MTH401 Solutions

MTH401 Tma Solutions

1. The Holder’s inequality states that: if \(1\leq p,q<\infty,\frac{1}{q}+\frac{1}{q}=1\) and if \(x_k,y_k,k=1,2,cdots,n\) are complex numnbers then

\(\left(\sum_{k=1}^{n}\left|x_k+y_k\right|^{p}\right)^{\frac{1}{p}\leq\left(\sum_{k=1}^{n}\left|x_k\right|^{p}\right)^{\frac{1}{p}+\left(\sum_{k=1}^{n}\left|y_k\right|^{p}\right)^{\frac{1}{p} \)

\(\left(\sum_{n=1}^{n}\left|x_k+y_k\right|^{p}\right)^{r}\leq\left(\sum_{k=1}^{n}\left|x_p\right|^{p}\right)^{\frac{1}{2}+\left(\sum_{k=1}^{n}\left|y_p\right|^{p}\right)^{\frac{1}{2} \)

—>> \(\left(\sum_{k=1}^{n}\left|x_k+y_k\right|^{p}\right)^{\frac{1}{p}\leq\left(\sum_{k=1}^{n}\left|x_k\right|^{p}\right)^{\frac{1}{p}+\left(\sum_{k=1}^{n}\left|y_k\right|^{q}\right)^{\frac{1}{q} \)

\(\left(\sum_{k=1}^{n}\left|x_k+y_k\right|^{p}\right)^{\frac{1}{p}\leq\left(\sum_{k=1}^{n}\left|x_p\right|\right)^{\frac{1}{p}+\left(\sum_{k=1}^{n}\left|y_p\right|\right)^{\frac{1}{q} \)

2. A sequence \({x_n}_{n=1}^{\infty}\) of points in a metric space \((E,d)\) is a Cauchy sequence if

for every \(\epsilon>0\), there exists a positive integer \(N\) such that \(x_n\in B(x,\epsilon)\) for all \(n\in N\) where \(B(x,r)={y\in E:d(y,x)<\epsilon}\)

\(X_n_k+1>X_n_k\) and \(n_k\geq k, k=1,2,cdots\) and \(n_k:\mathbb{N}\rightarrow\mathbb{N}

and only if its component sequence converges

—>> for any \(\epsilon>0\), there exists an integer \(N_0>0\) such that for all \(m,n>N_0\) we get that \(d(x_n,x_m)<epsilon\)

3. Let \((E,d)\) be a metric space and \(K\) a subset of \(E\). Then \(K\) is said to be connected if

and only if every subset of \(K\) are only closed

if and only if \(K\) is not the only nonempty set that is open and closed

—>> if and only if it is connected as a subspace

\(K\) is only open

4. If \((E,d)\)is metric space and \(x_0\in E\), then the open ball centred at \(x_0\) of radius \(r>0\) is given by the set

—>> \(B(x_0;r)={y\in E:d((x_0,y)<r}\)

\(B(x_0;r)={y\in E:d((x_0,y)\leq r}\)

\(S(x_0;r)={y\in E:d((x_0,y)\geq r}\)

\(S(x_0;r)={y\in E:d((x_0,y)=r}\)

5. The Euclidean metric on \(\mathbb{R}^n\) is defined as

\(d(x,y)=\sum_{i=1}^{n}\left|x_i-y_i\right|\)

—>> \(d_2(x,y)=\left(\sum_{i=1}^{n}\left|x_i-y_i\right|^2\right)^{\frac{1}{2}\)

\(d_{\infty}(x,y)=max_{1leq ileq n}\left{\left|x_i-y_i\right|\right}\)

\(d_{\infty}(x,y)=min_{1leq ileq n}\left{\left|x_i-y_i\right|^{\frac{1}{2}}\right}\)

6. let \(f:\mathbb{R}\rightarrow\mathbb{R}\) be defined by \(f(x)=\left{\begin{array}{rcl} x^2+1,&\mbox{if}&x\leq 0\\\frac{1}{2}(x+2),&\mbox{if}&x\geq 0\end{array}\right\), then \(f\) is

not continuous at \(x=0\)

not continuous on \(\mathbb{R}\)

—>> continuous on \(\mathbb{R}\)

continuous at \(x=0\)

7. Let \((X,d_x)\) and \((Y,d_Y)\) be arbitrary metric spaces. A mapping \(f:(X,d_x)\rightarrow (Y,d_Y)\) is called a strict contraction if

—>> there exist a constant \(k\in[0,1)\) such that \(d_Y(f(x),f(y))\leq kd_X(x,y)\) for all \(x,y\in X\)

there exist a constant \(k\in[0,1)\) such that \(d_Y(x,y)\leq kd_Y(x,y)\) for all \(x,y\in X\)

there exist a variable \(k\in\mathbb{N}\) such that \(d_X(f(x),f(y))\geq kd(x,y)\) for all \(x,y\in X\)

there exist a constant \(k\in\mathbb{N})\) such that \(d_Y(x,y)geq kd_X(x,y)\) for all \(x,y\in X\)

8. The discrete metric is defined as \(d_0:E\times E\rightarrow \mathbb{R}\) such that

\(d_0(x,y)=\left{\begin{array}{rcl} 1,&\mbox{if}&x\neq y\\-1,&\mbox{if}&x=y\end{array}\right\)

—>> \(d_0(x,y)=\left{\begin{array}{rcl}1,&\mbox{if}&x\neq y\\0,&\mbox{if}&x=y\end{array}\right\)

\(d_0(x,y)=\left{\begin{array}{rcl}0,&\mbox{if}&x\neq y\\-1,&\mbox{if}&x=y\end{array}\right\)

\(d_0(x,y)=\left{\begin{array}{rcl}1,&\mbox{if}&x\geq y\\-1,&\mbox{if}&x\leq y\end{array}\right\)

9. A metric space \((E,d)\) satisfies the following except

\(d(x,y)\leq d(x,z)+d(z,y)\) for all \(x,y,z\in E\)

\(d(x,y)= 0\)

—>> \(d(x,y)\leq 0\) for all \(x,y\in E\)

\(d(x,y)=d(y,x)\) for all \(x,y\in E\)

10. Let \((X,d_x)\) and \((Y,d_Y)\) be metric space and let \(f:D(f)\subset X\rightarrow Y\) where \(D(f)\) is the domain of \(f\), then \(f\) is continuous if

—>> given \(\epsilon>0\), there exist \(\delta>0\) such that if \(x\in D(f)\) and \(d_X(x,x_0)<\delta\), then \(d_Y(f(x),f(x_0))<\epsilon.

given \(\epsilon>0\), there exist \(\delta>0\) such that whenever \(d_2(x,a)<\delta\), it follows that \(|f(x)-f(a)|<\epsilon\)

given \(\epsilon>0\), there exist \(\delta>0\) such that whenever \(d_max(x,x_0)>\delta\), it follows that \(d_max_Y(f(x),f(x_0))<epsilon\)

given \(\epsilon>0\), there exist \(\delta>0\) such that \(d_x(x,x_0)<\epsilon\) then \(d_X(f9x),f(x_0))<\delta\)

JOIN OUR TELEGRAM ON VIP NOUN UPDATES – FOR FREE MTH401 PAST QUESTIONS AND EXAMS SUMMARIES

Leave a Reply

MEET OVER 2000 NOUN STUDENTS HERE. 

Join us for latest NOUN UPDATES and Free TMA answers posted by students on our Telegram. 

OUR ONLINE TUTORIAL CLASS IS NOW ON!!! JOIN US NOW. 
JOIN NOW!
close-link