A Bicycle, a Lesson, and Arithmetic Progression
We will begin our discussion with a story.
Ritesh Kumar approached his father, Mr. Kumar, requesting a bicycle. As a mathematician, Mr. Kumar told him he would need to earn it through effort. He proposed a daily allowance starting at one rupee, increasing by one rupee each day, contingent on completing his homework.
Initially, Ritesh thought this was unfair, realizing it would take too long to gather the 1770 rupees needed for the bicycle. However, he found a motivational book on his pillow the next morning, which sparked his interest in a chapter about "Arithmetic Progression." He soon realized that by completing his homework diligently for less than two months, he could earn the money, understanding the lesson his father intended to teach him.
Now, let us learn that lesson ourselves.
Arithmetic Progression
An Arithmetic Progression is nothing but a list of numbers (which can represent anything), in which each term is obtained by adding a fixed number, called the Common Difference, to the previous term(except for the first term). To construct an Arithmetic Progression, initiate with the specified first term, increment it by the designated common difference, and continue this procedure iteratively.
Some examples of Arithemetic progression are
- \(1,2,3,4,5,6,7,8,9,\cdots\)
First terms is 1 and common difference is 1.
- \(2,4,6,8,10,12,14,16,\cdots\)
First term is 2 and common difference is 2.
- \(3,9,15,21,\cdots\)
First term is 3 and common difference is 6.
We are going to analyze how Ritesh solved his problem by formulating this progression. For our discussion, we will denote the n-th term of the progression as \(a_n\), the common difference as \(d\), and the sum of the first n terms of the progression as \(S_n\). I will now present two important results.
I encourage readers to engage with these formulas personally and to attempt to prove them, even if it takes several weeks of dedicated effort. The time you invest is undoubtedly worthwhile. The inherent beauty of mathematics is that once you begin to appreciate the subject truly, it becomes a constant presence in your thoughts. You may find yourself contemplating mathematical problems while dining, navigating the streets, or during other daily activities. Therefore, pursuing mathematics is a valuable endeavor.
Only after you have made your best effort should you seek proof. This detailed proof proof may not align with your conclusions. Another beautiful aspect of mathematics is that there are often multiple solutions to every problem, yet every path ultimately leads to the same outcome.
Returning to the original problem, Ritesh needed a total of 1,770 rupees. He began earning 1 rupee on the first day, and his earnings increased by 1 rupee each subsequent day. To determine how many days it would take for him to accumulate 1,770 rupees, we can use the formula for the sum of an arithmetic sequence. This will help us find the number of days needed to reach that total. Using equation 2 we can write
\[\frac{n(n+1)}{2}=1770\]
This is a quadratic equation, which we will learn more about in this article in section titled Quadratic Equation . I recommend that anyone unfamiliar with solving quadratic equations read the relevant section. Solving the quadratic equation, the meaningful result we get is \(n=59\), which is less than two months.
Well, in another scenario, Mr. Kumar could also have said that he would double the amount every day, starting with giving 1 rupee on the first day. Then the progression would have looked something like this \[1,2,4,8,16,32,\cdots\]
This type of progression is known as geometric progression and closely resembles the process of deriving the formulas for the sum and the nth term of an arithmetic progression. Therefore, I will now transition to more complex and interesting progressions and sequences. I encourage readers to invest time in exploring geometric progression, attempting to formulate the relevant equations, and verifying them on their own. For those seeking a deeper understanding of Geometric progression or who want to check their derived formulas, additional resources are available for further study.
Harmonic Progression
I want to draw the reader’s attention to a more intricate concept. A Harmonic Progression is defined as a sequence in which the reciprocals of each term form an Arithmetic Progression. Or in mathematical terms, a general harmonic progression can be represented as follows. \[\frac{1}{a_1},\frac{1}{a_1+d},\frac{1}{a_1+2d},\frac{1}{a_1+3d},\cdots\] As one can see, the reciprocals of each term constitute an Arithmetic Progression, where the first term is denoted as \(a_1\) and the common difference is represented by \(d\). It must now be apparent that the general terms for the Harmonic progression is
where \(\frac{1}{a_1}\) is the first term of the Harmonic Progression.
In discussing harmonic progression, an intriguing aspect emerges when compared to arithmetic and geometric progressions. In the latter two cases, one can easily derive a formula for the sum of the first n terms. However, a general formula for computing the sum of n terms in a harmonic progression, given the first term and common difference, does not exist.
I have experienced numerous sleepless nights grappling with the realization that such a formula does not exist. I have dedicated extensive time and effort to this pursuit without achieving success. I encourage readers to engage in this exploration themselves, as one may uncover unexpected insights.
