Understanding the Infinitesimal
Cheenta Notes in Mathematics
Adding infinitely many positive quantities, you may end up having something finite. Greeks did not understand this very well. Archimedes had some idea. Kerala school of mathematics under the leadership of Madhavacharya made real progress in refining this notion. They created the necessary groundwork for the advent of calculus later in Germany and Britain.
Indian Statistical Institute’s 2017 entrance for B.Stat and B.Math had a very simple problem based on infinitesimals. Adding little things, can you end up having more than ‘little’?
Here is the problem:
Given f : ℝ → ℝ be a continuous function such that for any two real numbers x and y,
|f(x) – f(y)| ≤ 7|x-y|201
Then:
(A) f(101) = f(202) + 8;
(B) f(101) = f(201) +1;
(C) f(101) = f(200) + 2;
(D) None of the above;
Before we solve the actual problem, lets have a fun detour. What if f is differentiable (the problem does not say that) ? Then take y = x + δ where δ > 0 Clearly, by the given condition:
| f(x) – f(x + δ) | ≤ 7 |x – (x + δ) |201 = 7δ201
⇒ |f(x) – f(x+δ) |/δ ≤ 7 δ200
⇒ limδ → 0 |f(x) – f(x+δ) |/δ ≤ limδ → 0200 = 0
⇒ |f'(x)| = 0
That means, if f is differentiable, then it’s derivative is 0, or in other words it is constant function. In that case f(x) = f(y) = c (a constant for all x and y). Hence the none of the first three options would hold. (Interestingly enough this is sufficient to choose (D) as the correct option as differentiable functions are an important subclass of continuous functions).
However we cannot assume differentiability as it is not mentioned in the problem. But now we have a hunch! We are already guessing that may be f is not changing much.
Suppose we want to check:
| f(101) – f(202) |
Chop off the distance between 101 to 202 into intervals of 0.1 unit long. There are 1010 such intervals (this is the ‘adding the little’ part). Why did I choose 0.1 length? Well, that is because it is a fractional length and raising a fraction to large powers will make it even smaller.
Now note that:
|f(101) – f(202)|
= |f(101) – f(101.1) + f(101.1) – f(101.2) + … + f(201.9) – f(202)|
≤ |f(101) – f(101.1)| + |f(101.1) – f(101.2)| + … + |f(201.9) – f(202)|
≤ 7|101-101.1|201 + … + 7|201.9-202|201
= 7 ( 0.1201 + … + 0.1201 )
= 7 × 1010 × 0.1201
= 7070/10201

But that implies |f(101) – f(202)| is much smaller 1 let alone 8. Using this same technique you can make |f(101) – f(202| smaller than 1 and f(101) – f(200)| smaller than 2. Hence the answer is option (D).
More interestingly, can you make the difference between f(x) and f(y) arbitrarily small? If you can do that then f(x) will be a constant function! This is something for you to think about this week.
Here is a possible way to think about it:
Taking smaller intervals. For example, for integers x and y chop off |x-y| into intervals of length 1/n. There will be |x-y|/(1/n) = n|x-y| intervals. Then you can use the above algorithm to compute:
|f(x) – f(y)|
≤ 7 × n|x-y|/n201
= 7 × |x-y|/n200
No matter how large |x-y| is, (by archimedean property) we can find a positive integer k such that

|x-y| < k × n

Try to finish it off from here. Let me know if you get something.