Introduction

Recently I have been reading the book "Thinking Mathematically" by John Mason at al.

I demonstrated using the method from the book to solve problems to a friend. He thought the process is what some immature student who cannot perform abstract thinking would do, like I am in elementary school. He said I should think about theorems and definitions and try to brings all the conditions together with them.

But I've solved so many problems by applying the methods described in the book. Previously I would just give up trying and ended up in failure on those problems.

An successful example of problem solving by thinking concretely

The problem

For example, suppose knowing for $a \le x \le b$,

1. $0 \le f(x) \le \pi$
2. $f'(x) \ge m > 0$

Prove$$\left|\int_{a}^{b}\sin f(x) dx\right| \le {2 \over m}$$

The process

(Specialization)Following the advices from the book, I should do specialization first. I interpret this as trying some functions that satisfy the constraints. Specializing $f'(x)$ to make it always equal to $m$, $f'(x)=m$, solving for $f(x)$, I got $f(x)= mx + c$, with a (further specialization) $m=1, a=0, b=\pi$, $f(x)$ then equals to $x$, the problem reduces to proving$$\left|\int_0^\pi \sin x dx\right| \le 2$$which is $|2| \le 2$, what? What does this specialization give me?

(Generalization) $a$ and $b$ could be any values, like $a=0, b = 1000000000$, this inequalities could still hold. But the function is strictly increasing! (Conjucturing) This means $m$ is inversely proportional to $b-a$. Another interesting finding is when $a=0 \le f(x) \le b=\pi$, $m$ cannot got larger than $1$. Maybe this indicates that $1 \over b-a$ would give an upper bound for $m$. If I rewrite the integral as$$\left|\int_{a}^{b}\sin f(x) dx\right|=(b-a)\left|\sin f(\xi)\right| \le b-a, \xi \in (a,b)$$(AHA moment) till here i suddenly got another idea to generalize the function to make it work for any $a$ and $b$$$f(x)={x - a \over b - a}\pi={\pi \over b - a}x - {a\pi \over b - a}$$so $m={\pi \over b - a}$ in this specific case, which provided further evidence for the earlier assumption that m is inverseily propertional to $b-a$. However, this doesn't provide any insight of how to prove the inequality, like, what does$$\left|\int_a^b\sin {x - a \over b - a}\pi dx\right| \le {2\pi \over b - a}$$even mean? Huh... since whatever inside $\sin$ is always 0 to $\pi$ and increasingly grows from $0$ to $\pi$, (AHA moment) maybe a substituion will bring some hope:$$u={x - a \over b - a}\pi$$wait ... then $du = mdx$! I just introduced $m$ one the left side, which might be a good sign!$$\left|\int_a^b\sin {x - a \over b - a}\pi dx\right|={1 \over m}\left|\int_0^\pi \sin u du\right| = {2 \over m}$$and because of symmetry, the integral$${1 \over m}\left|\int_0^\pi \sin u du\right|={2 \over m}\left|\int_0^{\pi \over 2} \sin u du\right|$$see! magically $2$ also popped out!

The final solution

The above process gives me inspiration to substitute $f(x)$$$u=f(x), du = f'(x)dx, dx={1 \over f'(x)}du \le {1 \over m} du$$

Applying the substation (as well as symmetry) the inequality is proved:$$\left|\int_a^b\sin f(x) dx\right|\le{2 \over m}\left|\int_0^{\pi\over 2}\sin u du\right| = {2 \over m}$$

My questions

1. Will this type of thinking take more time than abstract thinking?

Because my feeling is that though I can solve harder problems on which previously I cannot make any meaningful progress, but it did take more time. I think I solved problems in 20 minutes , though I learnt much more from the problem than I could before.

2. How an abstract thinker like my friend would approach this problem?

3. Should I avoid this low-level (as my friend thought) thinking and try to think abstractly?

I'm not willing to give up it though, since it just feels like that it suits me. It suits the way my brain works.