Produced by Charles Wells Revised 2017-02-03 Introduction to this website website TOC website index blog Back to head of Doing Math chapter
CONTENTS |
When you are doing math, watch yourself doing it.
The idea is to install in your brain a Watcher who watches what you do without being judgmental. Math Ed people call this self-monitoring.
Self-monitoring enables you to discover what works and what doesn't work. When you practice self-monitoring successfully while doing math, then in the long run you will learn more about doing math than you will with any other practice suggested on abstractmath.org.
Abstractmath.org describes mistakes other people make, and sometimes reading about them may help you. But when you watch yourself doing math, you are learning about what you do, not about what other people do.
Suppose you are told that your college has 6 times as many students as professors and that the total number of students and professors is 1400. How many students does your college have?
You start working. Let $s$ be the number of students and $p$ be the number of professors. So \[6s=p\text{ and }s+p=1400\] Then $s+6s=1400$, so $7s=1400$, so $s=200$. So the college has $200$ students. Panic! That can’t be right! Let’s see, $6s=p$ and $s+p=1400$. So plug $p=6s$ into $s+p=1400$ and you get $s+6s=1400$, so $7s=1400$ but that's the same thing I got before!
But now the Watcher says: You did the same thing twice and got the same answer. Maybe it’s time to try something else. Like for example checking what you did!
Well, $6s=p$ is right, that’s $6$ students for every professor. No wait a minute, "$6s=p$" means that if there are $100$ students there are $600$ professors. Aargh. I got it backward. Rowrbazzle! I should have written $6p=s$…
Zooming and chunking is a technique that almost anyone with practice at dealing with abstract math will use to analyze a problem.
This example requires only first year calculus. It is also discussed in Representations of COntinuous Functions
Suppose your research has come up with the function \[f(x)=.0002{{\left( \frac{{{x}^{3}}-10}{3{{e}^{-x}}+1} \right)}^{6}}\]
defined for real numbers $x$. You want to understand how this function behaves. If I were looking at it, I would go through some process like this:
It is worth taking a very close look at what goes on in a mathematician’s head when they go through a process of analysis like this. What happens with me is that I zoom in and out and when I have zoomed out I see things in chunks:
I look at the big picture |
Chunking is a kind of psychological inverse to substitution. See this example.
In each step, I am selective in what details I pay attention to and lump the stuff I am not paying attention to into “something”, perhaps with a qualification such as “something $\geq0$”. The lumping part is called chunking or encapsulation in the math ed literature. (Computer science has a related idea of encapsulation.)
1) Zoom out. $f(x)=.0002{{\left( \frac{\text{something}}{\text{something }>0} \right)}^{6}}$ so it is always defined.
2) Zoom in on numerator: $f(x)=.0002{{\left( \frac{{{x}^{3}}-10}{\text{something }>1} \right)}^{6}}$ so $f(x)<.0002{{\left( {{x}^{3}}-10 \right)}^{6}}$.
3) Zoom out further (ignoring the fact that what is between the parentheses is a fraction): $f(x)=.0002{{\left( \text{something} \right)}^{6}}$, which is $.0002{{\left( \text{something else} \right)}^{2}}$, so it is always nonnegative.
4) Remember a math fact. Every real number has exactly one cube root, so ${{x}^{3}}-10=0$ when $x=\sqrt[3]{10}$ and nowhere else.
5) Blur out the denominator, keep everything else exact. Since \[f(x)=.0002{{\left( \frac{{{x}^{3}}-10}{\text{something nonzero}} \right)}^{6}}\] and the only time a fraction is zero when its numerator is zero, $x=\sqrt[3]{10}$ is the only $x$ for which $f(x)=0$.
And so on…
Not every mathematician would do this in exactly the way I did.
I have described a way of understanding something complicated by focusing on one detail and lumping all the other stuff into “something”, or “something with some property”. This is the zoom and chunk method.
Experts characteristically use the zoom and chunk method |
Tthe graph does not show you some aspects of the graph that the analysis above does show. You can’t tell just from looking at the graph that the function is asymptotic to $0$ in the negative direction and increases rapidly to the right of $\sqrt[3]{10}$. (It appears to do these things but you could be fooled.) You can’t tell it is zero only at one place, either.
Thanks to Olaf Stackleberg for corrections.
Every definition provides a method of proof:
METHOD: To prove
that a statement |
Definition: For any integer n:
This definition gives a precise meaning to the words positive, negative and nonnegative. Any question about whether a given integer is positive or negative must be answered by checking this definition.
$0$ is nonnegative.
The definition of “nonnegative” says $n$ is nonnegative if $n\ge 0$. So to prove $0$ is nonnegative, we must show that $0\ge 0$. But that is true because $0=0$.
Similarly, the statement “0 is positive” is false because “$0\gt0$” is false.
