My love for Mathematics really blossomed in the summer of 2017, when I first discovered the enigma that is Calculus. Or more precisely, Differential Calculus. The subject intrigued me for multiple reasons- its inherent ability to mesmerize its explorer while still being simple, its links with Algebra and Geometry, and most of all, the fact that it opens up beautiful avenues for the Math enthusiast and paves the way for a beautiful journey of discovery and enlightenment!

Now that’s a whole lot of generic talk. But it’s true nonetheless; from my high school classes that year onward, I became more interested in the discipline than I ever was. I would be engrossed by math-discussions during classes and once they got over, I would explore the generalizations of the problems we already discussed during class.

One of many such explorations was when I discovered how powerfully close the Fibonacci sequence (1,1,2,3,5,8,13,…) and its characteristic equation ( x²-x-1=0) are. I would find weird, objectively uninteresting (yet interesting for me) equations about this relationship and prove them after classes, and my Math teacher would repeatedly flash an uninterested/disinterested expression reminiscent of a purist’s reaction to the applications of his/her discovery.

My luck finally improved in the winter of the same year. My teacher would repeatedly ask me to find generalizations- not of the inelegant formulae I kept obtaining, but of the equation x²-x-1=0 itself. The idea wasn’t lucrative at first, but I only understood what he meant after a considerable while. During my winter vacation, I decided to spend a majority of time doing mathematical explorations, hoping they would lead to enlightening conjectures which I could prove later.

I decided to start with taking a higher-degree version of the quadratic (the characteristic equation). I took x³-x²-x-1=0 and did the same explorations I conducted with the quadratics.

I found some amazing generalizations. For example, the limit of the ratio of consecutive terms in the Fibonacci sequence (1/1, 2/1, 3/2, 5/3, 8/5, 13/8, 21/13, 34/21, and so on) starts to approach the golden ratio once the terms get very large. This golden ration also happens to be the largest magnitude root of x²-x-1=0. Similarly, if you take the limit of the ratio of consecutive terms in the sequence 1,1,2,4,7,13,24,44,81,… you end up getting the largest magnitude real root of x³-x²-x-1=0!

Notice how the first (Fibonacci) sequence had each successive term being obtained after adding the previous two terms. The second sequence was obtained by adding the previous three terms. This was when I took the third degree version of my equation. I found out on various internet sources that such a sequence was called the Tribonacci sequence.

I also found a third-degree equivalent for Binet’s formula. There was certainly some patterns brewing here!

So I decided to consider the relationship between the sequence 1,1,2,4,8,15,29,56,… and the equation x⁴-x³-x²-x-1=0 and lo and behold, I found exactly the fourth-degree equivalents for the conjectures I obtained earlier, although it must be mentioned that proving these would be much more difficult than proving the previous conjectures.

I decided to write about my findings. These were not just mere mathematical explorations- they were mind-blowing ideas to my eyes, and a tribute in my heart to our mathematical ancestors who dealt with similar sequences and found ground-breaking theorems. If I could even 0.0000001% of that, I would consider myself grateful to my moments of sweet discovery.

My resulting piece of work was a paper on what I called m-bonacci sequences. There were few research papers on the same (I found one published in Ithaca College and even referenced it in my paper; in fact, some papers even called these n-bonacci), but I decided to publish my findings in a good Math journal so I could present my ideas on a well-respected and reputed platform.

My experience in trying to find a good journal for publishing my research was a roller-coaster ride in itself, but I leave that narration for some other day.

I request all my readers to read my paper, for reading and attempting to understand them would mean a good deal of appreciation for my work, irrespective whether somebody disputes my methods or not.

You can find the journal issue in Volume 6 Issue 3 of International Journal of Mathematics And its Applications. And the link to my paper: http://ijmaa.in/v6n3/357-363.pdf

I would appreciate constructive feedback to my research.

Please clap, comment and share this article and follow my blog Cosmus Mathematicus for fun articles on Mathematics.

My Research Paper on M-Bonacci Sequences was originally published in Cosmus Mathematicus on Medium, where people are continuing the conversation by highlighting and responding to this story.

It’s been a long time since I’ve published something. However this article is really special to me and many others who discovered Calculus through sneaky explorations and fell in love with Math.

