To understand abstract concepts of mathematics, it is sometimes important to link abstraction to familiar situations from everyday life. In particular, it is often most helpful to visualise mathematical ideas, that is to associate them with pictorial perception. As an example of this idea, we will now envision the Pascal Triangle, which will also be used for later interpretation. Later, and because of it, we deduce, for example, the sum of the first hundred natural numbers or the first ten second powers of natural numbers.
The triangle is named after the 17th-century French mathematician Blaise Pascal, who studied e.g. liquids (you may know Pascal's Law from school), but as a proper enlightener he was also interested in many other branches of knowledge and the scientific method.
The Pascal triangle is not a geometric formation (in a true geometric sense). These are numbers arranged in a triangular shape, you may have seen it or even tried to draw it yourself when you were bored. It looks like this:
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1 1
1 2 1
1 3 3 1
1 4 6 4 1
...
You may have noticed the rule by which a person constructs a triangle. Starting with one, one place is created below each number on the right and left, and the empty spaces are filled with the sum of the upper numbers. So a two in the third row is e.g. the sum of two ones, etc.
Many peculiarities apply to this triangle. For example, it is obvious that the sum of the numbers in the $n$-th row is equal to $2^n$ (starting with a 1 and then adding all the numbers from one row twice to the numbers from the row below, thus doubling the sum). Less obvious is that the individual lines read as numbers are power 11 ($11^0=1$, $11^1=11$, $11^2=121$...), or that the totals of the diagonal of the triangle form the Fibonacci sequence.
Now let's look at a seemingly completely different topic, the exponents of the sum of $a+b$. The first power holds no secrets. $$(a+b)^1= a+b\,.$$
The second power has been memorized by many in school, but can be swiftly verified by multiplication. $$(a+b)^2 = a^2+2ab+b^2\,.$$
With a little diligence, pencil and paper, we can deduce what the formula will be for the sum of two numbers to the third: $$(a+b)^3 = a^3+3a^2b+3ab^2+b^2\,.$$
The rule we're referring to here is called the binomial theorem (English binomial theorem, German Binomischer Lehrsatz). Although we have run into this rule, we will not apologize to it and move on. There are more interesting things ahead in our journey.
Notice anything familiar? Indeed, the coefficients for the individual members of our expression are the same as some of the lines of the Pascal triangle. Obviously for $(a+b)$ raised to both zero and prime, we get coefficients from the zero and the first row, but this is also true for the higher powers. How so? I will not attach rigorous proof, for we are on our way to a differential count at the very beginning. As we go further and further, I will try to be more precise and accurate.
Notice how we go from the first power to the second power. The entire expression $a+b$ must be multiplied by $(a+b)$. Now, a little imagination: Imagine we're making a table in which we write the coefficients of members with a certain number of $a$ and $b$, with members with more than $a$ on the left. Then, when multiplying, the member that is multiplied by $a$, moves to the left, and the member that is multiplied by $b$ is moved to the right. Ultimately, we're going to get a member in the same way that we form the Pascal triangle. You can see the table for $(a+b)^2$, starting with the expression $(a+b)$, below.
|a^2| a | ab| b |b^2
----------------------------
(a+b) | 0 | 1 | 0 | 1 | 0
----------------------------
*a | 1 | 0 | 1 | 0 | 0
*b | 0 | 0 | 1 | 0 | 1
----------------------------
(a+b)^2| 1 | 0 | 2 | 0 | 1
If you think about this and visualize this, then hopefully you'll see why the coefficients of power $(a+b)$ are the coefficients from the Pascal triangle.
This is a fairly unsubstantiated story, but Gauss was indeed an unusually bright mathematician as a child (and even in adulthood).
It is said that when young mathematician Carl Friedrich Gauss was not listening to the teacher in class, he was tasked with calculating the sum of the first hundred natural numbers, $S_1=\sum_{i=1}^{100} i = 1+2+3+\dots +100$. Of course, young Gauss didn't want to perform this boring operation, so he thought about it and after a while simply said it was $5\,050, which turned out to be the right solution. Now let's see how you can shrewdly figure out the outcome. I will demonstrate two ways, one trick (probably used by Gauss) and then another that will seem even trickier. Nevertheless, the latter method unlocks the door for counting more complex rows later. I'll also explain the meaning of the $\sum$ symbol.