As one of the most prominent pure mathematicians of the 21st century, Manjul Bhargava remarked, "Some of the most prominent formulas in mathematics are found just because one was looking for the most beautiful mathematics out there." This statement underscores the intrinsic beauty that can be discovered within mathematical inquiry.
Though there are some practical uses of harmonic progression. The most famous one is the solution for block stacking problem.
The problem says -
"Place N identical rigid rectangular blocks in a stable stack on a table edge in such a way as to maximize the overhang."
Since this problem involves dealing with physical concepts like gravitational forces, centre of mass, and torques. It is not a mathematician's responsibility to do these; it's physicists.
All I can do is this- Proof of solution of single-wide variant .By clicking on the link provided, you will be directed to the relevant article on Wikipedia, where you can explore the application of harmonic progression in determining the general solution for this topic.
In forthcoming sections, we will explore the underlying reasons for the absence of a summation formula in harmonic progression. Additionally, it is noteworthy that while geometric progression permits a general formula for any infinite geometric series, neither arithmetic nor harmonic progression offers a similar capability.
Before jumping on to discovering more series and sequences, i recommend readers who are not much familiar with the definition of series, sequences, converging and diverging series to read this short page on A General Talk on Sequences and Series .And also be familiar with the concept infnity.
Some interesting infinite series
As we saw how infinty can not be treated as a number like we treat the other numbers. Before proceeding any further I want all the readers to first involve themselves with this.
The Question is -
What is the sum of all natural numbers. Or in other words
what is the value of S?
I want all the readers to atleast try to see the behavior and the properties of this series. We will return back to this after a little discussion on some other series.
Let's look at the following series :-
In the standard, rigorous definition of convergence, the series discussed herein are all divergent. The values assigned to them are derived from alternative summation methods that extend the concept of a sum to series that do not converge in the traditional sense. The following sections present the common heuristic, or non-rigorous, arguments used to arrive at these famous results. These manipulations are not valid under the standard rules of algebra for infinite series but provide an intuitive path to the values obtained through more advanced methods.
Though these series do not converge in the classical sense, using summation methods like Cesàro summation, Abel summation, or analytic continuation, we can assign finite values to them. For example, for the series \( S_2 \) , we have: \[ 1 - S_2 = S_2 \implies \boxed{S_2 = \frac{1}{2}} \] Next, let \( S \) denote the sum of natural numbers: \[ S = 1 + 2 + 3 + 4 + \cdots \] Adding \( S \) and \( S_5 \) term by term gives: \[ S + S_5 = (1+1) + (2-2) + (3+3) + (4-4) + \cdots = 2 + 0 + 6 + 0 + \cdots = 4 S \] Rearranging, we find: \[ S + S_5 = 4 S \implies S_5 = 3 S \] From analytic continuation, we assign: \[ \boxed{S = -\frac{1}{12}} \quad \] Thus, \[ \boxed{S_5 = \frac{1}{4}} \quad \text{(based on summation method conventions)} \] Finally, since \( S_3 \) is twice the sum of natural numbers, \[ S_3 = 2 S = 2 \times \left(-\frac{1}{12}\right) = \boxed{-\frac{1}{6}} \]
Try to find \(S_5\) yourself. There are a lot of stories related to these series. I can't tell them all here, but I encourage the readers to read and know about them.
Some interesting sequences
In this discussion, we will explore various sequences that manifest in nature and numerical patterns in unexpected ways.
Let's first start with the following question?
Question 7.1
Prove that if the \(n^\text{th}\) term of the sequence
\(\) is \(\frac{p}{q}\) , then
And also that the sequence contains all the rational number infinitely many times.
I encourage readers to explore this question on their own before reviewing any proof. If you discover an alternative proof, please share it—such discoveries are often enlightening!
Further details regarding the sequence in equation 10 will be discussed in the section on The Proof of Question 7.1.
Let’s check out a cool mathematical sequence that really shows off beauty in nature. You can find this sequence in all sorts of places, like the twists of seashells and the designs of flowers. It’s caught the attention of everyone from ancient Greek math whizzes to groundbreaking scientists, innovative architects from the 20th century, and even in a hit movie thriller.
What’s amazing about this sequence is that it pops up everywhere. It’s not just a neat math thing; It is found in the writings of an ancient Greek mathematician, the musings of a revolutionary cosmological scientist, the designs of a twentieth-century architect, and the intrigue of a best-selling thriller novel-turned-movie blockbuster. People love to debate its significance and how it can be used.
