Evaluate .

By (see “Integral: The CAS & I“),

*Exercise-1* Evaluate using a CAS.

Evaluate

For (see *Exercise-1*)

*Exercise-1* Show that for .

The proof in my post “A Computer Algebra Aided Proof in Plane Geometry” can be significantly simplified if complex numbers are used.

Consider complex numbers in FIg. 1.

Fig. 1

Since

we have,

i.e.,

Similarly,

and

give

Let denote the slope of line connecting and respectively.

From

we see that

lie on the same line.

Notice that since .

See also “Treasure Hunt with Complex Numbers“.

Given and two squares in Fig. 1. The squares are sitting on the two sides of and , respectively. Both squares are oriented away from the interior of . is an isosceles right triangle. is on the same side of . Prove that points and lie on the same line.

Fig. 1

Introducing rectangular coordinates show in Fig. 2:

Fig. 2

We observe that

Fig. 3

Solving system of equations (4), (5), (6), (7), we obtain four set of solutions:

Among them, only (10) truly represents the coordinates in Fig. 2.

Fig. 4

By Heron’s formula (see “Had Heron Known Analytic Geometryâ€¦“), the area of triangle with vortex is

.

From Fig. 4, we see that it is zero.

Therefore,

lie on the same line.

The reason we do not consider (9), (10), (11) is due to the fact that

(9) contradicts (2) since .

And,

by (1), (10) and (11) indicate which contradicts (3).

*Exercise-1* Prove “ lie on the same line” with complex numbers (hint: see “Treasure Hunt with Complex Numbers“).

* Problem:* and are squares.

** Solution**: Introducing a rectangular coordinate system and complex number

Let denote the area of triangle and respectively.

We have

Let

By Heron’s formula derived *without* *invoking trigonometric function *(see “An Algebraic Proof of Heronâ€™s Formula“),

We also have

(see “Treasure Hunt with Complex Numbers“) and,

Similarly,

Let

That is,

It is shown by Omega CAS Explorer that the expression under the square root of is the same as that of

Therefore,

*Exercise-1* The two squares with area 25 and 36 ( see figure below) are positioned so that Find the area of triangle TSC.

Besides “Wallis’ Pi“, there is another remarkable expression for the number as an infinite *product*. We derive it as follows:

From the trigonometric identity

,

we have

.

That is,

Dividing both sides by yields

or,

.

It follows that since ,

;

i.e.,

Let

(1) becomes

We know

Applying the half-angle formula

gives

Hence,

We compute the value of according to (3):

Fig. 1

*Exercise-1* Compute from (1) by letting

Solving

Even though ‘contrib_ode’, Maxima’s ODE solver choked on this equation (see “An Alternate Solver of ODEs“), it still can be solved as demonstrated below:

multiplied by

i.e.,

Integrate it, we have

By (see “Integration by Parts Done Right“),

As a result, (1-1) yields a new ODE

or,

Upon submitting (1-3) to Omega CAS Explorer in non-expert mode, the CAS asks for the range of

Fig. 1

Depending on the range provided, ‘ode2’ gives three different solutions (see Fig. 2, 3 and 4).

Fig. 2

Fig. 3

Fig. 4

Let’s also solve (1-3) manually:

If (1-3) has a constant solution

In fact, this solution can be observed from right away.

Otherwise (),

That is,

or

For let we have

Divide both numerator and denominator on the left side by ,

Write it as

Multiply both sides by

Integrate it,

we obtain

i.e.,

or,

where

For let

For

Notice when (1-2) becomes

This is a Bernoulli’s Equation with and Solving it (see “Meeting Mr. Bernoulli“),

Since we can verify (3-1) as follows:

substitute into (2-3),

Unsurprisingly, this is the same as (3-1).

*Exercise-1* Mathematica solves

But it only return one solution. Show that it is equivalent to (2-2).

*Exercise-2* Solving (1-2) using ‘contrib_ode’.

*Exercise-3* Show that (2-2), (2-3) and (2-4) are equivalent to results shown in Fig. 2, 3 and 4 respectively.

