# “Seek-Lock-Strike!” Again

We can derive a different governing equation for the missile in “Seek-Lock-Strike!“.

Fig. 1

Looking from a different viewpoint (Fig. 1), we see

$\frac{dx}{dy} = \frac{a+v_at-x}{b-y}.\quad\quad\quad(1)$

Solving (1) for $t$,

$t = -\frac{\frac{dx}{dy}y-b\frac{dx}{dy}-x+a}{v_a}.\quad\quad\quad(2)$

We also have

$v_m t = \int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2} \implies t = \frac{\int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2}\;dy}{v_m}.\quad\quad\quad(3)$

Equate (1) and (2) gives

$-\frac{\frac{dx}{dy}y-b\frac{dx}{dy}-x+a}{v_a}-\frac{\int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2}\;dy}{v_m} = 0.\quad\quad\quad(4)$

The governing eqaution emerges after differentiate (4) with respect to $x:$

$-\frac{d^2x}{dy^2}y+b\frac{d^2x}{dy^2}-\frac{v_a\sqrt{1+(\frac{dx}{dy})^2}}{v_m} = 0.\quad\quad\quad(5)$

We let $p = \frac{dx}{dy}$ so $\frac{d^2x}{dy^2} = \frac{d}{dy}\left(\frac{dx}{dy}\right)=\frac{dp}{dy}$ and express (5) as

$\frac{dp}{dy}(b-y) -\frac{v_a\sqrt{1+(\frac{dx}{dy})^2}}{v_m} = 0$

where $r = \frac{v_a}{v_m}.$

Fig. 2

Using Omega CAS Explorer, we compute the missile’s striking time $t_*$ (see Fig. 3). It agrees with the result obtained previously.

Fig. 3

# “Seek-Lock-Strike!” Animated

In “Seek-Lock-Strike!“, we obtained the missile’s trajectory. Namely,

$x = \frac{1}{2} \left(\frac{(b-y)^{r+1}}{b^r \cdot f \cdot (r+1)}-\frac{b^r \cdot f \cdot (b-y)^{1-r}}{1-r}\right) -k\quad\quad\quad(*)$

where

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

$k = \frac{b}{2}\left(\frac{1}{f \cdot (r+1)}-\frac{f}{1-r}\right).$

Since the fighter jet maintains its altitude ($y= b$), the missile must strike it at $(x_s, b)$. Setting $y=b$ in $(*)$ gives $x_s = -k.$

Hence, we can plot $(*):$

Fig. 1 $a = 0\;m, b=3000\;m, v_a=1500\;ms^{-1}, v_m=2000\;ms^{-1}$

We can also illustrate “Seek-Lock-Strike” in an animation:

Fig. 2 $a = 0\;m, b=3000\;m, v_a=1500\;ms^{-1}, v_m=2000\;ms^{-1}$

# “Seek-Lock-Strike!” Simplified

There is an easier way to derive the governing equation ((5), “Seek-Lock-Strike!“) for the missile.

Solving

$\frac{dy}{dx}(a+v_at-x) = b-y$

for $t,$ we have

$t = \frac{(x-a)\frac{dy}{dx}-y+b}{v_a\frac{dy}{dx}}.\quad\quad\quad(1)$

From

$v_m t = \int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx,$

we also have

$t = \frac{\int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx}{v_m}.\quad\quad\quad(2)$

Equate (1) and (2) gives

$\frac{\int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx}{v_m}-\frac{(x-a)\frac{dy}{dx}-y+b}{v_a\frac{dy}{dx}}=0.\quad\quad\quad(3)$

Differentiate (3) with repect to $x,$ we obtain

$(bv_m-v_my)\frac{d^2y}{dx^2} + v_a(\frac{dy}{dx})^2\sqrt{1+(\frac{dy}{dx})^2} = 0.\quad\quad\quad(4)$

(see Fig. 1)

Fig. 1

Let $r=\frac{v_a}{v_m}, p=\frac{dy}{dx} \implies \frac{d^2y}{dx^2} = p\frac{dp}{dy},$ (4) bcomes

$(b-y)p\frac{dp}{dy} + r p^2 \sqrt{1+p^2} = 0.$

Since $p = \frac{dy}{dx} \ne 0$, dividing $p$ through yields

$(b-y)\frac{dp}{dy} + rp\sqrt{1+p^2}=0,$

the governing equation for the missile.

