From proof to simpler proof

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Solve $\sin(x)=2x$ for $x$ .

By mere inspection, we have $x=0$ .

Visually, it appears that $0$ is the only solution (see Fig.1 or Fig. 2)

Fig. 1

Fig. 2

To show that $0$ is the only solution of $\sin(x)=2x$ analytically, let

$f(x) = \sin(x)-2x$

$0$ is a solution of $\sin(x)=2x$ means

$f(0) = 0\quad\quad\quad(1)$

Suppose there is a solution

$x^* \neq 0\quad\quad\quad(2)$

then,

$f(x^*)=\sin(x^*)-2x^*=0\quad\quad\quad(3)$

Since $f(x)$ is an function continuous and differentiable on $(-\infty, +\infty)$ ,

by Lagrange’s Mean-Value Theorem (see “A Sprint to FTC“), there $\exists c \in (0, x^*)$ such that

$f(x^*)-f(0)=f'(c)(x^*-0)$

$\overset{(1), (3)}{\implies} 0 = f'(c) x^*\quad\quad\quad(4)$

$\overset{(2)}{\implies} f'(c)=0\quad\quad\quad(5)$ .

We know

$f'(x) = \cos(x)-2$ .

From (5), we have

$\cos(c)-2=0$ .

i.e.,

$\cos(c)=2$ .

This is not possible since $\forall x \in R, -1 \le \cos(x) \le 1$ .

A simpler alternative without direct applying Lagrange’s Mean-Value Theorem is:

$f(x) = \sin(x)-2x \implies f'(x) = \cos(x) - 2 \overset{-1 \le \cos(x) \le 1}{\implies} f'(x) < 0\implies$

$f(x) =\sin(x)-2x$ is a strictly decreasing function.

Since $f(0) = 0, \forall x<0, f(x) > 0$ and $\forall x>0, f(x) < 0$ .

Therefore, $0$ is the only solution of $f(x)=0$ . i.e.,

$0$ is the only solution of $\sin(x) = 2x$ .

Analyzing Heron’s Algorithm

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Finding the value of $\sqrt{a}$ for $a>0$ is equivalent to seeking a positive number $x$ whose square yields $a$ : $x^2=a$ . In other words, the solution to $x^2-a=0$ .

We can find $\sqrt{4}$ by solving $x^2-4=0$ . It is $2$ by mere inspection. $\sqrt{9}$ can also be obtained by solving $x^2-9=0$ , again, by mere inspection. But ‘mere inspection’ can only go so far. For Example, what is $\sqrt{15241578750190521}$ ?

Method for finding the square root of a positive real number date back to the acient Greek and Babylonian eras. Heron’s algorithm, also known as the Babylonian method, is an algorithm named after the $1^{st}$ – century Greek mathematician Hero of Alexandria, It proceeds as follows:

Begin with an arbitrary positive starting value $x_0$ .
Let $x_{k+1}$ be the average of $x_k$ and $\frac{a}{x_k}$
Repeat step 2 until the desired accuracy is achieved.

This post offers an analysis of Heron’s algorithm. Our aim is a better understanding of the algorithm through mathematics.

Let us begin by arbitrarily choosing a number $x_0>0$ . If $x_0^2-a=0$ , then $x_0$ is $\sqrt{a}$ , and we have guessed the exact value of the square root. Otherwise, we are in one of the following cases:

Case 1: $x_0^2-a > 0 \implies x_0^2 > a \implies \sqrt{a} x_0 > a \implies \sqrt{a} > \frac{a}{x_0} \implies \frac{a}{x_0} < \sqrt{a} < x_0$

Case 2: $x_0^2-a <0 \implies x_0 < \sqrt{a} \implies \sqrt{a} x_0 < a \implies \sqrt{a} < \frac{a}{x_0} \implies x_0 < \sqrt{a} < \frac{a}{x_0}$

Both cases indicate that $\sqrt{a}$ lies somewhere between $\frac{a}{x_0}$ and $x_0$ .

Let us define $e_0$ as the relative error of approximating $\sqrt{a}$ by $x_0$ :

$e_0 =\frac{x_0-\sqrt{a}}{\sqrt{a}}$

The closer $e_0$ is to $0$ , the better $x_0$ is as an approximation of $\sqrt{a}$ .

Since $x_0>0, \sqrt{a}>0$ ,

$e_0 + 1 = \frac{x_0-\sqrt{a}}{\sqrt{a}} + 1 = \frac{x_0}{\sqrt{a}} > 0\quad\quad\quad(1)$

By (1),

$x_0 = \sqrt{a} (e_0+1)\quad\quad\quad(2)$

Let $x_1$ be the mid-point of $x_0$ and $\frac{a}{x_0}$ :

$x_1 = \frac{1}{2} (x_0 + \frac{a}{x_0})\quad\quad\quad(3)$

and, $e_1$ the relative error of $x_1$ approximating $\sqrt{a}$ ,

$e_1 = \frac{x_1-\sqrt{a}}{\sqrt{a}}\quad\quad\quad(4)$

We have

$x_1-\sqrt{a}\overset{(3)}{=}\frac{1}{2}(x_0+\frac{a}{x_0}) - \sqrt{a} \overset{(2)}{=}\frac{1}{2}(\sqrt{a}(e_0+1)+\frac{a}{\sqrt{a}(e_0+1)})-\sqrt{a}=\frac{\sqrt{a}}{2}\frac{e_0^2}{e_0+1}\overset{(1)}{>}0\quad(5)$

$e_1 = \frac{x_1-\sqrt{a}}{\sqrt{a}} \overset{(5)}{=} \frac{\frac{\sqrt{a}}{2}\frac{e_0^2}{e_0+1}}{\sqrt{a}} = \frac{1}{2}\frac{e_0^2}{e_0+1}\overset{(1)}{ > }0\quad\quad\quad(6)$

By (5),

$x_1 > \sqrt{a} \implies \sqrt{a} x_1 > a \implies \sqrt{a} > \frac{a}{x_1}\implies \frac{\sqrt{a}}{x_1} < \sqrt{a} < x_1$

i.e., $\sqrt{a}$ lies between $\frac{a}{x_1}$ and $x_1$ .