Important: The sum of the first $n$ of natural numbers is a very nice task, and I wouldn't want to deprive you of the joy of your own discovery. Therefore, try to figure out this task yourself before you start reading again. It may take 20 minutes, maybe a day, but I think it's worth it.
The origin of the $\sum$ symbol is that sigma represents the letter s, which stands for sum. Sum is another word for sum/summary and Latin for summa .
The $\sum$ symbol is the Greek capital letter sigma and is called sum. It marks an operation called a summation. Summation example: $$\sum_{i=1}^3 i+1 = (1+1) + (2+1) + (3+1) \,.$$
It means that we add several expressions starting with $i=1$ and ending with $i=3$. We get expressions by adding $i$ to the term loaded in sum, from the lower limit of $1 to the upper limit.
The sum of the first $n$ numbers is so clumsy, since all numbers are different. If you could somehow make the numbers equal to multiply them, it would be nicer. So why don't we try to take their average and multiply it by the number of censuses? That would work. However, how to calculate the average? Maybe you tried writing a special case of sum. For example, for the first three numbers, the mean is the middle — 2. But what if there is no middle at even number? Then you can try to draw the numbers on the number line -- the average lies in the middle between the first and the last number. Thus, for the average $a$ series of $n$ numbers, $a=(n+1)/2$. This average just needs to be multiplied by the number of censuses $n$. Thus, for the total: $$S_1=\frac{n(n+1)}{2}\,.$$
The second, perhaps more Gaussian way, is to write two sums down in reverse order. In case of $n=6$: $$1+2+3+4+5+6=S_1\,,$$
$$6+5+4+3+2+1=S_1\,.$$
It is now clear how to calculate the sum of $S_1+S_1$: $$6\cdot 7=2S_1\,.$$
So we generally get the same result $$S_1=\frac{n(n+1)}{2}\,.$$
The word telescope comes from the Greek tele (far) + skopos (observer). However, the method is called the Folding Telescope. We're going to come up with a sum that shortens and folds like the machine in question.
In the telescopic method, it begins boldly. Because if we want to calculate the formula for the sum of the first powers, we start by trying to calculate the formula for the sum of the second powers of the numbers. However, to take advantage of the Pascal triangle, we will not count $i^2$, but equal to $(i+1)^2$. We get: $$\sum_{i=1}^n (i+1)^2 = \sum_{i=1}^n i^2+2i+1 \,.$$
This expression indicates a sum from $i=1$ to $i=n$, where $n$ is some number that we specify later. The rule for summation is that the sum sum sum is the sum of summations. This rule comes from the way the census works, we can add members in any order we want. So we rewrite the right side: $$ \sum_{i=1}^n i^2+2i+1 = \sum_{i=1}^ni^2+\sum_{i=1}^n2i+\sum_{i=1}^n1 \,.$$
The last member on the right is actually the sum of $n$ times 1. That equals $n$. The $\sum_{i=1}^ni^2$ member can be subtracted from the entire equation: $$\sum_{i=1}^n (i+1)^2-i^2 = n+\sum_{i=1}^n 2i \,.$$
The sum on the left-hand side is the so-called telescopic sum. You can try to break down a number of its members or think about it, and you get a lot of them subtracted from each other, so the result is probably $(n+1)^2-1$. Try, for example, for $n=3$. If we take advantage of the fact that we can factor a 2 out of every sum addition on the right and make a few adjustments, we get: $$\begin{align*} (n+1)^2-1 &= n+2\sum_{i=1}^n i\\ n^2+2n -n &= 2\sum_{i=1}^n i\\ \sum_{i=1}^n i &= \frac{n^2+n}{2} = \frac{n(n+1)}{2}\,, \end{align*}$$
which is an old familiar result. Don't be fooled by the fact that we switched both sides of the equation in the last line, that's allowed. Subsequently, we also charged $n$. You can verify that for the first 100 numbers, the result is $100\cdot 101 /2 =5\,050$.