# Newton’s Pi Simplified

We know from “arcsin” :

$\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}}.$

Integrate from $0$ to $x\; (0

$\int\limits_{0}^{x}\frac{d}{dx}\arcsin(x)\;dx = \int\limits_{0}^{x}\frac{1}{\sqrt{1-x^2}}\;dx$

gives

$\arcsin(x)\bigg|_{0}^{x} = \int\limits_{0}^{x}\frac{1}{\sqrt{1-x^2}}\;dx.$

i.e.,

$\arcsin(x) = \int\limits_{0}^{x}\frac{1}{\sqrt{1-x^2}}\;dx.\quad\quad\quad(\star)$

Rewrite the integrand $\frac{1}{\sqrt{1-x^2}}$ as

$(1-x^2)^{-\frac{1}{2}}= (1+(-x^2))^{-\frac{1}{2}}$

so that by the extended binomial theorem (see “A Gem from Issac Newton“),

$\frac{1}{\sqrt{1-x^2}}\overset{(A-1)}{=} \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}1^{-\frac{1}{2}-i}(-x^2)^i.$

Hence,

$\frac{1}{\sqrt{1-x^2}} = \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i x^{2i}.$

And,

$\int\limits_{0}^{x}\frac{1}{\sqrt{1-x^2}}\;dx=\int\limits_{0}^{x}\sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i x^{2i}\;dx = \sum\limits_{i=0}^{\infty}\int\limits_{0}^{x}\binom{-\frac{1}{2}}{i}(-1)^i x^{2i}\;dx = \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i \frac{x^{2i+1}}{2i+1}.$

It follows that by $(\star)$,

$\arcsin(x) = \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i \frac{x^{2i+1}}{2i+1}.$

Let $x=\frac{1}{2},$ we have

$\frac{\pi}{6} = \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i \frac{(\frac{1}{2})^{2i+1}}{2i+1}.$

And so,

$\pi = 6 \sum\limits_{i=0}^{\infty}\binom{-\frac{1}{2}}{i}(-1)^i \frac{1}{2^{2i+1} 2i+1}.$

Fig. 1

Given $0 < x <1,$ prove:

$|-x^2| < 1.\quad\quad\quad(A-1)$

proof

Since $0

Exercise-1 Compute $\pi$ by applying the extended binomial theorem to $\frac{\pi}{6} = \int\limits_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+x^2}\;dx.$

Exercise-2 Can we compute $\pi$ by applying the extended binomial theorem to $\frac{\pi}{4}=\int\limits_{0}^{1}\frac{1}{1+x^2}\;dx?$ Explain.

# Newton’s Pi

Fig. 1

Shown in Fig. 1 is a semicircle centered at C $(\frac{1}{2}, 0)$ with radius = $\frac{1}{2}$. Its equation is

$(x-\frac{1}{2})^2+y^2=(\frac{1}{2})^2, 0 \le x \le 1, y \ge 0.$

Simplifying and solving for $y$ gives

$y = x^{\frac{1}{2}}\cdot(1-x)^{\frac{1}{2}}.\quad\quad\quad(1)$

We see that

Area (sector OAC) = Area (sector OAB) + Area (triangle ABC)$.\quad\quad(*)$

And,

$a = BC = \frac{1}{2}-\frac{1}{4} = \frac{1}{4}, \quad\quad h = AB = \sqrt{AC^2-BC^2} = \sqrt{\frac{1}{2} - \frac{1}{4}}=\frac{\sqrt{3}}{2}.$

It means

Area (triangle ABC) $= \frac{1}{2}ah =\frac{1}{2}\cdot\frac{1}{4}\cdot\frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{32}.\quad\quad\quad(2)$

Moreover,

$\cos(\theta) = \frac{BC}{AC} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \implies \theta = \frac{\pi}{3}.$

Since $\theta$ is one-third of the $\pi$ angle forming the semicircle, the sector is likewise a third of the semicircle. Namely,

Area (sector OAC) $= \frac{1}{3}$ Area (semicircle) = $\frac{1}{3} \cdot \frac{1}{2} \pi (\frac{1}{2})^2 = \frac{\pi}{24}.\quad\quad\quad(3)$

Area (sector OAB) is the area under the curve $y=x^{\frac{1}{2}}\cdot(1-x)^{\frac{1}{2}}$ from its starting point $0$ to the point $x=\frac{1}{4}.$ i.e.,

Area (sector OAB)

