The Get(topic, math_function) function returns the topic information on the function math_function. If the requested information is not available it returns NULL.

The topic argument must be one of:

To display the list of possible values for the math_function argument, use the FunctionAdvisor(known_functions) function. For more information, see FunctionAdvisor/known_functions.

The Get(topic, math_function) function is equivalent to FunctionAdvisor(topic, math_function), but does not attempt to match misspelled topic or math_function arguments to the correct names. For more information, see FunctionAdvisor.

The FunctionAdvisor command supports additional topics. For more information, see FunctionAdvisor/topics.

$\mathrm{with}\left(\mathrm{MathematicalFunctions}\right)$

$\left[\mathrm{\&Intersect}\,\mathrm{\&Minus}\,\mathrm{\&Union}\,\mathrm{Assume}\,\mathrm{Coulditbe}\,\mathrm{Evalf}\,\mathrm{Get}\,\mathrm{Is}\,\mathrm{SearchFunction}\,\mathrm{Sequences}\,\mathrm{Series}\right]$

$\mathrm{Get}\left(\mathrm{series}\,\arcsin \right)$

$\mathrm{series}\left(\arcsin \left(z\right)\,z\,4\right)=z+\frac{1}{6}{z}^3+O\left(z^5\right)$

$\mathrm{Get}\left(\mathrm{sum\_form}\,\tan \right)$

$\left[\tan \left(z\right)=\sum _{\mathrm{\_k1}=1}^{\mathrm{\infty }}\frac{\mathrm{bernoulli}\left(2\mathrm{\_k1}\right){\left(\mathrm{-1}\right)}^{\mathrm{\_k1}}{z}^{-1+2\mathrm{\_k1}}\left(4^{\mathrm{\_k1}}-{16}^{\mathrm{\_k1}}\right)}{\mathrm{\Gamma }\left(2\mathrm{\_k1}+1\right)}\&comma;\mathrm{And}\left(\mathrm{abs}\left(z\right)<\frac{1}{2}\mathrm{\&pi;}\right)\right]$

$\mathrm{Get}\left(\mathrm{special\_values}\&comma;\sec \right)$

$\left[\sec \left(\frac{\mathrm{\pi }}{6}\right)=\frac{2\sqrt{3}}{3}\&comma;\sec \left(\frac{\mathrm{\pi }}{4}\right)=\sqrt{2}\&comma;\sec \left(\frac{\mathrm{\pi }}{3}\right)=2\&comma;\sec \left(\mathrm{\infty }\right)=\mathrm{undefined}\&comma;\sec \left(\mathrm{\infty }\mathrm{I}\right)=0\&comma;\left[\sec \left(\mathrm{\pi }n\right)=\mathrm{-1}\&comma;\mathrm{And}\left(n::\mathrm{odd}\right)\right]\&comma;\left[\sec \left(\mathrm{\pi }n\right)=1\&comma;\mathrm{And}\left(n::\mathrm{even}\right)\right]\&comma;\left[\sec \left(\frac{\mathrm{\pi }n}{2}\right)=\mathrm{\infty }+\mathrm{\infty }\mathrm{I}\&comma;\mathrm{And}\left(n::\mathrm{odd}\right)\right]\right]$

$\mathrm{Get}\left(\mathrm{branch\_cuts}\&comma;\mathrm{arccot}\right)$

$\left[\mathrm{arccot}\left(z\right)\&comma;z\in \mathrm{ComplexRange}\left(-\mathrm{\infty }\mathrm{I}\&comma;\mathrm{-I}\right)\vee z\in \mathrm{ComplexRange}\left(\mathrm{I}\&comma;\mathrm{\infty }\mathrm{I}\right)\right]$

