Calculus – Page 4 – Physicspupil

May 25, 2022October 24, 2022

202205250911 Exercise 2.1A (Q1)

Exercises 1-8, show that the given function $y=f(x)$ is one-to-one over the given interval, then find the formulae for the inverse function $f^{-1}$ and its derivative.

1. $f(x)=x\quad \forall\, x$

2. $f(x)=3x\quad \forall\, x$
3. $f(x)=x^2\quad \forall\, x\ge 0$
4. $f(x)=\sqrt{x}\quad \forall\, x\ge 0$
5. $f(x)=\displaystyle{\frac{1}{x}}\quad \forall\, x>0$
6. $f(x)=\displaystyle{\frac{1}{x}}\quad \forall\, x<0$
7. $f(x)=\displaystyle{\frac{1}{x^2}}\quad \forall\, x>0$
8. $f(x)=x^5\quad \forall\, x$

_{Extracted from Michael Corral. (2020). Elementary Calculus.}

Background. (one-to-one; inverse; derivative of an inverse)

A one-to-one function (aka injection/injective function) is a function $f$ that maps distinct elements to distinct elements; i.e., $f(x_1)=f(x_2)\Rightarrow x_1=x_2$ , or equivalently in the contrapositive that $x_1\neq x_2\Rightarrow f(x_1)\neq f(x_2)$ . Not to be confused with one-to-one correspondence that refers to bijection.

_{Wikipedia on Injective function}

The inverse function of a function $f$ (aka inverse of $f$ ) is a function that undoes the operation of $f$ . The inverse of $f$ exists if and only if $f$ is bijective, and if it exists, is denoted by $f^{-1}$ . For a function $f:X\rightarrow Y$ , its inverse $f^{-1}:Y\rightarrow X$ admits an explicit description: it sends each element $y\in Y$ to the unique element $x\in X$ such that $f(x)=y$ .

_{Wikipedia on Inverse function}

If $y=f(x)$ is differentiable and has an inverse function $x=f^{-1}(y)$ , then $f^{-1}$ is differentiable and its derivative is

$\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}}\quad\textrm{ if }\enspace\frac{\mathrm{d}y}{\mathrm{d}x}\neq 0$ .

_{Text on pg. 38, Sec. 2.1, Ch. 2}

1.

$f(x)=x\quad \forall\, x$ :

$\begin{aligned} \forall\, x\in\mathbb{R},\quad & f(x_1)=x_1 \\ & f(x_2)=x_2 \\ \enspace & f(x_1)=f(x_2) \Rightarrow x_1=x_2 \\ \end{aligned}$

$\therefore$ $y=f(x)=x$ is injective/one-to-one.

In order for an inverse $f^{-1}(x)$ to exist, the original function $f(x)$ must be bijective (et injective et surjective). It remains to check whether or not $y=f(x)=x$ is surjective.

A function $f:X\rightarrow Y$ is said to be surjective if $\forall\, y\in Y,\enspace \exists\, x\in X\textrm{ s.t. }f(x)=y$ . In other words, every element of the function’s codomain is the image of at least one element of its domain.

_{Wikipedia on Surjective function}

Assume $f(x)=x$ is a real-valued function, i.e., $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $x\mapsto x$ . Obviously $f$ is surjective. Thus it is bijective enough to have an inverse $f^{-1}:\mathbb{R}\rightarrow\mathbb{R}$ . And apparently $f^{-1}$ is given by $x\mapsto x$ , such that $f^{-1}\circ f=f\circ f^{-1}=\textrm{id}_{\,\mathbb{R}}$ . Its derivative is $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\big( f^{-1}(x)\big) =\frac{\mathrm{d}}{\mathrm{d}x}(x)=1}$ .

The remaining questions are left the reader.

May 23, 2022October 24, 2022

202205231502 Exercise 1.3 A (Q1)

Exercises 1-9, let $\mathrm{d}x$ be an infinitesimal and prove the given formula.

