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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$