We can generate more values in stages by

$x_{k+1} = \frac{1}{2}(x_k + \frac{a}{x_k}), \;k=1, 2, 3, ...\quad\quad(8)$

Clearly,

$\forall k, x_k >0$ .

Let

$e_k = \frac{x_k-\sqrt{a}}{\sqrt{a}}$ .

We have

$x_k = \sqrt{a}(e_k+1)\quad\quad\quad(9)$

and

$e_{k+1} = \frac{x_{k+1}-\sqrt{a}}{\sqrt{a}}=\frac{\frac{1}{2}(\sqrt{a}(e_k+1) + \frac{a}{\sqrt{a}(e_k+1)})-\sqrt{a}}{\sqrt{a}}=\frac{1}{2}\frac{e_k^2}{e_k + 1}\quad\quad\quad(10)$

Consequently, we can prove that

$\forall k \ge 1, e_{k+1} > 0\quad\quad\quad(11)$

by induction:

When $k = 1, e_{1+1} \overset{(10)}{=} \frac{1}{2}\frac{e_1^2}{e_1+1} \overset{(6)} {>} 0$ .

Assume when $k = p$ ,

$e_{p+1} =\frac{1}{2}\frac{e_p^2}{e_p^2+1} >0\quad\quad\quad(12)$

When $k = p + 1, e_{(p+1)+1} \overset{(10)}{=}\frac{1}{2}\frac{e_{p+1}^2}{e_{p+1} + 1}\overset{(12)}{>}0$ .

Since (11) implies

$\forall k \ge 2, e_k > 0\quad\quad\quad(13)$

(10) and (13) together implies

$\forall k \ge 1, e_k > 0\quad\quad\quad(14)$

It follows that

$\forall k \ge 1, 0 \overset{(14)}{<} e_{k+1} \overset{(10)}{=} \frac{1}{2}\frac{e_k^2}{e_k+1} \overset{(14)}{<} \frac{1}{2}\frac{e_k^2}{e_k}=\frac{1}{2}e_k\quad\quad\quad(15)$

and

$\forall k \ge 1, 0 \overset{(14)}{<} e_{k+1} \overset{(10)}{=} \frac{1}{2}\frac{e_k^2}{e_k+1}\overset{(14)}{<}\frac{1}{2}\frac{e_k^2}{1}<e_k^2\quad\quad\quad(16)$

(15) indicates that the error is cut at least in half at each stage after the first.

Therefore, we will reach a stage $n$ that

$0 < e_n < 1\quad\quad\quad(17)$

It can be shown that

$\forall k \ge 1, 0< e_{n+k} < e_n^{2^k}\quad\quad\quad(18)$

by induction:

When $k = 1$ ,

$0 \overset{(14)}{<} e_{n+1} \overset{(10)}{=}\frac{1}{2}\frac{e_n^2}{e_n+1} \overset{}{<} e_n^2=e_n^{2^1}\quad\quad\quad(19)$

Assume when k = p,

$0 < e_{n+p} < e_n^{2^p}\quad\quad\quad(20)$

When $k = p+1$ ,

$0 \overset{(14)}{<} e_{n+(p+1)} = e_{(n+p)+1} \overset{(16)}{<}e_{n+p}^2\overset{(20)}{<}(e_n^{2^p})^2 = e_n^{2^p\cdot 2} = e_n^{2^{p+1}}$

From (17) and (18), we see that eventually, the error decreases at an exponential rate.

Illustrated below is a script that implements the Heron’s Algorithm:

/*
a   - the positive number whose square root we are seeking
x0  - the initial guess
eps - the desired accuracy
*/
square_root(a, x0, eps) := block(
    [xk],
    numer:true,
    xk:x0,
    loop,
        if abs(xk^2-a) < eps then return(xk),
        xk:(xk+a/xk)/2,
    go(loop)
)$

Suppose we want to find $\sqrt{2}$ and start with $x_0 = 99$ . Fig. 2 shows the Heron’s algorithm in action.

Fig. 1

Déjà vu!

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The stages of a two stage rocket have initial masses $m_1$ and $m_2$ respectively and carry a payload of mass $P$ . Both stages have equal structure factors and equal relative exhaust speeds. If the rocket mass, $m_1+m_2$ , is fixed, show that the condition for maximal final speed is

$m_2^2 + P m_2 = P m_1$ .

Find the optimal ratio $\frac{m_1}{m_2}$ when $\frac{P}{m_1+m_2} = b$ .

According to multi-stage rocket’s flight equation (see “Viva Rocketry! Part 2“), the final speed of a two stage rocket is

$v = -c \log(1-\frac{e \cdot m_1}{m_1 + m_2 +P})- c\log(1-\frac{e \cdot m_2}{m_2+P})$

Let $m_0 = m_1 + m_2$ , we have

$m_1 = m_0-m_2$

and,

$v = -c \log(1-\frac{e \cdot (m_0-m_2)}{m_0 +P})- c\log(1-\frac{e \cdot m_2}{m_2+P})$

Differentiate $v$ with respect to $m_2$ gives

$v' = \frac{c(e-1)e(2m_2 P-m_0 P +m_2^2)}{(P+m_2)(P-e m_2+m_2)(P+e m_2-e m_0+m_0)}= \frac{c(e-1)e(2m_2-m_0 P+m_2^2)}{(P+m_2)(P+(1-e)m_2)(P+e m_2 + (1-e)m_0)}\quad\quad\quad(1)$

It follows that $v'=0$ implies

$2m_2 P-m_0 P + m_2^2 =0$ .