$= \int\limits_{0}^{\frac{1}{4}}x^{\frac{1}{2}}\cdot(1-x)^{\frac{1}{2}}\;dx$

$= \int\limits_{0}^{\frac{1}{4}}x^{\frac{1}{2}}\cdot((-x)+1)^{\frac{1}{2}}\;dx$

By the extended binomial theorem: $(x+y)^r=\sum\limits_{i=0}^{\infty }x^iy^{r-i}, x, y, r \in R, |x|< |y|$ (see “A Gem from Isaac Newton“)

$= \int\limits_{0}^{\frac{1}{4}}x^{\frac{1}{2}}\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-x)^i\;dx$

$=\int\limits_{0}^{\frac{1}{4}}x^{\frac{1}{2}}\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-1)^ix^i\;dx$

$= \int\limits_{0}^{\frac{1}{4}}\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-1)^ix^{\frac{2i+1}{2}}\;dx$

$= \sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-1)^i\int\limits_{0}^{\frac{1}{4}}x^{\frac{2i+1}{2}}\;dx$

$=\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-1)^i\frac{2}{2i+3}x^{\frac{2i+3}{2}}\bigg|_{0}^{\frac{1}{4}}$

$x=\frac{1}{4} \implies x^{\frac{2i+3}{2}}$ simplifies beautifully: $\left(\sqrt{\frac{1}{4}}\right)^{2i+3}=\left(\frac{1}{2}\right)^{2i+3}.$

$=\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-1)^i\frac{2}{2i+3}(\frac{1}{2})^{2i+3}$

$=\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}\frac{1}{4(2i+3)}(-\frac{1}{4})^i.\quad\quad\quad(4)$

Expressing (*) by (3), (2) and (4), we have

$\frac{\pi}{24}=\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}\frac{(-1)^i}{4^{i+1}(2i+3)}+\frac{\sqrt{3}}{32}.$

Therefore,

$\pi=24\left(\sum\limits_{i=0}^{\infty}\binom{r}{i}\frac{(-1)^i}{4^{i+1}(2i+3)}+\frac{\sqrt{3}}{32}\right).\quad\quad\quad(5)$

Observe first that

$\sqrt{3} = \sqrt{4-1} = \sqrt{4(1-\frac{1}{4})}=2\sqrt{1-\frac{1}{4}}=2\sqrt{\frac{-1}{4}+1}=2(\frac{-1}{4}+1)^{\frac{1}{2}}$

and so we replace $(\frac{-1}{4}+1)^{\frac{1}{2}}$ by its binomial expansion. As a result,

$\sqrt{3} = 2\sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}(-\frac{1}{4})^i.\quad\quad\quad(6)$

Substituting (6) into (5) then yields

$\pi = \sum\limits_{i=0}^{\infty}\binom{\frac{1}{2}}{i}\left(-\frac{1}{4}\right)^i\left(\frac{1}{4(2i+3)} + \frac{2}{32}\right).$

Fig. 2 shows that with just ten terms (0 to 9) of the binomial expression, we have found $\pi$ correct to seven decmal places.

Fig. 2

# A Gem from Isaac Newton

The Binomial Theorem (see “Double Feature on Christmas Day“, “Prelude to Taylor’s theorem“) states:

$(x+y)^n = \sum\limits_{i=0}^{n} \binom{n}{i}x^{n-i}y^i, \quad n \in \mathbb{N}.$

Provide $x$ and $y$ are suitably restricted, there is an Extended Binomial Theorem. Namely,

$(x+y)^r = \sum\limits_{i=0}^{\infty} \binom{r}{i}x^{r-i}y^i, \underline{|\frac{x}{y}|<1}, r \in \mathbb{R}$

where $\binom{r}{i} \overset{\Delta}{=} \frac{r(r-1)(r-2)...(r-i+1)}{i!}.$

Although Issac Newton is generally credited with the Extended Binomial Theorem, he only derived the germane formula for any rational exponent (i.e., $r \in \mathbb{Q}$).

