# Computer Algebra

# Beauty and the Beast [2]

# Beauty and the Beast [1]

# A Gift That Keeps On Giving

We see from “Seek-Lock-Strike!” Again that given the missile’s position

where and are themselves functions of time

It means

That is, let

We also have (see “Seek-Lock-Strike!”)

Since

Substitute (2) into (1) yields

It follows that , the position of the missile satisfies the initial-value problem

To obtain the missile’s trajectory, we solve (4) numerically using the Runge-Kutta algorithm. It integrates (4) from to (see “Seek-Lock-Strike!”).

Fig. 1

The missile strike is illustrated in Fig. 1 and 2.

Fig. 2

Fig. 3

The trajectories shown are much smoother than those in “Seek-Lock-Strike!” Animated.

# “Seek-Lock-Strike!” Animated

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

where

Since the fighter jet maintains its altitude (), the missile must strike it at . Setting in gives

Hence, we can plot

Fig. 1

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

Fig. 2

Fig. 3

# “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

Solving (1) for ,

We also have

Equate (1) and (2) gives

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

We let so and express (5) as

where

Fig. 2

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

Fig. 3

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

# “Seek-Lock-Strike!” Simplified

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

Solving

for we have

From

we also have

Equate (1) and (2) gives

Differentiate (3) with repect to we obtain

(see Fig. 1)

Fig. 1

Let (4) bcomes

Since , dividing through yields

the governing equation for the missile.

# Newton’s Pi Simplified

We know from “arcsin” :

Integrate from to

gives

i.e.,

Rewrite the integrand as

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

Hence,

And,

It follows that by ,

Let we have

And so,

Fig. 1

See also “Newton’s Pi“.

Given prove:

*proof*

Since

*Exercise-1* Compute by applying the extended binomial theorem to

*Exercise-2* Can we compute by applying the extended binomial theorem to Explain.

# Newton’s Pi

Fig. 1

Shown in Fig. 1 is a semicircle centered at C with radius = . Its equation is

Simplifying and solving for gives

We see that

Area (sector *OAC*) = Area (sector *OAB*) + Area (triangle *ABC*)

And,

It means

Area (triangle *ABC*)

Moreover,

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

Area (sector* OAC*) Area (semicircle) =

Area (sector *OAB*) is the area under the curve from its starting point to the point i.e.,

Area (sector *OAB*)

By the extended binomial theorem: (see “A Gem from Isaac Newton“)

simplifies beautifully:

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

Therefore,

Observe first that

and so we replace by its binomial expansion. As a result,

Substituting (6) into (5) then yields

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

Fig. 2

# Have we a new proof ?

For a right triangle:

On one hand, its area is

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

where

Hence,

Squaring it gives

Using a CAS, we obtain

from which the Pythagorean theorem emerges: