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Some series on Fibonacci and Lucas numbers

March 16, 2011 7:32 am / Leave a comment

In this post, we shall see several curious summation formulas of the Fibonacci numbers $F_n$ and the related Lucas numbers $L_n$ that involves the Bell polynomials $B_n(x)$ and the incomplete gamma function $\Gamma(a,x)$ . The Fibonacci numbers are defined as $F_0=0, F_1=1, F_{n+2}=F_{n+1} + F_n, n \ge 0$ and the Lucas numbers are defined as $L_0=2, L_1=1, L_{n+2}=L_{n+1} + L_n, n \ge 0$ .

Bell polynomials

The Bell polynomials $B_n(x)$ of polynomials are defined by the exponential generating function

$\displaystyle{\sum_{n=0}^{\infty}\frac{B_n(x)t^n}{n!}=e^{(e^t-1)x}}$ .

The first few Bell polynomials are

$B_0(x) = 1$

$B_1(x) = x$

$B_2(x) = x^2 + x$

$B_3(x) = x^3 + 3x^2 + x$

$B_4(x) = x^4 + 6 x^3 + 7x^2 + x$

$B_5(x) = x^5 +10x^4 + 25 x^3 + 15x^2 + x$ .

Incomplete gamma function

The incomplete gamma function $\Gamma(a,x)$ is defined as

$\displaystyle{\Gamma(a,x) = \int_{x}^{\infty}t^{a-1}e^{-t}dt}$ .

Entry 1-6 involve summation identities involving the Bell polynomials or the incomplete gamma function.

As usual, $\displaystyle{\phi=\frac{1+\sqrt5}{2}}$ denotes golden ratio. We have the following results:

Entry 1.

$\displaystyle{\sum_{r=1}^{\infty}\frac{F_r r^n}{r!} = \frac{e^{-1/\phi}}{\sqrt5}\{e^{\sqrt5} B_n(\phi)-B_n(-1/\phi)\}}$ .

Entry 2.

$\displaystyle{\sum_{r=1}^{\infty}\frac{L_r r^n}{r!} = e^{-1/\phi}\{e^{\sqrt5} B_n(\phi)+B_n(-1/\phi)\}}$ .

Entry 3.

$\displaystyle{\sum_{k=0}^{n-1}\frac{F_{k+m}x^k}{k!} = \frac{(-1)^me^{-x/\phi}}{\phi^m\sqrt5(n-1)!}\{(-\phi^2)^m e^{\sqrt5x}\Gamma(n,\phi x) - \Gamma(n,-x/\phi)\}}$

Entry 4.

$\displaystyle{\sum_{k=0}^{n-1}\frac{F_{k+m}(-x)^k}{k!} = \frac{(-1)^m e^{-x\phi}}{\phi^m\sqrt5(n-1)!}\{(-\phi^2)^m \Gamma(n,-\phi x) - e^{\sqrt5x}\Gamma(n,x/\phi)\}}$

Entry 5.

$\displaystyle{\sum_{r=0}^{\infty}\frac{F_{r+m} x^r}{r!} = \frac{e^{-x/\phi}}{\sqrt5}\{\phi^m e^{\sqrt5x}-(-\phi)^{-m}\}}$

Entry 6.

$\displaystyle{\sum_{r=0}^{\infty}\frac{F_{r+m} (-x)^r}{r!} = \frac{e^{-x\phi}}{\sqrt5}\{\phi^m -e^{\sqrt5x}(-\phi)^{-m}\}}$

Entry 7.

$\displaystyle{\sum_{r=0}^{\infty}\frac{F_r x^r}{r!} = -e^x \sum_{r=1}^{\infty}\frac{F_r (-x)^r}{r!}}$

Entry 8.

$\displaystyle{\sum_{k=0}^{n-1}\frac{L_{k+m}x^k}{k!} = \frac{(-1)^m e^{-x/\phi}}{\phi^m (n-1)!}\{(-\phi^2)^m e^{\sqrt5x}\Gamma(n,\phi x) + \Gamma(n,-x/\phi)\}}$

Entry 9.

$\displaystyle{\sum_{k=0}^{n-1}\frac{L_{k+m}(-x)^k}{k!} = \frac{(-1)^m e^{-x\phi}}{\phi^m (n-1)!}\{(-\phi^2)^m \Gamma(n,-\phi x) + e^{\sqrt5x}\Gamma(n,x\phi)\}}$

Entry 10.