We offer a complete proof as follows:

Let

$f(t) = (1+t)^r,\;|t|<1, r \in \mathbb{R}.\quad\quad\quad(1)$

Differentiate (1) with respect to $t$ yields

$f'(t) = r(1+t)^{r-1}.$

We have

$(1+t)f'(t) = (1+t)\cdot r(1+t)^{r-1} = r(1+t)^r.$

That is,

$(1+t)f'(t) = rf(t).$

From

$f(0) = (1+0)^r =1,$

we see that $f(t) = (1+t)^r$ is a solution of initial-value problem

$\begin{cases} (1+t)z'(t)=rz(t) \\ z(0) = 1 \end{cases}\quad\quad\quad(2)$

By $(A-1),$ we also have

$g(t) = \sum\limits_{i=0}^{\infty}\binom{r}{i}\cdot t^{i}, |t|<1, r \in \mathbb{R}.\quad\quad\quad(3)$

Express (3) as

$g(t) = \binom{r}{0}t^0 + \sum\limits_{i=1}^{\infty}\binom{r}{i}t^i=1+ \sum\limits_{i=1}^{\infty}\binom{r}{i}t^i$

and then differentiate it with respect to $t:$

$g'(t) = \sum\limits_{i=0}^{\infty}\binom{r}{i}\cdot i\cdot t^{i-1}\overset{(A-2)}{=} r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^{i-1}$

gives us

$(1+t)g'(t) = (t+1)\cdot r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^{i-1}$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^{i-1}$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r\sum\limits_{k=0}^{\infty}\binom{r-1}{k}t^{k}$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r\left(\binom{r-1}{0}t^0+\sum\limits_{k=1}^{\infty}\binom{r-1}{k}t^k\right)$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r\left(1+\sum\limits_{i=1}^{\infty}\binom{r-1}{i}t^i\right)$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r + r\sum\limits_{i=1}^{\infty}\binom{r-1}{i}t^i$

$= r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}t^i + r\sum\limits_{i=1}^{\infty}\binom{r-1}{i}t^i+r$

$= r\sum\limits_{i=1}^{\infty}\left(\binom{r-1}{i-1} + \binom{r-1}{i}\right)t^i+r$

$\overset{(A-3)}{=} r\sum\limits_{i=1}^{\infty}\binom{r}{i}t^i+r$

$= r\sum\limits_{i=1}^{\infty}\binom{r}{i}t^i+r\binom{r}{0}t^0$

$= r\left(\sum\limits_{i=1}^{\infty}\binom{r}{i}t^i+\binom{r}{0}t^0\right)$

$= r\sum\limits_{i=0}^{\infty}\binom{r}{i}t^i$

$\overset{(3)}{=} r\cdot g(t).$

i.e.,

$(1+t)g'(t) = rg(t).$

Since

$g(0) = 1+\sum\limits_{i=1}^{\infty}\binom{r}{i}0^i = 1,$

we see that $g(t)$ is also a solution of initial-value problem (2).

Hence, by the Uniqueness theorem (see Coddington: An Introduction to Ordinary Differential Equations, p. 105),

$f(t) = g(t).$

And so,

$(1+t)^r = \sum\limits_{i=0}^{\infty}\binom{r}{i}t^i.$

Consequently, for $|\frac{x}{y}| <1,$

$(1+\frac{x}{y})^r = \sum\limits_{i=0}^{\infty}\binom{r}{i}(\frac{x}{y})^i.$

Multiply $y^r$ throughout, we obtain

$(x+y)^r = y^r\sum\limits_{i=0}^{\infty}\binom{r}{i}x^iy^{-i}.$

i.e.,

$(x + y)^r = \sum\limits_{i=0}^{\infty}\binom{r}{i}x^iy^{r-i}.$

Prove $(A-1)$ is convergent:

$\sum\limits_{i=1}^{\infty}\binom{r}{i}\cdot t^{i}, |t|<1\quad\quad\quad(A-1)$

Proof

$\frac{\binom{r}{i+1}t^{i+1}}{\binom{r}{i}t^i}$

$= \frac{\frac{r(r-1)(r-2)...(r-i+1)(r-(i+1)+1)}{(i+1)!}}{\frac{r(r-1)(r-2)...(r-i+1)}{i!}}\cdot t$

$=\frac{r(r-1)(r-2)...(r-i+1)(r-(i+1)+1)}{(i+1)!}\cdot \frac{i!}{r(r-1)(r-2)...(r-i+1)}\cdot t$

$= \frac{r-(i+1)+1}{(i+1)i!}i!\cdot t$

$=\frac{r-(i+1)+1}{i+1}\cdot t$

$\lim\limits_{i\rightarrow \infty}\bigg|\frac{\binom{r}{i+1}t^{i+1}}{\binom{r}{i}t^i}\bigg|=\lim\limits_{i\rightarrow \infty}|\frac{r-(i+1)+1}{i+1}\cdot t|=\lim\limits_{i\rightarrow \infty}|\frac{i-r}{i+1}|\cdot |t|=|t|\overset{}{<}1$