1. $(\mathrm{d}x+1)^2=2\,\mathrm{d}x+1$

_{Extracted from Michael Corral. (2020). Elementary Calculus.}

Background. (Infinitesimal)

A number $\delta$ is an infinitesimal if the conditions (a) – (d) hold: (a) $\delta\neq 0$ ; (b) If $\delta >0$ then $\delta$ is smaller than any positive real number; (c) If $\delta <0$ then $\delta$ is larger than any negative real number; (d) $\delta^2=0$ (and hence all higher powers of $\delta$ , such as $\delta^3$ and $\delta^4$ , are also $0$ ) N.b. Any infinitesimal multiplied by a nonzero real number is also an infinitesimal, while $0$ times an infinitesimal is $0$ .

Proof.

Suppose the contrary is true:

$(\mathrm{d}x+1)^2\neq 2\,\mathrm{d}x+1$ .

$\begin{aligned} \textrm{LHS} & = (\mathrm{d}x+1)^2 \\ & = (\mathrm{d}x)^2+2(\mathrm{d}x)(1)+(1)^2 \\ & \stackrel{(\textrm{d})}{=} 0+2\,\mathrm{d}x+1 \\ & = 2\,\mathrm{d}x+1 \\ & = \textrm{RHS} \qquad \perp\\ \end{aligned}$

Thus converse is the case.

September 10, 2021October 24, 2022

202109101556 Exercises 1.2.C (Q16)

For Exercises 16-21, assuming that $f'(x)$ exists, prove the given formula.

$f'(x)=\displaystyle{\lim_{h\to 0}\frac{f(x+2h)-f(x-2h)}{4h}}$

Proof.

Renaming by dummy variables.

Let $y=x-2h$ , then $x+2h=(x-2h)+4h=y+4h$ .

Rewrite it as

$f'(x)=\displaystyle{\lim_{h\to 0}\frac{f(y+4h)-f(y)}{4h}}$ .

Note that

$\displaystyle{\lim_{h\to 0}}[\,\cdots ]\Rightarrow \displaystyle{\lim_{4h\to 0}}[\,\cdots ]$ .

So,

$\begin{aligned} f'(x) & = \lim_{4h\to 0}\frac{f(y+4h)-f(y)}{4h} \\ & = \lim_{\Delta y\to 0}\frac{f(y+\Delta y)-f(y)}{\Delta y} \\ & = \lim_{\Delta y\to 0}\frac{\Delta f}{\Delta y}\\ & = \frac{\mathrm{d}f}{\mathrm{d}y}\\ & = \dots\enspace \textrm{(discontinued)}\enspace \dots \\ \end{aligned}$

Do you spot the flaw in the ~~Proof~~?

(revised)

As left-hand limit and right-hand limit are equivalent,

i.e., $f'(x)=\displaystyle{\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(x)-f(x-h)}{h}}$ ,

in our scenario, do write

$\begin{aligned} f'(x) & = \lim_{2h\to 0}\frac{f(x+2h)-f(x)}{2h}=\lim_{2h\to 0}\frac{f(x)-f(x-2h)}{2h} \\ \frac{1}{2}f'(x) & =\lim_{2h\to 0}\frac{f(x+2h)-f(x)}{4h}=\lim_{2h\to 0}\frac{f(x)-f(x-2h)}{4h}\\ \end{aligned}$

Then

$\begin{aligned} & \quad \lim_{h\to 0}\frac{f(x+2h)-f(x-2h)}{4h} \\ & = \lim_{h\to 0}\frac{\big( f(x+2h)-f(x)\big) + \big( f(x)-f(x-2h) \big) }{4h} \\ & = \lim_{h\to 0}\frac{f(x+2h)-f(x)}{4h} + \lim_{h\to 0}\frac{f(x)-f(x-2h)}{4h} \\ & = \lim_{2h\to 0}\frac{f(x+2h)-f(x)}{4h} + \lim_{2h\to 0}\frac{f(x)-f(x-2h)}{4h} \\ & = \frac{1}{2}\cdot f'(x)+\frac{1}{2}\cdot f'(x) \\ & = f'(x) \\ \end{aligned}$

QED

September 10, 2021October 24, 2022

202109101417 Exercises 1.1.A (Q5)

By equation (1.1), $\pi =4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\cdots )$ , where the $n^{\textrm{th}}$ term in the sum inside the parenthesis is $\frac{(-1)^{n+1}}{2n-1}$ (starting at $n=1$ ). So the first approximation of $\pi$ using this formula is $\pi\approx 4(1)=4.0$ , and the second approximation is $\pi\approx 4(1-\frac{1}{3})=8/3\approx 2.66667$ . Continue like this until two consecutive approximations have $3$ as the first digit before the decimal point. How many terms in the sum did this require? Be careful with rounding off in the approximations.