When you are starting to learn about a part of abstract math, the method of rewriting what you want to prove using the definitions is the first thing you should do when faced with a statement to be proved. In most cases, that will not be enough to give the proof! But it is a start, and it may suggest what to do next.
The proof above that $0$ is nonnegative is a baby example. Abstractmath.org uses baby examples a lot. Their value is that you can concentrate on what is being illustrated without getting stuck in unfamiliar math. Some students sneer at baby examples. “I already knew that!” But the example above does not exist to teach you something about $0$. It exists to teach you how, when you come across a definition new to you, you use examples to understand it.
Do not diss baby examples. |
There are many different kinds of examples in math, discussed in Varieties of Mathematical Prose, section 2.3.1.
Whether they use rewrite by definition (at least in their heads) or not is a sharp dividing line between newbies to abstract math and those with even some successful experiences with it. Those with more experience do it automatically and even unconsciously, and some of those who read the description I give here will react like:
But that’s trivial!! Why did you devote so much space to that?? It’s OBVIOUS.
See ratchet effect.
As math educators who pay attention to their students know, it is not obvious to some beginners. Rewriting according to the definition may seem trivial, but:
In beginning abstract math courses, |
When performing a calculation to solve a problem, you may want to
Look ahead to the form the solution must take |
Suppose you have a right triangle with legs $a$ and $b$ and hypoteneuse $c$. The Pythagorean Theorem says that
\[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]Derive the trig identity
\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]from the Pythagorean Theorem.
This is what you are given: \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]
Look ahead to what you want to prove: ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
By using the definition of sine and cosone, this means you want to prove
\[\frac{{{a}^{2}}}{{{c}^{2}}}+\frac{{{b}^{2}}}{{{c}^{2}}}=1\]You can do that this way: Take the given equation \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\] and divide through by ${{c}^{2}}$, getting
\[\frac{{{a}^{2}}}{{{c}^{2}}}+\frac{{{b}^{2}}}{{{c}^{2}}}=\frac{{{c}^{2}}}{{{c}^{2}}}\]But if $c\neq0$, then $\frac{{{c}^{2}}}{{{c}^{2}}}=1$, so the proof is finished. (What happens if $c=0$?)
How did I know to
divide by ${{c}^{2}}$?
Answer: I looked ahead to see what I needed to prove!
Dividing both sides of a correct equation by a nonzero number gives you another correct equation.
Look ahead to see what you want to get |
See method addiction and rabbits.
I learned this example from David A. Olson.
One characteristic of people that are good at math is that when they are faced with a new concept, they immediately start thinking of examples:
When you learn about a new concept |
Suppose you just learned the definition of prime number:
A positive integer $n\gt 1$ is prime if its only positive divisors are $1$ and $n$. (See definition of prime.)
You might have thoughts like this go through your head:
It is vital to generate example of any new concept (if you can!) because that is the fastest road to understanding the concept. See [Selden & Selden]
Suppose you need to know the largest integer n for which $n!\lt109$. One way to do it is to calculate: $4! = 24$, $5! = 120$, so the answer is $n=4$. When I gave a problem that came down to this calculation in my discrete math class, most students solved it correctly, but several wrote apologies on their paper for doing it by trial and error. Apologies are not necessary!
Trial and error is a valid method for doing math. |
In the case of this problem it is probably the easiest way to do it. Even if you need to know the largest integer for which $n\lt 70,000,000$ it makes sense to do it using trial and error with a calculator or a program such as Mathematica or Maple.
You don't have to start at $5!$ and work your way up.
This experiment may have made you realize how fast the factorial grows..
Experimenting with a computer may |
Guessing at the answer to a problem and then using a theorem to prove it is correct is legitimate. Some students don't believe this!
You need to find $\int{\sin x \cos x\,dx}$. You vaguely remember that it was $\sin^2 x$. But that is not a proof. So you differentiate $\sin^2 x$ using the chain rule and you get $2\sin x\cos x$. Oops. That's twice what it should be. So the answer must be $\frac{1}{2}\sin^2 x$. Check again (always check again!). The derivative of $\frac{1}{2}\sin^2 x$ is $\frac{1}{2}\cdot 2\cdot \sin x\cos x$ which is $\sin x\cos x$. The fundamental theorem of calculus tells you that $\frac{1}{2}\sin^2 x$ must be the correct answer.
This is a perfectly respectable way to do math.
It is good math behavior |
You need not use a specific method (for example substitution or integration by parts) to get an integral. Any way of coming up with the answer is OK as long as you can check it out (in this case by using differentiation).
A teacher might insist that you solve an integral problem by using substitution. Of course, if you practice doing five integrals by substitution you will get better at it. That didn’t stop me from being irritated by such practices when I was a student.
This work is licensed under a Creative Commons Attribution-ShareAlike 2.5 License.