Open Textbook Library – Page 2

September 16, 2021October 24, 2022

202109161108 Exercises 3.3.A (Q2)

For Exercises 1-8, evaluate the given triple integral.

$\displaystyle{\int_{0}^{1}\int_{0}^{x}\int_{0}^{y}xyz\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x}$

Solution.

$\begin{aligned} &\qquad \int_{0}^{1}\int_{0}^{x}\int_{0}^{y}xyz\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{1}\int_{0}^{x} \bigg( \frac{xyz^2}{2}\bigg|_{z=0}^{z=y} \bigg)\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{1}\int_{0}^{x} \frac{xy^3}{2}\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{1} \bigg( \frac{xy^4}{8}\bigg|_{y=0}^{y=x} \bigg) \,\mathrm{d}x \\ & = \int_{0}^{1} \frac{x^5}{8}\,\mathrm{d}x \\ & = \bigg[ \frac{x^6}{48} \bigg]_{0}^{1} \\ & = \frac{1}{48} \\ \end{aligned}$

September 16, 2021October 24, 2022

202109161055 Exercises 3.3.A (Q1)

For Exercises 1-8, evaluate the given triple integral.

$\displaystyle{\int_{0}^{3}\int_{0}^{2}\int_{0}^{1}}xyz\,\mathrm{d}x\, \,\mathrm{d}y\,\mathrm{d}z$

Solution.

$\begin{aligned} & \qquad \int_{0}^{3}\int_{0}^{2}\int_{0}^{1}xyz\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \\ & = \int_{0}^{3}\int_{0}^{2} \bigg(\frac{x^2yz}{2}\bigg|_{x=0}^{x=1}\bigg)\,\mathrm{d}y\,\mathrm{d}z \\ & = \int_{0}^{3}\int_{0}^{2}\bigg(\frac{yz}{2}\bigg)\,\mathrm{d}y\,\mathrm{d}z \\ & = \int_{0}^{3}\bigg( \frac{y^2z}{4}\bigg|_{y=0}^{y=2} \bigg) \,\mathrm{d}z \\ & = \int_{0}^{3} z\,\mathrm{d}z \\ & = \bigg[\frac{z^2}{2}\bigg]_{0}^{3} \\ & = 4\frac{1}{2} \\ \end{aligned}$

September 16, 2021October 24, 2022

202109161046 Exercises 1.1.A (Q2)

For the points $P=(1,-1,1)$ , $Q=(2,-2,2)$ , $R=(2,0,1)$ , $S=(3,-1,2)$ , does $\overrightarrow{PQ}=\overrightarrow{RS}$ ?

Solution.

$\begin{aligned} \overrightarrow{PQ} & = (2,-2,2) - (1,-1,1) \\ & = (1,-1,1) \\ \overrightarrow{RS} & = (3,-1,2) - (2,0,1) \\ & = (1,-1,1) \\ \end{aligned}$

$\therefore\enspace \overrightarrow{PQ}=\overrightarrow{RS}$

September 16, 2021October 24, 2022

202109161039 Exercises 1.1.A (Q1)

Calculate the magnitudes of the following vectors:

(a) $\mathbf{v}=(2,-1)$
(b) $\mathbf{v}=(2,-1,0)$
(c) $\mathbf{v}=(3,2,-2)$
(d) $\mathbf{v}=(0,0,1)$
(e) $\mathbf{v}=(6,4,-4)$

Solution.

(a)

$\begin{aligned} v & = |\mathbf{v}| \\ & = \sqrt{(2)^2+(-1)^2} \\ & = \sqrt{5} \end{aligned}$

(b)

$\begin{aligned} v & = |\mathbf{v}| \\ & = \sqrt{(2)^2+(-1)^2+(0)^2} \\ & = \sqrt{5} \end{aligned}$

(c)

$\begin{aligned} v & = |\mathbf{v}| \\ & = \sqrt{(3)^2+(2)^2+(-2)^2} \\ & = \sqrt{17} \end{aligned}$

(d)

$\begin{aligned} v & = |\mathbf{v}| \\ & = \sqrt{(0)^2+(0)^2+(1)^2} \\ & = \sqrt{1} \\ & = 1 \end{aligned}$

(e)

$\begin{aligned} v & = |\mathbf{v}| \\ & = \sqrt{(6)^2+(4)^2+(-4)^2} \\ & = \sqrt{68} \\ & = 2\sqrt{17} \\ \end{aligned}$

September 16, 2021October 24, 2022

202109160920 Exercises 5.3.B (Q21)

Show that for all $x>0$ ,

$\displaystyle{\int_{0}^{x}\frac{\mathrm{d}t}{1+t^2}+\int_{0}^{1/x}\frac{\mathrm{d}t}{1+t^2}}$

is independent of $x$ .

Solution.