Attempts.

$1^{\textrm{st}}$ approximation:

$\pi\approx 4(1)=4.0$

$2^{\textrm{nd}}$ approximation:

$\pi\approx 4(1-\frac{1}{3})=\frac{8}{3}\approx 2.66667$

$3^{\textrm{rd}}$ approximation:

$\pi\approx 4(1-\frac{1}{3}+\frac{1}{5})=\frac{52}{15}\approx 3.466667\quad (\textrm{5 d.p.})$

$4^{\textrm{th}}$ approximation:

$\pi\approx 4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7})=\frac{304}{105}\approx 2.89523\quad (\textrm{5 d.p.})$

$5^{\textrm{th}}$ approximation:

$\pi\approx 4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9})=\frac{1052}{315}\approx 3.33968\quad (\textrm{5 d.p.})$

$6^{\textrm{th}}$ approximation:

$\pi\approx 4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11})=\frac{10312}{3465}\approx 2.97605\quad (\textrm{5 d.p.})$

$7^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}\bigg) \\ & =\frac{147916}{45045} \\ & \approx 3.28374\quad (\textrm{5 d.p.}) \end{aligned}$

$8^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}\bigg) \\ & =\frac{135904}{45045} \\ & \approx 3.01707\quad (\textrm{5 d.p.}) \end{aligned}$

$9^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}\bigg) \\ & =\frac{2490548}{765765} \\ & \approx 3.25237\quad (\textrm{5 d.p.}) \end{aligned}$

$10^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}\bigg) \\ & =\frac{44257352}{14549535} \\ & \approx 3.04184\quad (\textrm{5 d.p.}) \end{aligned}$

$11^{\textrm{th}}$ approximation:

$12^{\textrm{th}}$ approximation:

$13^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}\bigg) \\ & =\frac{5385020324}{1673196525} \\ & \approx 3.21840\quad (\textrm{5 d.p.}) \end{aligned}$

$14^{\textrm{th}}$ approximation:

$15^{\textrm{th}}$ approximation:

$16^{\textrm{th}}$ approximation:

$17^{\textrm{th}}$ approximation:

$18^{\textrm{th}}$ approximation:

$19^{\textrm{th}}$ approximation:

$20^{\textrm{th}}$ approximation:

$21^{\textrm{st}}$ approximation:

$22^{\textrm{nd}}$ approximation:

$23^{\textrm{rd}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\bigg) \\ & =\frac{937558224301357108}{294362129962575675} \\ & \approx 3.18505\quad (\textrm{5 d.p.}) \end{aligned}$

$24^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}\bigg) \\ & =\frac{42887788022313481376}{13835020108241056725} \\ & \approx 3.09994\quad (\textrm{5 d.p.}) \end{aligned}$

$25^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}\bigg) \\ & =\frac{308120241932332116332}{96845140757687397075} \\ & \approx 3.18158\quad (\textrm{5 d.p.}) \end{aligned}$

$26^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}\bigg) \\ & =\frac{300524544618003693032}{96845140757687397075} \\ & \approx 3.10315\quad (\textrm{5 d.p.}) \end{aligned}$

$27^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}\bigg) \\ & =\frac{16315181427784945318996}{5132792460157432044975} \\ & \approx 3.17862\quad (\textrm{5 d.p.}) \end{aligned}$

$28^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}\bigg) \\ & =\frac{15941887430682586624816}{5132792460157432044975} \\ & \approx 3.10589\quad (\textrm{5 d.p.}) \end{aligned}$