That is, $2 m_2 P - (m_1+m_2) P + m_2^2 = (m_2-m_1) P + m_2^2=0$ . i.e.,

$m_2^2 + P m_2 = P m_1\quad\quad\quad(2)$

It is the condition for an extreme value of $v$ . Specifically, the condition to attain a maximum (see Exercise-2)

When $\frac{P}{m_1+m_2} = b$ , solving

$\begin{cases} (m_2-m_1)P+m_2^2=0\\ \frac{P}{m_1+m_2} = b\end{cases}$

yields two pairs:

$\begin{cases} m_1=\frac{(\sqrt{b}\sqrt{b+1}+b+1)P}{b}\\m_2= -\frac{\sqrt{b^2+b}P+bP}{b}\end{cases}$

and

$\begin{cases} m_1= - \frac{(\sqrt{b}\sqrt{b+1}-b-1)P}{b}\\ m_2=\frac{\sqrt{b^2+b}P-bP}{b}\end{cases}\quad\quad\quad(3)$

Only (3) is valid (see Exercise-1)

Hence

$\frac{m_1}{m_2} = -\frac{(\sqrt{b}\sqrt{b+1}-b-1)P}{\sqrt{b^2+b}P-b P} = \frac{\sqrt{b+1}}{\sqrt{b}} = \sqrt{1+\frac{1}{b}}$

The entire process is captured in Fig. 2.

Fig. 2

Exercise-1 Given $b>0, 0<e<1, P>0$ , prove:

$- \frac{(\sqrt{b}\sqrt{b+1}-b-1)P}{b}>0$
$\frac{\sqrt{b^2+b}P-bP}{b}>0$

Exercise-2 From (1), prove the extreme value attained under (2) is a maximum.

Boosting rocket flight performance without calculus (Viva Rocketry! Part 2.1)

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Fig. 1

The stages of a two-stage rocket have initial masses $m_1$ and $m_2$ respectively and carry a payload of mass $P$ . Both stages have equal structure factors $e$ and equal relative exhaust speed $c$ . The rocket mass, $m_1+m_2$ is fixed and $\frac{P}{m_1+m_2}=b$ .

According to multi-stage rocket’s flight equation (see “Viva Rocketry! Part 2“), the final speed of a two-stage rocket is

$v = -c \log(1-\frac{em_1}{m_1+m_2+P}) - c\log(1-\frac{em_2}{m_2+P})$

Let $a = \frac{m1}{m_2}$ , it becomes

$v = -c\log(1-\frac{ea}{a+1+b(a+1)})-c \log(1-\frac{e}{1+b(a+1)})$

where $a>0, b>0, c>0, 0< e < 1$ . We will maximize $v$ with an appropriate choice of $a$ .

That is, given

$v = c\log(\frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1})$

where $a>0, b>0, c>0, 0<e<1$ . Maximize $v$ with an appropriate value of $a$ .

The above optimization problem is solved using calculus (see “Viva Rocketry! Part 2“). However, there is an alternative that requires only high school mathematics with the help of a Computer Algebra System (CAS). This non-calculus approach places more emphasis on problem solving through mathematical thinking, as all symbolic calculations are carried out by the CAS (e.g., see Fig. 2). It also makes a range of interesting problems readily tackled with minimum mathematical prerequisites.

The fact that

$\log$ is a monotonic increasing function $\implies v_{max} = c\log(w_{max})$

where

$w = \frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)}$

$w(1-e+(a+1)b)((1-e)a+(a+1)b+1) - (a+1)(b+1)((a+1)b+1)=0\quad\quad\quad(1)$

(1) can be written as

$A_1 a^2 + B_1 a +C_1= 0$

where

$A_1 = -bew+b^2w+bw-b^2-b$

$B_1 = e^2w-2bew-2ew+2b^2w+3bw+w-2b^2-3b-1$ ,

$C_1 = -bew-ew+b^2w+2bw+w-b^2-2b-1$ .

Since $A_1 = 0$ means

$-b e w + b^2 w + b w - b^2 - b =0$ .

That is

$w = -\frac{b+1}{e-b-1}$ .

Solve

$\frac{(a+1)(b+1)((a+1)b+1)}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = -\frac{b+1}{e-b-1}$

for $a$ gives $a = 0 \implies A_1 \neq 0$ if $a > 0$ .

Hence, (1) is a quadratic equation. For it to have solution, its discriminant $B_1^2-4A_1C_1$ must be nonnegative, i.e.,

$(e^2-2be-2e+b+1)^2 w^2-2(b+1)(2be^2+e^2-2be-2e+b+1) w +(b+1)^2 \geq 0\quad(2)$

Consider

$(e^2-2be-2e+b+1)^2 w^2-2(b+1)(2be^2+e^2-2be-2e+b+1)w+(b+1)^2 = 0\quad(3)$

If $e^2-2be-2e+b+1 \neq 0$ , (3) is a quadratic equation.

Solving (3) yields two solutions

$w_1 = -\frac{(b+1)(2\sqrt{b(b+1)}e^2-2be^2-e^2-2\sqrt{b(b+1)}e+2be+2e-b-1)}{(e^2-2be-2e+b+1)^2}$ ,

$w_2=\frac{(b+1)(2\sqrt{b(b+1)}e^2+2be^2+e^2-2\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}$ .

Since $0 < e < 1$ ,

$w_1 - w_2 = -\frac{4(b+1)\sqrt{b(b+1)}(e-1)e}{(e^2-2be-2e+b+1)^2} > 0\quad(4)$

(4) implies

$w_1 > w_2$

and, the solution to (2) is

$w \leq w_2$ or $w \ge w_1$

i.e.,

$w \leq \frac{(b+1)(2\sqrt{b(b+1)}e^2+2be^2+e^2-2\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}\quad\quad\quad(4)$

$w \ge -\frac{(b+1)(2\sqrt{b(b+1)}e^2-2be^2-e^2-2\sqrt{b(b+1)}e+2be+2e-b-1)}{(e^2-2be-2e+b+1)^2}\quad\quad\quad(5)$

We prove that (4) is true by showing (5) is false:

Consider $w - w_1= 0$ :

$\frac{(b+1)(e-1) e \cdot f(a)}{(1-e+ab+b)(a(1-e)+ab+b+1)(e^2-2be-2e+b+1)^2} = 0\quad\quad\quad(6)$

where

$f(a) = 2a\sqrt{b(b+1)}e^2+a^2be^2+be^2+e^2-2a^2b\sqrt{b(b+1)}$

$-4ab\sqrt{b(b+1)}e - 2b\sqrt{b(b+1)}e - 4a\sqrt{b(b+1)}e$

$-2\sqrt{b(b+1)}e-2a^2b^2e-4ab^2e-2a^2be-4abe-4be$

$-2e+2a^2b^2\sqrt{b(b+1)}+4ab^2\sqrt{b(b+1)} + 2b^2\sqrt{b(b+1)} +2a^2b\sqrt{b(b+1)}$

$+6ab\sqrt{b(b+1)} + 4b\sqrt{b(b+1)} + 2a\sqrt{b(b+1)} + 2\sqrt{b(b+1)}$

$+2a^2b^3 + 4ab^3 + 2b^3 +3a^2b^2 + 8ab^2 +5b^2+a^2b+4ab+4b+1$ .

It can be written as

$A_2a^2 + B_2a + C_2\quad\quad\quad(7)$

where

$A_2 = be^2-2b\sqrt{b(b+1)}e-2b^2e-2be+2b^2\sqrt{b(b+1)}+2b\sqrt{b(b+1)}$

$+2b^3+3b^2+b$ ,

$B_2 = 2\sqrt{b(b+1)}e^2-4b\sqrt{b(b+1)}e -4\sqrt{b(b+1)}e$

$-4b^2e-4be+4b^2\sqrt{b(b+1)}+6b\sqrt{b(b+1)}+2\sqrt{b(b+1)}+4b^3+8b^2+4b$ ,

$C_2=be^2+e^2-2b\sqrt{b(b+1)}e-2\sqrt{b(b+1)}e-2b^2e-4be$

$-2e+2b^2\sqrt{b(b+1)}+4b\sqrt{b(b+1)}+2\sqrt{b(b+1)}+2b^3+5b^2+4b+1$ .

Since $A_2 > 0$ (see Exercise 1) and,

solve (7) for $a$ yields

$a = -\sqrt{1+\frac{1}{b}}$ .

It follows that for $a > 0, f(a) > 0$ .

Consequently, $w-w_1$ is a negative quantity. i.e.,

$w-w1 < 0$

which tells that (5) is false.

Hence, when $e^2-2be-2e+b+1 \neq 0$ , the global maximum $w_{max}$ is $w_2$ .

Solving $w = w_2$ for $a$ :

$\frac{(a+1)(b+1)((a+1)b+1}{(1-e+(a+1)b)((1-e)a+(a+1)b+1)} = \frac{(b+1)(2\sqrt{b(b+1)}e^2+2be^2+e^2-2\sqrt{b(b+1)}e-2be-2e+b+1)}{(e^2-2be-2e+b+1)^2}$ ,

we have

$a = \sqrt{1+ \frac{1}{b}}$ .

Therefore,

$e^2-2be-2e+b+1 \neq 0 \implies w$ attains maximum at $a = \sqrt{1+ \frac{1}{b}}$ .

In fact, $w$ attains maxima at $a = \sqrt{1+\frac{1}{b}}$ even when $e^2-2be-2e+b+1 = 0$ , as shown below:

Solving $e^2-2be-2e+b+1 = 0$ for $e$ , we have

$e_1= -\sqrt{b(b+1)}+b+1$ or $e_2 = \sqrt{b(b+1)} + b + 1$ .

Only $e_1$ is valid (see Exercise-2),

When $e = e_1$ ,

$w(\sqrt{1+\frac{1}{b}} )- w(a) = - \frac{(b+1)g(a)}{4 \sqrt{b(b+1)} (\sqrt{b(b+1)}+ab) (a\sqrt{b(b+1)}+b+1)}\quad(8)$

where

$g(a) = (2a^2b+4ab+2b+2a+2)\sqrt{b(b+1)}-2a^2b^2-4ab^2-2b^2-a^2b-4ab-3b-1$

Solve quadratic equation $g(a) = 0$ for $a$ yields

$a = \sqrt{1+\frac{1}{b}}$ .

The coefficient of $a^2$ in $g(a)$ is $2b\sqrt{b(b+1)}-2b^2-b$ , a negative quantity (see Exercise-3).

The implication is that $g(a)$ is a negative quantity when $a \neq \sqrt{1 + \frac{1}{b}}$ .