$\mathrm{Get}\left(\mathrm{identities}\&comma;\mathrm{BesselK}\right)$

$\left[\left[\mathrm{BesselK}\left(a\&comma;\mathrm{I}z\right)=-\frac{\mathrm{\pi }\mathrm{BesselY}\left(a\&comma;z\right)}{2{\mathrm{I}}^a}+\frac{\mathrm{BesselJ}\left(a\&comma;z\right)\left(\ln \left(z\right)-\ln \left(\mathrm{I}z\right)\right)}{{\mathrm{I}}^a}\&comma;\mathrm{And}\left(a::\mathrm{integer}\right)\right]\&comma;\left[\mathrm{BesselK}\left(a\&comma;\mathrm{I}z\right)=-\frac{\mathrm{\pi }{z}^a\mathrm{BesselY}\left(a\&comma;z\right)}{2{\left(\mathrm{I}z\right)}^a}+\frac{\mathrm{\pi }\mathrm{BesselJ}\left(a\&comma;z\right)\left(-\frac{{\left(\mathrm{I}z\right)}^a}{z^a}+\frac{z^a\cos \left(\mathrm{\pi }a\right)}{{\left(\mathrm{I}z\right)}^a}\right)\csc \left(\mathrm{\pi }a\right)}{2}\&comma;\mathrm{And}\left(a::\left(\mathrm{Not}\left(\mathrm{integer}\right)\right)\right)\right]\&comma;\left[\mathrm{BesselK}\left(a\&comma;-z\right)={\left(\mathrm{-1}\right)}^a\mathrm{BesselK}\left(a\&comma;z\right)+\mathrm{BesselI}\left(a\&comma;z\right)\left(\ln \left(z\right)-\ln \left(-z\right)\right)\&comma;\mathrm{And}\left(a::\mathrm{integer}\right)\right]\&comma;\left[\mathrm{BesselK}\left(a\&comma;-z\right)=\frac{z^a\mathrm{BesselK}\left(a\&comma;z\right)}{{\left(-z\right)}^a}+\frac{\mathrm{\pi }\left(\frac{z^a}{{\left(-z\right)}^a}-\frac{{\left(-z\right)}^a}{z^a}\right)\mathrm{BesselI}\left(a\&comma;z\right)\csc \left(\mathrm{\pi }a\right)}{2}\&comma;\mathrm{And}\left(a::\left(\mathrm{Not}\left(\mathrm{integer}\right)\right)\right)\right]\&comma;\left[\mathrm{BesselK}\left(a\&comma;b{\left(c{z}^q\right)}^p\right)=\frac{{\left(b{c}^p{z}^{pq}\right)}^a\mathrm{BesselK}\left(a\&comma;b{c}^p{z}^{pq}\right)}{{\left(b{\left(c{z}^q\right)}^p\right)}^a}-\frac{\mathrm{\pi }\csc \left(\mathrm{\pi }a\right)\mathrm{BesselI}\left(a\&comma;b{c}^p{z}^{pq}\right)\left(\frac{{\left(b{\left(c{z}^q\right)}^p\right)}^a}{{\left(b{c}^p{z}^{pq}\right)}^a}-\frac{{\left(b{c}^p{z}^{pq}\right)}^a}{{\left(b{\left(c{z}^q\right)}^p\right)}^a}\right)}{2}\&comma;a::\left(\neg \mathrm{integer}\right)\wedge \left(2p\right)::\mathrm{integer}\right]\&comma;\left[\mathrm{BesselK}\left(a\&comma;b{\left(c{z}^q\right)}^p\right)={\left(\frac{{\left(c{z}^q\right)}^p}{c^p{z}^{pq}}\right)}^a\left(\mathrm{BesselK}\left(a\&comma;b{c}^p{z}^{pq}\right)-{\left(\mathrm{-1}\right)}^a\mathrm{BesselI}\left(a\&comma;b{c}^p{z}^{pq}\right)\left(\ln \left(b{\left(c{z}^q\right)}^p\right)-\ln \left(b{c}^p{z}^{pq}\right)\right)\right)\&comma;a::\mathrm{integer}\wedge \left(2p\right)::\mathrm{integer}\right]\&comma;\mathrm{BesselK}\left(a\&comma;z\right)=\frac{2\left(a-1\right)\mathrm{BesselK}\left(a-1\&comma;z\right)}{z}+\mathrm{BesselK}\left(a-2\&comma;z\right)\&comma;\mathrm{BesselK}\left(a\&comma;z\right)=-\frac{2\left(a+1\right)\mathrm{BesselK}\left(a+1\&comma;z\right)}{z}+\mathrm{BesselK}\left(a+2\&comma;z\right)\right]$

$\mathrm{Get}\left(\mathrm{calling\_sequence}\&comma;\mathrm{Zeta}\&comma;\mathrm{all}\right)$

$\mathrm{\zeta }\left(s\right),\mathrm{\zeta }\left(n\&comma;s\right),\mathrm{\zeta }\left(n\&comma;s\&comma;a\right)$

$\mathrm{Get}\left(\mathrm{definition}\&comma;\mathrm{JacobiAM}\right)$

$\left[z\&equals;\mathrm{JacobiAM}\left(\mathrm{Int}\left(\frac{1}{\sqrt{1-k^2{\sin \left(\mathrm{\&theta;}\right)}^2}}\&comma;\mathrm{\&theta;}\&equals;0..z\right)\&comma;k\right)\&comma;z::\left(\mathrm{RealRange}\left(-\frac{3}{2}\&comma;\frac{3}{2}\right)\right)\right]$

$\mathrm{Get}\left(\mathrm{definition}\&comma;\mathrm{InverseJacobiAM}\right)$

$\left[\mathrm{InverseJacobiAM}\left(\mathrm{\&phi;}\&comma;k\right)\&equals;\mathrm{Int}\left(\frac{1}{\sqrt{1-k^2{\sin \left(\mathrm{\_\&theta;1}\right)}^2}}\&comma;\mathrm{\_\&theta;1}\&equals;0..\mathrm{\&phi;}\right)\&comma;\mathrm{with\; no\; restrictions\; on}\left(\mathrm{\&phi;}\&comma;k\right)\right]$