With a telescopic method at hand (though it may seem trivial), we can calculate more totals, like $$S_2 = \sum_{i=1}^n i^2\,.$$
I shall mention only the procedure in which the past result is used. However, I suggest you try yourself first: $$\begin{align*} \sum_{i=1}^n (i+1)^3 &= \sum_{i=1}^n i^3 + 3i^2+3i+1\\ \sum_{i=1}^n (i+1)^3-i^3 &= \left( 3\sum_{i=1}^n i^2 \right) + 3\frac{n(n+1)}{2}+ n \\ n^3+3n^2+3n &=3\frac{n(n+1)}{2}+ n + 3\sum_{i=1}^n i^2\\ n^3+\frac{3}{2}n^2+\frac{1}{2}n &= 3\sum_{i=1}^n i^2\\ \frac{2n^3+3n^2+n}{6} &= \sum_{i=1}^n i^2\\ \frac{n(2n^2+3n+1)}{6} &= \sum_{i=1}^n i^2\\ S_3=\sum_{i=1}^n i^2&=\frac{n(2n+1)(n+1)}{6}\,. \end{align*}$$
For a specific case of the top ten squares, we get e.g.: $$\sum_{i=1}^{10} i^2 = 385 \,.$$
The calculus is all about infinity and neglect. So far in this chapter, however, we have only been discussing discrete (discontinuous) examples, and infinity has not appeared anywhere. I'll try to correct that.
Imagine if we wanted to calculate $S_1$ for the first billion numbers. Let's look at the formula: $$S_1=\frac{n(n+1)}{2}=\frac{n^2+n}{2}\,.$$
Suppose we neglect the $n$ member if it is some number of orders of magnitude less than $n^2$. Then for each number of orders, the member $n$ we always neglect from some number up.
We have $n^2+n$ in the denominator. But when it's a billion, we get $10^{18}+10^9$. Ten to the ninth is a big number, but it's nine orders of magnitude lower than $10^{18}$, which is already quite negligible. What if we wanted to calculate the sum of the first $10^{100}$ numbers? By then, the two members would be on the order of $10^{100}$, an unimaginable difference. As you know, there are infinite numbers, so we might want to calculate the sum of even more numbers, and then the second term would appear smaller and smaller. In fact, we might neglect it. Then for the big numbers, we would get: $$S_1 \approx \frac{n^2}{2} \,.$$
A $\approx$ sign means approximately equal. While this seems to be a mathematically inaccurate term, it will come in very handy when it comes to building intuition.
We can also make a counter estimate for the higher $S_l$. It's easy to imagine that with the big numbers, the largest power of $n$ will remain dominant, the rest we can neglect. We get that for a big $n$: $$S_2 \approx \frac{n^3}{3} \,.$$
How about a bigger $n$? We need to look at how $S_l$ is made by the telescopic method. To avoid confusion in marking, let's calculate the sum of the first $n$ numbers raised to $l$. We always get an expression similar to this: $$\sum_{i=1}^n (i+1)^{l+1}-i^{l+1} = K\sum_{i=1}^n i^l + \dots $$
Here $K$ is the second coefficient of the $n$-series Pascal triangle, and the expressions behind the dots are always in the form of $C\cdot i^m$, where $C$ is some number and $m$ is less than $n$.
The assumption that the totals of $\sum_{i=1}^n i^m$ are in the order of no more than $n^{m+1}$ should correctly be proved by mathematical induction. But we will rely on intuition and trust.
We will now edit the left side and leave all members that are less than $n^{l+1}$. Also, let's disregard any members that have been marked with the three dots above -- their total is in the order of $n^l$. That's why we get: $$\begin{align*} n^{l+1} &\approx K\sum_{i=1}^n i^l\\ \sum_{i=1}^n i^l &\approx \frac{n^{l+1}}{K}\\ S_l = \sum_{i=1}^n i^l &\approx \frac{n^{l+1}}{l+1}\,. \end{align*}$$
I took one more step $K=l+1$, which can be seen in the Pascal Triangle. So we have an expression for the sum of the first $n$ of natural numbers to $l$ for the big $n$. Later, we'll show how to use this insight to count the contents of shapes under a curve, Archimedes-style.