Hence, (8) is a positive quantity, i.e.,

$e^2-2be-2e+b+1 = 0, a \neq \sqrt{1+\frac{1}{b}} \implies w(\sqrt{1+\frac{1}{b}})-w(a) > 0$

We therefore conclude

$\forall 0 < e < 1, b > 0, w$ attains its maximum at $a = \sqrt{1+\frac{1}{b}}$ .

Fig. 2

Exercise-1 Prove: $0<e<1, b>0 \implies$

$be^2-2b\sqrt{b(b+1)}e-2b^2e-2be+2b^2\sqrt{b(b+1)}+2b\sqrt{b(b+1)}+2b^3+3b^2+b > 0$

Exercise-2 Prove: $b > 0 \implies 0 <-\sqrt{b(b+1)} + b +1 <1$

Exercise-3 Prove: $b > 0 \implies 2b\sqrt{b(b+1)}-2b^2-b < 0$

Prelude to Taylor’s theorem

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As an application of derivative, we may prove the Binomial theorem that concerns the expansion of $(1+x)^n$ as a polynomial. Namely,

$(1+x)^n = 1+ a_1 x + a_2 x^2 + ... + a_n x^n\quad\quad\quad(1)$

where $a_k = \frac{n(n-1)(n-2)...(n-k+1)}{k!}, 1 \leq k \leq n, n \in N$ .

There are two steps:

Step 1) Prove $(1+x)^n$ can be expressed as a polynomial $1+a_1 x + a_2 x^2 + ... + a_n x^n$ , i.e.,

$(1+x)^n = 1 + \sum\limits_{i=1}^{n}a_i x^i$

where $a_k$ s are constants.

Step 2) Show that

$a_k = \frac{n(n-1)(n-2)...(n-k+1)}{k!}$

We use mathematical induction first.

When $n = 1$ ,

$(1+x)^1 = 1+x = 1 +a_1 x$

where $a_1 = 1$ .

Assume when $n=k, (1+x)^k$ is a polynomial:

$(1+x)^k = 1+b_1 x + b_2 x + ... +b_k x^k\quad\quad\quad(2)$

When $n = k+1$ ,

$(1+x)^{k+1} = (1+x)^k (1+x)$ .

By (2), it is

$(1+b_1 x+b_2 x^2+ ...+ b_k x^k) (1+x)$

$= (1+\sum\limits_{i=1}^{k} b_i x^i)(1+x)$

$= 1+ x + \sum\limits_{i=1}^{k}b_i x^i (1+x)$

$= 1+x + \sum\limits_{i=1}^{k}(b_i x^i +b_i x^{i+1})$

$= 1+x +\sum\limits_{i=1}^{k}b_i x^i + \sum\limits_{i=1}^{k} b_i x^{i+1}$

$= 1+x + (b_1 x + b_2 x^2 +... +b_k x^k) + (b_1 x^2 + ... + b_{k-1} x^k + b_k x^{k+1})$

$= 1+ (b_1+1)x + (b_2 + b_1) x^2 + ... + (b_k + b_{k-1}) x^k + b_k x^{k+1}$

$= 1 +a_1 x + a_2 x^2 + ... + a_k x^k + a_{k+1} x^{k+1}$

where $a_1 = b_1+1, a_2=b_2+b_1, ..., a_k = b_k + b_{k-1}, a_{k+1} = b_k$ .

Once (1) is established, we proceed to step 2) to construct $a_k$ :

From (1),

$\frac{d^k}{dx^k}(1+x)^n = \frac{d^k}{dx^k}(1+a_1 x + a_2 x^2 + ... + a_k x^k + ... +a_n x^n)$ .

That is

$n(n-1)(n-2)...(n-k+1)(1+x)^{n-k} = k(k-1)(k-2)...1 \cdot {a_k} + \sum\limits_{i=1}^{n-k} c_i x^i\quad\quad\quad(3)$

where $c_i$ s are constants.

Let $x = 0$ , (3) becomes

$n(n-1)(n-2)...(n-k+1) \cdot 1= k(k-1)(k-2)...1 \cdot {a_k} + 0$

i.e.,

$n(n-1)(n-2)...(n-k+1) = k!\;a_k\quad\quad\quad(4)$

Solving (4) for $a_k$ gives

$a_k = \frac{n(n-1)(n-2)...(n-k+1)}{k!}$ .

Since

$\frac{n(n-1)(n-2)...(n-k+1)}{k!} = \frac{n(n-1)(n-k+1)\boxed{(n-k)(n-k-1)...1}}{\boxed{(n-k)(n-k-1)...1}\;k!}=\frac{n!}{(n-k)!\;k!}$ ,

$a_k$ is often expressed as

$a_k = \frac{n!}{(n-k)!\;k!}$

Piece of Pi

1 Reply

A while back, we deemed that it is utterly impractical to calculate the value of $\pi$ using the partial sum of Leibniz’s series due to its slow convergence (see “Pumpkin Pi“)

Fig. 1

As illustrated in Fig. 1, in order to determine each additional correct digit of $\pi$ , the number of terms in the summation must increase by a factor of 10.

What we need is a fast converging series whose partial sum yields given number of correct digits with far fewer terms.