If you’ve never done any integral calculus problems before, I suggest you look through Thomas’ Calculus and Apostol’s Calculus and get the prerequisites for this article.

Now let me cut to the chase and show you the first problem.

The problem at first looks really difficult. But once you look at the idea you may feel the same way I did: this problem is one of the easiest integrals!

Now the idea involves an understanding of Riemann’s integration. You can watch the video below to get an understanding-

Now once you think of the definite integral as a Riemann sum, things become much easier. But before I start doing the problem directly I’ll show you a result related to even/odd functions:

Let f(x) be any function. We can write f(x) in an alternate way-

And use this to compute our integral above. Letting f(x) be the integrand-

And Voila! We have found what we set out to. This is why I said that it would become one of the easiest integrals ever if you just did the necessary manipulations. The idea was to represent a function as the sum of an even and odd function- that you can do such a manipulation to any function is a testament to the beauty of Math. And you could do the above trick for any f(x) instead of the e^(1/x) in the denominator as long as f(x)f(-x) = 1. Also if we had integrate the same integrand from -A to A for all A in the domain of f we would always get sin(A), which is something fantastic in itself.

Now let’s look at another seemingly ‘difficult’ integration problem that becomes easy with some manipulations.

Wow! Such a complicated function to nothing. Literally nothing. You could look at the graph and get the intuition that the areas bounded by the x-axis and graph below and above the point (1,0) are equal. Still wonderful!

We can generalize this by applying the same trick to the more general case-

I suggest you try this definite integral problem with a cool trick of its own-

(Note the singularities of the integrand- asymptotes in graph of 1/(x³-1)- try taking limits of the indefinite integral before integrating it over the interval)

Now that we have done some really fascinating but easy definite integrals, let me show you something you can do with just the knowledge of integration techniques.

First consider integration by parts. We use this technique when the integrand appears to be the product of two functions whose derivatives we know. But we also use the technique in the case when the integrand is difficult to integrate.

Consider this: Wikipedia- ln x integration

But what if we do a similar thing for any function f(x)? Integration by parts states-

I leave proving the conjecture up to you. I proved it using First principle of Finite induction since it is conjectured to be true for all positive integers n.

But what happens if we continue integrating by parts again and again? Let us limit the above expression on right-hand side with n approaching infinity-

And we have shown the integral as a power series! Just note the integration constant C that I added at last. It is necessary for the indefinite integral for a general answer.

This brings me to the conclusion of this article. Please go check out my other articles on the main website and go read ‘Inside Interesting Integrals’ by Paul Nahin.

Clap, comment, share and follow my blog Cosmus Mathematicus!

Wonders of Integral Calculus was originally published in Cosmus Mathematicus on Medium, where people are continuing the conversation by highlighting and responding to this story.

Leonhard Euler was the most distinguished and popular 18th century mathematician. His contributions to Mathematics are as profound as they are diverse- he worked on number theory, algebra, analysis, geometry, statistics, linear algebra, applied mathematics, and so many other subjects within the discipline. Just to give you an idea of how prolific Euler was: He started writing for the journal of St. Petersburg academy from 1727, and continued doing so till he died in 1783; however, the journal continued to publish Euler’s articles for 48 years after his death!

Euler is too tall a giant for me, or for most other math enthusiasts to stand on the shoulders of. Of him the great French scholar and mathematician Pierre-Simon Laplace said: “Read Euler, read Euler, he is the master of us all”.

Like my last article, this one too is inspired by a book I am reading currently. It’s called “Euler: The Master of Us All” written by Professor William Dunham. I recommend everybody reading this tiny article read the book too, because the book contains many formulas and theorems Euler derived, although even these “many” are a small proportion of Euler’s achievements.

Euler ‘s biggest achievements are in analytic number theory and infinite series. Let me start with the definition of a the most well-known function in the subject, known as the Riemann zeta (or Euler-Riemann zeta) function:

Where ‘s’ is any complex variable (if imaginary part is zero it is real) . Euler was interestingly the first one to study such types of functions (for even positive integers). His association with it started with the famous “Basel’s problem”, where one was asked to find the sum of the function when s=2. This problem had challenged the Bernoulli's even. But alas, Euler found out the following result, amazing and yet completely non-intuitive:

Euler’s first proof for this was a totally non-rigorous one and at that time was not completely accepted by mathematicians, although he did receive a lot of acclaim. I’ll explain the non-rigorous proof because it is easier than Euler’s rigorous proofs to the result. To prove this, let us introduce two infinite expansions for sine of an angle:-

Now using the two Euler did something amazing. He first expanded the second infinite product and equated it to the first divided by x:

By equating the coefficients for x² from the two sides of this equation he got:

And from this Leonhard Euler concluded in dramatic fashion-

Similarly Euler went ahead and proved results for the function for 4,6,8,etc.