$29^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}\bigg) \\ & =\frac{16302083392798897645516}{5132792460157432044975} \\ & \approx 3.17607\quad (\textrm{5 d.p.}) \end{aligned}$

$30^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}-\frac{1}{59}\bigg) \\ & =\frac{941291750334505232905544}{302834755149288490653525} \\ & \approx 3.10827\quad (\textrm{5 d.p.}) \end{aligned}$

$31^{\textrm{st}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}-\frac{1}{59}+\frac{1}{61}\bigg) \\ & =\frac{58630135791001973169852284}{18472920064106597929865025} \\ & \approx 3.17384\quad (\textrm{5 d.p.}) \end{aligned}$

$32^{\textrm{nd}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}-\frac{1}{59}+\frac{1}{61}-\frac{1}{63}\bigg) \\ & =\frac{57457251977407903460019584}{18472920064106597929865025} \\ & \approx 3.11035\quad (\textrm{5 d.p.}) \end{aligned}$

$33^{\textrm{rd}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}-\frac{1}{59}+\frac{1}{61}-\frac{1}{63}+\frac{1}{65}\bigg) \\ & =\frac{4507234389098153984166548}{1420993851085122917681925} \\ & \approx 3.17189\quad (\textrm{5 d.p.}) \end{aligned}$

$34^{\textrm{th}}$ approximation:

$\begin{aligned} \pi & \approx 4\bigg( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23} \\ & \qquad\quad +\frac{1}{25}-\frac{1}{27}+\frac{1}{29}-\frac{1}{31}+\frac{1}{33}-\frac{1}{35}+\frac{1}{37}-\frac{1}{39}+\frac{1}{41}-\frac{1}{43}+\frac{1}{45}\\ & \qquad\qquad\quad -\frac{1}{47}+\frac{1}{49}-\frac{1}{51}+\frac{1}{53}-\frac{1}{55}+\frac{1}{57}-\frac{1}{59}+\frac{1}{61}-\frac{1}{63}+\frac{1}{65}-\frac{1}{67}\bigg) \\ & =\frac{296300728665235825268431016}{95206588022703235484688975} \\ & \approx 3.11219\quad (\textrm{5 d.p.}) \end{aligned}$

Summing without aim, I forgot my purpose. Where am I?

(discontinued)

(refreshed)

Please scroll up to the $7^{\textrm{th}}$ and $8^{\textrm{th}}$ approximation.

This required seven or eight terms in the sum for having $3$ as the first digit before the decimal point.

September 10, 2021October 24, 2022

202110091141 Exercises 1.1.A (Q1-Q4)

For Exercises 1-4, suppose that an object moves in a straight line such that its position $s$ after time $t$ is the given function $s=s(t)$ . Find the instantaneous velocity of the object at a general time $t\ge 0$ . You should mimic the earlier example for the instantaneous velocity when $s=-16t^2+100$ .

1. $s=t^2$

2. $s=9.8t^2$

3. $s=-16t^2+2t$

4. $s=t^3$

Ans.

1. $2t$

2. $19.6t$

3. $-32t+2$

4. $3t^2$

Solution.

1.

The average velocity of the object over the interval $[t,t+\Delta t]$ is $\frac{\Delta s}{\Delta t}$ , so since $s(t)=t^2$ :

$\begin{aligned} \frac{\Delta s}{\Delta t} & = \frac{s(t+\Delta t)-s(t)}{\Delta t} \\ & = \frac{(t+\Delta t)^2 - t^2}{\Delta t} \\ & = \frac{(t^2+2t\Delta t+(\Delta t)^2)-(t^2)}{\Delta t} \\ & = \frac{2t\Delta t+(\Delta t)^2}{\Delta t} \\ & = \frac{\Delta t(2t+\Delta t)}{\Delta t} \\ & = 2t + \Delta t \end{aligned}$

Now let the interval $[t,t+\Delta t]$ get smaller and smaller indefinitely—that is let $\Delta t$ get closer and closer to $0$ . Then the average velocity $\frac{\Delta s}{\Delta t}=2t+\Delta t$ gets closer and closer to $2t+0=2t$ . Thus, the object has instantaneous velocity $2t$ at time $t$ . This calculation can be interpreted as taking the limit of $\frac{\Delta s}{\Delta t}$ as $\Delta t$ approaches $0$ , written as follows:

$\begin{aligned} & \qquad \textrm{instantaneous velocity at }t \\ & = \textrm{limit of average velocity over }[t,t+\Delta t]\textrm{ as }\Delta t\textrm{ approaches to }0 \\ & = \lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t} \\ & = \lim_{\Delta t\to 0}(2t+\Delta t) \\ & = 2t+(0) \\ & = 2t \end{aligned}$

2.