Looking back, we see that the origin of Leibniz’s series is the definite integral

$\frac{\pi}{4} = \int\limits_{0}^{1}\frac{1}{1+x^2} dx\quad\quad\quad(1)$

To find the needed new series, we consider a variation of (1), namely,

$\frac{\pi}{6} = \int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+x^2} dx\quad\quad\quad(2)$

Given the fact (see “Pumpkin Pi“) that

$\frac{1}{1+x^2} = \sum\limits_{k=1}^{n}(-1)^{k+1}x^{2k-2}+\frac{(-1)^n x^{2n}}{1+x^2}\quad\quad\quad(3)$

We proceed to integrate (3) with respect to $x$ from $0$ to $\frac{1}{\sqrt{3}}$ ,

$\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+x^2} dx=\int\limits_{0}^{\frac{1}{\sqrt{3}}}\sum\limits_{k=1}^{n}(-1)^{k+1}x^{2k-2} dx + \int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{(-1)^n x^{2n}}{1+x^2}dx$

$= \sum\limits_{k=1}^{n}(-1)^{k+1}\int\limits_{0}^{\frac{1}{\sqrt{3}}}x^{2k-2} dx + (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

$= \sum\limits_{k=1}^{n}(-1)^{k+1}\frac{x^{2k-1}}{2k-1}\bigg|_{0}^{\frac{1}{\sqrt{3}}}+ (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

As result, (2) becomes

$\frac{\pi}{6}=\sum\limits_{k=1}^{n}(-1)^{k+1}\frac{(\frac{1}{\sqrt{3}})^{2k-1}}{2k-1}+(-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

$= \sum\limits_{k=1}^{n}(-1)^{k+1}\frac{{(\frac{1}{3}})^k}{(2k-1)\frac{1}{\sqrt{3}}} + (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

$= \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)} + (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx$

i.e.,

$\frac{\pi}{6} - \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)} = (-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx\quad\quad\quad(4)$

By (4),

$|\frac{\pi}{6} - \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}| = |(-1)^n\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx|=\int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{x^{2n}}{1+x^2}dx<\int\limits_{0}^{\frac{1}{\sqrt{3}}}x^{2n}dx$

$=\frac{x^{2n+1}}{2n+1}\bigg|_0^{\frac{1}{\sqrt{3}}}$

$=\frac{1}{3^n \sqrt{3} (2n+1)}$

which gives

$|\sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)} - \frac{\pi}{6}| < \frac{1}{3^n \sqrt{3} (2n+1)}$ .

And so

$-\frac{1}{3^n \sqrt{3} (2n+1)}<\sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}-\frac{\pi}{6}<\frac{1}{3^n \sqrt{3} (2n+1)}\quad\quad\quad(5)$

Since $\lim\limits_{n\rightarrow \infty}\frac{1}{3^n \sqrt{3} (2n+1)}=0$ , (5) implies

$\lim\limits_{n\rightarrow \infty} \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}-\frac{\pi}{6}= 0$ .

Hence,

$\lim\limits_{n\rightarrow \infty} \sqrt{3}\sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{3^k(2k-1)}=\frac{\pi}{6}$ .

It follows that the value of $\pi$ can be approximated by the partial sum of a new series

$6\sqrt{3}\sum\limits_{k=1}^{\infty}\frac{(-1)^{k+1}}{3^k(2k-1)}$

Let’s compute it with Omega CAS Explorer (see Fig. 2, 3)

Fig. 2

Fig. 2 shows the series converges quickly. The sum of the first 10 terms yields the first 6 digits!

Fig. 3

Totaling the first 100 terms of the series gives the first 49 digits of $\pi$ (see Fig. 3)

Exercise 1. Show that $\lim\limits_{n\rightarrow \infty}\frac{1}{3^n \sqrt{3} (2n+1)}=0$ .

Exercise 2. Can we use $\frac{\pi}{3} = \int\limits_{0}^{\sqrt{3}}\frac{1}{1+x^2}dx$ to compute the value of $\pi$ in a similar fashion? Explain.

Viva Rocketry! Part 2

Screen Shot 2018-12-24 at 5.13.14 AM.png

Fig. 1

A rocket with $n$ stages is a composition of $n$ single stage rocket (see Fig. 1) Each stage has its own casing, instruments and fuel. The $n$ th stage houses the payload.

This image has an empty alt attribute; its file name is screen-shot-2019-01-09-at-6.24.56-pm.png

Fig. 2

The model is illustrated in Fig. 2, the $i^{th}$ stage having initial total mass $m_i$ and containing fuel $\epsilon_i m_i (0 < \epsilon_i <1, 1 \leq i \leq n)$ . The exhaust speed of the $i^{th}$ stage is $c_i$ .

The flight of multi-stage rocket starts with the $1^{st}$ stage fires its engine and the rocket is lifted. When all the fuel in the $1^{st}$ stage has been burnt, the $1^{st}$ stage’s casing and instruments are detached. The remaining stages of the rocket continue the flight with $2^{nd}$ stage’s engine ignited.

Generally, the rocket starts its $i^{th}$ stage of flight with final velocity achieved at the end of previous stage of flight. The entire rocket is propelled by the fuel in the $i^{th}$ casing of the rocket. When all the fuel for this stage has been burnt, the $i^{th}$ casing is separated from the rest of the stages. The flight of the rocket is completed if $i=n$ . Otherwise, it enters the next stage of flight.

When all external forces are omitted, the governing equation of rocket’s $i^{th}$ stage flight (see “Viva Rocketry! Part 1” or “An alternate derivation of ideal rocket’s flight equation (Viva Rocketry! Part 1.3)“) is

$0 = m_i(t) \frac{dv_i(t)}{dt} + c_i \frac{dm_i(t)}{dt}$

It can be written as

$\frac{dv_i(t)}{dt} = -\frac{c_i}{m_i(t)}\frac{dm_i(t)}{dt}\quad\quad\quad(1)$

Integrate (1) from $t_0$ to $t$ ,

$\int\limits_{t_0}^{t}\frac{dv_i(t)}{dt}\;dt = -c_i \int\limits_{t_0}^{t}\frac{1}{m_i(t)}\frac{dm_(t)}{dt}\;dt$

gives

$v_i(t) - v_i(t_0) = -c_i(\log(m_i(t))-\log(m_i(t_0)))=-c_i\log(\frac{m_i(t)}{m_i(t_0)})$ .

