ending

(noun)

A termination or conclusion.

Examples of ending in the following topics:

  • Examples

    • $\displaystyle{\left( \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \end{array} \right) \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \end{array} \right). }$
    • $\left\{ \left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right), \left( \begin{array}{c} 0 \\ 2 \\ 3 \end{array} \right), \left( \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right), \left( \begin{array}{c} 3\\ 6 \\ 6 \end{array} \right) \right\}$
    • $x \left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right) + y \left( \begin{array}{c} 0 \\ 2 \\ 3 \end{array} \right) + z \left( \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right) = \left( \begin{array}{c} 3\\ 6 \\ 6 \end{array} \right)$
    • $\left( \begin{array}{ccc} 1 & 0 & 1 \\ 1 & 2 & 2 \\ 0 & 3 & 3 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 3 \\ 6 \\ 6 \end{array} \right)$
    • $2 \left( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right) + 1 \left( \begin{array}{c} 0 \\ 2 \\ 3 \end{array} \right) + 1 \left( \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right) = \left( \begin{array}{c} 3\\ 6 \\ 6 \end{array} \right)$

  • Negotiating the End of Apartheid

  • Cramer's Rule

    • $\displaystyle \left\{\begin{matrix} 3x+2y & = 10\\ -6x+4y & = 4 \end{matrix}\right.$
    • $\displaystyle \begin{aligned} x&=\frac{\begin{vmatrix}10&2\\4&4\end{vmatrix}}{\begin{vmatrix}3&2\\-6&4\end{vmatrix}}\\&=\frac{10\cdot 4-2 \cdot 4}{(3 \cdot 4) -[2 \cdot (-6)]}\\&=\frac{32}{24}=\frac{4}{3}\end{aligned}$ 
    • $\displaystyle \begin{aligned} y&=\frac{\begin{vmatrix}3&10\\-6&4\end{vmatrix}}{\begin{vmatrix}3&2\\-6&4\end{vmatrix}}\\ &=\frac{(3 \cdot 4)-[10 \cdot(-6)]}{(3 \cdot 4)-[2 \cdot (-6)]}\\ &=\frac{72}{24}=3 \end{aligned}$

  • Binomial Expansion and Factorial Notation

    • $\displaystyle \begin{pmatrix} n \\ k \end{pmatrix} = \frac { n!
    • $\displaystyle \begin{aligned} (x+y)^4&={ \begin{pmatrix} 4 \\ 0 \end{pmatrix} } { x }^{ 4-0}{y}^{0} + { \begin{pmatrix} 4 \\ 1 \end{pmatrix} } { x }^{ 4-1}{ y }^{1} + { \begin{pmatrix} 4 \\ 2 \end{pmatrix} } { x }^{ 4-2 }{ y }^{ 2 } + { \begin{pmatrix} 4 \\ 3 \end{pmatrix} } { x }^{ 4-3 }{ y }^{ 3 } + { \begin{pmatrix} 4 \\ 4 \end{pmatrix} } { x }^{ 4-4 }{ y }^{ 4 } \\ &={ \begin{pmatrix} 4 \\ 0 \end{pmatrix} } { x }^{ 4} + { \begin{pmatrix} 4 \\ 1 \end{pmatrix} } { x }^{ 3}{ y } + { \begin{pmatrix} 4 \\ 2 \end{pmatrix} } { x }^{ 2 }{ y }^{ 2 } + { \begin{pmatrix} 4 \\ 3 \end{pmatrix} } { x }^{ 1 }{ y }^{ 3 } + { \begin{pmatrix} 4 \\ 4 \end{pmatrix} } { y }^{ 4 } \end{aligned}$
    •  $\displaystyle = { x }^{ 4} + { \begin{pmatrix} 4 \\ 1 \end{pmatrix} } { x }^{ 3}{ y } + { \begin{pmatrix} 4 \\ 2 \end{pmatrix} } { x }^{ 2 }{ y }^{ 2 } + { \begin{pmatrix} 4 \\ 3 \end{pmatrix} } { x }{ y }^{ 3 } + { y }^{ 4 } $
    • $\displaystyle \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \frac { 4!
    • $\displaystyle \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \frac { 4!

  • Additional Reading

    • "The End of Gender?

  • Examples of Least Squares

    • $\left( \begin{array}{c} 1 \\ 1 \\ 1 \\ \vdots \\ 1 \end{array} \right) x = \left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n \end{array} \right)$
    • $\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ 0 & 2 \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} \alpha \\ \beta \\ \gamma \end{array} \right)$
    • $\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ 0 & 2 \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right)$
    • $A^T A = \left( \begin{array}{cc} 1 & 1 \\ 1 & 6 \end{array} \right) .$
    • $\mathbf{x_{ls}} = \left( \begin{array}{ccc} 1 & -1/5 & -2/5 \\ 1 & 1/5 & 2/5 \end{array} \right) \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) = \left( \begin{array}{c} -2/5 \\ 2/5 \end{array} \right) .$

  • Ending Punctuation

    • Ending punctuation comprises symbols that indicate the end of a sentence.
    • A period (.) is the punctuation mark that indicates the end of a sentence.
    • replaces a period at the end of a sentence that asks a direct question.
    • Periods are used at the end of declarative or imperative sentences.
    • Periods can also be used at the end of an indirect question.

  • Telomere Replication

    • Linear chromosomes have an end problem.
    • After DNA replication, each newly synthesized DNA strand is shorter at its 5' end than at the parental DNA strand's 5' end.
    • This produces a 3' overhang at one end (and one end only) of each daughter DNA strand, such that the two daughter DNAs have their 3' overhangs at opposite ends
    • Telomerase adds complementary RNA bases to the 3' end of the DNA strand.
    • Once the 3' end of the lagging strand template is sufficiently elongated, DNA polymerase adds the complementary nucleotides to the ends of the chromosomes; thus, the ends of the chromosomes are replicated.

  • A few examples

    • $A = \left[ \begin{array}{cc} 3 & 1 \\ 0 & 3 \\ \end{array} \right]$
    • $\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right]?
    • $\left[ \begin{array}{ccc} -2 & 1& 0 \\ 1 &-2& 1 \\ 0 &1& -2 \end{array} \right] \left[ \begin{array}{c} a \\ b \\ c \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right]$
    • $\left[ \begin{array}{cc} a & b\\ b & d \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \end{array} \right]$
    • $\left[ \begin{array}{cc} 1 & -1 \\ 0 & 0 \end{array} \right] \mbox{ and } \left[ \begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & 0 \end{array} \right]$

  • A Matrix Appears

    • $\displaystyle{ \mathbf{u} = \left[ \begin{array}{c} A e ^{i \omega t} \\ B e ^{i \omega t} \end{array} \right] = e^ {i \omega t} \left[ \begin{array}{c} A \\ B \end{array} \right]. }$
    • $\displaystyle{ \mathbf{u} = e^{i \omega _ 0 t} \left[ \begin{array}{c} 1 \\ 1 \end{array} \right], }$
    • $\displaystyle{ \mathbf{u} = e^{i \sqrt{3} \omega _ 0 t} \left[ \begin{array}{c} 1 \\ -1 \end{array} \right]. }$
    • $\displaystyle{ \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] \hbox{ and } \left[ \begin{array}{c} 1 \\ -1 \end{array} \right] }$
    • $\left( \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] \cdot \left[ \begin{array}{c} 1 \\ -1 \end{array} \right] \equiv \left[ 1,1 \right] \left[ \begin{array}{c} 1 \\ -1 \end{array} \right] = 1\cdot 1 - 1 \cdot 1 = 0.