$\begin{aligned} &\qquad \textrm{instantaneous velocity at }t\\ & = \lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t}\\ & = \lim_{\Delta t\to 0}\frac{s(t+\Delta t)-s(t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{9.8(t+\Delta t)^2 - 9.8t^2}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{9.8(t^2+2t(\Delta t)+(\Delta t)^2) - 9.8t^2}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{19.6t(\Delta t)+9.8(\Delta t)^2}{\Delta t} \\ & = \lim_{\Delta t\to 0}19.6t+9.8(\Delta t) \\ & = 19.6t+9.8(0) \\ & = 19.6t \end{aligned}$

3.

$\begin{aligned} &\qquad \textrm{instantaneous velocity at }t\\ & = \lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t}\\ & = \lim_{\Delta t\to 0}\frac{s(t+\Delta t)-s(t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{\big( -16(t+\Delta t)^2+2(t+\Delta t)\big) - (-16t^2+2t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{-16(t^2+2t(\Delta t)+(\Delta t)^2)+2(t+\Delta t)+16t^2-2t}{\Delta t} \\ & = \lim_{\Delta t\to 0}\frac{-32t(\Delta t)-16(\Delta t)^2+2(\Delta t)}{\Delta t} \\ & = \lim_{\Delta t\to 0}(-32t-16(\Delta t)+2) \\ & = -32t-16(0)+2 \\ & = -32t+2 \end{aligned}$

4.

$\begin{aligned} s(t) & = t^3 \\ s(t+\Delta t) & = (t+\Delta t)^3 \\ & = t^3+3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3 \\ s(t+\Delta t)-s(t) & = \big( t^3+3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3 \big) - (t^3) \\ & = 3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3 \\ \frac{s(t+\Delta t)-s(t)}{\Delta t} & = \frac{3t^2(\Delta t)+3t(\Delta t)^2+(\Delta t)^3}{\Delta t} \\ & = 3t^2+3t(\Delta t)+(\Delta t)^2 \\ \lim_{\Delta t\to 0}\frac{s(t+\Delta t)-s(t)}{\Delta t} & = 3t^2+3t(0)+(0)^2 \\ \frac{\mathrm{d}s}{\mathrm{d}t} & = 3t^2\\ \end{aligned}$

January 14, 2021October 24, 2022

202101141227 Exercise 1 (Q10)

If $f(x)=2x^2$ , find expressions for:

1. $f(x+h)$ .

2. $f(x+h)-f(x)$ .

3. $\displaystyle{\frac{f(x+h)-f(x)}{h}}$ .

Solution.

1.

$\begin{aligned} f(x+h) & = 2(x+h)^2 \\ & = 2(x^2+2hx+h^2) \\ & = 2x^2+4hx+2h^2 \end{aligned}$

2.

$\begin{aligned} f(x+h)-f(x) & = 2(x+h)^2 - 2(x)^2 \\ & = 2(x^2+2hx+h^2) - 2x^2 \\ & = 4hx+2h^2 \\ \end{aligned}$

3.

$\begin{aligned} \frac{f(x+h)-f(x)}{h} & = \frac{2(x+h)^2-2(x)^2}{h} \\ & = \frac{2(x^2+2hx+h^2)-2x^2}{h} \\ & = \frac{4hx+2h^2}{h} \\ & = 4x+2h \\ \end{aligned}$

Remark.

The first step to derivation from first principles, is calculate the slope of tangent

$\displaystyle{\frac{f(x+h)-f(x)}{h}}$ ,

and put the limit $h\to 0$ , i.e.,

$\begin{aligned} f'(x) & =\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}x}} \\ & = \displaystyle{\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}} \end{aligned}$

a right-hand limit.

