# Infinity Product Equality.

Let $\{I_n\}_{n\in\mathbb{N}}$ be a sequence of intervals in the form

$$I_n = \Big [ \frac{q_n}{b_n}, \frac{q_n + 1}{b_n} \Big),$$

where $q_{n}$ is some integer, for all $n\in\mathbb{N}$. Define the sequences of real numbers $\{ y_n \}_{n\in\mathbb{N}}$ and $\{ z_n \}_{n\in\mathbb{N}}$ by

$$y_n = \frac{q_n + 1}{b_n}\;\;\;\;\text{ and }\;\;\;\; z_n = \frac{q_n + 3/2}{b_n}.$$

I want to to find a upper bound for

$$\Delta_n = \Big | \prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi z_n) -\prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi y_n)- \sigma_i]\Big|$$

Where $\{a_n\}_{n\in\mathbb{N}}$ an $\{b_n\}_{n\in\mathbb{N}}$ are real number sequences such that

$$\sum a_n < \infty\;\;\;\;\text{and }\;\;\;\; b_n = \prod_{i=0}^{n}p_i,$$

$\{p_n\}_{n\in\mathbb{N}}$ is a prime sequence such that

$$\lim_{n\to\infty}\frac{2^n}{a_n p_n} = 0.$$

Johan Thim proved [2] that $$\Delta_n \leq \frac{b\pi}{p_n}2^{n-2},$$

by the following: Define the constants

$$a = \prod_{i=0}^{\infty}(1 - a_i)\;\;\;\;\text{ and }\;\;\;\; b=\prod_{i=0}^{\infty}(1+a_i).$$

and the partial products $$w_n(x) = \prod_{i=0}^{n}[1+a_i\sin(b_i\pi x)].$$

Given a natural $k>n$, the quotient $b_k/b_n$ is a even integer. Thus, for some $r_k\in\mathbb{Z}$,

$$\sin(b_k\pi y_n) = \sin(2r_k\pi(q_n+1)) = 0 \;\text{ and }\; \sin(b_k\pi z_n) = \sin(2r_k\pi(q_n + 3/2)) = 0.$$

And $k=n$, we have $$\sin(b_k \pi y_n) = 0\;\;\text{ and }\;\; \sin(b_k\pi z_n) = -(-1)^{q_n}.$$

Therefore, $$\Delta_n = W_{n-1}(z_n)[(1+a_n\sin(b_n\pi z_n))] - W_{n-1}(y_n)[(1+a_n\sin(b_n\pi y_n))]=$$ $$= W_{n-1}(z_n) - W_{n-1}(y_n) - (-1)^{q_n}(z_n)[1+a_n\sin(b_n \pi z_n)]$$

I understood everything until here. The trouble comes with the following steps:

Thus, there is a $\sigma_k\in\mathbb{R}$ such that

$$a_k\sin(b_k\pi z_n) = a_k\sin(b_k\pi y_n) + \sigma_k.$$

and for $|W_{n-1}(z_n) - W_{n-1}(y_n)|$, it's true that

$$|W_{n-1}(z_n) - W_{n-1}(y_n)| = \Big | \prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi z_n) -\prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi y_n)- \sigma_i]\Big| =$$ $$= \Big | \sum_{i=0}^{2^{n-1}-1}\sigma_ {l_i}\big( \prod_{j\in I_i}\sigma_j\big)\big( \prod_{j\in J_i}[1+a_j\sin(b_j\pi y_n)]\big)\Big | \leq \frac{\pi}{2p_n} \sum_{i=0}^{2^{n-1}-1} \Big (\prod_{j\in J_i}|1+a_j|\Big)\leq \frac{b\pi}{2p_n}(2^{n-1}-1)\leq\frac{b\pi}{p_n}2^{n-2}$$

For $I_i, J_i\subseteq \mathbb{N}$ and some natural $l_i$. Therefore $$|\Delta_n|\leq a_n\Big(a - \frac{2^{n-2}}{a_n p_n}b\pi\Big)$$

My question is concerning the step in which Thim states the equality

$$\Big | \prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi z_n) -\prod_{i=0}^{n-1}[1+a_i\sin(b_i\pi y_n)- \sigma_i]\Big| =$$ $$= \Big | \sum_{i=0}^{2^{n-1}-1}\sigma_ {l_i}\big( \prod_{j\in I_i}\sigma_j\big)\big( \prod_{j\in J_i}[1+a_j\sin(b_j\pi y_n)]\big)\Big |.$$

Why is that true?

[1] - Wen, Liu : A Nowhere Differentiable Continuous Function Constructed by Infinite Products - The American Mathematical Monthly, Vol. 109, No. 4 (Apr., 2002), pp. 378-380.

[2] - Thim, Johan: Continuous Nowhere Differentiable Functions - December 2003, Departament of Mathematics, LuleĆ„ University of Technology.

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This is not an answer, but you might like to look at this handout (math.cmu.edu/~bobpego/21131/nowhdiff.pdf) slightly simplifying Liu Wen's nice construction, which I prepared for a basic analysis class. – Bob Pego Mar 22 '12 at 16:57
Dear @BobPego, the page is off. – Paulo Henrique Mar 23 '12 at 12:19
Sorry! Try this one. – Bob Pego Mar 24 '12 at 15:04