At $t=t^*$ when $i^{th}$ stage’s fuel has been burnt, we have

$v_i(t^*) - v_i(t_0) = -c_i\log(\frac{m_i(t^*)}{m_i(t_0)})\quad\quad\quad(2)$

where

$m_i(t_0) = \sum\limits_{k=i}^{n} m_k +P$

and,

$m_i(t^*) = \sum\limits_{k=i}^n m_k - \epsilon_i m_i + P$ .

Let $v^*_i = v_i(t^*), \; v^*_{i-1}$ the velocity of rocket at the end of ${i-1}^{th}$ stage of flight.

Since $v_i(t_0) = v^*_{i-1}$ , (2) becomes

$v^*_i - v^*_{i-1} = -c_i \log(\frac{ \sum\limits_{k=i}^n m_k - \epsilon_i m_i+P}{\sum\limits_{k = i}^{n} m_k + P}) = -c_i\log(1-\frac{\epsilon_i m_i}{\sum\limits_{k = i}^{n} m_k + P})$

i.e.,

$v^*_i = v^*_{i-1}-c_i\log(1-\frac{\epsilon_i m_i}{\sum\limits_{k = i}^{n} m_k + P}), \quad\quad 1\leq i \leq n, v_0=0\quad\quad\quad(3)$

For a single stage rocket ( $n=1$ ), (3) is

$v_1^*=-c_1\log(1-\frac{\epsilon_1}{1+\beta})\quad\quad\quad(4)$

In my previous post “Viva Rocketry! Part 1“, it shows that given $c_1=3.0\;km\;s^{-1}, \epsilon_1=0.8$ and $\beta=\frac{1}{100}$ , (4) yields $4.7 \;km\;s^{-1}$ , a value far below $7.8\;km\;s^{-1}$ , the required speed of an earth orbiting satellite.

But is there a value of $\beta$ that will enable the single stage rocket to produce the speed a satellite needs?

Let’s find out.

Differentiate (4) with respect to $\beta$ gives

$\frac{dv_1^*}{d\beta} = - \frac{c_1 \epsilon_1}{(\beta+1)^2 (1-\frac{\epsilon_1}{\beta+1})} = - \frac{c_1\epsilon_1}{(\beta+1)^2(\frac{1-\epsilon_1+\beta}{\beta+1})} < 0$

since $c_1, \beta$ are positive quantities and $0< \epsilon_1 < 1$ .

It means $v_1^*$ is a monotonically decreasing function of $\beta$ .

Moreover,

$\lim\limits_{\beta \rightarrow 0}v_1^*=\lim\limits_{\beta \rightarrow 0}-c_1\log(1-\frac{\epsilon_1}{1+\beta}) = - c_1 \log(1-\epsilon_1)\quad\quad\quad(5)$

Given $c_1=3.0\;km\;s^{-1}, \epsilon_1=0.8$ , (5) yields approximately

$4.8\;km\;s^{-1}$

This image has an empty alt attribute; its file name is screen-shot-2019-01-21-at-4.56.14-pm.png

Fig. 3

This upper limit implies that for the given $c_1$ and $\epsilon_1$ , no value of $\beta$ will produce a speed beyond (see Fig. 4)

Let’s now turn to a two stage rocket ( $n=2$ )

From (3), we have

$v_2^* = -c_1\log(1-\frac{\epsilon_1 m_1}{m_1+m_2+P}) - c_2\log(1-\frac{\epsilon_2 m_2}{m_2+P})\quad\quad\quad(6)$

If $c_1=c_2=c, \epsilon_1 = \epsilon_2 = \epsilon, m_1=m_2$ and $\frac{P}{m_1+m_2} = \beta$ , then

$P = \beta (m_1+m_2) = 2m_1\beta = 2m_2\beta$ .

Consequently,

$v_2^*=-c \log(1-\frac{\epsilon}{2(1+\beta)}) - c\log(1-\frac{\epsilon}{1+2\beta})\quad\quad\quad(7)$

When $c=3.0\;km\;s^{-1}, \epsilon=0.8$ and $\beta = \frac{1}{100}$ ,

$v_2^* \approx 6.1\;km\;s^{-1}$

This image has an empty alt attribute; its file name is screen-shot-2019-01-21-at-10.30.21-am.png

Fig. 5

This is a considerable improvement over the single stage rocket ( $v^*=4.7\; km\;s^{-1}$ ). Nevertheless, it is still short of producing the orbiting speed a satellite needs.

In fact,

$\frac{dv_2^*}{d\beta} = -\frac{2c\epsilon}{(2\beta+2)^2(1-\frac{\epsilon}{2\epsilon+2})}-\frac{2c\epsilon}{(2\beta+1)^2(1-\frac{\epsilon}{2\beta+1})}= -\frac{2c\epsilon}{(2\beta+2)^2\frac{2\beta+2-\epsilon}{2\beta+2}}-\frac{2c\epsilon}{(2\beta+1)^2\frac{2\beta+1-\epsilon}{2\beta+1}} < 0$

indicates that $v_2^*$ is a monotonically decreasing function of $\beta$ .

In addition,

$\lim\limits_{\beta \rightarrow 0} v_2^*=\lim\limits_{\beta \rightarrow 0 }-c\log(1-\frac{\epsilon}{2+2\beta})-c\log(1-\frac{\epsilon}{1+2\beta})=-c\log(1-\frac{\epsilon}{2})-c\log(1-\epsilon)$ .