Afterword.

$\begin{aligned} f(x) & =2x^2 \\ f'(x) & = \frac{\mathrm{d}f}{\mathrm{d}x} \\ & = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ & \stackrel{\textrm{(3)}}{=} \lim_{h\to 0} 4x+2h \\ & = 4x+2(0) \\ & = 4x \end{aligned}$

November 11, 2020July 25, 2022

202011110613 Sidenote of Dummy Variables

Q: What is a dummy variable?

A: $f(x)=x^2$ and $g(y)=y^2$ are dummy variables because they describe the same pattern.

The indefinite integrals $\int x\,\mathrm{d}x=x^2/2+C$ and $\int y\,\mathrm{d}y=y^2/2+C$ are $\textrm{\scriptsize \textbf{NOT}}$ dummy variables because they are functionals admitting of different functions.

But if we put an upper and a lower limit to make it a definite integral,

i.e.,

$\begin{aligned} \int_0^1x\,\mathrm{d}x=\bigg[ \frac{x^2}{2} \bigg]_0^1=\frac{1}{2} \\ \int_0^1 y\,\mathrm{d}y=\bigg[ \frac{y^2}{2} \bigg]_0^1=\frac{1}{2} \\ \end{aligned}$ ,

they are dummy variables as the structure preserves the value.

November 9, 2020October 24, 2022

202011091423 Exercise 1 (Q2)

If $f(x)=(x-1)(x+5)$ , find the values of $f(2)$ , $f(1)$ , $f(0)$ , $f(a+1)$ , $f\bigg(\displaystyle{\frac{1}{a}}\bigg)$ , $f(-5)$ .

Solution.

$\begin{aligned} f(2) & = (2-1)(2+5)=7 \\ f(1) & = (1-1)(1+5)=0 \\ f(0) & = (0-1)(0+5)=-5 \\ f(a+1) & = ((a+1)-1)((a+1)+5) = a(a+6) \\ f\bigg( \frac{1}{a}\bigg) & = \Bigg( \bigg( \frac{1}{a}\bigg) -1\Bigg) \Bigg( \bigg( \frac{1}{a}\bigg) +5 \Bigg) \\ f(-5) & = (-5-1)(-5+5) = 0 \end{aligned}$

On reflection.

Suppose you are given an unknown quadratic function $f(x;a_0,a_1,a_2)=a_0+a_1x+a_2x^2$ such that

$\begin{aligned} 0 & \mapsto -5 \\ 1 & \mapsto 0 \\ 2 & \mapsto 7 \\ \end{aligned}$

and you are asked to make polynomial interpolation.

$\begin{cases} a_0 + a_1 (0) + a_2(0)^2 = f(0) = -5 \\ a_0 + a_1 (1) + a_2(1)^2 = f(1) = 0 \\ a_0 + a_1 (2) + a_2(2)^2 = f(2) = 7 \\ \end{cases}$
$\begin{cases} a_0=-5 \\ a_0 + a_1 + a_2 = 0\\ a_0 +2a_1 + 4a_2 = 7\\ \end{cases}$
$\begin{cases} a_0 = -5 \\ a_1 + a_2 = 5 \\ a_1 + 2a_2 = 6 \\ \end{cases}$
$\begin{cases} a_0 = -5 \\ a_1 = 4 \\ a_2 = 1 \\ \end{cases}$

And then is the interpolated polynomial $f(x)=x^2+4x-5\qquad \textrm{\scriptsize OR}\qquad (x+5)(x-1)$ .

November 5, 2020October 24, 2022

202011051527 Exercise 1 (Q1)

If $f(x)=2x^2-4x+1$ , find the values of $f(1)$ , $f(0)$ , $f(2)$ , $f(-2)$ , $f(a)$ , $f(x+\delta x)$ .

Solution.

Given $f(x)=2x^2-4x+1$ .