$\displaystyle{\sum_{r=0}^{\infty}\frac{L_{r+m} x^r}{r!} = e^{-x/\phi}\{\phi^m e^{\sqrt5x}+(-\phi)^{-m}\}}$

Entry 11.

$\displaystyle{\sum_{r=0}^{\infty}\frac{L_{r+m} (-x)^r}{r!} = e^{-x\phi}\{\phi^m + e^{\sqrt5x}(-\phi)^{-m}\}}$

Entry 12.

$\displaystyle{\sum_{r=0}^{\infty}\frac{L_r x^r}{r!} = e^x \sum_{r=1}^{\infty}\frac{L_r (-x)^r}{r!}}$

Identities on the zeta function and harmonic numbers

March 14, 2011 11:23 am / Leave a comment

Leonard Euler gave a general identity involving the harmonic number $H_n$ and the Riemann zeta function $\zeta(n)$ for natural numbers $n$ .

$\displaystyle{2\sum_{n=1}^{\infty}\frac{H_n}{n^m} = (m+2)\zeta(m+1)-\sum_{k=1}^{m-2}\zeta(m-k)\zeta(k+1)}$ .

A special case of this is the identity

$\displaystyle{\sum_{n=1}^{\infty}\frac{H_n}{n^2} = 2\zeta(3)}$ .

I first saw the above identity in Alex Youcis’s blog Abstract Nonsense and in course of further investigation, I was able to find several identities involving the Riemann zeta function and the harmonic numbers. While it is practically impossible to go through the entire mathematical literature to see if a formula is new or a rediscovery, in this post we shall see a few identities which are not given in the above links. Also if I encounter an identity elsewhere at any point of time, I have taken care to delete it from this post.

For $n \ge 1$ the following relations hold.

Entry 1.

$\displaystyle{\sum_{r=1}^{\infty}H_r \Big\{\frac{1}{r^{2n}} - \frac{1}{(r+1)^{2n}}\Big\}=\zeta(2n+1)}$ .

Entry 2.

$\displaystyle{\sum_{r=1}^{\infty}H_r \Big\{\frac{1}{r^{4n-1}} - \frac{1}{(r+1)^{4n+1}}\Big\}=\zeta(4n)}$ .

Entry 3.

$\displaystyle{\sum_{r=1}^{\infty}H_r \Big\{\frac{1}{r^{4n+1}} - \frac{1}{(r+1)^{4n+1}}\Big\}=\zeta(4n+2)}$ .

Entry 4.

$\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+1)^{2n}} = n\zeta(2n+1) - \sum_{k=2}^{n}\zeta(k)\zeta(2n+1-k)}$ .

Entry 5.

$\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+1)^{4n-1}} = \frac{4n-3}{4}\zeta(4n) - \sum_{k=1}^{n-1}\zeta(2k+1)\zeta(4n-2k-1)}$ .

Entry 6.

$\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+1)^{4n+1}}= \frac{4n-1}{4}\zeta(4n+2)-\frac{\zeta(2n+1)^2}{2}}$

$\displaystyle{-\sum_{k=1}^{n-1}\zeta(2k+1)\zeta(4n-2k+1)}$ .

Entry 7.

$\displaystyle{\sum_{r=1}^{\infty}\frac{H_r}{(r+2)^{2n}} = n\zeta(2n+1) - 2n - \sum_{k=2}^{n}\zeta(k)\zeta(2n+1-k) + \sum_{k=2}^{2n}\zeta(k)}$ .

Entry a. If $|x| < 1$ then

$\displaystyle{\sum_{n=1}^{\infty}\zeta(4n)x^{4n} = \frac{1}{2}-\frac{\pi x}{4}(\cot(\pi x) +\coth(\pi x))}$ .

On the real roots of ζ(x) = a

March 9, 2011 7:07 am / Leave a comment

Let $\mathbb N$ be the set of natural numbers and $\mathbb Q^+$ be the set of positive rational numbers. For every positive integer $n \ge 2$ , there exists a unique real number $s_n$ such that $\zeta (s_n) = n$ . The question is does there exists an $n \in \mathbb N$ such that $s_n \in \mathbb Q^+$ be rational? In other words, does there exist a positive rational $\frac{p}{q}$ such that $\zeta(\frac{p}{q}) \in \mathbb N$ ? At present, we don’t even know if $s_n$ is always rational, let alone irrational or transcendental. Investigation seems to suggest that $\zeta(\frac{p}{q}) \notin \mathbb N$ for any $\frac{p}{q} \in \mathbb Q^+$ . However this has not been completely established. In this post, we shall look at some on the results in this line of research.

