Books For Understanding Books

I couldn't help myself.

I've bought another book.

The reason I bought it is to help me with the book I'm currently reading... but it's not as bad as it sounds.

The last Analysis query I had, I posted to the OU forums. As usual, my question was answered almost instantly in a concise and understandable way. Awesome!

However, there was another post a few days later suggesting that I read "How To Think About Analysis", another book by Lara Alcock.

howToThinkAboutAnalysis

Initially I thought it would be a bad idea to drop my current book about Analysis, and pick up a new one, but it turns out that after the first 50 pages, this book can be used as a companion to learning different sub-topics of Analysis. Specifically, after the first 50 pages, the remainder of the book is split into sections about Real Numbers, Sequences, Series, Continuity, Differentiability, and Integrability. Each sub-topic appears to be about 40-ish pages long, and can be read just before or during the actual study of each. Bite-size!

Although I've not quite finished the first section yet,  I've already added several new strategies to my set of learning techniques. Despite the fact that the knock-on effect is that the learning process may become slower, the idea is that my knowledge and understanding of the material will become much much deeper.

I'll try and give another review once I've completed the first part, the section on real numbers, and related it all to where I currently am in my larger Analysis text book.

Reading Mathematics

Still working through my book on Analysis (which I will be until July 2016, so I should probably stop mentioning it...), but I recently came across another proof that I had difficulty understanding. I had to reach out to the mathematics forums in the end, but after getting a reply and working through some further steps that they mentioned, everything fell into place.

I felt really good that I'd finally understood the proof, but I also tried to work out what I could've done to push for that answer myself (so hopefully next time, I won't have to post a question to the forum).

I realised that I might not be reading through the mathematics effectively enough. After reading through Lara Alcock's book I realised how important it is to make effective notes whilst reading through "all the symbols".  When reading through proofs I have this nasty habit of reading them like a novel, keeping this story of logic in my head... and then very quickly becoming confused because I didn't see how you could logically progress from one sentence to another.

This sounds really simple, and almost obvious, but I think all it really takes is to sit down with a pen and go through the mathematics of the proof in gritty detail, liberally re-arranging things as you go. It's worth mentioning that this is perhaps quite a different act than just "taking notes".

Being completely confounded by something only to solve the issue entirely on your own is enormously satisfying. The hope is that if I stay mindful and remain aware as to when to write the right kinds of notes, not only will I be able to solve more complicated problems on my own, but also in time these seemingly large logical steps will become second nature.

Anyway, doing independent reading of mathematics is proving, generally, to be really satisfying. Let's see if I can become a more effective reader...

 

 

Proof of Inequalities by Mathematical Induction

Still reading though my book in Analysis, I've come across a section on proving inequalities. I'm glad to say that all of this made sense... until I reached a sub-section on proving an inequality by mathematical induction.

As  I've written previously, I find that proofs are notoriously unintuitive. In the past however, I have been particularly puzzled by the logical steps involved in proving an inequality by mathematical induction.

To explain my difficulties, let's have a look at the example provided in the book:

 

Prove that 2^n \geq n^2, for n \geq 4.

If we're proving this by mathematical induction, we generally follow these steps:

  1. Let P(n) be the statement 2^n \geq n^2.
  2. Show that P(4) is true.
  3. Assume that P(k) is also true for k \geq 4.
  4. Show that P(k) \Rightarrow P(k+1). Or rather, show that if P(k) is true, then P(k+1) is also true.

Step 4 is the key step here in the proof as it shows that if any number is true, and the next number is also true, then you can apply this rule forever, and your original statement must be true for all numbers!

Anyway, lets jump to step 2. Show that P(4) is true. Well if n=4, then 2^4 = 16 and 4^2 = 16. So P(4) is true! Easy.

Let's look at step 3. Let's ASSUME that P(k) is true for some k \geq 4.

Now, this is the part that caught me by surprise... At this step, the text in the book reads as follows:

"So, we are assuming that 2^k \geq k^2 . Multiplying this inequality by 2 we get:

2^{k+1} \geq 2k^2,

so it is therefore sufficient to prove that 2k^2 \geq (k+1)^2."

Wait, what? How is it that all of a sudden, all we need to prove is that 2k^2 \geq (k+1)^2? This isn't explained explicitly in the text so I had to close the book and do a bit more thinking.

First thing I had to realise here is that the "Step 4" I've listed above requires a bit more detail... What you're actually trying to do is show that you can progress naturally from P(k) to P(k+1). ie: We should be able to show that we can progress naturally from:
2^k \geq k^2
to:
2^{k+1} \geq (k+1)^2.

Now, if we multiply 2^k \geq k^2 by 2, as mentioned in the text, we do arrive at:
2^{k+1} \geq 2k^2

This is good, as we've managed to get the 2^{k+1} we were looking for on the left-hand side of the inequality. But the right-hand side looks nothing like the right-hand side of P(k+1) ie: (k+1)^2.

Here's the key though... It doesn't matter they they're not the same. We only need to see how 2k^2 and (k+1)^2 relate to each other. Look back at Step 3. Part of this assumption is that k \geq 4. Just as a test, let's try k=4:

2k^2 = 2 \times 4^2 = 32
and
(k+1)^2 = (4+1)^2 = 25

Well this is interesting. It's looking as if 2k^2 \geq (k+1)^2. This is exactly what was written in the text!

But to really ram it home, what we really have now is the following:

2^{k+1} \geq 2k^2 \geq (k+1)^2

So.... this show us that IF we can prove that last bit (2k^2 \geq (k+1)^2 ) is true for all k \geq 4, and not just k=4 we have managed to prove that we can get from P(k) to P(k+1)!!! This is exactly why the text in the book said "so it is therefore sufficient to prove that 2k^2 \geq (k+1)^2."

I'm sure in future I'll jump on this immediately and say "oh yes, of course that's all we need to do now", but working through the derivation of why it was sufficient was extremely useful. Long-winded... but useful.

Ah, the learning process...