The Presenting Problem

Recently I came across a video on YouTube that asked the question, “Which is bigger: 99! or 50⁹⁹?” There are so many such questions on YouTube and I wonder why. After giving it some thought, I realize the truth in the saying, “Give a woman a fish and you feed her for a day. Show a woman to fish and you feed her for a lifetime.” Ok, fine, the original had ‘man’ instead of ‘woman’! But you get the point. When we teach people how to solve a particular problem rather than show them the principles that would enable them to solve problems in general, we are giving them a fish that will last them a day.

So I’m going to treat problems like this under a single umbrella. You see, the question could have been, “Which is bigger: 101! or 51¹⁰¹?” or “Which is bigger: 199! or 100¹⁹⁹?” and the approach would have been exactly the same. Since I’m saying this, try to determine a pattern in the numbers involved in each of the questions before you proceed.

The Pattern

What we should be able to see in the pairs {99! 50⁹⁹}, {101! 51¹⁰¹}, and {199! 100¹⁹⁹} is that the number whose factorial forms the first part of the pair shows up as the exponent in the second part of the pair. And these are odd numbers. Also, the base in the second part is half of one more than the exponent (50 = (99 + 1)/2, 51 = (101 + 1)/2, and 100 = (199 + 1)/2). So we can generalize the question to, “Which is bigger (2n-1)! or n^(2n-1).

Breakdown

We can recognize that (2n-1)! and n^(2n-1) are each the products of 2n-1 numbers. Hence,

Now, the middle number in each product is n. Moreover, the sum of the first and last terms in each product is 2n. And the sum of the second and second last terms is also 2n. In fact, every pair made of numbers equidistant from the ends (or the center) will add up to 2n.

Let us consider the first pair from the ends. From (2n – 1)! we have 2n – 1 and 1 and from n^{2n – 1} we have n and n. Now, while the sum is the same for both pairs, the products are not the same.

In fact

Now, from the previous post (and elsewhere, of course!) we know that the product of two positive numbers or two negative numbers is a positive number. Hence,

as long as n is not 1. We can expand the square to get

Generalization

Now, if we chose a general pair (say, the k^th pair), we would have 2n – (2k – 1) and (2k – 1) from (2n – 1)! and n and n from n^{2n – 1}. Now if we take the products we get

Once again, let us consider the product of two numbers as below

Expanding the square we get

In other words, no matter which pair we pick, the product of the pair from n^{2n – 1}, in which all terms are n, will be greater than the product of the pair from (2n – 1)! It follows, therefore that

Hence, we can conclude that 99! < 50⁹⁹, that 101! < 51¹⁰¹, and that 199! < 100¹⁹⁹.

Stepping Back

What we did in looking at the above was identify patterns between as many numbers involved in the question as possible. Here, specifically four things were crucial. First, both sides consisted of products of n numbers. Second, even though I did not explicitly mention it, the factorial contains all natural numbers from 1 to a particular higher number (n). However, on the other side we had n copies of the same number, the base. Third, the sum of numbers equidistant from the ends was a constant. And fourth, the base happened to be half this sum.

Recognizing these patterns allowed us to frame the issue in terms of the product of pairs of numbers whose sum is constant. From there it was an issue of recognizing that we can use something we knew about products, namely that the product of two numbers having the same sign is always positive, to produce the inequality that would allow us to proceed to a solution.

Now, it must be noted that not all solutions will involve identical steps. However, for most problems pattern recognition is crucial. Further, as I have shown here and here, mathematics is a coherent body of knowledge. This means that things we learn in one area can be reframed to be applicable in another area. Hence, in this post, something that a student might learn in a chapter on inequalities turns out to be applicable to algebra in general and number theory in particular.

The willingness to ‘borrow’ knowledge from another area of mathematics to make it applicable in a new area is something that we realize is important as mathematics becomes more complex. And speaking of that which is complex, I will be starting a new series on complex numbers next week. But next week’s post will only be a beginning. Till then, try proving that, for all n > 3, n^{n + 1} > (n + 1)ⁿ. If you need something to point the way, the series on e might be helpful.

Inane Phrases

If you read my motivations page and the previous post, you will realize that my experience of teaching grade 6 students had an immense impact on me. While the previous post focused on BEDMAS (or BODMAS or PEMDAS, etc.), in the motivations page I also reveal that the students came to me with what I said were “some inane phrases to remember while performing operations.” Unfortunately, I have found these inane phrases all to prevalent even among students in high school.