Expanding the Riemann zeta function about $s=1$ gives

$\displaystyle{\zeta(s) = \frac{1}{s-1} + \gamma_0 + \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\gamma_n (s-1)^n}$

where the constants

$\displaystyle{\gamma_n = \lim_{m \rightarrow \infty}\Bigg [\sum_{k=1}^{m}\frac{(\ln k)^n}{k} - \frac{(\ln m)^{n+1}}{n+1}\Bigg]}$

are known as Stieltjes constants. Inverting the above series we obtain:

Theorem 1. Let $x > 1$ and $\zeta(x) = a$ ; then,

$\displaystyle{x = 1 + \frac{1}{a-\gamma _0} - \frac{\gamma _1}{1!(a-\gamma _0)^2} + \frac{\gamma _2}{2!(a-\gamma _0)^3}- \frac{\gamma _3 - 12\gamma _1}{3!(a-\gamma _0)^4} + O\Big(\frac{1}{a^5}\Big)}$ .

Proof. The proof follows by inverting the Stieltjes series expansion of $\zeta (x)$ . $\Box$

We can rewrite the Stieltjes series expansion of the Riemann zeta function in the following form which we shall use regularly in our analysis:

$\displaystyle{\zeta \Big (1+\frac{1}{s}\Big ) = s + \gamma_0 + \sum_{n=0}^{\infty}\frac{(-1)^n\gamma_n}{n!^n}}$ .

We want to find an interval where the function $\zeta(1+\frac{1}{s})$ is strictly decreasing. Clearly $\zeta(1+\frac{1}{s})$ is continuous and strictly decreasing in the interval $(0, \infty)$ . The trivial zeros of $\zeta(s)$ occur at $s=-2n$ where $n \in N$ . In other words, the trivial zeros of $\zeta(1+\frac{1}{s})$ occur at $s = \frac{-1}{2n+1}$ and all these zeros lie in the interval $(-1/3, 0)$ . Hence in this interval, $\zeta(1+\frac{1}{s})$ will have infinitely many points of inflexion. Excluding the interval $(-1/3, 0)$ , we can divide the real line in to two parts $(-\infty, -1/3)$ and $(0,\infty)$ ; and in each of these two intervals, $\zeta(1+\frac{1}{s})$ in strictly decreasing. For positive rationals $\frac{p}{q}$ , the scope of our current study is the interval $(0,\infty)$ .

Lemma 2. For every finite real $a>1$ , there exists a unique real number $c_a, 0< c_a < 1-\gamma _0$ , such that

$\displaystyle{\zeta\Big(1+\frac{1}{a-1 +c_a}\Big) = a}$ .

Proof. Let lemma follows form the fact that in our interval of interest $(0,\infty)$ , $\zeta (1+\frac{1}{s}) - s$ is strictly decreasing. $\Box$

Theorem 3. Let $c_a$ be as defined in Lemma 2. The asymptotic expansion of $c_a$ is

$\displaystyle{c_a = 1-\gamma_0 + \frac{\gamma_1}{a-1} - \frac{\gamma_2 + \gamma_1-\gamma_1\gamma_0}{(a-1)^2}}$

$\displaystyle{+ \frac{\gamma_3 + 2\gamma_2- 2\gamma_2\gamma_0 + \gamma_1 - 2\gamma_1\gamma_0 + \gamma_1\gamma_0^2 - \gamma_1^2}{(a-1)^3}+ O\Big(\frac{1}{a^4}\Big)}.$

Proof. This is obtained by inverting the above Stieltjes series expansion of $\zeta (1+\frac{1}{a-1+c_a})$ . $\Box$

The next result gives us a method to compute $c_a$ numerically. We have

Lemma 4. Let $k_r$ be defined recursively as

$\displaystyle{k_{r+1} = a + k_r - \zeta \Big (\frac{a+k_r}{a-1+k_r}\Big )}$

where $k_0 = 1-\gamma_0$ . Then $\lim_{r \rightarrow \infty}k_r = c_a$ .

Marek Wolf, has sent me the computed values of $c_n$ for $2 \le c_n \le 1000$ . Some of these numbers were used in the proof of the following theorem.

Proof. Trivial. $\Box$

Theorem 5. If $1 \le p-q \le 40$ then $\zeta (\frac{p}{q}) \notin \mathbb N$ .