Therefore, there is an upper limit to the speed a two stage rocket can produce. When $c=3.0\;km\;s^{-1}, \epsilon=0.8$ , the limit is approximately

$6.4\;km\;s^{-1}$

This image has an empty alt attribute; its file name is screen-shot-2019-01-24-at-2.07.48-pm.png

Fig. 6

In the value used above, we have taken equal stage masses, $m_1 = m_2$ . i.e., the ratio of $m_1 : m_2 = 1 : 1$ .

Is there a better choice for the ratio of $m_1:m_2$ such that a better $v_2^*$ can be obtained?

To answer this question, let $\frac{m_1}{m_2} = \alpha$ , we have

$m_1 = \alpha m_2\quad\quad\quad(8)$

Since $P = \beta (m_1+m_2)$ , by (8),

$P = \beta (\alpha m_2 + m_2)\quad\quad\quad(9)$

Substituting (8), (9) into (6),

$v_2^* = -c\log(1-\frac{\epsilon \alpha m_2}{\alpha m_2+m_2+\beta(\alpha m_2+m_2))}) - c\log(1-\frac{\epsilon m_2}{m_2 + \beta(\alpha m_2 + m_2)})$

$= -c\log(1-\frac{\epsilon \alpha}{\alpha+1+\beta(\alpha+1)})-c\log(1-\frac{\epsilon}{1+\beta(\alpha+1)})$

$= c\log\frac{(\alpha+1)(\beta+1)((\alpha+1)\beta+1)}{(1-\epsilon+(\alpha+1)\beta)((1-\epsilon)\alpha + (\alpha+1)\beta+1)}\quad\quad\quad(10)$

a function of $\alpha$ . It can be written as

$v_2^*(\alpha) = c\log(w(\alpha))$

where

$w(\alpha) = \frac{(\alpha+1)(\beta+1)((\alpha+1)\beta+1)}{(1-\epsilon+(\alpha+1)\beta)((1-\epsilon)\alpha + (\alpha+1)\beta+1)}$

This is a composite function of $\log$ and $w$ .

Since $\log$ is a monotonic increasing function (see “Introducing Lady L“),

$(v_2^*)_{max} = c\log(w_{max})$

Here, $(v_2^*)_{max}, w_{max}$ denote the maximum of $v_2^*$ and $w$ respectively.

To find $w_{max}$ , we differentiate $w$ ,

$\frac{dw}{d\alpha} = \frac{(\beta+1)(\alpha^2\beta-\beta-1)(\epsilon-1)\epsilon}{(\epsilon-\alpha \beta-\beta-1)^2(\alpha\epsilon-\alpha \beta-\beta-\alpha-1)^2}=\frac{(\beta+1)(\alpha^2\beta-\beta-1)(\epsilon-1)\epsilon}{(1-\epsilon+\alpha \beta+\beta)^2(\alpha(1-\epsilon)+\alpha \beta+\beta)^2}\quad\quad\quad(11)$

Solving $\frac{dw}{d\alpha} = 0$ for $\alpha$ gives

$\alpha = - \sqrt{1+\frac{1}{\beta}}$ or $\sqrt{1+\frac{1}{\beta}}$ .

This image has an empty alt attribute; its file name is screen-shot-2019-01-29-at-11.44.15-am.png

Fig. 7

By (8), the valid solution is

$\alpha = \sqrt{1+\frac{1}{\beta}}$ .

It shows that $w$ attains an extreme value at $\sqrt{1+\frac{1}{\beta}}$ .

Moreover, we observe from (11) that

$\alpha < \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} > 0$

and

$\alpha > \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} < 0$ .

i.e., $w$ attains maximum at $\alpha=\sqrt{1+\frac{1}{\beta}}$ .

It follows that

$v_2^*$ attains maximum at $\alpha = \sqrt{1+\frac{1}{\beta}}$ .

Therefore, to maximize the final speed given to the satellite, we must choose the ratio

$\frac{m_1}{m_2} = \sqrt{1+\frac{1}{\beta}}$ .

With $\beta =\frac{1}{100}$ , the optimum ratio $\frac{m_1}{m_2}=10.05$ , showing that the first stage must be about ten times large than the second.

Using this ratio and keep $\epsilon=0.8, c=3.0\;km\;s^{-1}$ as before, (10) now gives

$v_2 = 7.65\;km\;s^{-1}$ ,

a value very close to the required one.

This image has an empty alt attribute; its file name is screen-shot-2019-01-23-at-4.14.53-pm.png

Fig. 8

Setting $\beta = \frac{1}{128}$ , we reach the goal:

$v_2^* = 7.8\;km\;s^{-1}$

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Fig. 9

This image has an empty alt attribute; its file name is screen-shot-2019-01-30-at-10.00.04-pm-1.png

Fig. 10

At last, it is shown mathematically that provided the stage mass ratios ( $\frac{P}{m_1+m_2}$ and $\frac{m_1}{m_2}$ )are suitably chosen, a two stage rocket can indeed launch satellites into earth orbit.

Exercise 1. Show that $\alpha < \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} > 0$ and $\alpha > \sqrt{1+\frac{1}{\beta}} \implies \frac{dw}{d\alpha} < 0$ .

Exercise 2. Using the optimum $\frac{m_1}{m_2} = \sqrt{1+\frac{1}{\beta}}$ and $\epsilon=0.8, c=3.0\;km\;s^{-1}$ , solving (10) numerically for $\beta$ such that $v_2^* = 7.8$ .

Vroom

A Laboratory for Advanced Computing

From proof to simpler proof

Analyzing Heron’s Algorithm

Déjà vu!

Boosting rocket flight performance without calculus (Viva Rocketry! Part 2.1)

Prelude to Taylor’s theorem

Piece of Pi

Viva Rocketry! Part 2

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