# back substitution

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##### 1: 3.2 Linear Algebra
βΊWith $\mathbf{y}=[y_{1},y_{2},\dots,y_{n}]^{\rm T}$ the process of solution can then be regarded as first solving the equation $\mathbf{L}\mathbf{y}=\mathbf{b}$ for $\mathbf{y}$ (forward elimination), followed by the solution of $\mathbf{U}\mathbf{x}=\mathbf{y}$ for $\mathbf{x}$ (back substitution). … βΊIn solving $\mathbf{A}\mathbf{x}=[1,1,1]^{\rm T}$, we obtain by forward elimination $\mathbf{y}=[1,-1,3]^{\rm T}$, and by back substitution $\mathbf{x}=[\frac{1}{6},\frac{1}{6},\frac{1}{6}]^{\rm T}$. … βΊand back substitution is $x_{n}=y_{n}/d_{n}$, followed by …
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##### 3: 3.6 Linear Difference Equations
βΊ(This part of the process is back substitution.) …
##### 4: Philip J. Davis
βΊDavis left NBS in 1963 to become a faculty member in the Division of Applied Mathematics at Brown University, but during the early development of the DLMF, which started in 1998, he was invited back to give a talk and speak with DLMF project members about their plans. …
##### 5: 27.13 Functions
βΊThis conjecture dates back to 1742 and was undecided in 2009, although it has been confirmed numerically up to very large numbers. …
##### 6: 8.15 Sums
βΊ
8.15.2 $a\sum_{k=1}^{\infty}\left(\frac{{\mathrm{e}}^{2\pi\mathrm{i}k(z+h)}}{\left(2% \pi\mathrm{i}k\right)^{a+1}}\Gamma\left(a,2\pi\mathrm{i}kz\right)+\frac{{% \mathrm{e}}^{-2\pi\mathrm{i}k(z+h)}}{\left(-2\pi\mathrm{i}k\right)^{a+1}}% \Gamma\left(a,-2\pi\mathrm{i}kz\right)\right)=\zeta\left(-a,z+h\right)+\frac{z% ^{a+1}}{a+1}+\left(h-\tfrac{1}{2}\right)z^{a},$ $h\in[0,1]$.
##### 7: 29.10 Lamé Functions with Imaginary Periods
βΊThe substitutions