This was later generalized for even positive integers using Bernoulli numbers-

For all positive integers or natural ‘n’ and where Bernoulli numbers are explicitly defined as follows:-

You can also try out some other really cool stuff Euler proved in his papers, like an expression for the Zeta function using primes(which is divergent for s=1, first shown by Jacob Bernoulli) and the famous Wallis product for pi:

Now let us try another of Euler’s adventurous proofs- one in the field of Number Theory, something , much easier, that related perfect numbers to Mersenne primes, known as the Euclid-Euler theorem.

Euclid has proved in his proved the following result (Euclid, Prop.IX.36):-

And from this Alhazen conjectured that every even perfect number is of this form. It was only Leonhard Euler who proved this conjecture and in his proof he used the divisor function for positive integers which is defined as follows:

Suppose N is even and perfect. Factor out all powers of 2 to write N as follows:

Since N is even k > 1. Also because N is a perfect positive integer we know:

At the same time since 2^k-1 & b are co-prime, using a property of co-primes:

Now using the previous two equations for the divisor function of N we get:

The left-hand side is in its simplest form (or in lowest terms) but the right-hand side is not yet clear, so we can write that for some c ≥1 we have:

Now what Euler decided to do was to consider the two cases for c(> or =1):

c > 1

Each of the whole numbers 1,b,c,-1+2^k divides b. Now considering pairwise equalities for these numbers we get:

1. 1 ≠ b for otherwise N is 2^(k-1) which is not possible because no power of 2 can be perfect.
2. 1 ≠ c for we have stipulated c to be c > 1.
3. 1 ≠ (2^k)-1 for otherwise N = b which makes N odd but N is even.
4. b ≠ c for if these were equal then 1 = (2^k)-1 which is not true.
5. b ≠ (2^k)-1 for otherwise b=cb which implies c=1.
6. c ≠ (2^k)-1 for otherwise b=c² but the sum of divisors of b is c² + c and in this case the sum is more than or equal to c² + c +1 which is absurd.

Therefore the numbers 1,b,c,(2^k)-1 are four different divisors of b. Thus:

Therefore this case is entirely false since this is a contradiction. Then considering c=1-

c = 1

This implies the only divisors of b are 1 and b, or in other words, b is prime.

Thus we have shown that if N is an even perfect number, then N = (2^(k-1))b

⇒ N=(2^(k-1))*((2^k)-1) where (2^k)-1 is prime. The necessity of Euclid’s condition is thus established and hence Euler proved it.

Euler did so many things in every field of Mathematics that there is no space big enough to fill in all of his achievement. I suggest everybody reading my article to not only read Prof. Dunham’s book but also try catching hold of a translated version of Introductio in Analysin Infinitorium and also Opera Omnia which according to me summarize who Euler really was- an adventurous scientist and carefree explorer. He gave non-rigorous proofs to so many infinite expansions without the use of Calculus and used those in his papers on Calculus thus independently proving a lot of great Math theorems.

I really wish somebody gets inspired by this article like I did and reads his work. Truly, Euler is the “Master of us all”!

Please Clap, Comment, and Follow my blog. It would mean a lot to me if you could share this article. If you love Mathematics, you will love Cosmus Mathematicus!

Ode to Euler was originally published in Cosmus Mathematicus on Medium, where people are continuing the conversation by highlighting and responding to this story.

A lot of Math-enthusiasts dislike doing too much of geometry and perhaps that is justified because before high school,geometry is mostly drawing and measurements. However, what makes trigonometry different is the amount of algebra involved, encountered by students mostly in high school- right from mundane sine-cosine identities to infinite series and calculus. It is this fusion of algebra and geometry that makes trigonometry so interesting. This article discusses few contributions of great Mathematicians, namely François Viète, Leonhard Euler and Brahmagupta- all of whom proved some really “cool” formulas and identities in Trigonometry.