Artists, architects, and many others get inspired by this idea, using it to make their work more beautiful and appealing. Let’s dive into this fascinating world of math and see how it influences art, science, and nature!
I understand that many individuals may be inquiring about the nature of this sequence and how it can enhance their ability to create with greater artistry.
Fibonacci Sequence
This sequence is named after the famous Italian mathematician (Fibonacci) . However, Leonardo Fibonacci did not discover the Fibonacci sequence. Although he popularized it in Western Europe, the sequence was known to Indian mathematicians centuries earlier. Fibonacci introduced it to the Western world in his 1202 book "Liber Abaci", but it was studied in India as early as the 6th century BC by Pingala and later by Hemachandra.
The sequence is described as follows.
Can u tell me the formula for the general term of this sequence? Yes, that's right. The general terms of this sequence are related by the formula
The general term of the fibonacci series only in terms of n can be written as follows:
We will prove the above equation in two ways: First here we show the derivation and second proof by induction is given in the proof section.
The easiest derivation is by constructing a function as follows,
Let,
where \(a_n\) is the n-th term in the fibonacci sequence.
Since we know from equation 13 that \( a_{n+2} = a_{n+1} + a_n \text{ and } a_0 = 1 \text{ and } a_1 = 1\). Multiplying both sides by \(x^n\) and using a summation from zero to infinity, we get,
\[ \begin{align*} & \sum_{n=0}^{\infty} x^n a_{n+2} = \sum_{n=0}^{\infty} x^n a_{n+1} + \sum_{n=0}^{\infty} x^n a_n \\ \implies & \frac{1}{x^2}( f(x) - a_0 - a_1 x ) = \frac{1}{x} ( f(x) - a_0 ) + f(x) \\ \end{align*} \]Which gives us,
Let's assume
where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(x^2+x-1=0\).
Let \(\alpha = \frac{-1 + \sqrt{5}}{2}\) and \(\beta = \frac{-1 - \sqrt{5}}{2}\). We now solve for \(A\) and \(B\) using the equation 17 our assumed partial fraction form.
\begin{align*} \text{For } x = \alpha:& \quad -1 = A(\alpha - \beta) = A\sqrt{5} \implies A = -\frac{1}{\sqrt{5}} \\ \text{For } x = \beta:& \quad -1 = B(\beta - \alpha) = B(-\sqrt{5}) \implies B = \frac{1}{\sqrt{5}} \end{align*}Substituting \(A\) and \(B\) back into our assumed form for \(f(x)\):
\[ f(x) = \frac{1}{\sqrt{5}}\left(\frac{1}{x-\beta} - \frac{1}{x-\alpha}\right) \]To use the geometric series expansion, we rearrange the terms into the form \(\frac{C}{1-rx}\), noting that \(\frac{1}{\alpha} = \phi\) and \(\frac{1}{\beta} = 1 - \phi\), where \(\phi = \frac{1+\sqrt{5}}{2}\) is the golden ratio.
\[ f(x) = \frac{1}{\sqrt{5}} \left( \frac{1/\alpha}{1-x/\alpha} - \frac{1/\beta}{1-x/\beta} \right) = \frac{1}{\sqrt{5}}\left(\frac{\phi}{1 - \phi x} - \frac{1-\phi}{1 - (1-\phi)x}\right) \]Using the geometric series formula \(\frac{1}{1-y} = \sum_{n=0}^{\infty} y^n\), we can expand the fractions:
\[ f(x) = \frac{1}{\sqrt{5}} \left( \sum_{n=0}^{\infty} \phi^{n+1} x^n - \sum_{n=0}^{\infty} (1-\phi)^{n+1} x^n \right) \]By equating the coefficients of \(x^n\) with \(a_n\), we arrive at the closed-form expression for the sequence starting with \(a_0=1, a_1=1\):
\[ a_n = \frac{1}{\sqrt{5}}\left[ \left(\frac{1 + \sqrt{5}}{2}\right)^{n+1} - \left(\frac{1 - \sqrt{5}}{2}\right)^{n+1} \right] \]This formula allows you to calculate the \(n\)-th term of the sequence directly.
The first interesting thing about the Fibonacci sequence is: Almost every flower follow this pattern, as the number of petals they have always happens to be one of the terms of this Fibonacci sequence.
🌸 A Task for the Readers
Please visit a nearby garden and pick a flower of one of the following types:
🌼 Happy flower picking!
Finding the value of golden ratio
In the sequence, it seems that there is no clear pattern, except for what is indicated in equation 7. However, mathematicians are inherently curious and enjoy exploring mathematical concepts. They continuously dig deeper until they find elegant relationships and patterns. They have noticed that, as we move further along in the series, the ratio of two preceding terms approaches a fascinating value.
