4.1 Basic requirements

\begin{eqnarray*}\vec{x}&=&\vect{x_1\\\vdots\\x_N}\in\mathbb{R}^N\;\mbox{vector and $N$ dimensional space}\\ (\vec{x}_k)&=&\vec{x}_0,\vec{x}_1,\vec{x}_2,\ldots\;\;\mbox{vector sequence}\\ &=&\vect{x_{1,0}\\x_{2,0}\\\vdots\\x_{N,0}},\vect{x_{1,1}\\x_{2,1}\\\vdots\\x_{N,1}},\ldots\end{eqnarray*}

Definition 36 vector sequence \((\vec{x}_k)\in\mathbb{R}^N\) converges to \(\vec{a}\in\mathbb{R}^N\) if \begin{eqnarray*} x_{1,k}&\rightarrow&a_1\\ x_{2,k}&\rightarrow&a_2\\ \vdots&&\vdots\\ x_{N,k}&\rightarrow&a_N \end{eqnarray*}

each single component does converge in the sense of 1D convergence

\[\lim_{k\to\infty}\vec{x}_k=\vec{a}\Leftrightarrow\,\vect{x_{1,k}\\\vdots\\x_{N,k}}\;\rightarrow\,\vect{a_1\\\vdots\\a_N}\]

sum with vector series:

\[\sum_{k=0}^\infty \vec{x}_k=\vect{\sum x_{1,k}\\\sum x_{2,k}\\\vdots\\\sum x_{N,k}}\]

Distance between points:

\[\left|\vec{x}-\vec{y}\right|=\left(\sum_{j=1}^N\left(x_j-y_j\right)^2\right)^{\frac{1}{2}}\]
\[\lim_{k\to\infty}\vec{x}_k=\vec{a}\Rightarrow\left|\vec{x}_k-\vec{a}\right|^2=\sum_{j=1}^N\left(x_{j,k}-a_j\right)^2\rightarrow0\quad\mbox{if $\quad k\to\infty$}\]


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© J. Carstensen (Math for MS)