Showing below is the abstract of my talk at ACA 2018. Stay tuned for the complete presentation after the conference.

# Six of one, half a dozen of the other

We have defined the derivative of function in “Inching towards Definite Integral” as

An equivalent definition of is

We will prove the equivalency below:

Let , then , (1) becomes

Similarly,

Let , then , (2) becomes

# Every dog has its day

On a school day in the late 1700s, a teacher, for unknown reason, ordered her pupils to add up all the numbers from 1 up to 100.

The children immediately started adding except for the future prince of mathematics.

It took him a few seconds before he proudly announced the result.

What a glorious day for the little Carl Friedrich Gauss!

Fast forward to a hot summer day in the late 1970s. I was then a high school student. On that day, factoring was taught first, then an assignment was given:

“Factor all expressions on page 18, *only* 88 of them”

The students went to work immediately except for me who got a funny idea of doing something else.

While pretending to be working, I was peeking though the illustrated book “Ring of Nibelungs” by Arthur Rackham. I brought it to school and hid it under the dull textbooks.

What a sixteen-year-old would rather do? Slaying dragon or factoring algebraic expressions?

Just as I was enthralled by the adventures of Siegfried, my hero, a voice boomed from behind:

“What are you doing rascal ?!”

It was the teacher.

The punishment – Factoring , for is the additional homework!

I managed to complete this before staggering into the bed:

As I was factoring, I made an observation:

The non-vanishing coefficients in any factor are either -1 or 1.

I wondered: Could this be true for all values of ?

As soon as I finished factoring , I wrote down a conjecture named after Siegfried:

*The absolute value of all coefficients appear in the polynomial factor of is always 1. *

That night I had a lucid dream, where I proved “Siegfried conjecture.”

Reviewing the proof in the morning, however, to my dismay, it was a pile of junk.

Still, for several weeks I continued my effort fervently, but I could neither prove nor refute “Siegfried conjecture”.

All to no avail. I had to give up.

Fast forward again to a mid summer day a few years later.

I visited Joe, a high school classmate who was studying at MIT to become an aeronautical engineer.

While walking on campus, I teased my friend: “How many basketball championships does MIT have? Do you even have any sports team here?”

Joe shrugged: “Haha, but we have created MACSYMA, the world’s first program that does symbolic algebra! Come, let me show you its power.”

That evening, I sat quietly in front of a VT100 tube, getting acquainted with MACSYMA.

It was fantastic. I felt as though I was brought out of the hold of a ship where I had lived all my life, onto the deck, into the fresh air.

I saw the good –

Then the better –

Finally the best –

I grabbed the keyboard and commanded MACSYMA:

It responded:

At is the coefficient of term .

In a blink of an eye, MACSYMA disproved “Siegfried conjecture”.

And I, while fully clothed, jumped up and shouted

Thanks to MACSYMA, the day was finally mine!

Fast forward once more to the summer of 2005, the spark MACSYMA lit in me twenty years ago gave life to Omega CAS Explorer:

Hooray! Every dog has its day!

Recently, as I was preparing for my talk at ACA 2017 (See “Little Bird and a Recursive Generator“), I looked at the formula

and wondered:

For what value(s) of , (1) produces a perfect square ?

A search shows when , (1) yields , a perfect square:

But are there any other such ‘s?

Further search yields no result:

It leads me to think:

For is the only solution to Diophantine equation

Can you prove or refute my new conjecture?

# Inching towards Definite Integral

In a blog titled “Introducing Lady L“, we showed that

In light of the fact that is a monotonic function on , i.e.,

,

we can prove that

Let is *continuous* and *monotonic*

The proof is simple, rigorous and similar to what we have done in “Introducing Lady L“.

Let be a monotonically increasing function,

.

and

.

Since

It follows that

.

The fact that is continuous tells us

.

In addition,

.

Therefore,

.

The case for can be handled in a similar fashion.

(2) becomes more general when the condition of being a monotonic function is removed:

Let is *continuous*

Let’s prove it.

By definition, is continuous at means

.

It implies .

For , we have

That is

Since ,

it follows that

or,

By (3), we have

i.e.,

As a result,

.

For , since ,

.

That is

or,

Divide throughout, and express as , we arrived at

as before.

We are now poised to define the derivative of a function:

Let be a function on an opensubset of . Let . We say that is differentiable at if

exists. If exists, this limit, commonly denoted by or , is called the derivative of at .

For function , the difference of two differentiable functions,

.

By definition,

We have

or,

With this definition, we can also re-state (4) as:

Let is *continuous*

From (7), it is clear that is a solution of the following equation:

where is the unknown function.

In fact, for *any* function that satisfies (8), we have

.

Therefore by (6),

That is, is a function whose derivative is everywhere zero.

Geometrically, if the curve of a function is horizontally directed at every point, it represents a constant function.

It is even more obvious if one considers a function that describes the position (on some axis) of a car at time . Then the derivative of the function, is the instantaneous velocity of the car. If the derivative is zero for some time interval (the car does not move) then the value of the function is constant (the car stays where it is).

Hence, we assert

A function on an open interval has derivative zero at each point , a constant.

From (9) and above assertion, whose rigorous proof we postpone until later in “Sprint to FTC“, it follows that

or,

where c is a constant.

We know

.

It implies

.

i.e.,

and, (10) becomes

Let , we have

This is the well known Newton-Leibnitz formula. It expresses an algorithm for evaluating the definite integral :

Find *any* function whose derivative is , and the difference gives the answer.

# A Sprint to FTC

Our sprint starts with **Lagrange’s Mean-Value theorem** which states:

A function is

(1) continous on closed interval

(2) differentiable on open interval

.

Let’s prove it.

The area of a triangle with vortices and is the absolute value of

where (See “Had Heron Known Analytic Geometry“)

Let

,

Since is differentiable, we have

.

Clearly,

Therefore, by Rolle’s Theorem (See “Rolle’s theorem”) , i.e.,

or

.

The geometric meaning of Lagrange’s Mean-Value theorem is illustrated for several functions in Fig. 1. It shows that the graph of a differentiable function has at least one tangent line parallel to the cord connecting and .

Fig. 1

The bottom curve falsifies the theorem due to its missing differentiability at one point.

Following Lagrange’s Mean-Value theorem are two corollaries. We have encountered and accepted the first one without proper proof in the past (See “Inching towards Definite Integral“)

Let’s state and prove them now.

**Corollary 1**. A function on an open interval has derivative zero at each point , a constant.

It is true due to the fact that is both continous on and differentiable on . By Lagrange’s Mean-Value theorem, . Since , We have . i.e., . Hence , a constant on .

**Corollary 2**. Two functions, and have the same derivative at each point on an open interval , a constant.

For . By corollary 1, , a constant.

Next, we define a set as follows:

i.e.,

is a set of function such that for all ‘s.

is certainly not empty for we have proved that by showing (See “Inching towards Definite Integral“). It follows that ,

By Corollary 2,

.

Let , we have

.

Hence

or

.

Let’s summarize the result of our exploration in a theorem:

On an open interval containing , a function is differentiable and, ,

This is FTC, **The Fundamental Theorem of Calculus**.

Let , (1) becomes the Newton-Leibnitz formula

,

our old friend for evaluating (See “Inching towards Definite Integral”)

# Knock Euler’s Line

*my imagination*

* is a piece of board*

* my sole instrument*

* is a wooden stick*

*I strike the board*

* it answers me*

* yes—yes*

* no—no*

*“A Knocker” by Zbigniew Herbert*

Euler’s line theorem states

In every triangle

the intersection of the medians

the intersection of the heights

and the center of the circumscribed circle

are on a straight line

Let’s prove it

with the aid of Omega CAS Explorer

We know

is a circle centered at with radius

provide (2) is positive

**Find d, e, f from triangle’s vertices :**

*ceq:x^2+y^2+d*x+e*y+f=0;
eq1:ev(ceq, x=x1, y=0);
eq2:ev(ceq, x=-x1, y=0);
eq3:ev(ceq, x=x2, y=y2);
sol:linsolve([eq1, eq2, eq3], [d,e,f]);*

**Evaluate (2):**

*ev(d^2 + e^2-4*f, sol);*

always positive for

**Find the center of the circumscribed circle :**

*xc:ev(-d/2, sol)$*

yc:ev(-e/2, sol)$

**Find the intersection of the medians :**

*eq1:y*((x1+x2)/2+x1)=y2/2*(x+x1)$
eq2:y*((x2-x1)/2-x1)=y2/2*(x-x1)$
sol:solve([eq1, eq2], [x,y])$
xm:ev(x,sol);*

*ym:ev(y, sol);*

*is(ev(x2*y=y2*x, x=xm, y=ym));
*

**Find the intersection of the heights :**

*eq1:y2*y=-(x2+x1)*(x-x1)$
eq2:y2*y=-(x2-x1)*(x+x1)$
sol:solve([eq1, eq2], [x,y])$
xh:ev(x, sol);*

*yh:ev(y, sol);*

**Compute the area of triangle with vertices :**

*m:matrix([xc, yc,1], [xm, ym, 1], [xh, yh,1]);
determinant(m)$
ratsimp(%);*

Indeed

and are on a straight line.

# A Computer Algebra Aided Proof of Feuerbach’s Nine-Point Circle Theorem

Feuerbach’s Nine-Point Circle Theorem states that a circle passes through the following nine significant points of any triangle can be constructed:

1. The midpoint of each side of the triangle

2. The foot of each altitude

3. The midpoint of the line segment from each vertex of the triangle to the orthocenter

Let’s prove it with the aid of Omega CAS Explorer.

**Step-1 ****Set up the circle equation**:

is a circle centered at with radius

provide (2) is positive.

**Step-2** **Find d,e,f using p1, p2, p3**:

*ceq: x^2+y^2+d*x+e*y+f=0;*

*eq1: ev(ceq, x=(x1+x2)/2, y=y2/2); *

*eq2: ev(ceq, x=0, y=0);*

*eq3: ev(ceq, x=(x2-x1)/2, y=y2/2);*

*sol: linsolve([eq1, eq2, eq3], [d,e,f]);*

The new circle equation is

*nceq: ev(ceq, sol);*

,

Evaluate (2)

*ev(d^2+e^2-4*f, sol);*

,

always positive for .

**Step-3** **Show p5, p6, p4 are on the circle**:

*p5:ev(nceq, x=x2, y=0);*

*p4: linsolve([(x2-x1)*y=y2*(x-x1), y2*y=-(x2-x1)*(x+x1)], [x,y]);*

*ratsimp(ev(nceq, p4));*

*p6: linsolve([(x2+x1)*y=y2*(x+x1), y*y2=-(x2+x1)*(x-x1)], [x,y]);*

*ratsimp(ev(nceq, p6));*

**Step-4** **Find the orthocenter**

*o: linsolve([x=x2, y*y2=-(x2+x1)*(x-x1)], [x,y]);*

**Step-5** **Show p7, p8, p9 are on the circle**:

*xo: rhs(o[1]);*

*yo: rhs(o[2]);*

*p7: ratsimp(ev(nceq, x=(xo+x2)/2, y=(yo+y2)/2));*

*p8: ratsimp(ev(nceq, x=(xo-x1)/2, y=(yo+0)/2));*

*p9: ratsimp(ev(nceq, x=(xo+x1)/2, y=(yo+0)/2));*

This post is essentially my presentation at 21st Conference on Applications of Computer Algebra, July 20-23, 2015, Kalamata, Greece.