You may be wondering what these inane phrases are. Well, for instance, you may hear a student say, “Minus and minus equals plus” or “Minus and plus equals minus.” Another student may say, “Positive and positive gives positive” or “Positive and negative gives negative.” Whatever in the world is this nonsense?

Most of you might, however, know what they are trying to say. However, without any proper context within which the phrases are to be interpreted, I could conclude, as many students do, that – 2 – 3 = +5. After all, “minus and minus equals plus,” right? Or I could insist that – 7 + 10 = – 3 since “minus and plus equals minus.”

Need for Precision and Clarity

What the phrases actually intend to communicate are “the product of two positive numbers is a positive number,” “the product of two negative numbers is a negative number,” and “the product of a positive number and a negative number is a negative number.” Of course, my phrases have 10, 10, and 14 words, while the original inaccurate phrases have 5 words each. The inaccurate phrases are certainly more economical in terms of the number of words they contain and the time it takes to say them. And don’t get me wrong. I’m all for saving time and effort.

However, I am not for saving time and effort at the cost of precision and accuracy and clarity. If the price exacted by the shorter phrases is the confusion among students about whether these are applicable only for multiplication or also for addition and subtraction, then I think the price is too high to pay.

Basic BEDMAS

But you may ask, “If these phrases are applicable only for multiplication, what rules do we follow when it comes to addition and subtraction?” This is a valid question because here too I have found students floundering. It’s not just that they have got the order of operations wrong, but also that they do not understand what is happening, especially when we are asked to add and subtract multiple numbers, some of which are positive and others of which are negative.

Let us consider one problem. Suppose we are asked to evaluate

Equipped with our understanding that the convention is to perform division and multiplication first, we obtain

Now what do we do? We begin by performing the calculations from left to right. What we first do is orient ourselves so we imagine we are on the number line facing the right (i.e. the positive direction). Now, we locate ourselves where the first number is. So we land on the position for -7.

Now all we have to do is move forward or backward the appropriate number of steps according to whether we encounter an addition or subtraction respectively. So, we first move forward 10 spaces, landing on +3, then back 3 spaces, landing on 0, and then forward 48 spaces, landing on +48.

BEDMAS with the Brackets

But what if we have something more complicated, say something that involves brackets? Consider the following

Here we have two separate parentheses to deal with. Since both are unrelated, in that there is a subtraction operator between the two, we can deal with both simultaneously. So we can first obtain

We can now complete our dealings with the parentheses to get

Now the –8 represents the negation of -8, which would give +8. Hence, we have

Now, the initial multiplication involves one positive number and one negative number, meaning that the product will be negative. The division involves two positive numbers and hence will yield a positive number. So we have

This gives the final answer as -2.

In the above example, we had obtained –8 and I stated that this was the negation of -8. We could have also considered

Here, instead of the negation of a negative number, we are considering the product of -1, which is inherent in the standalone negative sign, and -8, which gives us +8 since it is the product of two negative numbers.

BEDMAS ‘Brain Teasers’

Now, almost everyday I get some BEDMAS ‘brain teaser’ in my social media feeds. I don’t know why. Maybe my interest in mathematics related posts makes the idiotic algorithm ‘think’ that I would like such ‘brain teasers’. The reason I’m placing ‘brain teaser’ in inverted commas is because I do not think these problems qualify as brain teasers. They only check whether you know the rules of BEDMAS.

‘Brain Teaser’ 1

Anyway, one such ‘brain teaser’ is

We know we need to deal with the brackets first. To do this, we need to calculate 8 – 6 + 3, which equals 5. Hence, the problem reduces to

Since we only have multiplication and division with no grouping, we can proceed from left to right. This gives us

It is important to note that in the expression 5(8 – 6 + 3), we are not instructed to group the 5 and the value of the parentheses. Hence, the 5, which comes after the division sign, becomes the divisor, while the 5, which is the value of the parentheses, is a normal multiplicand.

‘Brain Teaser’ 2

How would we calculate the value of

Here, we observe that there is a parentheses that we need to deal with first, yielding

Now, we have an exponent, which we need to address next, giving us

Now we have to perform the multiplication to get

This leaves just addition and subtraction, which we can perform from left to right to get

which yields 14 as the final answer.