$\begin{aligned} f(1) & =2(1)^2-4(1)+1=-1 \\ f(0) & = 2(0)^2 - 4(0) +1 = 1 \\ f(2) & = 2(2)^2-4(2)+1 =1 \\ f(-2) & = 2(-2)^2-4(-2)+1=17\\ f(a) & = 2a^2 - 4a +1 \\ f(x+\delta x) & = 2(x+\delta x)^2 - 4 (x+\delta x) +1 \end{aligned}$

This exercise is done.

On reflection.

Suppose you are given the following conditions:

$\begin{aligned} x_0 = 0 & \qquad f(x_0) = 1 \\ x_1 = 1 &\qquad f(x_1) = -1 \\ x_2 = 2 &\qquad f(x_2) =1 \end{aligned}$

and you are asked to interpolate by Lagrange polynomials over the range $[0,2]$ .

$\begin{aligned} \mathcal{L}(x) & = (1)\bigg( \displaystyle{\frac{x-1}{0-1}} \bigg)\bigg( \displaystyle{\frac{x-2}{0-2}} \bigg) + (-1)\bigg( \displaystyle{\frac{x-0}{1-0}} \bigg) \bigg( \displaystyle{\frac{x-2}{1-2}} \bigg) + (1)\bigg( \displaystyle{\frac{x-0}{2-0}} \bigg) \bigg( \displaystyle{\frac{x-1}{2-1}} \bigg) \\ & = \displaystyle{\frac{(x-1)(x-2)}{2}} + x(x-2) + \displaystyle{\frac{x(x-1)}{2}} \\ & = \displaystyle{\frac{(x-1)(x-2)+2x(x-2)+x(x-1)}{2}} \\ & = \displaystyle{\frac{x^2-3x+2+2x^2-4x+x^2-x}{2}} \\ & = \displaystyle{\frac{4x^2-8x+2}{2}} \\ & = 2x^2-4x+1\\ \end{aligned}$

The interpolating polynomial $\mathcal{L}(x)$ checks with the original function $f(x)$ .

September 26, 2020October 24, 2022

202009260344 Exercise 2.5.1

Show that $x=e^{2\theta}\sin\theta$ satisfies the equation $x''-4x'+5x=0$ .

Solution.

(bottom-up)

Let $x'$ denote $\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}\theta}}$ , $x''$ denote $\displaystyle{\frac{\mathrm{d}^2x}{\mathrm{d}\theta^2}}$ .

$\begin{aligned} x' & = \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \sin\theta \big) \\ & = e^{2\theta} \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( \sin\theta \big) + \sin\theta \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \big) \\ & = e^{2\theta} \cos\theta + 2e^{2\theta}\sin\theta \\ & = e^{2\theta}(\cos\theta + 2\sin\theta ) \\ \end{aligned}$

$\begin{aligned} x'' & = \frac{\mathrm{d}}{\mathrm{d}\theta} \big( x' \big) \\ & = e^{2\theta} \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} (\cos\theta +2\sin\theta ) + (\cos\theta + 2\sin\theta )\cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \big) \\ & = (e^{2\theta})(-\sin\theta + 2\cos\theta ) + (\cos\theta + 2\sin\theta )(2e^{2\theta }) \\ & = e^{2\theta}(3\sin\theta + 4 \cos\theta ) \\ \end{aligned}$

To know whether or not $x=e^{2\theta}\sin\theta$ is a solution, I simply do substitution in the equation $x''-4x'+5x=0$ .

$\begin{aligned} \textrm{LHS} & = x''-4x'+5x \\ & = e^{2\theta}(3\sin\theta +4\cos\theta )-4\big[ e^{2\theta}(\cos\theta +2\sin\theta )\big] + 5\big(e^{2\theta}\sin\theta \big) \\ & = \dots \\ & = 0 \\ & = \textrm{RHS} \\ \end{aligned}$

Revision.

(top-down)

We are given a second-order linear homogeneous ordinary differential equation (ODE):

$x''-4x'+5x = 0$

with some independent variable $\theta$ and some dependent variable $x(\theta )=e^{2\theta }\sin\theta$ , the coefficients of $x''$ , $x'$ , and $x$ being constants $1$ , $-4$ , and $5$ .

