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\[x=Ax+b\ \ \ (A,\ b\ \ \text{given})\]
_of \(n\) linear equations in \(n\) unknowns, \((x_{1},x_{2},\dots,x_{n})\), satisfies_
\[\alpha=\max_{1\leq i\leq n}\sum_{j=1}^{n}|a_{ij}|<1, \tag{2.5} \]\]
_then there is precisely one solution \(x\). This solution can be obtained as the limit of the iterative sequence \((x^{(0)},x^{(1)},\dots)\), where \(x^{(0)}\) is arbitrary and_
\[x^{(m+1)}=Ax^{(m)}+b,\ \ m=0,1,2,\dots\]
_with error bounds of_
\[d(x^{(m)},x)\leq\frac{\alpha^{m}}{1-\alpha}d(x^{(0)},x^{(1)}).\]
Now we will work through a concrete example. Consider the following set of algebraic equations,
\[\begin{cases}\frac{4}{5}x_{1}-\frac{1}{2}x_{2}-\frac{1}{4}x_{3}=-1\\ -\frac{1}{3}x_{1}+\frac{4}{3}x_{2}+\frac{1}{4}x_{3}=2\\ -\frac{1}{4}x_{1}-\frac{2}{15}x_{2}+\frac{3}{4}x_{3}=3.\end{cases}\]
To apply the Row Sum Criterion, we first rewrite the equations in the form \(x=Ax+b\); we have
\[\begin{cases}x_{1}=\frac{1}{5}x_{1}+\frac{1}{2}x_{2}+\frac{1}{4}x_{3}-1\\ x_{2}=\frac{1}{3}x_{1}-\frac{1}{3}x_{2}-\frac{1}{4}x_{3}+2\\ x_{3}=\frac{1}{4}x_{1}+\frac{2}{15}x_{2}+\frac{1}{4}x_{3}+3\end{cases}\]which can also be written as \[\begin{pmatrix}x_{1}\\ x_{2}\\ x_{3}\end{pmatrix}=\begin{pmatrix}\frac{1}{5}&\frac{1}{2}&\frac{1}{4}\\ \frac{3}{4}&-\frac{1}{2}&-\frac{1}{4}\\ \frac{1}{4}&\frac{2}{15}&\frac{1}{4}\end{pmatrix}\begin{pmatrix}x_{1}\\ x_{2}\\ x_{3}\end{pmatrix}+\begin{pmatrix}-1\\ 2\\ 3\end{pmatrix}.\] We also know that \[\alpha = \max_{1\leq i\leq 3}\sum_{j=1}^{3}|a_{ij}|=\max\left\{\frac{1}{5}+ \frac{1}{2}+\frac{1}{4},\frac{1}{3}+\frac{1}{3}+\frac{1}{4},\frac{1}{4}+\frac {2}{15}+\frac{1}{4}\right\}\] \[= \max\left\{\frac{19}{20},\frac{11}{12},\frac{19}{30}\right\}= \frac{19}{20}<1.\] So we can begin with an iteration scheme, \(x^{(0)}=(0,0,0)\), and we can iterate to find \[x^{(1)} = (-1,2,3),\] \[x^{(5)} = (0.507535,0.942640,4.303294),\] \[x^{(10)} = (0.641658,0.835535,4.360547),\] \[x^{(19)} = (0.639519,0.841889,4.362843),\] \[x^{(20)} = (0.639559,0.841833,4.362843).\] This leads to the exact solution of \[x=(0.639548,0.841853,4.362845).\] * Suppose in \((\mathbb{R}^{n},d)\) the metric \(d(\cdot,\cdot)\) is given by \[d(x,z)=\sum_{i=1}^{n}|x_{i}-z_{i}|.\] (2.6) Using this metric, in the following we obtain a condition (the column sum condition) under which \(T\) is a contraction. \[d(y^{\prime},y^{\prime\prime}) = \sum_{i=1}^{n}|y^{\prime}_{i}-y^{\prime\prime}_{i}|=\sum_{i=1}^{ n}\left|\sum_{j=1}^{n}a_{ij}x^{\prime}_{j}+b_{i}-(\sum_{j=1}^{n}a_{ij}x^{\prime \prime}_{j}+b_{i})\right|\] \[= \sum_{i=1}^{n}\left|\sum_{j=1}^{n}a_{ij}(x^{\prime}_{j}-x^{ \prime\prime}_{j})\right|\leq\sum_{i=1}^{n}\sum_{j=1}^{n}|a_{ij}||x^{\prime}_ {j}-x^{\prime\prime}_{j}|\] \[\leq \max_{1\leq j\leq n}\sum_{i=1}^{n}|a_{ij}|\sum_{j=1}^{n}|x^{ \prime}_{j}-x^{\prime\prime}_{j}|\] \[= \max_{1\leq j\leq n}\sum_{i=1}^{n}|a_{ij}|\,d(x,x^{\prime\prime}).\]
**Theorem 85** (Column Sum Criterion).: _If_
\[\beta=\max_{1\leq j\leq n}\sum_{i=1}^{n}|a_{ij}|<1, \tag{2.7} \]\]
_then \(T\) is a contraction and the system \(x=Ax+b\) has precisely one solution._
By the same concrete exercise as before we have
\[\beta=\max_{j}\sum_{i=1}^{3}|a_{ij}| = \max\left\{\frac{1}{5}+\frac{1}{3}+\frac{1}{4},\quad\frac{1}{2} +\frac{1}{3}+\frac{2}{15},\quad\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right\}\] \[= \frac{29}{30}<1.\] 3. If we use the Euclidean metric on \(\mathbb{R}^{n}\) \[d(x,z)=\left(\sum_{i=1}^{n}|x_{i}-z_{i}|^{2}\right)^{\frac{1}{2}},\] we then see \[d(y^{\prime},y^{\prime\prime})^{2} = \sum_{i=1}^{n}|y^{\prime}_{i}-y^{\prime\prime}_{i}|^{2}=\sum_{i =1}^{n}\left|\sum_{j=1}^{n}(a_{ij}x^{\prime}_{j}+b_{i})-\sum_{j=1}^{n}(a_{ij} x^{\prime\prime}_{j}-b_{i})\right|^{2}\] \[= \sum_{i=1}^{n}\left|\sum_{j=1}^{n}a_{ij}(x^{\prime}_{j}-x^{ \prime\prime}_{j})\right|^{2}\leq\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}^{2}\sum_{j =1}^{n}(x^{\prime}_{j}-x^{\prime\prime}_{j})^{2}\] \[= \sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}^{2}d(x^{\prime},x^{\prime \prime})^{2}.\] So we have if \[\gamma=\left(\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}^{2}\right)^{\frac{1}{2}}<1,\] then \(T\) is a contraction and the system has precisely one solution. In our exercise we obtain that \[\gamma = \left[\left(\frac{1}{25}+\frac{1}{4}+\frac{1}{16}\right)+\left( \frac{1}{9}+\frac{1}{9}+\frac{1}{16}\right)+\left(\frac{1}{16}+\frac{4}{225}+ \frac{1}{16}\right)\right]^{\frac{1}{2}}\] \[= \left(\frac{39}{50}\right)^{\frac{1}{2}}<1.\] Note the iterative sequence turns out to converge faster than in part a).
### 2.2 Fixed Point Theorems and Applications
#### Applications to Differential Equations
Consider the Cauchy initial value problem for ordinary differential equations of the first order
\[\begin{cases}x^{\prime}=f(t,x)\\ x(t_{0})=x_{0}\end{cases}. \tag{2.8} \]\]
We examine the solution of this initial value problem with the help of the Contraction Mapping Theorem.
**Definition 71** (Lipschitz Condition).: We say \(f\) satisfies a **Lipschitz Condition** if there exists \(L>0\) such that
\[|f(x)-f(y)|\leq L|x-y|.\]
**Theorem 86** (Picard Existence and Uniqueness Theorem, 1890).: _Let \(f\) be continuous on the rectangle,_
\[R=\{(t,x):|t-t_{0}|\leq a,|x-x_{0}|\leq b\}=[t_{0}-a,t_{0}+a]\times[x_{0}-b,x_{ 0}+b].\]
_Suppose that \(f\) satisfies a Lipschitz condition with respect to the second argument which implies that there exists \(L>0\) such that_
\[|f(t,x_{1})-f(t,x_{2})|\leq L|x_{1}-x_{2}|\ \ \forall(t,x_{1}),(t,x_{2})\in \mathbb{R}^{2}.\]
_Then the Cauchy initial value problem (2.8) has a unique solution (Figure 2.11). This solution exists on an interval \([t_{0}-\beta,t_{0}+\beta]\) where_
\[\beta<\min\left\{a,\frac{b}{M},\frac{1}{L}\right\},\ \ M=\sup_{(t,x)\in \mathbb{R}}|f(t,x)|.\]
The solution passes through \((t_{0},x_{0})\).
Proof
: We will proceed in steps.
