$$
It will help , hopefully!!!
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