Let the primed $[ *]'$ be the function derived wrt. to $\theta$ . Following the routine procedures,

$\begin{aligned} x & = e^{r\theta} \\ x' & = re^{r\theta }\\ x'' & = r\cdot ( e^{r\theta })' + r' \cdot ( e^{r\theta }) \\ & = r^2e^{r\theta} \end{aligned}$

rewrite it,

$\begin{aligned} x''-4x'+5x & = 0 \\ r^2e^{r\theta} - 4re^{r\theta} + 5e^{r\theta} & = 0 \\ \dots \textrm{\quad excepting\quad} & x=e^{r\theta}=0\textrm{\quad \dots} \\ r^2 -4r+5 & = 0 \\ \end{aligned}$

and we shall obtain the auxiliary equation (aka. the characteristic equation) on the very last line.

$r=\displaystyle{\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(5)}}{2(1)}}=2\pm \textrm{i}$ .

Note.

If the roots of the auxiliary equation $ar^2+br+c=0$ are the complex numbers $r_1=\alpha +\textrm{i}\beta$ and $r_2=\alpha -\textrm{i}\beta$ , the general solution of $ay''+by'+cy=0$ is

$y=e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x)$

_{(please refer to J. Stewart’s Calculus, Second-Order Linear Differential Equations)}

Now that $r_1=2+\textrm{i}$ and $r_2=2-\textrm{i}$ and $c_1$ , $c_2$ are arbitrary constants, to this $2^{\textrm{nd}}$ -order ODE, the general solution is

$x=e^{2\theta}(c_1\cos\theta +c_2\sin\theta )$ .

Examination.

It is a good practice, no matter how much time one would allow oneself, to countercheck one’s solution…

Here it goes,

$\begin{aligned} x & =e^{2\theta}(c_1\cos\theta +c_2\sin\theta ) \\ & = (c_2e^{2\theta })(\sin\theta ) + (c_1e^{2\theta })(\cos\theta ) \\ \end{aligned}$ .

$\begin{aligned} x' & = e^{2\theta}(-c_1\sin\theta +c_2\cos\theta ) + 2e^{2\theta}(c_1\cos\theta +c_2\sin\theta ) \\ & = (2c_2e^{2\theta }-c_1e^{2\theta})(\sin\theta ) + (2c_1e^{2\theta} + c_2e^{2\theta} )(\cos\theta ) \\ \end{aligned}$

$\begin{aligned} x''& = \big[ (2c_2e^{2\theta }-c_1e^{2\theta})(\sin\theta ) \big]' + \big[ (2c_1e^{2\theta} + c_2e^{2\theta} )(\cos\theta ) \big]' \\ & = \big[ (2c_2e^{2\theta}-c_1e^{2\theta })(\cos\theta )+(4c_2e^{2\theta }-2c_1e^{2\theta })(\sin\theta ) \big] \\ & \qquad\quad + \big[ (2c_1e^{2\theta }+c_2e^{2\theta })(-\sin\theta )+(4c_1e^{2\theta }+2c_2e^{2\theta })(\cos\theta ) \big] \\ & = (-4c_1e^{2\theta }+3c_2e^{2\theta })(\sin\theta ) + (3c_1e^{2\theta }+4c_2e^{2\theta })(\cos\theta ) \\ \end{aligned}$

Then,

$\begin{aligned} \textrm{LHS} & = x'' - 4x' + 5x \\ & = \big( (-4c_1e^{2\theta }+3c_2e^{2\theta })-4(2c_2e^{2\theta}-c_1e^{2\theta})+5c_2e^{2\theta}\big)(\sin\theta ) \\ & \qquad\quad + \big( (3c_1e^{2\theta}+4c_2e^{2\theta}) - 4(2c_1e^{2\theta}+c_2e^{2\theta})+5c_1e^{2\theta} \big)(\cos\theta ) \\ & = 0 \\ & = \textrm{RHS} \end{aligned}$

In conclusion, $x=e^{2\theta}\sin\theta$ is a particular solution, the general solution being $x=e^{2\theta}(c_1\cos\theta +c_2\sin\theta )$ .