MATH114 Integration and Differentiation

Written Assessment 2 – Solutions

[2.5ex]

• Prove Proposition 2.1.4: If a series is convergent/divergent then so is any other series formed from it by altering only finitely many summands. [4]

  Consider a series ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n$ and denote (as always) the sequence of partial sums by ${(A_n)}_{n\in \mathbb{N}}$, where $A_n={\sum }_{i=1}^n{a}_n$.

  Let us now alter a finite number of summands $a_n$; more precisely, let $a_n^{\prime }$ be numbers such that $a_n^{\prime }=a_n$ for all $n$ except for a finite number of $n$ where $a_n^{\prime }\ne a_n$. Then there will be $N\in \mathbb{N}$ such that $a_n^{\prime }=a_n$, for all $n>N$. [1]

  We are interested in the series ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^{\prime }$. Let us look at the difference of partial sums:

  $$A_n^{\prime }-A_n=\sum _{i=1}^n{a}_n^{\prime }-\sum _{i=1}^n{a}_n=\sum _{i=1}^n(a_n^{\prime }-a_n).$$

  Now all the summands with $n>N$ are $0$. Thus there is a number $c\in ℝ$ such that $A_n^{\prime }-A_n=c$, for all $n>N$. [1]

  Therefore, if ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n$ converges, meaning that ${(A_n)}_{n\in \mathbb{N}}$ converges, then ${(A_n^{\prime })}_{n\in \mathbb{N}}$ converges as well with

  $$A_n^{\prime }\to c+\sum _{n=1}^{\mathrm{\infty }}{a}_n,n\to \mathrm{\infty },$$

  (according to general statements from MATH113), so that ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^{\prime }$ converges, too. [1]

  Since, the other way round, ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n$ is obtained from ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^{\prime }$ by altering finitely many summands, we find that, if ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^{\prime }$ converges then ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n$ converges; in summary, ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^{\prime }$ converges if and only if ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n$ converges.

  The statement of divergence is just the contrapositive of this one, namely ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n$ diverges if and only if ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^{\prime }$ diverges. [1]

• No model solutions provided for essays. [6]

Quiz 2 – Solutions

• Which of the following statements is false?

  (C) The product of two uniformly continuous functions is not necessarily uniformly continuous. Here is a counter-example: consider the function $f:ℝ\to ℝ$ defined by $f(x)=x$, and let $g=f$. Then $f$ is uniformly continuous as, for every $\epsilon >0$, we can choose $\delta =\epsilon$ to satisfy the criterion of uniform continuity in Definition 1.1.7. However, $f\cdot g:ℝ\to ℝ$ is given by $(f\cdot g)(x)=x^2$, which is continuous but not uniformly continuous, see W1.1 Solutions.

• Given two continuous functions $f,g:I\to ℝ$ on a compact interval $I$. Which of the following relations is true?

  (A) ${\max }_{x\in I}(f(x)+g(x))\le {\max }_{x\in I}f(x)+{\max }_{x\in I}g(x)$. In fact, from the definition of the maximum we get

  $$f(x)+g(x)\le \underset{y\in I}{\max }f(y)+\underset{z\in I}{\max }g(z),x\in I,$$

  which can be rewritten with a constant function on the right:

  $$f(x)+g(x)\le (\underset{y\in I}{\max }f(y)+\underset{z\in I}{\max }g(z)){\mathrm{𝟏}}_I(x),x\in I.$$

  Taking the maximum on both sides gives

  $$\underset{x\in I}{\max }(f(x)+g(x))\le \underset{x\in I}{\max }\left((\underset{y\in I}{\max }f(y)+\underset{z\in I}{\max }g(z)){\mathrm{𝟏}}_I(x)\right)=\underset{y\in I}{\max }f(y)+\underset{z\in I}{\max }g(z).$$

• Let $f:I\to ℝ$ be a continuous function and $I$ compact. Let ${\mathrm{\ell }}_n^{(f)}$ and $u_n^{(f)}$ be the integrals of the lower and upper, respectively, approximating step function of $f$ after $n$ bisections. Then which of the following statements is true?

  (C) ${(u_n^{(f)})}_{n\in \mathbb{N}}$ and ${({\mathrm{\ell }}_n^{(f)})}_{n\in \mathbb{N}}$ converge to the same limit point. This is part of Theorem 1.2.12. The limit defines the integral of $f$.

• Consider the function $g:[-1,1]\to ℝ,g(x)=x^3.$ What are the values of the integrals of the approximating steps functions after $n=2$ bisections, ${\mathrm{\ell }}_2^{(g)}$ and $u_2^{(g)}$?

  (A) $-\frac{1}{2}$ and $\frac{1}{2}$. In fact, after $n=2$ bisections of the interval $[-1,1]$, we have $4$ intervals of length $\frac{1}{2}$, and we find for the lower approximating step function according to Definition 1.2.11:

  $$\mathrm{\ell }(f,[-1,1],2,x)=-1\cdot {\mathrm{𝟏}}_{[-1,-\frac{1}{2})}(x)-\frac{1}{8}\cdot {\mathrm{𝟏}}_{[-\frac{1}{2},0)}(x)+0\cdot {\mathrm{𝟏}}_{[0,\frac{1}{2})}(x)+\frac{1}{8}\cdot {\mathrm{𝟏}}_{[\frac{1}{2},1)}(x).$$

  Then

  $${\int }_{-1}^1\mathrm{\ell }(f,[-1,1],2,x)dx=-1\cdot (-\frac{1}{2}+1)-\frac{1}{8}\cdot (0+\frac{1}{2})+0\cdot (\frac{1}{2}-0)+\frac{1}{8}\cdot (1-\frac{1}{2})=-\frac{1}{2}.$$

  Analogously, the upper approximating step function is

  $$u(f,[-1,1],2,x)=-\frac{1}{8}\cdot {\mathrm{𝟏}}_{[-1,-\frac{1}{2})}(x)-0\cdot {\mathrm{𝟏}}_{[-\frac{1}{2},0)}(x)+\frac{1}{8}\cdot {\mathrm{𝟏}}_{[0,\frac{1}{2})}(x)+1\cdot {\mathrm{𝟏}}_{[\frac{1}{2},1)}(x).$$

  This implies

  $${\int }_{-1}^{1}u(f,[-1,1],2,x)dx=\frac{1}{2}.$$

• Consider the function $f:[-\pi ,\pi ]\to ℝ,f(x)={\cos }^2(x).$ What are the values of the integrals of the approximating steps functions after $n=3$ bisections, ${\mathrm{\ell }}_3^{(f)}$ and $u_3^{(f)}$?

  (D) $\frac{\pi }{2}$ and $\frac{3\pi }{2}$. This is calculated by the same procedure as Q2.4.