Tom Apostol Calculus One, Archimedes Method of Exhaustion

Question

I am trying to find a way to move from pre-calculus to learning calculus by reading Tom Apostol Calculus Volume One. Since this is the first time I am opening up a calculus book I am struggling to grasp the topic.
In Tom Apostol's book he is talking about Archimede's Method of Exhaustion. Tom Apostol shows a bunch of steps on how Archimedes and other mathematicians may have contributed to the idea of finding the area of a shape that is unknown or hard to find. I was looking for some help on one of his steps. I found this page on Math Stack Exchange but was looking to see if I understand what Tom Apostol is talking about. Here is the link: YouTube Video showing how to cancel equations

The step I have questions on starts with an identity $$(K+1)^3 = K^3+3K^2+3K+1$$ Equation is rewritten as $$3K^2+3K+1=(K+1)^3-K^3$$ I am good up to here, Tom Apostol subtracted the $K^3$ from both sides. The next step then is to start plugging in integers. The identity up to this point should be satisfied for every integer $n\ge 1$. Tom Apostol sets up this step like this: $$\begin{array}{L}3* 1^2 + 3 * 1 + 1 = 2^3 -1^3 \\3 * 2^2 +3 * 2 + 1 = 3^3-2^3\\ \phantom2 \vdots \\ 3(n-1)^2+3(n-1)+1 = n^3-(n-1)^3\end{array}$$

I have an idea of what Tom Apostol comes up with the right side of the equation. $$\begin{array}{L} = \phantom9 \phantom1 \require{cancel}\cancel{2^3}- 1^3 \\ = -( \cancel{3^3}-\cancel{2^3} )\\ = -( 4^3 - \cancel{3^3}) \\ = - (n-1)^3 + 1^3 \end{array}$$ Am I going in the right direction? I feel like I'm close to understanding the right side, but can see my method is not exact if it is in the right direction. Thanks for any help.

Answer 1

What Tom Apostol want to do in that part of the book is to show that for any $n\geq 1$ you have: $$ 1^2+2^2\dots+n^2=\frac{n^3}{3} + \frac{n^2}{2}+\frac{n}{6} $$ As you pointed out, he starts with the identity $$ 3k^2+3k+1 = (k+1)^3-k^3 $$ from which the following equations follow, and proceeds to add all them up: $$ \begin{array}{lrl} &3\cdot 1^2+3\cdot 1+1&=2^3-1^3 \\ &3\cdot 2^2+3\cdot 2+1&=3^3-2^3 \\ &\vdots & \\ +&3(n-1)^2+3(n-1)+1&=n^3-(n-1)^3\\ \hline\\ &3[1^2+2^2+\dots+(n-1)^2]+3[1+2+\dots+(n-1)]+(n-1) &=n^3-1^3 \end{array} $$ Let's see how he got all the terms one by one. First, the left-hand side is composed of three terms: $3[1^2+2^2+\dots+(n-1)^2]$ plus $3[1+2+\dots+n]$ plus $(n-1)$. These terms are obtained by adding the first, second and third terms of the right hand side of all $(n-1)$ equations independently: $$ \begin{aligned} 3\cdot 1^2 + 3\cdot 2^2+\cdots+3\cdot(n-1)^2 &= 3[1^2+2^2+\cdots+(n-1)^2]\\ 3\cdot 1 + 3\cdot 2+\cdots+3\cdot(n-1) &= 3[1+2+\cdots+(n-1)]\\ 1+1+\cdots+1&=(n-1) \end{aligned} $$ Now, lets look at the right hand side. Similarly we add the first and second terms of the $(n-1)$ by separate and then add them at the end: $$ \begin{array}{cccccccc} & &+2^3&+3^3&+\cdots&+(n-1)^3&+n^3\\ +&-1^3&-2^3&-3^3&-\cdots&-(n-1)^3\\ \hline\\ &-1^3+&+0&+0&+\cdots&+0&+n^3 \end{array} $$ The next step is to substitute the well known formula for $1+2+\dots+(n-1) = \frac{n(n-1)}{2}$ to obtain: $$ 3[1^2+2^2+\dots+(n-1)^2]+\frac{3n(n-1)}{2}+(n-1) =n^3-1^3 $$ and rearrange to obtain: $$ \begin{aligned} 1^2+2^2+\dots+(n-1)^2 &= \frac{1}{3}\left(n^3-1^3 - \frac{3n(n-1)}{2}-(n-1)\right)\\ & = \frac{n^3}{3} -\frac{n^2}{2} + \frac{n}{6} \end{aligned} $$ Finally, just add $n^2$ to both sides of the previous equation to obtain the desired result: $$ \begin{aligned} 1^2+2^2+\dots+(n-1)^2 +n^2&= \frac{n^3}{3} -\frac{n^2}{2} + \frac{n}{6}+n^2\\ &=\frac{n^3}{3} + \frac{n^2}{2}+\frac{n}{6} \end{aligned} $$ Hopefully this will help you in addition to what Tom Apostol already explains in the book. Good luck!

