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 $\dfrac{1}{x} - \dfrac{x}{3}$.

Feb 10, 2025 · The (real) cotangent function has a Taylor series expansion: ∞ ∑ n = 0(− 1)n22nB2nx2n − 1 (2n)! ∑ n = 0 ∞ (− 1) n 2 2 n B 2 n x 2 n − 1 (2 n)! where B2n denotes the Bernoulli numbers. This converges for 0 <|x| <π. i + 1 x ∞ ∑ n = 0Bn(2ix)n n! i + 1 x ∑ n = 0 ∞ B n (2 i x) n n! i + 1 x1 + −1 2(2ix) 1! + ∑n= 2∞ Bn(2ix)n n!)

Elementary Functions Cot: Series representations. Generalized power series. Expansions at z==z 0. For the function itself. Expansions at z==0. For the function itself. Expansions at z==Pi/2. For the function itself. q-series. Dirichlet series. Asymptotic …

Mar 8, 2021 · $\ds \pi \cot \pi z = \frac 1 z - 2 \sum_{k \mathop = 1}^\infty \frac z {k^2} \cdot \sum_{n \mathop = 1}^\infty \paren {\frac {z^2} {k^2} }^{n - 1}$ from which:

Jan 18, 2015 · Remove the singularity by expanding coth x − 1/x instead. There is a simple way of approximating coth by noticing that it is a logarithmic derivative. Since: sinh z z = ∏n=1+∞(1 + z2 π2n2) (1) by the Weierstrass product for the (hyperbolic) sine function, we have: log sinh z − log z =∑n=1+∞ log(1 + z2 π2n2), (2) so, by differentiating both sides:

cot numerically evaluates these units automatically: radian, degree, arcmin, arcsec, and revolution. Show this behavior by finding the cotangent of x degrees and 2 radians. You can calculate cotf by substituting for x using subs and then using double or vpa.

Oct 4, 2016 · I'm looking for clarification on how to compute a Laurent series for $\cot z$ I started by trying to find the $\frac{1}{\sin z}$. I've found multiple references that go from an Taylor expansion f...

Feb 16, 2011 · Remove the singularity by multiplying through by $$x (you should know that \cot(x)\sim x^{-1} for small $$x), so take the series expansion of x \cot(x) (though taking the derivatives of this is rather tedious)

Feb 10, 2025 · The hyperbolic cotangent function has a Taylor series expansion: ∞ ∑ n = 022nB2nx2n − 1 (2n)! ∑ n = 0 ∞ 2 2 n B 2 n x 2 n − 1 (2 n)! where B2n denotes the Bernoulli numbers. This converges for 0 <|x| <π. 1 + 1 x ∞ ∑ n = 0Bn(2x)n n! 1 + 1 x ∑ n = 0 ∞ B n (2 x) n n! 1 x + 1 x ∞ ∑ n = 2Bn(2x)n n! 1 x + 1 x ∑ n = 2 ∞ B n (2 x) n n!

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