By (see “Integral: The CAS & I“),
Exercise-1 Evaluate using a CAS.
By (see “Integral: The CAS & I“),
Exercise-1 Evaluate using a CAS.
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.
Let denote the slope of line connecting and respectively.
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.
Introducing rectangular coordinates show in Fig. 2:
We observe that
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.
By Heron’s formula (see “Had Heron Known Analytic Geometry…“), the area of triangle with vortex is
From Fig. 4, we see that it is zero.
lie on the same line.
The reason we do not consider (9), (10), (11) is due to the fact that
(9) contradicts (2) since .
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. Without invoking trigonometric functions, show that the area of triangle equals that of
Solution: Introducing a rectangular coordinate system and complex number
Let denote the area of triangle and respectively.
By Heron’s formula derived without invoking trigonometric function (see “An Algebraic Proof of Heron’s Formula“),
We also have
(see “Treasure Hunt with Complex Numbers“) and,
It is shown by Omega CAS Explorer that the expression under the square root of is the same as that of
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
Dividing both sides by yields
It follows that since ,
Applying the half-angle formula
We compute the value of according to (3):
Exercise-1 Compute from (1) by letting
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:
Integrate it, we have
By (see “Integration by Parts Done Right“),
As a result, (1-1) yields a new ODE
Upon submitting (1-3) to Omega CAS Explorer in non-expert mode, the CAS asks for the range of
Depending on the range provided, ‘ode2’ gives three different solutions (see Fig. 2, 3 and 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.
For let we have
Divide both numerator and denominator on the left side by ,
Write it as
Multiply both sides by
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.
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.
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.
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
(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
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.
(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“).
From “Deriving Two Inverse Functions“:
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:
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.
we have i.e.,
It follows that
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
Substituting it back into (1-6) yields
This is a depressed cubic with By Cardano’s formula,
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
 Has every real number as its member.
 The arithmetic operations () are so defined that the commutative, associative and distributive law are obeyed.
 There is a member such that i.e.,
Bombelli sanity checked his idea by consider any quadratic equation
where which can be written as
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
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
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:
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