Fig. 1

In his popular book “One, Two, Three … Infinity“, physicist George Gamow told a story:

Once upon a time, there was a young man who found among his great grandfather’s papers a piece of parchment that revealed the location of a hidden treasure. It read:

“Sail to ____ North Latitude and ____ West longitude where you will find a deserted island. There is a large meadow on the north shore of the island where stand an oak and a pine. You will see also an old gallows on which we once used to hang traitors. Start from the gallows and walk to the oak counting your steps. At the oak, you must turn *right* by a right angle and take the same number of steps. Put a spike in the ground there. Now you must return to the gallows and walk to the pine counting your steps. At the pine, you must turn *left* by a right angle and see that you take the same number of steps, and put another spike into the ground. Dig half way between the spikes; the treasure is there.”

So the young man charted a ship and sailed to the South Seas. He found the island, the meadow, the oak and the pine, but to his great sorrow the gallows was gone. Unlike the living trees, the gallows has long since disintegrated in the weather, and not a trace of it or its location remains.

Unable to carry out the rest of the instructions (or so he believes), the young man fell into despair. In an angry frenzy he began to dig at random all over the field. But all his efforts were in vain; the island was too big!

Needless to say, the young man sailed back empty handed. And the treasure is still there.

This is a sad story, but what is sadder still is the fact that the young man might have gotten the treasure, if only he had known some mathematics, and specifically the use of complex numbers.

How come?

Consider the island as a plane of complex numbers; Place the origins of three rectangular coordinate systems at the location of oak(), pine () and half way between them (). and are complex numbers. Notably, is the half way point between the spikes.

Fig. 2

From Fig. 2, we see that

By the fact (see *Exercise-1*) that

(1) The multiplication by is geometrically equivalent to a *counterclockwise *rotation by a right angle

and

(2) The multiplication by is geometrically equivalent to a *clockwise* rotation by a right angle,

Since the treasure is halfway between the spikes, we have

Subtracting (6) from (5) gives

It means

Therefore,

We see that the unknown position of gallows denoted by fell out in (7), and (8) tells regardless where the gallows stood, the treasure must be located at the point of rectangular coordinate system with origin .

And so, had the young man done the simple math shown above, he could have looked for the treasure at the point indicated by the cross in Fig. 1 and found it there.

*Exercise-1 *Prove

(1) The multiplication by is geometrically equivalent to a *counterclockwise *rotation by a right angle.

(2) The multiplication by is geometrically equivalent to a *clockwise* rotation by a right angle.

*Exercise-2* Locate the treasure using a computer algebra system (hint: see “A Computer Algebra Aided Proof in Plane Geometry“).

Evaluate

Let

we have

and

Consequently,

From “Deriving Two Inverse Functions“:

Therefore,

Had we written as we would have

the same as (*).

**In Memory of Johann Weilharter (1953-2021)**

Girolamo Cardano (1501-76) was an Italian intellect whose interests and proficiencies ranged through those of mathematician, physician, biologist, physicist, chemist, astrologer, astronomer, philosopher, writer, and gambler.

While conducting research on solving algebraic equations, Cardano discovered that by means of a suitable substitution, the general cubic equation

can be simplified. His substitution is , which yields

Upon expanding and rearrange the terms, this becomes

a depressed cubic (without the term) where

Cadano applied this substitution in solving cubic equation

Substituting into the cubic in , he obtained a depressed cubic in namely,

Without a formula for this simplified equation, Cardano proceeded to solve it by way of ad hoc factoring:

Clearly, is a solution to

Applying the quadratic formula to gave But this expression was immediately dismissed (for Cardano knew has no real solution).

Therefore, is the only solution to the original cubic equation.

has no solution

Cardano also solved in a similar fashion:

Obtaining first the depressed cubic (with )

Next is the ad hoc factoring again:

Surely,

is a solution.

Furthermore, two additional solutions: were obtained by applying the quadratic formula to

has three solutions:

But Cardano did not like the ad hoc factoring. He wanted a *formula* that readily solves the depressed cubic , one that resembles the formula for the quadratics (see “Deriving the quadratic formula without completing the square“).

His relentless search for such a formula took many years (see William Dunham’s “Journey through genuis“) but, lo and behold, he found one:

To be clear, (*) is not Cardano’s own making. The formula bears the name ‘Cardano’s formula’ today only because Cardano was the one who published it in his 1545 book “Ars Magna” but without its derivation. However, in a chapter titled “On the Cube and First Power Equal to the Number”, Cardano did give acknowledgement to Scipio del Ferro and Niccolo Fontana, who had independently derived (*) around 1515, but had kept the knowledge away from the public.