* Let \(X\) be the Banach space \(C(J)\), \(J=[t_{0}-\beta,t_{0}+\beta]\), with the sup norm and \[\tilde{C}=\{x\in C(J):x_{0}-M\beta\leq x(t)\leq x_{0}+M\beta\ \ \forall t\in J\},\] where \(M\beta Figure 2.11: Rectangle in Picard’s theorem* Consider the initial value problem given in 2.8; we form the following integral equation \[x(t)=x(t_{0})+\int_{t_{0}}^{t}f(s,x(s))ds\quad\text{ and }\quad Tx(t)=x_{0}+\int_{t_{0}}^{t}f(s,x(s))ds.\] **Corollary 14**.: _The Contraction Mapping Theorem also implies that the unique solution of (2.8) is the limit of the sequence, \(\{x_{0},x_{1},x_{2},\dots\}\) obtained by the Picard iteration_ \[x_{n+1}(t)=x_{0}+\int_{t_{0}}^{t}f(s,x_{n}(s))ds.\] Note that a bigger interval than \([t_{0}-\beta,t_{0}+\beta]\) can be obtained by using a different metric on \(C(J)\). Also, with such a metric we can solve the Cauchy problem in the strip \[S=\{(t,x):|t-t_{0}|\leq a,\,x\in\mathbb{R}\}.\] ### 2.2 Fixed point theorems and applications #### Applications to Integral Equations A little more than a hundred years ago, Fredholm made a study of integral equations of the type \[x(t)=\mu\int_{a}^{b}k(t,s)x(s)\,ds+v(t).\] These arose from questions of theoretical physics and the aim was to solve the function \(x\) in terms of the function \(v\). Since one can think of an integral as a limit of a finite sums, one can view the Fredholm integral equation as an infinite dimensional counterpart of finite dimensional linear systems. There is also a close connection between differential and integral equations and some problems may be formulated either way. In this section, we examine an application of the Contraction Mapping Theorem to integral equations. We can consider integral equations in various function spaces and we will consider them here in \(C[a,b]\), the space of continuous functions on \([a,b]\). We start by defining the Fredholm integral equation. **Definition 72**.: The Fredholm Integral equation of the second kind is an equation of the form \[x(t)=\mu\int_{a}^{b}k(t,s)x(s)ds+v(t), \tag{2.9} \]\] where \(k,v\) are given functions, \(\mu\in\mathbb{R}\) is a parameter, and \(x\) is an unknown function on \([a,b]\), \(k(x,y)\in C([a,b]\times[a,b])\) is called the _kernel_ of this integral equation. Note that a Fredholm equation of the first kind is given by \[\int_{a}^{b}k(t,s)x(s)ds=x(t).\] **Theorem 87** (Fredholm Integral Equation).: _Suppose \(k\in C([a,b]\times[a,b])\), \(v\in C[a,b]\) and_ \[|\mu|<\frac{1}{M(b-a)},\ \ M=\sup_{s,t\in[a,b]}|k(t,s)|.\] _Then (2.9) has a unique solution \(x\in C[a,b]\). This function \(x\) is the limit of the iterative sequence \((x_{0},x_{1},\dots),\) where \(x_{0}\) is any function \(C[a,b]\) and_
* Consider \(T:\tilde{C}\to\tilde{C}\), and notice that \[|Tx(t)-x_{0}| = \left|\int_{t_{0}}^{t}f(s,x(s))ds\right|\] \[\leq \int_{t_{0}}^{t}|f(s,x(s))|ds\,\leq M|t-t_{0}|\leq M\beta.\] This implies that \(Tx(t)\in\tilde{C}\).
* Now we will show that \(T\) is a contraction on \(\tilde{C}\) using the hypothesis that \(T\) is Lipschitz in the second variable: \[|Tx(t)-Ty(t)| = \left|\int_{t_{0}}^{t}[f(s,x(s))-f(s,y(s))]ds\right|\] \[\leq \int_{t_{0}}^{t}|f(s,x(s))-f(s,y(s))|ds\leq L\int_{t_{0}}^{t}|x(s )-y(s)|ds\] \[\leq L||x-y||_{\infty}|t-t_{0}|\leq L\beta||x-y||_{\infty}\quad\forall t \in J.\] This implies that \[||Tx-Ty||_{\infty}\leq L\beta||x-y||_{\infty}.\]
* The Banach Fixed Point Theorem implies \(T\) has a unique fixed point \(x\in\tilde{C}\), that is, a continuous function \(x\) on \(J\) satisfying \(x=Tx\) which implies \[x(t)=x_{0}+\int_{t_{0}}^{t}f(s,x(s))ds,\] and if \((s,x(s))\in\mathbb{R}^{2}\) and \(f\) is continuous we may differentiate and this satisfies the Cauchy initial value problem (2.8).
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