“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) -r\sqrt{1+(\frac{dx}{dy})^2} = 0\quad\quad\quad(*)

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


Exercise-1 Obtain the missile’s trajectory from (*).

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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<x<1):

\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

See also “Newton’s Pi“.


Given 0 < x <1, prove:

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

proof

Since 0<x<1 \implies 0< x^2 < 1, |-x^2|=|x^2|=x^2 < 1.


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┬áDay“, “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}.