Let's determine the value of \(\frac{a_{n+1}}{a_n}\) as n approaches infinity.{ In calculus, we commonly use the symbol "lim" to address these types of problems. However, in this instance, we will proceed without the concept of limits and use a more intuitive approach to compute our value.}
Using equation 7, we can write:
And when \( n \) is large enough, we can write:
Where \( r \) is the ratio of two preceding terms as \( n \) approaches infinity.
This gives us a quadratic equation:
Solving this quadratic equation yields:
We give this irrational number a new symbol and name: the golden ratio, denoted by \( \phi \). Approximating:
\[ \begin{equation}\tag{19} \phi = 1.61803398875\cdots \end{equation} \]Let’s see if this is correct:
\[ \begin{align*} \frac{2}{1} &= 2 \\ \frac{3}{2} &= 1.5 \\ \frac{5}{3} &= 1.666\cdots \\ \frac{8}{5} &= 1.6\cdots \\ \frac{13}{8} &= 1.625 \\ \frac{21}{13} &= 1.615\cdots \\ \frac{34}{21} &= 1.619\cdots \\ \frac{55}{34} &= 1.617 \end{align*} \]
And so on. As we progress further, we get closer to the value of \( \phi \). It takes the ratio of the 11th and 10th terms to achieve the third digit correct after the decimal of \( \phi \) , which is \( \frac{88}{55} = 1.61818\cdots \). Additionally, it takes the ratio of the 20th and 19th terms to achieve 10 correct digits of \( \phi \) after the decimal point.
The golden ratio is used in every field and if you are wondering why this page is looking beautiful in terms of dimensions? The answer is because i have used the golden ratio throughout the designing this website.
Pascal's Triangle
Blaise Pascal developed Pascal's triangle, which is useful in many cases; however, we will discuss only one application here. It is also worthwhile to know the history of this triangle.
The pascal's triangle looks something like this.
The rule is simple: for each row, start with 1, then add the two terms just above it from the preceding row, and end with 1.
Blaise Pascal was clever enough to notice that this triangle follows the very same pattern as the coefficients of the expansion of \((x+y)^n\).
Let's see the pattern ourself.
\[ \begin{align*} (x+y)^0 &= 1 \\ (x+y)^1 &= x+y \\ (x+y)^2 &= x^2+2xy+y^2 \\ (x+y)^3 &= x^3+3x^2y+3xy^2+y^3 \\ (x+y)^4 &= x^4+4x^3y+6x^2y^2+4xy^3+y^4 \\ (x+y)^5 &= x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5 \\ (x+y)^6 &= x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6 \\ (x+y)^7 &= x^7+7x^6y+21x^5y^2+35x^4y^3+35x^3y^4+21x^2y^5+7xy^6+y^7 \\ (x+y)^8 &= x^8+8x^7y+28x^6y^2+56x^5y^3+70x^4y^4+56x^3y^5+28x^2y^6+8xy^7+y^8 \\ (x+y)^9 &= x^9+9x^8y+36x^7y^2+84x^6y^3+126x^5y^4+126x^4y^5+84x^3y^6+36x^2y^7+9xy^8+y^9 \end{align*} \]As one can see that the coefficient of expansion of \((x+y)^n\) resembles the nth row of the Pascal's triangle.
There's two nice way to formulate the binomial expansion. One is given here and the other is talked in the calculus section.
We define a function for the rth column and nth row of the pascals triangle as \(F(r,n)\). This funtion also tells the rth term coefficient in the expansion of \((x+y)^n\). By observing the pascals triangle, one can easily write the following:
Why we can write this. Because thats basically how we define the rule for writing out the pascal's triangle( see the rule above).
Again i would definitely encourage the reader to try and work out a well defined function that behaves and satisfies all the conditions in equation 20. But everyone familiar with the notation of \(\binom{n}{r}\) or sometimes also written as \({}^{n}C_{r}\), might know this is one that satisfies the above requirement in equation 19. And is defined as
The following property about the \(\binom{n}{r}\) can be confirmed by the readers:
As it can be seen that the equation 22 satisfies the property or condition in equation 20.
Thus using the above equations and pascal's triangle, one can write the general expansion of \((x+y)^n\).
Quadratic Equation
Now we are going to discuss quadratic equations. As seen in the article until now, it is that one concept that can pop up while dealing with other concepts as well.
So a quadratic equation is just a special kind of polynomial (to know more about polynomials, go to the function section) with degree 2. Or in simpler words, any equation that can be written in the following form is a quadratic equation.
This equation must have seemed quite similar to most of the readers, as anyone who has passed the eighth standard all around the world knows this equation and also the solution of the roots of this equation. By roots of the equation, we mean to find such values of x that make the value of the equation zero, that is, y = 0.
The equation in equation (24) can also be written in terms of its roots. But how?
Let \(\alpha\) be one of the roots of the equation (24) . Thus, dividing the equation (24) by \(x-\alpha\) gives us the following equation,
Since \(\alpha\) is a root of the equation (24) , that means \(c+(b\alpha+a\alpha^2)=0\) . This tells us that
If we assume \(\beta\) to be the second root of the equation, then we can write, using the above equation, the following relation between roots of the quadratic equation,
Squaring both sides gives,
\[ \begin{align*} \alpha^2+2\alpha\beta+\beta^2 &= \frac{b^2}{a^2} \\ (a\alpha^2+b\alpha+c)+2a\alpha\beta+(a\beta^2+b\beta+c) &= \frac{b^2}{a}+b(\alpha+\beta)+2c \\ \end{align*} \]Using equation (27) equation (27) and the fact that \(\alpha,\beta\) are the roots of the equation, we get,
Note:
This is intended to motivate and capture the reader's attention. I'm deliberately skipping steps to keep readers engaged. The focus should remain on the solutions, preventing any distractions.
It should be noticed that a quadratic equation can have at most two roots as \(\alpha \text{ and } \beta\). Their difference can be given by as follows
Good for us that we already know the value for \(\alpha+\beta\) and \(\alpha\beta\). Putting them in the above equation from equation (27) and (28), we get,
Adding and then subtracting equation (28) and (30) with each other, we get the roots of the equation in equation (24),
Playing with the quadratic equation
We are now going to see some interesting results that we can come up with using the quadratic equation.
So we know that,
\[ \begin{align*} a\alpha^2+b\alpha+c &= 0 \quad \text{and} , \quad a\beta^2+b\beta+c = 0 \end{align*} \]Now we are gonna do some simple algebra that ought to be understood by the readers.
\[ \begin{align*} & a\alpha^{n+2}+b\alpha^{n+1}+c\alpha^n = 0 \quad \text{ and } \quad a\beta^{n+2}+b\beta^{n+1}+c\beta^n = 0 \\ \implies & \quad a(\alpha^{n+2}+\beta^{n+2})+b(\alpha^{n+1}+\beta^{n+1})+c(\alpha^n+\beta^n) = 0 \end{align*} \]Now if we let, \((\alpha^n+\beta^n)=S_n\) , we can write the following equation,
The relation above found is sometimes called newton's identities. There are a lot more interesting relations and equation that can be seen by just by doing a little bit of algebra to the equation. Below is given such identity and its for the readers to find a derivation for these. (I encourage readers to find a derivation and not a general proof, though it might work as well to checck if the identity is correct or not. But i want the readers to go through the journey a mathematician goes through whilst finding out these identities and remember that every individual has got to choose his own unique journey).
Let, \(G(x)=\sum_{n=0}^{\infty} S_nx^n\) , where \(S_n\) , has the same meaning as in equation (31). Then,
Quadratic Graphs in Different Cases:
We are now going to see how the quadratic functions behave on the real plane. We write a quadratic function as follows,
Now if you notice in equation (30) the roots of the equation depends on all the coefficients. But if u notice there's a term \(\sqrt{b^2-4ac}\) . This can be considered as the most powerful term in the formula if we put it that way, not only because of the fact that it is the term that decides if the equation will have a real root or not, but also because it is contains all the terms of the equation. So if we change even a single coefficient, the behaviour and the roots changes.
So we will start of by letting, \((b^2-4ac)=D\) .
It is very easy to check that when,
Case 1: D>0, we get two distinct real roots.
Case 2: D<0, we can't get two distinct real roots, instead we get two distinct complex roots. (for those without the knowlege of imaginary numbers, i suggest you to read a little bit of the complex analysis section.)
Case 3: D=0, we get one real root.
\(y = ax^2 + bx + c, \text{ when } a>0, D>0\)
\(y = ax^2 + bx + c, \text{ when } a<0, D>0\)
\(y = ax^2 + bx + c, \text{ when } a>0, D=0\)
\( y = ax^2 + bx + c, \text{ when } a<0, D=0 \)
\(y = ax^2 + bx + c, \text{ when } a>0, D<0\)
\(y = ax^2 + bx + c, \text{ when } a<0, D<0 \)
I expect the readers to justify these graphs themselves.