# Tom Apostol Calculus One, Archimedes Method of Exhaustion

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.

• It's difficult to determine what your question is. Can you clarify or narrow down what it is you are asking? Commented Aug 4, 2021 at 14:53

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!

• I think I am going to have to find another book that goes between a pre-calculus and Tom Apostol. I am not sure how all these steps take place. Thanks. Commented Aug 4, 2021 at 20:43
• Do you have trouble with some step in my answer. I can elaborate. Besides, i agree. Apostol may not be te best book for someone starting with calculus. Commented Aug 4, 2021 at 21:33
• Hi, I can see much better how the equations are being added together. I was wondering how the third term in the polynomial. $3(n-1)^2+3(n-1)+1$. Goes from adding $1+1+1… to (n-1)$? I think this is a polynomial. I did just study them but have not added the terms in this way. Commented Aug 5, 2021 at 10:53
• The thing is that you are adding $n-1$ ones. I mean, if you add two ones, you get 2. If you add 3 ones you get 3. In the case of 1+1+1... but $n-1$ times, you get $n-1$. Commented Aug 5, 2021 at 12:13
• Thanks for the response. I didn’t grasp that. I think I can move forward! The variable K is the only variable in the identity. When integers start to plug into the equation I thought (n-1) needs a K. Almost like a substitution. I forgot or didn’t realize (n-1) was the number of integers we plugged in for each term. I think I know it a little bit better! Thanks! Commented Aug 5, 2021 at 13:38

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.

• Thanks, much gratitude I have for your time. I am learning a lot by reading your example and I like that you showed me the identity. I guess now that I have taken the time to learn trig identities I actually see that both sides of the equations are equal. Commented Aug 3, 2022 at 1:36
• I have a question; if I put 2 in for $i$ I do not think I get $n^2$ $$2^2 - (2-1)^2$$ $$4-1=3$$. I’d have to look closer at that unless I am missing the sum notation. Commented Aug 3, 2022 at 2:03
• @Benp404 You need to sum all the terms to have $n^2$. If you let $n=2$, then... $$\sum_{i=1}^2{i^2-(i-1)^2}=1-0+4-1=4$$ Commented Aug 3, 2022 at 9:10
• @Benp404 No problem. I encourage you to learn that notation because is very useful in linear algebra, taylor series, etc., and then you leter can jump to the Einstein convention, which is more useful. You can think of the $\Sigma$ notation as a Foor loop, if you know basic coding. Commented Aug 6, 2022 at 0:55
• @Benp404 You can use Riemann Sum, which is the definition of an Riemann integral, or you can use a Taylor series aproximation of your curve and integrate that. Given that you are using a computer, try to use the first. The integration of Taylor series converge more quick, but You have to calculate the Taylor coefficient, so I would use the last if I will do by hand. Commented Aug 6, 2022 at 21:26