The LinearDependency(v,opts) command finds an integer relation between the numbers in v - if they are linearly dependent. Given a list (or a Vector) of $n$ real or complex numbers, LinearDependency outputs a list (or a Vector) $u$ of $n$ integers such that $\sum _{i=1}^n{u}_i{v}_i$ is close to zero.

By default, Bailey and Ferguson's PSLQ (Partial Sum of Least Squares) algorithm is used if the numbers in v are real.

The optional argument method=LLL specifies that the LLL (Lenstra-Lenstra-Lovasz) lattice basis reduction algorithm be used, which is the default if v contains non-real values.

The internal working precision of the LinearDependency command corresponds to the value of Digits. For best results, the same value of Digits should be used with which the input approximation was obtained.

$\mathrm{with}\left(\mathrm{IntegerRelations}\right)\:$

$r≔\mathrm{sqrt}\left(2\right)+\mathrm{sqrt}\left(3\right)$

$r≔\sqrt{2}+\sqrt{3}$

$v≔\mathrm{expand}\left(\left[\mathrm{seq}\left(r^i\,i=0..4\right)\right]\right)$

$v≔\left[1\,\sqrt{2}+\sqrt{3}\,5+2\sqrt{2}\sqrt{3}\,11\sqrt{2}+9\sqrt{3}\,49+20\sqrt{2}\sqrt{3}\right]$

$v≔\mathrm{evalf}\left(v\,12\right)$

$v≔\left[1.\,3.14626436994\,9.89897948556\,31.1448064542\,97.9897948556\right]$

$v≔\mathrm{evalf}\left(v\right)$

$v≔\left[1.\,3.146264370\,9.898979486\,31.14480645\,97.98979486\right]$

$u≔\mathrm{LinearDependency}\left(v\right)$

$u≔\left[1\,0\,\mathrm{-10}\,0\,1\right]$

$\mathrm{add}\left(u\left[i\right]v\left[i\right]\,i=1..5\right)$

$0.$

$m≔\mathrm{add}\left(u\left[i\right]z^{i-1}\,i=1..5\right)$

$m≔z^4-10z^2+1$

$\mathrm{simplify}\left(\mathrm{eval}\left(m\,z=r\right)\right)$

$0$

$r≔1+{\left(-2\right)}^{\frac{1}{3}}$

$r≔1+{\left(\mathrm{-2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$v≔\mathrm{Vector}\left(\mathrm{expand}\left(\left[\mathrm{seq}\left(r^i\,i=0..4\right)\right]\,12\right)\right)$

$v≔\left[\begin{array}{c}1\\ 1+{\left(\mathrm{-2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ 1+2{\left(\mathrm{-2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+{\left(\mathrm{-2}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ -1+3{\left(\mathrm{-2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+3{\left(\mathrm{-2}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ -7+2{\left(\mathrm{-2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6{\left(\mathrm{-2}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\end{array}\right]$

$v≔\mathrm{evalf}\left(v\,12\right)\:$$v≔\mathrm{evalf}\left(v\right)$

$v≔\left[\begin{array}{c}1.\\ 1.629960525+1.091123636\mathrm{I}\\ 1.466220524+3.556976909\mathrm{I}\\ \mathrm{-1.491220003}+7.397559819\mathrm{I}\\ \mathrm{-10.50228211}+10.43062509\mathrm{I}\end{array}\right]$

$u≔\mathrm{LinearDependency}\left(v\,\mathrm{method}=\mathrm{LLL}\right)$

$u≔\left[\begin{array}{c}\mathrm{-1}\\ \mathrm{-2}\\ 6\\ \mathrm{-4}\\ 1\end{array}\right]$

$\mathrm{add}\left(u\left[i\right]v\left[i\right]\,i=1..5\right)$

$0.-1.\times {10}^{\mathrm{-8}}\mathrm{I}$

$m≔\mathrm{add}\left(u\left[i\right]z^{i-1}\,i=1..5\right)$

$m≔z^4-4z^3+6z^2-2z-1$

$\mathrm{simplify}\left(\mathrm{eval}\left(m\,z=r\right)\right)$

$0$

$\mathrm{solve}\left(m=0\,z\right)$

$1,-2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+1,\frac{2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{2}-\frac{\mathrm{I}\sqrt{3}{2}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{2}+1,\frac{2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{2}+\frac{\mathrm{I}\sqrt{3}{2}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{2}+1$

$\mathrm{evalc}\left(r\right)$

$\frac{2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{2}+\frac{\mathrm{I}\sqrt{3}{2}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{2}+1$