Prove

$\sum\limits_{i=1}^{\infty}\binom{r}{i}\cdot i\cdot t^{i-1} = r\sum\limits_{i=1}^{\infty}\binom{r-1}{i-1}\cdot t^{i-1}\quad\quad\quad(A-2)$

Proof

$\binom{r}{i} = \frac{r(r-1)(r-2)...(r-i+1)}{i!}$

$= \frac{r}{i}\cdot\frac{(r-1)(r-1-1)...(r-1+1-i+1)}{(i-1)!}$

$= \frac{r}{i}\cdot\frac{(r-1)(r-1-1)...(r-1-i+1+1)}{(i-1)!}$

$= \frac{r}{i}\cdot\frac{(r-1)(r-1-1)(r-1-2)...(r-1-(i-1)+1)}{(i-1)!}$

$= \frac{r}{i}\binom{r-1}{i-1}$

that is,

$\binom{r}{i} = \frac{r}{i}\binom{r-1}{i-1}.$

Therefore,

$\sum\limits_{i=1}^{\infty}\boxed{\binom{r}{i}}\cdot i\cdot t^{i-1} =\sum\limits_{i=1}^{\infty}\boxed{\frac{r}{i}\binom{r-1}{i-1}}\cdot i\cdot t^{i-1}=r\sum\limits_{k=1}^{\infty}\binom{i-1}{i-1}t^{i-1}.$

Prove

$\binom{r-1}{i-1} + \binom{r-1}{i} = \binom{r}{i}\quad\quad\quad(A-3)$

Proof

$\binom{r-1}{i-1} + \binom{r-1}{i}$

$= \frac{(r-1)((r-1)-1)...((r-1)-(i-1)+1)}{(i-1)!} + \frac{(r-1)((r-1)-1)...((r-1)-i+1)}{i!}$

$= \frac{(r-1)((r-1)-1)((r-1)-2)...(r-(i-1)+1)}{(i-1)!} + \frac{(r-1)((r-1)-1)((r-1)-2)...((r-1)-(i-1)+1)((r-1)-i+1)}{(i)!}$

$= (r-1)((r-1)-1)((r-1)-2)...((r-1)-(i-1)+1)\cdot \left(\frac{1}{(i-1)!} + \frac{(r-1)-i+1}{i!}\right)$

$= \frac{(r-1)(r-1-1)(r-1-2)...(r-1-(i-1)+1)}{(i-1)!}\cdot \left(1+\frac{(r-1)-i+1}{i}\right)$

$= \frac{(r-1)(r-1-1)(r-1-2)...(r-1-(i-1)+1)}{(i-1)!}\cdot \left(\frac{i + (r-1)-i+1}{i}\right)$

$= \frac{(r-1)(r-1-1)(r-1-2)...(r-1-(i-1)+1)}{(i-1)!}\cdot\frac{r}{i}$

$= \frac{r(r-1)(r-2)...(r-1-i+1+1)}{i(i-1)!}$

$= \frac{r(r-1)(r-2)...(r-i+1)}{i!}$

$= \binom{r}{i}.$

# A Mind Unleashed

In 1665, following an outbreak of the bubonic plague in England, Cambridge University closed its doors, forcing Issac Newton, then a college student in his 20s, to go home.

Away from university life, and unbounded by curriculum constraints and tests, Newton thrived. The year-plus he spent in isolation was later referred to as his annus mirabilis, the “year of wonders.”

First, he continued what he had begun at Cambridge: “forging the sword” in mathematical problem solving; Within a year, he gave birth to differential and integral calculus.

Next, he acquired a few glass prisms and made a hole in his window shutter so only a small beam could come through. What he saw after placing a prism in the sunbeam sprung his theories of optics.

And yes, there was an apple tree in the garden! One fateful day in 1666, while contemplating celestial body movements under that tree, Newton was bonked by a falling apple. It dawned on him that the force pulling the apple to the ground might be the same force that holds celestial bodies in orbit. The epiphany led him to discover the law of universal gravitation.

Newton returned to Cambridge in 1667 after the plague had ended. Within six months, he was made a fellow of Trinity College; two years later, the prestigious Lucasian Chair of Mathematics.

# Have we a new proof ?

For a right triangle:

On one hand, its area is

$\frac{1}{2}ab.$

On the other hand, according to Heron’s formula (see “An Algebraic Proof of Heron’s Formula“),

$\sqrt{s(s-a)(s-b)(s-c)}\quad\quad\quad(*)$

where $s=\frac{a+b+c}{2}.$

Hence,

$\frac{1}{2} ab=\sqrt{s(s-a)(s-b)(s-c)}.$

Squaring it gives

$\frac{1}{4}a^2b^2=s(s-a)(s-b)(s-c).$

Using a CAS, we obtain

$\frac{(c^2-b^2-a^2)^2}{16} = 0 \implies c^2-b^2-a^2=0$

from which the Pythagorean theorem emerges:

$a^2+b^2=c^2$

# Seek-Lock-Strike!

A guided missile is launched to destroy a fighter jet (Fig. 1).

Fig. 1

We introduce a coordinate axes such that at $t=0,$ the missile is at origin $(0, 0)$ and the jet at $(a,b)$. The jet flies parallel to the x-axis with constant speed $v_a$. The missile has locked onto the jet so it is always pointing at the jet as it moves. Its speed is $v_m.$

Find the time and position the missile strikes its target.

“The missile has locked onto the jet so it is always pointing at the jet as it moves” means that the tangent to the missile’s path at any point $(x, y)$ will pass through the position of the jet. The equation of the tangent is

$\frac{dy}{dx}\cdot(a + v_a t -x) = b-y.\quad\quad\quad(1)$

$(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2=v_m^2.\quad\quad\quad(2)$

Notice $x, y$ are functions of time $t : x=x(t), y=y(t).$

Differentiate (1) with respecte to $t:$

$\frac{d}{dt}(\frac{dy}{dx})\cdot(a+v_at-x) + \frac{dy}{dx}\cdot\frac{d}{dt}(a+v_at-x) = -\frac{dy}{dt},$

$\frac{d}{dt}(\frac{dy}{dx}) = \frac{d}{dx}(\frac{dy}{dx})\cdot\frac{dx}{dt}=\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}$

$\left(\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}\right)\cdot(a+v_at-x) + \frac{dy}{dx}\cdot(v_a -\frac{dx}{dt}) = -\frac{dy}{dt},$

$\left(\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}\right)\cdot(a+v_at-x) + \frac{dy}{dx}\cdot v_a -\underbrace{\frac{dy}{dx}\cdot\frac{dx}{dt}}_{\frac{dy}{dt}}= -\frac{dy}{dt},$

$\frac{d^2y}{dx^2}\cdot\frac{dx}{dt}\cdot(a+v_at-x) + \frac{dy}{dx}\cdot v_a = 0,$

$\frac{d^2y}{dx^2}= \frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dy}{dx}$

$\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dy}{dx}\cdot\frac{dx}{dt}(a+v_at-x) + \frac{dy}{dx}v_a=0,$

$\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dx}{dt}\cdot\underline{\frac{dy}{dx}(a+v_at-x)} + \frac{dy}{dx}v_a=0.$

By (1), substituting $b-y$ for $\frac{dy}{dx}\cdot(a + v_a t -x)$,

$\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{dx}{dt}\cdot(b-y) + \frac{dy}{dx}v_a=0.\quad\quad\quad(3)$

Let $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt},$ we express (2) as

$(\frac{dx}{dt})^2 + (\frac{dy}{dx}\cdot\frac{dx}{dt})^2=v_m^2,$

$(\frac{dx}{dt})^2\cdot(1+(\frac{dy}{dx})^2)= v_m^2.$

Solving for $\frac{dx}{dt}$ gives

$\frac{dx}{dt} = \frac{v_m}{\sqrt{1+(\frac{dy}{dx})^2}}.\quad\quad\quad(4)$

Submitting (4) into (3),

$\frac{d}{dy}(\frac{dy}{dx})\cdot\frac{v_m}{\sqrt{1+(\frac{dy}{dx})^2}}\cdot(b-y) + \frac{dy}{dx}\cdot v_a=0,$

$\frac{d}{dy}(\frac{dy}{dx})\cdot(b-y) + \frac{v_a}{v_m}\cdot\frac{dy}{dx}\cdot\sqrt{1+(\frac{dy}{dx})^2}=0,\quad\quad\quad(5)$

