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A mathematician’s mind-2

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Link to Chapter 1.



Section 2.1. Method of Decomposing N

As the name of the chapter say, here I will write about how to see things that are not visible. This is best explained by the following example. Recall Lemma 1.2 from Chapter 1 which states that

Lemma 1.2. If the sequence a_n is uniformly distributed modulo 1 then

\displaystyle{\lim _{n \rightarrow \infty} \frac{1}{n}\sum_{r=1}^{n} f(a_r)  = \int_{0}^{1}f(x)dx. }

Now let us rewrite the above result as an asymptotic formula

\displaystyle{\sum_{r=1}^{n} 1. f(a_r) \sim  n \int_{0}^{1}f(x)dx. }

This form looks pretty much normal and fails to reveal any hidden pattern. But if we break the above asymptotic form one step further as shown below;

\displaystyle{\sum_{r=1}^{n} 1. f(a_r) \sim  \sum_{r=1}^{n}1. \int_{0}^{1}f(x)dx }

then it begins to look more promising and starts revealing inner hidden patterns. Such a pattern should should immediately ignite a mathematical mind to ask oneself to ask how would the formula look if I replace the coefficient of f(a_r); which is currently 1; by a general term say d_r. Once we reach this question, we know that we are on the right track; we have found possible clues to a hidden pattern that could lead to a missing mathematical formula.

At this point we have no idea how the actual formula; if at all it exists; would look like. So the best approach at this point would be to a reasonable guess based on intuition and logic. In the above example, it is not hard to see that a reasonable guess should be

\displaystyle{\sum_{r=1}^{n} d_r f(a_r) \sim  \sum_{r=1}^{n}d_r \int_{0}^{1}f(x)dx. }

If it happens that more than one guess seems reasonable, take all of them for consideration for the time begin. Later we can eliminate the incorrect models in the verification step. A good mathematician should be a great guesser. If you can make ten good guess, probably one is right and you have found something to work on. Ramanujan himself relied very strongly on his intuition and made discoveries that probably would never have been made had he not possessed this ability.

Section 2.2. More on the method of decomposing N

Let us see one more example of the method of decomposing N this time slightly more advanced then in our previous example. The following result is well known.

Lemma 2.1If f is monotonic and continuous and defined in [1; n] then

\displaystyle{\sum_{r=1}^{n}f(n) = \int_{1}^{n}f(x)dx + O(|f(n)| + |f(1)|.}

Take a good look at this result and try to find out if there is any possibility of improvisation. Observe that f(n) and the integral on the RHS both contain an n. Let us decompose this as the sum of n 1’s. We write

\displaystyle{\sum_{r=1}^{n} 1.f(1+...+1) = \int_{1}^{1+...+1}f(x)dx + O(|f(1+...+1)| + |f(1)|).}

Now lets us work with our intuition and guess. Let d_r be a sequence of positive reals and let \sum_{k=1}^{r} d_k = S_r. The above decomposition of n should at one point or another lead us to guess that there could be a result which would look like

\displaystyle{\sum_{r=1}^{n} d_r f(S_r) = \int_{S_1}^{S_n}f(x)dx + O(|f(S_n)| + |f(S_1)|).}

Later we shall see that we have guessed the integral on the RHS is correctly while the error term is slightly incorrect. Nonetheless we have made a good guess based on which we can begin working.

Section 2.3. Verify your formula numerically

Now that our observation and intuition have led us to guess some new mathematical results, we should verify or test our guesses with a few numerical examples. The reason for doing this is that in case of explicit formulas like in our examples, it is advisable to gain some confidence in our guesses before we attempt to prove them rigorously. Rigorous proof are usually harder and more time consuming and we want attempt a proof only when we have sufficient confidence in our guesses. Remember that while verification for any number of cases cases does not prove a formula, one counter example is sufficient to disprove the formula.

I verified my guesses using simple d_r = 1/r, r^2 and f(a_r) = 1/(1+a_r ^2) where a_r is the sequence of the fractional part of r\sqrt 2 which is uniformly distributed modulo one as shown my the Equidistribution Theorem. In the age of computers we can verify a formula easily by writing a program (PARI, Mathematica or even using MS Excel).

Section 2.4. Prove your formula rigorously

After successful verification of the formula, we can begin to construct a rigorous mathematical proof. Yes there is absolutely no guarantee that we will indeed find a proof. But there is bright side of this. The harder it is to prove something, the more likely it is that we have stumbled upon something important. If we are unable to find a rigorous proof but our formula passes all verification tests thenwe can conjecture the result and hope that someone else finds a rigorous proof of disproof of our result. This is how mathematicians usually come up with conjectures. Perhaps the most example of this method is the Prime Number Theorem. Based on numerical observation, young Gauss conjectured that the number of primes not exceeding x is approximately equal to Li(x). Almost a century later, this conjecture was proved true by Hadamard and Charles de la Vallee Poussin.

Continuing with our example, after verifying the formulas, I became more confident that the formulas are indeed true and I began looking for a mathematical proof. This stage is the most important, not only because we might end up proving a new result but also because we may encounter many unforeseen cases that needs to be taken care of, especially the boundary conditions, error terms, exceptions etc. For example when I was working on the proof of the first formula (or first guess), I found that the result is not true if d_r = \sin (r), 1/r^2 etc. It was then then I realized that the series \sum_{r=1}^{\infty}d_r should be divergent for the first formula. Similarly in the second formula, I had guessed the error term incorrectly and it was only during the proof stage that I could find the correct error term. Such intricate details might escape the verification stage and unless we are careful in the proof stage, our formula would be incomplete. Finally I could prove the following results rigorously; the details of which can be found in the paper ‘On a unified theory of numbers‘.

Lemma 2.2. If d_r is a divergent series of positive terms and a_r is uniformly distributed modulo 1 then

\displaystyle{\sum_{r=1}^{n} d_r f(a_r) \sim  \sum_{r=1}^{n}d_r \int_{0}^{1}f(x)dx. }

Notice that with Lemma 2.1, we have generalized Lemma 1.2.

Lemma 2.3If f is monotonic and continuous and defined in [1; n] and d_r is a series of positive terms such that \sum_{k=1}^{r} d_k = S_r; then

\displaystyle{\sum_{r=1}^{n} d_r f(S_r) = \int_{S_1}^{S_n}f(x)dx + O(\delta (n))}


\displaystyle{\delta (n) = d_n |f(S_n)| + d_1 |f(S_1)| + \int _{1}^{n} \frac{d(d_y)}{dy} S(y)dy.}


\displaystyle{\lim_{n \rightarrow \infty} \sum_{r < n} \frac{d_r}{S_n\sqrt{\ln S_n - \ln S_r}} = \sqrt{\pi}}.

\displaystyle{\sum_{r \le n}{p_r}^{a}({p_1}^{a}+{p_2}^{a} + \ldots + {p_r}^{a})^{b}\sim \frac{n^{b+1} p_n^{ab+a}}{(b+1)(a+1)^{b+1}}}.

Notes at the end of Chapter 2

In this chapter we saw how breaking numbers using the method of decomposition throws light on how well know results in mathematics could be improved. Writing n as \sum_{r=1}^{n} 1 is an explicit example of this. This does not mean that this is the only trick. What the reader should grasp here is the idea of decomposition. A mathematical explorer should always try to visualize results by improvising and by breaking the results into simpler forms in one or more ways.


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