We derive (*) as follows:

Consider an algebraic identity that reminiscent of the depressed cubic Namely,

It suggests that if we can determine the quantity and in terms of and from

then is a solution to

Asume , (1-1) gives

Substituting this into (1-2) yields

Multply both sides by and rearrange terms, we have a sixth-degree equation:

But it is also quadratic in.

Therefore, using the formula for quadratics,

There are two cases to consider.

For

we have i.e.,

It follows that

For

the same as (1-5).

If we see that on the one hand,

On the other hand, letting in (*) yields

Cardano first tested (*) on cubic by letting . The formula yields

It came as a surprise to Cardano initially. But he quickly realized that this sophisticated looking expression is nothing more than “2”, the unique solution of , in disguise.

Today, this is easily checked by a CAS:

For a mathematical *proof*, see “A Delightful Piece of Mathematics“.

Cardano then tested the formula on cubic Substituting into it gave

He was startled by the result!

The presence of alone did not surprise him for he had seen negative number under the square root before (while solving a quadratic clearly has no solution). What really perplexed Cardano this time was the fact that square root of negative number appearing in the result for a cubic that has three real solutions!

Cardano thus sought the value of to see which solution, amongst and it represents.

He started with . At once, Cardano noticed that is a number in the form of

And he speculated that the result of calculating has the same manner.

So Cardano wanted to find and such that

He proceeded as follows:

Cubing both sides gives

Equating the similar parts on both sides yields a system of nonlinear algebraic equations

Squaring both (1-6) and (1-7) gives:

and subtracting (1-9) from (1-8) results in

or,

Substituting it back into (1-6) yields

And so,

This is a depressed cubic with By Cardano’s formula,

That is,

So solving for by Cardano’s formula resulting in having to calculate another square root of a negative number. Cardano was put right back to where he had started. With the frustration he called the cubic “irreducible” and pursued the matter no further.

It would be another generation before Rafael Bombelli (1576-72) took upon the challenge of calculating again.

Bombelli’s was an engineer who knew how to drain the swampy marshes, and only between his engineering projects was he actively engaged in mathematics. Being practical and sound minded, he read the near-mystical not as* the square root of a negative number* but a symbolic representation for a new type of number that *extends *the real number. He imagined a set for a new type of number that

[1] Has every real number as its member.

[2] The arithmetic operations () are so defined that the commutative, associative and distributive law are obeyed.

[3] There is a member such that i.e.,

Bombelli sanity checked his idea by consider any quadratic equation

That is

where which can be written as

or,

if is positive, then it has a square root, and is a solution of the equation (so is the number If is not positive, then is, and therefore has a square root

Let

The left side of (2-2) becomes

That is, (2-3) is a solution of (2-2).

Bombelli was elated as it suggested that by considering his set which contains the real numbers and *all *quadratic equations have solutions!

It also gave him much needed confidence in showing what really is.

Right away, Bombelli saw

so he replaced in with

He then anticipated that the value of is a new type of number where and are real numbers. i.e.,

And finally, he proceeded to find and from (3-1).

As an illustration, we solve (3-1) for as follows:

Cubing it gives

Since

this is

Equating similar terms on both sides yields a system of nonlinear equations:

After factoring, it becomes

Assuming and are both integers, then and on the left side of (3-2) are two integer factors of Since has only two factors, namely, and If then from (3-1), a contradiction. However, yields or While contradicts (3-3), gives Therefore, is the solution to (3-2, 3-3). i.e.,

It is also easy to see that (3-4) is true as follows:

And so

Similarly, Bombelli obtained (see *Exercise-1*)

By (3-4) and (3-5) Bombelli was able to reproduce the solution to cubic :

Thus, with and the ordinary rules of real numbers’ arithmetic, Bombelli broke the mental logjam concerning negative number under the square root.

Satisfied with his work that unlocked what seemed to be an impassable barrier, Bombelli moved on without constructing his set for the new type of number in a logically unobjectionable way. The world had to wait another two hundred years for that (see “Mr. Hamilton does complex numbers”). Still, Bombelli deserves the credit for not only recognizing numbers of a new type have a role to play in algebra, but also giving its initial impetus and now undisputed legitimacy.

*Exercise-1 *Show that