Let $p = \frac{dy}{dx}, r=\frac{v_a}{v_m},$

(5) becomes

$\frac{dp}{dy}(b-y) + r\cdot p\cdot\sqrt{1+p^2}=0.$

We solve this non-linear differential equation as follows:

For $y < b$,

$\frac{1}{p\sqrt{1+p^2}}\frac{dp}{dy} = \frac{-r}{b-y},$

$\int (\frac{\sqrt{1+p^2}}{p}-\frac{p}{\sqrt{1+p^2}})\frac{dp}{dy} dy = \int \frac{-r}{b-y} dy,$

$\int\frac{\sqrt{1+p^2}}{p}dp-\int\frac{p}{\sqrt{1+p^2}}dp = r\log(b-y)+k_2.\quad\quad\quad(6)$

Since $\int\frac{\sqrt{1+p^2}}{p}dp=\sqrt{1+p^2} - \mathrm{arcsinh}\left(\frac{1}{|p|}\right)$ (See “I vs. CAS“), $\int\frac{p}{\sqrt{1+p^2}}dp=\sqrt{1+p^2},$

(6) gives

$-\mathrm{arcsinh}\left(\frac{1}{|p|}\right) = r\log(b-y)+k_1.\quad\quad\quad(7)$

At $t=0, y=0, p=\frac{dy}{dx}=\frac{b}{a},$

(7) yields

$k_1 = -\mathrm{arcsinh}\left(\frac{a}{b}\right) - r\log(b).$

Moreover, since $p = \frac{dy}{dx} > 0$, we have

$\frac{dy}{dx} = \frac{1}{-\sinh(r\log(b-y)+k_1)},$

$-\sinh(r\log(b-y)+k_1)\cdot \frac{dy}{dx} = 1,$

$\displaystyle\int -\sinh(r\log(b-y) + k_1) \cdot \frac{dy}{dx}\;dx = \displaystyle \int 1\; dx = x +k_2.$

Using Omega CAS Explorer:

we obtain

$\frac{1}{2}\left(\frac{e^{k_1}(b-y)^{r+1}}{r+1} - \frac{e^{-k_1}(b-y)^{1-r}}{1-r}\right) = x + k_2.\quad\quad\quad(8)$

Since $y=0, x=0$, (8) gives

$k_2 = \frac{1}{2}\left(\frac{e^{k_1}b^{r+1}}{r+1} - \frac{e^{-k_1}b^{1-r}}{1-r}\right).\quad\quad\quad(9)$

Suppose when $t = t_*,$ the missile hits the target. Then the striking coordinates $(x_*, y_*)$ are the same as that of the fighter jet, i.e.,

$x_* = a + v_a t_*, \;\;y_*=b.\quad\quad\quad(10)$

Hence,

$\lim\limits_{t \rightarrow t_*} x = x_* \overset{(10)}{=} a+v_a t_*, \;\;\lim\limits_{t \rightarrow t_*} y = y_* \overset{(10)}{=}b \implies \lim\limits_{t \rightarrow t_*} b-y = 0.\quad\quad\quad(11)$

For $r \le 1$ (i.e., $v_a \le v_m$), as $t \rightarrow t_*,$ (8) gives (see Exercise-1)

$0 = a +v_a t_* + k_2 \implies -k_2-a = v_a t_*.$

That is,

$-\underbrace{\frac{1}{2}\left(\frac{e^{k_1}b^{1+r}}{1+r}-\frac{e^{-k_1}b^{1-r}}{1-r}\right)}_{(9):\;\;k_2}-a=v_a t_*.$

By $(\star)$ (see below),

$-\frac{1}{2}\left(\frac{1}{b^r\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}\cdot\frac{b^{1+r}}{1+r}-b^r\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)\cdot\frac{b^{1-r}}{1-r}\right) - a = v_a t_*,$

$-\frac{1}{2}\left(\frac{1}{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}\cdot\frac{b}{1+r}-\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)\cdot\frac{b}{1-r}\right) - a = v_a t_*,$

$\frac{\frac{-b}{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)(1+r)} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{1-r}}{2} -a = v_a t_*,$

$\frac{\frac{-b\left(\frac{a}{b}-\sqrt{1+(\frac{a}{b})^2}\right)}{-1(1+r)} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{1-r}}{2} -a = v_a t_*,$

$\frac{\frac{b\left(\frac{a}{b}-\sqrt{1+(\frac{a}{b})^2}\right)}{1+r} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{1-r}}{2} -a = v_a t_*,$

$\frac{b\left(\frac{a}{b}-\sqrt{1+(\frac{a}{b})^2}\right)}{2(1+r)} + \frac{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b}{2(1-r)} -a = v_a t_*,$

$\frac{a-\sqrt{b^2+a^2}}{2(1+r)} + \frac{a+\sqrt{b^2+a^2}}{2(1-r)} -a = v_a t_*,$

$\frac{(1-r)(a-\sqrt{b^2+a^2}) +(1+r)(a+\sqrt{b^2+a^2})-2a(1-r^2)}{2(1+r)(1-r)} = v_a t_*,$

$\frac{a-\sqrt{b^2+a^2}-ar+r\sqrt{b^2+a^2}+ar+r\sqrt{b^2+a^2}+a+\sqrt{b^2+a^2}-2a + 2ar^2}{2(1-r^2)}=v_a t_*,$

$\frac{2r\sqrt{b^2+a^2} + 2ar^2}{2(1-r^2)} = v_a t_*,$

$\frac{ar^2+\sqrt{b^2+a^2}r}{1-r^2} = v_a t_*,$

$\frac{r(ar + \sqrt{b^2+a^2})}{v_a(1-r^2)} = t_*.$

Since $r = \frac{v_a}{v_m}$, we have

$t_* = \frac{v_a}{v_m}\cdot\frac{ar + \sqrt{b^2+a^2}}{v_a(1-r^2)}=\frac{1}{v_m}\cdot\frac{ar + \sqrt{b^2+a^2}}{1-r^2}.$

Namely,

$t_* = \frac{\sqrt{b^2+a^2} + a\cdot r}{(1-r^2)\cdot v_m}, \quad\quad r=\frac{v_a}{v_m}.$

It follows that the guided missile strikes the fighter jet at $(a+r\cdot\frac{\sqrt{b^2+a^2}+a\cdot r}{1-r^2}, b).$

$e^{k_1} = e^{-\mathrm{arcsinh}(\frac{a}{b})-r\log(b)}$

$= \frac{1}{e^{\mathrm{arcsinh}(\frac{a}{b})+r\log(b)}}$

$= \frac{1}{e^{\mathrm{arcsinh}(\frac{a}{b})}\cdot e^{r\log(b)}}$

$\mathrm{arcsinh}(\blacksquare) = \log\left(\blacksquare + \sqrt{1+\blacksquare^2}\right)$ (see “Deriving Two Inverse Functions“)

$= \frac{1}{e^{\log\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}\cdot e^{r\log(b)}}$

$e^{r\log(b)}=e^{\log(b^r)}=b^r$ (see “Introducing Lady L” and “Two Peas in a Pod, Part 3“)

$= \frac{1}{\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)b^r}.$

i.e.,

$e^{k_1} = \frac{1}{b^r\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)}.\quad\quad\quad(\star)$

Exercise-1 Show that for $r < 1$, (8) gives $0 = a +v_a t_* + k_2 \implies -k_2-a = v_a t_*.$ (Hint: (10))

Exercise-2 For $r < 1$, what is the total distance traveled by the missile when it strikes the fighter jet? (hint: Don’t make things harder than they are)

Exercise-3 Show that if $r \ge 1$ (i.e., $v_a \ge v_m$), the missile will not strike the fighter jet. Explain.

# Faster Pi

There is a more astonishing algorithm than what is described in “Fast Pi” for rapidly calculating $\pi.$

Let us consider the three-term iteration with initial values

$a_0 = \sqrt{2}, \quad b_0 =0, \quad \pi_0 = 2+\sqrt{2}$

given by

$a_n = \frac{1}{2}\left(\sqrt{a_{n-1}} + \frac{1}{\sqrt{a_{n-1}}}\right), \quad b_{n} = \sqrt{a_{n-1}}\left(\frac{b_{n-1}+1}{b_{n-1} + a_{n-1}}\right), \quad \pi_{n} = \pi_{n-1}b_n\left(\frac{1+a_{n}}{1+b_{n}}\right).$

Then $\pi_n$ converges exponentially to $\pi$. In fact,

$|\pi-\pi_n|< \frac{1}{10^{2^n}}.$

Implemented in Omega CAS Explorer, $4$ iterations yield $40$ digits of $\pi$:

The $8$th iteration gives $\pi$ correctly to $694$ digits:

Exercise-1 Show that $20$ iterations will provide over $2$ million digits of $\pi.$