‘Brain Teaser’ 3

But what if we have something like

We recognize that the first term has a nested bracket. Hence, we need to first deal with the innermost bracket. This bracket contains an exponent, which we need to perform first. The second term has an exponent, which we will perform first in that term. The third and fourth terms have division, which we will perform at this stage. This gives us

Now, we have to perform the division in the innermost bracket of the first term. At the same time, we can evaluate the brackets of the second, third, and fourth terms. This gives

Now we can evaluate this innermost bracket of the first term. At the same time, we can evaluate the exponent on the second term. We now have

Evaluating the first bracket, we get

Now evaluating the exponent gives us

Observe that we have two division operators one after the other. When faced with this situation, we need to perform the operations one by one from left to right. So we first get

because the first division was 3÷1, which equals 3. Now we can perform the remaining division to get

Now we have only addition and subtraction, which we can perform from left to right to give us 9 as the final answer.

It’s a Wrap

What we can see is that following the BEDMAS rules carefully will yield an unambiguous answer. As mentioned in the previous post, the is nothing sacrosanct about BEDMAS. It is not inherent to any of the operations. However, this is the convention that has been adopted by mathematicians so as to remove confusion and to ensure that any particular arithmetic calculation yields one and only one answer. In these two posts I hope I have communicated the rules of BEDMAS well enough for anyone at or above grade 4 to understand. I hope the examples I have given serve as good illustrations of the process. But more than that, I hope I have communicated the fact that BEDMAS is a convention that could only have been agreed upon if there was a spirit of collaboration and camaraderie among mathematicians.

Order! Order!

As a mathematics teacher, I primarily teach students in high school, preferring, within this group, to teach grades 11 and 12. There are two reasons for this. First, mathematics taught in grades 11 and 12 is complex enough to allow for interesting nuancing of the ideas and also blending of the various areas like algebra, geometry, trigonometry, statistics and probability, and calculus. Prior to that the concepts are too superficial to allow for such unrestrained exploration. Second, I really do not know how to handle students who are younger than 14.

Anyway, one year I was feeling somewhat adventurous and agreed to teach students in grade 6. This was the academic year 2016-2017. The kids were, as expected, difficult to handle. However, what I was surprised with was the fact that most of them were very weak with order of operations. For those of you who are confused about this terminology, perhaps BODMAS or PEMDAS or some such six letter variation might prove to be a reminder.

In the above, ‘M’ stands for ‘multiplication’, ‘D’ for ‘division’, ‘A’ for ‘addition’, and ‘S’ for ‘subtraction’. Those are the letters that do not change, even though D and M often swap positions. We will see why shortly. However, the first two letters have some variety. The first letter varies between ‘B’ for ‘brackets’, ‘P’ for ‘parentheses’, and ‘G’ for ‘grouping’. The second letter varies between ‘O’ for ‘orders’, ‘E’ for ‘exponents’, and ‘I’ for ‘indices’. So, we can get 18 variations of these letters, all of which essentially give the same sequence for performing operations. However, whether we call it BODMAS or PEMDAS or GIDMAS, whatever does it mean? And, more importantly, why do we have such an ordering?

Mathematics aims for consistency. In other words, if 2 people perform the same set of operations accurately, we expect them to obtain the same answer. We will see how this desire for consistency is at the root of BEDMAS. I will use BEDMAS because I think ‘O’ for ‘orders’ is weird since we never call the exponentiation operation by that name anywhere else. Also, ‘G’ for ‘grouping’ is vague and is also not common terminology. So how does BEDMAS give us the consistency we aim for?

Testing for Consistency

We begin with the most basic operation – addition. And we know that 2 + 3 and 3 + 2 gives us 5. We call this property the commutative property. Hence, we say that addition is commutative. This means that, given any two numbers, say a and b, a + b = b + a. When we consider subtraction, however, we get a different result because 2 – 3 is not the same as 3 – 2. In fact, the first is -1 while the second is 1. We can actually generalize this to say that a – b and b – a are negatives of each other. We often say that subtraction is anticommutative.

When we move to multiplication and division, we see quite similar things. For example, we realize that multiplication is commutative because a × b = b × a. However, we say that division is noncommutative because a ÷ b and b ÷ a are not negatives of each other but reciprocals of each other.

We have, therefore, seen how the four basic operations function. Now we have to ensure that our sequence of operations maintains this feature.

So let us consider 2 + 5 × 3. Since we normally perform operations from left to right, let us do this here. Let us also give preference to the addition operation and perform it before multiplication. With this approach, we will first add 2 and 5 to get 7 and them multiply by 3 to get 21. However, we know that addition is commutative. Hence, if it was true that we should do the addition first, we should get the same answer if we did 5 + 2 × 3. This does prove to be the case. However, we know that multiplication is commutative. So swapping the 3 and the 7 (from 2 + 5 or 5 + 2) should give us the same answer. Now we have 3 × 2 + 5. But this gives us 6 + 5 = 11. And if we swap the 2 and 5 we have 3 × 5 + 2, which gives us 17.

Since we are getting different answers with different approaches, let us now give precedence to multiplication over addition. Hence, if we are given 2 + 5 × 3, then we need to perform the multiplication first, to get 2 + 15, which then gives us 17. But again, we recognize that addition is commutative. So, we could write 5 + 2 × 3, which gives 11 if we perform the multiplication first.

What we can see is that there is no sequence that is inherent to the operations themselves. After all, multiplication itself is ‘repeated addition’! So one would not expect anything inherent to the two operations than can actually make a distinction between them.

An Alternate Convention

But what this means is that we need to decide upon a ‘convention’ that all of us will follow, which will remove the ambiguity concerning which operations need to be done first and which ones last.

For example, we could propose an alternate order, namely SAMD, just for the four primary operations. Then if we have to calculate 4 – 5 × 6 + 8, we would perform the subtraction first, to get -1 × 6 + 8. Then we would perform the addition, resulting in -1 × 48. Finally, we would perform the multiplication to get -48. Similarly, if we had 6 – 2 × 3 ÷ 4, we could get the following steps 4 × 3 ÷ 4 = 12 ÷ 4 = 3. As long as everyone followed the convention we would all get the same answer all the time.

The Need for Conventions

Conventions are crucial in any area of knowledge. For example, in the sentence, “The cat climbed the tree” it is only by convention that we accept that ‘the cat’ is the subject of the verb ‘to climb’, which ‘the tree’ is the object. There is nothing inherent to the order of the words that tell us this. That is why in many languages, the adjective comes after the noun it qualifies, while in English it comes before.

Similarly, in chemistry we may come across the symbol Na₂CO₃. Only convention tell us that Na represents Sodium, C represents Carbon and O represents Oxygen. Only convention tells us that the subscripted 2 and 3 indicate the number of atoms of the element preceding it that constitutes the molecule.

The same is true about mathematics. The symbols do not interpret themselves. We need to accept conventions that everyone agrees to follow in order to communicate mathematical knowledge reliably.

Unfortunately, most of us mathematics teachers do not recognize the complete arbitrariness of the order of operations since we have become so used to the convention that we cannot see it as arbitrary. However, as I have attempted to show, mathematicians could have adopted another convention for the order of operations without there being any confusion. I think that, if mathematics teachers spent some time asking students to arrive at a new convention they could follow for a time, it might be a good exercise in helping students realize that there is a consensus that has been adopted, which everyone needs to follow so that we can communicate mathematical ideas reliably and without confusion.

Down Memory Lane

Just recently, while introducing me for a talk, someone stated that I have two loves – theology and mathematics. This is a reasonably accurate statement and I blog about these two areas regularly. I have been quite open about both of these with my students too. Hence, even though I am their mathematics teacher, they all know that I am also a pastor and that I think about theological issues. As a result many students often ask me questions related to issues outside the immediate sphere of mathematics.

Hence, way back in 2006, shortly after the movie was released, some students asked me about my take on The Da Vinci Code. Some of them, inspired by the movie, had started reading the novel by Dan Brown and had observed some differences between the book and the movie. They wanted to know which version was correct.

I had to disabuse – or at least attempt to disabuse – them of the notion that either the book or the movie had anything that could be considered historically reliable. Secret societies and hidden genealogies are all well and good for a novel, but they hardly hold up to rigorous scrutiny.

In 2006 I had neither seen the movie nor read the book. My work schedule only permitted time for one ‘distraction’ and, at that time, J. K. Rowling was regaling me with the adventures at Hogwarts, both on the page and on the screen. Having had a late start with the adventures of Harry, I was attempting to catch up and finish reading till the Half Blood Prince before the Deathly Hallows was scheduled to be published the next year.

One of my student expressed disbelief that I had not read the novel by Brown. She showed me the illustration of The Vitruvian Man in the novel and asked me how I, as a mathematics teacher, was not intrigued by the proportions of the human body that the drawing indicated. I had to tell her that, since humans come in all shapes and sizes, there is no ideal proportions for the body. After reading a bit about the Vitruvian man, I had to tell her that this was some idealized set of proportions that might have held some appeal for da Vinci, but that this does not mean it was some universal principle.

The Golden Ratio

She also expressed some interest in learning about the Golden Ratio, which she believed to reside in the drawing. But I had to tell her that this was impossible since da Vinci was working with ratios of whole numbers, which meant that he was not working with the Golden Ratio since the ratio is an irrational number. In this blog, I have dealt with two other irrational numbers at length – e and π. The Golden Ratio is a different sort of irrational number. Unlike e and π, which are both transcendental irrational numbers, the Golden Ratio is an algebraic irrational number.

Of course, here we come across something very strange. There are people who claim that the Golden Ratio has some inherent aspect of beauty to it and that, therefore, many artists use it in their art. Let me be very blunt. Even if it is the case that the work of some artists might seem to indicate that there are some ratio similar to the Golden Ratio, and even if some artists today might be intentionally trying to include the Golden Ratio in their work, there is absolutely no justification, from a mathematical perspective, to think that there is any truth to such claims. In order to justify my assertion, let us consider what the Golden Ratio is.

Suppose we have a rectangle with smaller side of length a and larger side having a length a + b. Now suppose we cut off a square of side length a from the rectangle. This would leave a rectangle with sides of length a and b. This is shown in the diagrams below

Now if the ratio of the lengths of the sides of the original rectangle is the same as the ratio of the lengths of the sides of the smaller rectangle, then we say that the ratio of the lengths of the sides is the Golden Ratio. This gives us the equation

Dividing the numerator and denominator of the left fraction by b we get

Denoting a/b as ɸ we get

This can be rearranged to give the quadratic equation

Using the quadratic formula, we can obtain

Since ɸ is a ratio of lengths, it cannot be negative. Hence, ignoring the negative root above we get

Using a calculator, we can obtain ɸ = 1.618033989…. The presence of the square root of 5 in the expression for ɸ indicates that it is an irrational number, as I claimed earlier.

An Oft-Repeated Claim

However, the irrationality of the ratio cannot be the only reason for which I reject the claims about its prevalence in art. After all, the diagonal of a unit square is √2, which is also irrational. Similarly, the ratio of the distance between parallel sides of a regular hexagon to the length of the sides is √3. While artists, in general, may not use geometric figures in their art, it would be fallacious for me to discount the presence of some ratio simply on the grounds that it is irrational.

The claims about art do not come on their own. Rather, there is the prior claim that the Golden Ratio occurs often in nature. This is a bald-faced lie. After all, what can we observe in nature but discrete occurrences of phenomena? That is, we can count certain things. For example, it is claimed that, if we count bands of fruitlets on pineapples, we will count 5, 8, or 13 bands for small pineapples or 8, 13, and 21 bands for larger ones. So what?

Well, these numbers are consecutive numbers in the Fibonacci sequence. And if we take the ratio of consecutive terms in the sequence, the ratio will turn out to more closely approximate ɸ. For example, see the table below

I have collapsed many of the rows so that the table could fit and be readable in a single screen. However, the table shows that the terms in the fourth column converge to the value of ɸ indicated earlier. It is crucial, however, to note that, while the ratio in the fourth column do converge to the value of ɸ, they will never be equal to ɸ because the ratios are rational numbers while ɸ is irrational.

Debunking the Claim

But that’s not my beef with the earlier claims. Let us start the sequence with different numbers. Instead of the first two terms being 1 and 1, let us start with 1 and 4. Note that 4 does not appear in the original sequence. With these two as the starting point, we will get the following table

Since the starting terms are different, all the terms in the new sequence differ from the original one. However, the ratios quickly converge to the same value!

Now let us change the second term quite drastically. Let’s make it 1,000,000! Now we get the following table

We observe the same thing, namely that, while the individual terms are the ratios converge to the same value.

We could take a different first term also and see what we get. Choosing a = 4 and b = 9 we get the following

There is no change to the ratio to which the terms of this sequence converge. Indeed, we can start with a value of a that is greater than b and get something like

In other words, no matter what numbers we choose to start with, the resulting ratio of consecutive terms converses to the value of ɸ. Hence, if you give me any two starting numbers, I can confidently assert that they are part of a sequence where the ratios of consecutive terms converges to ɸ. For example, if you gave me 11 and 19, I could form the sequence as below

Here, I have shown the third row in red so you can see where in the sequence the given numbers lie. And since the sequence of ratios converges to ɸ, I can confidently state that, if we find the numbers 11 and 19 in nature, then this must be an approximation to ɸ. But the careful reader will recognize this as quite spurious reasoning. After all, every pair of starting numbers will yield a sequence of ratios that converges to ɸ. Hence, there is nothing extraordinary in finding any such pairs of numbers anywhere.

Freedom to Explore

Over my career as a teacher, which spans more years than I would care to admit, I have come across the claim that the Golden Ratio appears in nature many more times than I would consider acceptable. Most of these are from excited young students, who have just been introduced to the Golden Ratio or the Fibonacci sequence. However, a fair number are from teachers, some very experienced in terms of years. They point to the fruitlets of the pineapple and to the petals of various flowers to give their students examples of these presumed occurrences. However, as I have shown in this post, any starting point, as long as both numbers are positive, will result in a sequence of ratios that converges to ɸ. Here is one that starts with the transcendental number e and π.

By pointing to numbers in nature and asking their students to draw conclusions that nature has examples of the Golden Ratio, such teachers are stunting the mathematical acumen of their students because they make the students think that something is special when in case it is nothing more than run of the mill. Instead of asking their students to wonder why every sequence of numbers defined as

has a corresponding sequence of ratios defined as

that converges to ɸ, the teachers make students think that the example of the Fibonacci sequence is unique in that regard.

Don’t get me wrong. I am not saying that the Fibonacci sequence does not represent something insightful about mathematics. What I am saying is that we should not claim any distinction for it that it actually does not have. But for this to happen the teachers should have the time to explore various ideas both individually and in groups. However, given how content heavy the high school mathematics curriculum is, teachers in the lower grades also have a lot to cover.

As a result, most mathematics teachers are perpetually scrambling to complete the syllabus for the year, with hardly any time for the kind of exploration that would result in robust learning for the students. Because of this, most teachers are forced to use prepackaged material, such as from a textbook or online platforms. But prepackaged materials can never cater to the unique requirements of a class of students, each of whom has a uniquely expressed curiosity, unless, of course, we wish to quench that curiosity!

If we are serious about teaching students to be adept at mathematics, we must reduce the content of our curriculums and syllabuses and focus instead on the development of a small set of skills that can be transferred not just to different areas within mathematics but also to other disciplines. And we must allow our students – and teachers – the freedom to explore the subtleties of mathematics without which their understanding of mathematics would be superficial at best.

Note: I will be taking a break next week and will return with new posts in January 2025. Have a happy holiday season.

The Presenting Problem

In this post, I wish to continue with some geometry along with some insights from sequences and series. Consider the figure below

It is given that ∠AOB = 60°. Also, the radius of the largest circle is 1 unit. The successive smaller circles are tangential to the circles on either side. This continues indefinitely. In other words, there are infinitely many circles. We are asked to find the total area of all the infinite circles.

If you are inclined to solve this, please pause here before proceeding.

Introducing Infinite Geometric Series

Ok. I hope those who attempted the solution have obtained a satisfactory answer. Now, some of you may be wondering how the sum of the area of infinite circles could be determined. Should the finite area of infinite circles also be infinite?

Let us take a brief detour into the marvelous world of infinite series. Consider the series

The ellipses (i.e. ‘…’) at the end indicate that the pattern continues. We encountered an expanded version of this series in the post Naturally Bounded. There we considered the infinite series

After some nifty algebraic manipulation we showed that S = 2. Since the sum we are considering in this post only lacks the starting 1, we can conclude that the sum of the infinite series for this post is 1.

However, let us spend some time understanding why this is the case. What in the world is the pattern in our series? We can recognize that all the values in the denominator are powers of 2. Let us designate the sum with S. Then we can write

Now let us try to see why the sum turns out to be 1. As in the earlier post, we can multiply both sides by 2 to get

Let us make this colorful. We can write the above two equations one below the other as follows

We can see that, if we subtract the first equation from the second, the LHS will yield S_n, while the RHS will reduce to 1 since all the terms in red, green, blue, and purple will cancel out. This gives us S = 1.

But why does this work. We can see that, in the series we are considering, every term is obtained by multiplying the preceding term by 1/2. Such a series is known as a geometric series.

In order to understand geometric series, let us start with a general case. Let’s say that we have a starting term, a, and that each subsequent term is obtained by multiplying by r. The multiplier, r, is called the common ratio because it is the ratio of any term to the term before it. Anyway, we can see that our geometric series up to n terms will be

Note that the n^th term is ar^n-1 because the first term, a, is actually ar⁰ and as the power of a starts from 0, the count after n terms will be n – 1. We can multiply the above equation by r and arrange a similarly color coordinated set of two equations one below the other as follows

Now, when we subtract the second equation from the first, the red, green, blue, and purple terms will vanish, leaving us with

Both sides can be factorized to give

Now, if we divide the entire equation by 1 – r we will obtain

The first thing we can observe is the denominator 1 – r. Since the denominator cannot be zero, we can conclude that this formula will not work for r = 1. Further, we can take a look at the rⁿ term. When the value of r is either greater than 1 or less than -1, multiplying by r results in a number that has a greater magnitude. In this case, it is obvious that we cannot find a sum of infinite terms of the series since subsequent terms have larger and larger magnitudes.

But what happens when the value of r is between -1 and 1? In this case, multiplying by r results in a number that has a smaller magnitude. Hence, the more times we multiply by r, the closer the result gets to zero. We can see this in the original series

We can see that each term is closer to zero than the one before it. So we can potentially consider an infinite series for these values of r and write

Once again we can multiply this equation by r and stack the two equations as follows

Now, when we subtract the second equation from the first, the red, green, and blue terms will vanish giving us

Dividing the equation by 1 – r we get

Now, we can return to our original series. For this series, we observe that a = 1/2 and r = 1/2. Plugging these values into the above formula gives us

Return to the Problem

This matches with our earlier result and puts us in a position where we can actually attempt the problem with which I started the post. Let me reproduce the figure so we can recall what the problem was.

Starting with the largest and second largest circles, we can obtain the following figure.

Here, X and Y are the centers of the larger and smaller circles respectively. XP and YQ are perpendiculars drawn from X and Y respectively to AO. YZ is a perpendicular drawn from Y to XP. Since we were given that the largest circle has a radius of 1, this means that XP = 1. Also, R is the point of tangency of the two circles, which is also the point where the line OX intersects both circles.

Now, we are given that ∠AOB = 60°. From symmetry, it follows that ∠AOX = 30°. Using some basic geometry, we can see that OX = 2 XP. Similarly, since the triangles OXP, PYQ, and YXZ are similar, we can obtain OY = 2YQ and XY = 2XZ.

Suppose we say that YQ = r. This gives us OY = 2r. Also, since XP = 1, we can conclude that OX = 2. But OX = OY + YR + RX. This gives us 2 = r + 2r + 1, yielding r = 1/3. However, r is the radius of the smaller circle. Since, with the use of geometry, it turns out to be a constant (i.e. 1/3), this means that the radius of each successive circle in the series will be 1/3 times the radius of the previous circle. In other words, the radii of the circles form the following sequence

Now, the area of a circle is πr². This gives us the infinite series for the area is

Taking π common we can express this as

Here, the terms in red form an infinite geometric series with a = 1 and r = 1/9. Using the formula we earlier derived for the sum of an infinite geometric series we can obtain

Reflection

While solving the problem we have used both algebra and geometry. In an earlier post I have bewailed the unfortunate trend in some countries of offering distinct mathematics subjects like Algebra 1, Geometry, and Pre-Calculus. In another post I demonstrated that one’s understanding of mathematics is furthered when we consider algebra and geometry together. I have done this in this post too. Mathematics is a coherent body of knowledge. By introducing artificially defined segments of mathematics as stand-alone bodies of knowledge, we convey the idea that the whole is the sum of its parts. However, mathematics is much greater than the sum of its branches. And I will continue to harp on this till my dying breath!