François Viète

François Viète was a French Mathematician who gave the world some really beautiful proofs. He is mentioned extensively in Trigonometric Delights by Eli Maor (a book I just finished reading recently). One of his biggest achievements was to provide a trigonometric solution to special cubic equations. I had given a challenge problem in my last article related to the same, and let me describe the method shortly.

Consider the depressed cubic equation x³+Hx+G=0

Let x = u*cos θ (u is some constant we don’t know yet)

=> u³cos³ θ+Hucos θ+G=0

Dividing both sides of the equation by u³/4 gives:

4cos³ θ + (4H/u²)cos θ + (4G/u³) = 0 (we assume that u is not zero since G is some non-zero constant in the case)

Using this we can find u if we attempt to coincide this with the identity -

cos(3θ)=4cos³θ- 3cosθ, we equate 4H/u² and minus 3 to get:

u = ±2 √-H/3 => cos (3θ) =(3G√3)/2H √H

Thus we can solve for θ and hence find x.(However here the method only works if -1 <(3G√3)/2H √H < 1)

Viete was so adept at such transformations of algebraic equations to trigonometric equations that once when the Dutch Ambassador to France challenged King Henry IV with a 45th degree equation, Viete came up with 23 solutions to the problem. This served him well especially because the ambassador boasted that not a single Frenchman could solve it. Let me describe the method below-

The equation was:

x⁴⁵-45x⁴³+945x⁴¹-12300x³⁹+……..+95634x⁵-3795x³+45x=c for some constant ‘c’.

Now let c=2sin(45θ), z=2sin(15θ), y=2sin(5θ), x=2sinθ

Using the sin3θ= 3sinθ-4sin³θ identity, we get c=6z-8z³(let it be equation 1) and then using the same identity z=6y-8y³ (let it be equation 2).

Also since sin5θ = 16sin⁵θ-20sin³θ+5sinθ, we get y=x⁵-5x³+5x (Equation 3)

Substituting equation 3 for y in equation 2 to get z in terms of x and then substituting this new equation for z in equation 1 to get c in terms of x gives us exactly the original equation! (Unfortunately, back then trigonometric ratios were taken to be positive so Viete could only find 23 solutions). The complete set of solutions is given by:

Another magnificent achievement of Viete was the discovery of the remarkable formula:

But how did Viete come up with this?

Viete discovered this in 1593 and in proving it he used a geometric argument based on the ratio of areas of regular polygons of n and 2n sides inscribed in the same circle. This marks a milestone in mathematical history, as noted by Eli Maor in his book, because it was the first time “ an infinite process was explicitly written as a succession of algebraic operations”.

Leonhard Euler

Not much can be said of Euler, who undoubtedly is one of the greatest of geniuses to have stepped on planet earth. The only person comparable to Euler is Carl Gauss, but Euler’s achievements are nothing short of transcendental. Notable achievements include the discovery of the number e and also the formula e^(π*i) = -1. Some of his less known work includes a unique trigonometric proof of Heron’s formula for Area of triangle.

This proof is in fact unique to Euler- the very idea of taking tangents of all half-angles in a triangle and taking radius of in-circle to derive one of the most popular mathematical formulas is just ingenious. Euler still has a lot to be credited for- this article is just too small a space to even include another of his humongous feats.

Brahmagupta

Brahmagupta was the first person to compute rules for dealing with zero and also one of the first people to provide a general solution (although incomplete) to quadratic equations. His work on Diophantine equations are well noted by mathematicians the world over. One of his remarkable achievements was the discovery of the formula for the area of a cyclic quadrilateral. What really makes it remarkable is its similarity to the Heron’s formula (reason being that Heron’s formula is actually a special case of Brahmagupta’s formula), and perhaps trigonometry provides with the easiest, if not the most elegant method to prove it. The formula goes as follows:

Where S is the semi perimeter and a,b,c,d the sides. Now let’s try to prove this using simple trigonometric identities. What makes a cyclic quadrilateral different from other shapes? Well, the obvious answer would be that the opposite angles add up to 180 degrees, and in fact, that is the most fundamental idea to proving this. Let’s build on this-

Notice how very simple identities of trigonometry like cosine rule and sine rule (for ares of triangles) lead to such an interesting result. As you will see in the challenge problem, this can be generalized for all quadrilaterals too.

Challenge Problems

Conclusion

Trigonometry has developed as field of Mathematics over the years. Initially it was limited to spherical shapes during Ancient Greece but later the world saw the development of flat-plane trigonometry. With this mathematicians began to find links between trigonometry and algebraic equations. Today, it has deep applications in engineering and astronomy. You’ll even encounter sines and cosines in complex analysis, differential geometry, etc.

Hope you liked this article. Please clap, share and comment and please follow my publication for more interesting articles on Mathematics!

Viète, Euler and Brahmagupta: Beauty of Trigonometry was originally published in Cosmus Mathematicus on Medium, where people are continuing the conversation by highlighting and responding to this story.

Algebra is said to be the link between most of Mathematics, and dates back to the Ancient Babylonians and Egyptians. It was treated as a formal science for the first time by the Greeks, of whom the great Mathematician Diophantus (3rd century AD) is sometimes called the “Father of Algebra”. The study later flourished at the hands of the Indians, the Chinese, the medieval Arabs and later the Europeans of the Renaissance Era and Modern Era.

Most of Algebra is concerned with expressions, identities and equations, and it is the latter we are interested with at the moment. One of the special forms of equations are equations in the third degree, more commonly known as cubic equations. The early Greeks and Indians made great advances while dealing with these, but most of what we know of cubic equations is from 15th century onward, when the general solution to these equations was discussed in great detail. Our job is to do just that: attempt at solving a cubic equation of any kind.

What is a Cubic Equation?

Any equation of the form ax³+bx²+cx+d=0 where a,b,c,d are all real numbers such that a is not zero are considered as cubic equations. Thus the only difference between a cubic equation and a quadratic equation (equations in 2nd degree) is that extra term ax³ which really does change the equation entirely. You may realize this from the graphs we shall study later of cubic functions and try comparing them with any quadratic function using your graphing calculator.

What earlier mathematicians found strange was their inability to solve these equations, primarily because it is nearly impossible to solve them using simple Euclidean geometry, unlike quadratic equations, for which the Persian Mathematician Al-Khwarizmi found a general solution to by completion of squares. His method was used later by René Descartes to arrive at:

However if one tries to find a general solution to a cubic, she may meet dead ends because it requires prerequisite knowledge of complex numbers. Amazingly, the Persian poet Omar Khayyam came up with a method to solve a cubic equation using geometrical analysis, but it was only in the 16th century that an Italian genius named Scipione del Ferro was able to solve for cubic equations of a special type known as depressed cubics, which we shall discuss shortly.

After del Ferro’s death, it was Niccolò Tartaglia who took forward the study of cubic equations by arriving at a general solution. Another Italian named Girolamo Cardano is claimed to have pressed Tartaglia into revealing the secret method (in around 1530), and promised him that he would never publish it. Alas Cardano broke the promise, and although it still remains a controversy as to whether whose method it actually was, today the method to solve a cubic equation is known as Cardano’s method.

Cardano’s Method

Consider the general cubic ax³+bx²+cx+d=0. In order to solve it easily, lets try to take the bx² term out. Transforming the left-hand side of the equation as a function on the Cartesian plane tells us that the second and third terms are responsible for horizontal transformation of the graph. If we can convert this to the form x³+3Hx+g=0, things become much easier. This form of a cubic is called a depressed cubic equation.

Perhaps one of Del-Ferro, Cardano and Tartaglia realized this and thus transformed the cubic by replacing x with the x-coordinate of the point of inflection of the cubic (shown below).

If you know some differential calculus, maybe you’d know this is the zero of the 2nd derivative of any cubic.

(The fact that the x-coordinate was the zero of the 2nd derivative wasn’t known to either of the Italians back then since Calculus was only developed by Isaac Newton and Gottfried Leibniz later in the 17th century. So if you don’t know Calculus, it’s not a problem at all!)

Thus the x-coordinate was found out to be x=(-b/3a) for any cubic. Using this they replaced x: x= y- (b/3a). When this substitution was used in ax³+bx²+cx+d=0, the equation was transformed into: y³+3Hy+G=0 where H and G are: H= (c/3a)-(b²/9a²) and G=(d/a)+(2b³/27a³)-(bc/3a²).

Let y=s-t for some numbers s and t such that H=st. This implies:

G= -(s-t)³-3st(s-t) => G=t³-s³+3st(s-t)-3st(s-t) => G= t³-s³

This may take some time to get comprehended because the step of taking y as the difference between any two numbers is just arbitrary. Perhaps somebody noticed a similarity between the depressed cubic equation and identity used to find (a+b)³ and randomly thought of this. Mathematicians therefore are artists in their own right!

Now H=st => s=H/t => G= t³- (H/t)³ => t⁶-Gt³-H³=0

This is a quadratic equation in t³ and can be solved using the quadratic formula:

Thus from the values of t³ and s³ we can find s and t and solve for y. Therefore, we can solve for x.

Discriminant of a Cubic

In Cardano’s Method discussed above, we haven’t really gone further to find the formula to find the solutions to the general cubic since that requires an understanding of complex numbers (If you are aware of primitive roots of unity then go ahead!). However this tells us that a cubic equation must have complex roots, also because not all cubic functions plotted on the Cartesian plane intersect the x-axis.

Consider the depressed cubic x³-3Hx-G=y

So solutions to the cubic can be solved for by finding the points of intersection of the curve y=x³-Hx and the line y=2Hx+G. When the former is sketched, it would look like the graph on right.

It could look like other similar curves as well, but let’s just stick to this one.

Call curve y=x³-Hx as C and line y=2Hx+G as L. C and L meet at least once, so depressed cubic must have at least one real root.

When L passes through one of C’s stationary points or vertices, that is the maximum or minimum point (look at the point where C is about to turn or change direction in the graph), the cubic has a repeated root.

So when:

(If you do not understand the notation, which uses differential calculus, you need not worry. All it means is that the slope of the line tangent to the curve at a stationary point equals to plus or minus square root of H by 3. If you still seem unconvinced or seem interested you are welcome to explore derivatives of functions.)

So if L meets C at one of the stationary points,

This implies that if depressed cubic has three real solutions, L is bound between lines passing through the two stationary points of C. Thus,

And for C to meet L at one of the stationary points, 64H³-27G²=0

Therefore, x³-3Hx-G=0 has:

1) 1 real root and 2 complex/imaginary roots if Δ<0

2) 1 real root and 2 real, equal roots if Δ=0

3) 3 distinct real roots if Δ>0

Where Δ=64H³-27G² is the discriminant. Since the nature of the roots of the general cubic also depend on the nature of the roots of the depressed cubic, we can determine nature of roots of any cubic equation using this discriminant. This method that we used falls under a Mathematical discipline called Analytic Geometry.

An Alternate Representation

Can we represent a cubic expression as the sum of two cubes? This thought may have struck your head after you noticed a similarity between a depressed cubic and the identity for (a+b)³. Let’s try and find this out.

This implies that: y = A(x- λ)³+B(x- μ)³

Thus we have represented the cubic expression as a sum of two cubes. Also,

Therefore, we also conclude that λ and μ are the roots of the equation:

Challenge Question:

Conclusion

Mathematics is as old as human civilisation itself. Human beings have come across so many different patterns in nature and while measuring quantities in recorded history. Cubic equations are just one of the many beautiful things we have stumbled upon and luckily we know how to solve such types of equations.

It is wonderful to know that Cardano’s method is not the only method that exists which can be used to solve a cubic equation. Rafael Bombelli, François Viète, Lodovico Ferrari and so many others found out new and innovative methods to solve cubic equations. So it is always important to look at the same problem in more than one way so that Mathematics is explored with rigor. We have only discussed a few aspects of cubic equations here. There are in fact so many research papers and articles on them that the internet is filled with, but what matters here is how much one is willing to go through to enjoy the processes involved in Mathematics.

References

Please Like (or Clap), Comment and Follow this page if you love Math like me. Also if you’re interested, please do check out Vieta’s trigonometric solution to certain types of cubic equations and Euler’s solution too. (Geniuses they are, aren’t they?)

Cubic Equations: Beauty of Algebra was originally published in Cosmus Mathematicus on Medium, where people are continuing the conversation by highlighting and responding to this story.