Answer 2

Maybe instead of follow the book, if you can't understand something, try to prove it by yourself. Calculus 1 is not a very hard topic for doing self-learning, with the use of Wikipedia.

The first time I solved a sum of a geometric series, I didn't even know what they are, but I knew that calculus start with a discrete case, and then goes to the limit.

In your case, the autor tried to explain the sum of square with a very unintuitive method. In my approach, I first would take sum of difference, because maybe, it can lead to something, just like the Riemann sum, so I would first define the difference of something, in this case, the difference $\Delta$ of $i^2$ just because you are trying to get the sum of squares:

$$\Delta i^2 = i^2-(i-1)^2 $$

suming $\Delta i^2$ gives:

$$\sum_{i=1}^n{\Delta i^2} = \sum_{i=1}^n{i^2-(i-1)^2} = (1 - 0) + (2 - 1) + ... + (n^2-(n-1)^2)$$

Note that every term are cancelled except $0$ and $n^2$. so the sum is:

$$ \sum_{i=1}^n{i^2-(i-1)^2} = n^2$$ $$ \sum_{i=1}^n{i^2-(i^2-2i+1)} = n^2$$ $$ \sum_{i=1}^n{2i-1} = n^2$$ $$ \sum_{i=1}^n{2i}-\sum_{i=1}^n{1} = n^2$$ $$ \sum_{i=1}^n{2i}-n = n^2$$ $$ \sum_{i=1}^n{2i} = n^2+n$$ $$ \sum_{i=1}^n{i} = {n^2+n\over 2}$$

Now we didn't get the sum of square, instead, we get the sum of integers, but, we think: "this approach may be lead to something using $\Delta i^3$, instead of $\Delta i^2$, because we got the sum of $i^1$, instead of $i^2\ $":

$$ \sum_{i=1}^n{\Delta i^3} = \sum_{i=1}^n{i^3-(i-1)^3} = \sum_{i=1}^n{i^3-(i^3-3i^2+3i-1)} = \sum_{i=1}^n{3i^2-3i+1} = n^3$$

$$ \sum_{i=1}^n{3i^2}-\sum_{i=1}^n{3i}+\sum_{i=1}^n{1} = n^3$$ $$ \sum_{i=1}^n{3i^2}-3\left({n^2+n\over 2}\right )+ n = n^3$$ $$ \sum_{i=1}^n{3i^2} = n^3+3\left({n^2+n\over 2}\right )- n = n^3 +{3n^2\over 2}+{n\over 2} $$ $$ \sum_{i=1}^n{i^2} = {n^3\over 3} +{n^2\over 2}+{n\over 6}$$

So, What's the difference between approach? The author start from an identity that he doesn't explain why it's use, then he does $n-1$ equation, which for me is insane, and then arrives to the conclution. For last, that method doesn´t seems generalizable for a student at first. Using $\Delta i^k$ seems more intuitive for me, even when $k=1$. Also works for $\Delta a^i$ to calculate the geometric sum: $\sum{a^i}$, which is where I first use it.

Obviously, you have to understand $\Sigma$ notation to use it, but just reading the Wikipedia article and doing some examples are all you need in my opinion.