$\begin{aligned} & \qquad\int_{0}^{x}\frac{\mathrm{d}t}{1+t^2}+\int_{0}^{1/x}\frac{\mathrm{d}t}{1+t^2} \\ \dots &\enspace\textrm{as }\sin^2u +\cos^2u =1 \enspace\dots\\ \dots &\enspace\textrm{so }\tan^2u +1 = \sec^2u\enspace\dots \\ \dots &\enspace\textrm{let }t=\tan u \textrm{ then }\mathrm{d}t=\sec^2u\,\mathrm{d}u\enspace\dots \\ & = \int_{0}^{\arctan x}\frac{\sec^2u\,\mathrm{d}u}{1+\tan^2u} + \int_{0}^{\arctan (\frac{1}{x})}\frac{\sec^2u\,\mathrm{d}u}{1+\tan^2u} \\ & = \int_{0}^{\arctan x}\frac{\sec^2u\,\mathrm{d}u}{\sec^2u} + \int_{0}^{\arctan (\frac{1}{x})}\frac{\sec^2u\,\mathrm{d}u}{\sec^2u} \\ & = [u]|_{0}^{\arctan x} + [u]|_{0}^{\arctan (1/x)}\\ & = \arctan x + \arctan\bigg(\frac{1}{x}\bigg) \\ & \stackrel{(*)}{=} \frac{\pi}{2} \end{aligned}$

Lemma (*)

Let $y=\arctan x$ and $w=\arctan \displaystyle{\bigg(\frac{1}{x}\bigg)}$ ,

then $x=\tan y$ and $\displaystyle{\frac{1}{x}}=\tan w$ .

Thus,

$\begin{aligned} \tan y & = \frac{1}{\tan w} \\ \tan y & = \cot w \\ \tan y & = \tan \bigg(\frac{\pi}{2}-w\bigg) \\ y & = \frac{\pi}{2} - w \\ y + w & = \frac{\pi}{2} \\ \end{aligned}$

$\therefore\enspace \displaystyle{\arctan x + \arctan\bigg(\frac{1}{x}\bigg) = \frac{\pi}{2}}$

(independent of $x$ )

September 16, 2021July 25, 2022

202109160908 Exercises 5.1.A (Q18)

Use the free fall motion equation for position to show that the maximum height reached by an object launched straight up from the ground with an initial velocity $v_0$ is $\frac{v_0^2}{2g}$ .

Proof.

$\begin{aligned} v^2 & = u^2+2as \\ \dots \textrm{ upward} & \textrm{ taken to be positive }\dots \\ 0 & = v_0^2 + 2(-g)s \\ s & = \frac{v_0^2}{2g} \end{aligned}$

September 16, 2021October 24, 2022

202109160857 Exercises 5.1.A (Q17)

Integrate both sides of the equation

$\displaystyle{\frac{\mathrm{d}P}{P}+\frac{\mathrm{d}M}{M}=\frac{\mathrm{d}T}{2T}}$

to obtain the ideal gas continuity relation:

$\displaystyle{\frac{PM}{\sqrt{T}}=\textrm{constant}}$ .

Attempts.

$\begin{aligned} \int\frac{\mathrm{d}P}{P} + \int\frac{\mathrm{d}M}{M} & = \int\frac{\mathrm{d}T}{2T} \\ \ln |P| + \ln |M| - \frac{1}{2}\ln |T| & = C \\ \ln \bigg|\frac{PM}{T^{1/2}}\bigg| & = C \\ \frac{PM}{\sqrt{T}} & = \textrm{Const.} \\ \end{aligned}$

September 10, 2021July 25, 2022

202109101713 Exercises 1.2.1

Let $T:\mathbb{R}^2\rightarrow \mathbb{R}^3$ be the function defined by $T(x_1,x_2)=(x_1,x_2,0)$ . Show that $T$ is a linear transformation.

Definition. A function $T:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is called a linear transformation if:
(1) $T(\mathbf{x}+\mathbf{y})=T(\mathbf{x})+T(\mathbf{y})$ ; and
(2) $T(c\mathbf{x})=cT(\mathbf{x})$ .
The conditions must be satisfied for all $\mathbf{x},\mathbf{y}$ in $\mathbb{R}^n$ and all $c$ in $\mathbb{R}$ .

Proof.

Let $\mathbf{x}=(x_1,x_2)$ and $\mathbf{y}=(y_1,y_2)$ .

Then

$\begin{aligned} T(\mathbf{x})& =T(x_1,x_2)=(x_1,x_2,0) \\ T(\mathbf{y})&=T(y_1,y_2)=(y_1,y_2,0) \\ T(\mathbf{x})+T(\mathbf{y}) & = (x_1,x_2,0) + (y_1,y_2,0) \\ \Rightarrow\enspace\textrm{RHS} & = (x_1+y_1,x_2+y_2,0)\\ \mathbf{x}+\mathbf{y} & = (x_1,x_2) + (y_1,y_2) \\ & = (x_1+y_1,x_2+y_2) \\ T(\mathbf{x}+\mathbf{y}) & = T(x_1+y_1,x_2+y_2) \\ \Rightarrow\enspace\textrm{LHS} & = (x_1+y_1,x_2+y_2,0) \\ \therefore\enspace \textrm{LHS} & = \textrm{RHS} \\ \end{aligned}$

Condition (1) is thus satisfied.

$\forall\, c\in\mathbb{R}$ and $\forall\, \mathbf{x}=(x_1,x_2)\in\mathbb{R}^2$ ,

$\begin{aligned} cT(\mathbf{x}) & = cT(x_1,x_2) \\ & = c(x_1,x_2,0) \\ \Rightarrow\enspace\textrm{RHS} & = (cx_1,cx_2,0) \\ T(c\mathbf{x}) & = T(c(x_1,x_2)) \\ & = T(cx_1,cx_2) \\ \Rightarrow \enspace\textrm{LHS} & = (cx_1,cx_2,0) \\ & = \textrm{RHS} \\ \end{aligned}$

Condition (2) is also satisfied.

In conclusion, $T$ is a linear transformation.

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.