Proof. We have $c_2 \approx 0.37240621590$ . There are exactly twenty three positive fractions $\frac{m}{n}$ with $n \le 40$ such that

$\displaystyle{c_2 \le \frac{m}{n} < 1- \gamma_0}$ .

For each of these fractions, we find (by actual computation) that there exists a integer $r \le 43$ such that

$\displaystyle{c_r \le \frac{m}{n} < c_{r+1}}$ .

Since $c_n$ is strictly increasing, it implies that if $c_n \in \mathbb Q^+$ then the denominator of $c_n$ has to be $\ge 41$ . Hence the theorem follows. $\Box$

If we continue working with the idea given the proof of Theorem 5, we can prove $p-q > n_0$ for some large natural number $n_o$ . But this method gives no hope of proving that $\zeta (\frac{p}{q}) \notin \mathbb N$ . Therefore we must look for some other ideas.

Theorem 6. Let $\zeta (1+\frac{1}{x_i}) \in \mathbb N$ , $x_i \in \mathbb R^+ , i=1, 2, \ldots , k$ and $a_j \in \mathbb N$ such that

$\sum a_j = \{1, 2, 3, 4, 5, 7, 8, 10, 12 , 15\}$ .

If $y=m+\sum a_k x_k$ where is any non-negative integer; then $\zeta (1+\frac{1}{y}) \notin \mathbb N$ .

Some applications of Theorem 6 are given below.

Example 7. If $x \in \mathbb R^+$ and $2 \le n \le 5, n \in \mathbb N$ then at least one of the two numbers

$\zeta (1+\frac{1}{x})$ and $\zeta (1+\frac{1}{nx})$

is not a integer.

Example 8. If $x, y \in \mathbb R^+$ then at least one of the three numbers

$\zeta (1+\frac{1}{x}), \zeta (1+\frac{1}{y})$ and $\zeta (1+\frac{1}{x+y})$

is not a integer.

Dividing an angle into an arbitrary ratio

March 8, 2011 9:04 am / Leave a comment

Back in my school days, probably in the year 1997, I first read about a famous and ancient geometrical problem, the problem of trisecting an angle using only a compass and an unmarked ruler. It took me some time to understand that a general method for trisecting all angles is impossible arbitrary angle $\theta$ . Of course with a marked ruler, we can always trisect an arbitrary angle, and moreover, in case of some special angles such as $\pi$ , $\pi/2$ or some non-constructible angles such as $3\pi/7$ , we can trisect even with an unmarked ruler. I found some approximate trisection methods, each of which gave a dominant term $\theta/3$ and an error terms that was much smaller than the dominant term. At that time, I thought is it possible to find a method that gives an approximate division of an arbitrary angle into an arbitrary ratio. In formal terms, this problem can be stated as:

Problem: Let $n$ be positive integer and $x$ be any positive real. Given an angle $\theta$ and a line that divides it in the ratio $1/x$ find a method with finite number of steps to construct an line that approximately divides the angle in the ratio $1/(x+n)$ using only a compass and an unmarked ruler.

Solution: We shall use the notation $[A]$ to denote the angle $A$ . In the diagram let $[COB]=\theta$ be an acute angle and let $[IOB]=\theta/x$ be given.

Steps:

With O as center and radius OC, draw an arc to cut OB at D.
Join CD and draw DJ perpendicular to OB.
Let OI intersect DJ at I and the arc CD at K.
Produce DK to intersect OC at G.
Let OI intersect CD at E.
Extent GE to intersect OB at H.
Join IH to intersect GD at F.
Draw a line OA through F. $[AOB] \approx \theta/(x+1)$ .
Repeating this method $n$ times, to construct a line that approximately divides $\theta$ in the ratio $1/(x+n)$ .

As $\theta$ increases, the lines OC and DG tend to become parallel to each other and therefore the paper size required for construction increases. To overcome this problem, we can consider an obtuse angle as the sum of a right angle and an acute angle and apply our method on each part separately and then add up the final angle resulting form each of the two parts.

A conjecture on consecutive primes

March 7, 2011 6:24 pm / Leave a comment

In 1982, Farideh Firoozbakht made an interesting conjecture that the sequence $p_n ^{1/n}$ decreases i.e. for all $n \ge 1$ ,

$p_n ^{\frac{1}{ n}} > p_{n+1} ^{\frac{1}{n+1}}$ .

I present a stronger form of Firoozbakht’s conjecture.

Conjecture: Let $p_n$ be the n-th prime and let $r(n) = p_n^{1/n}$ .

$\displaystyle{\lim_{n \rightarrow \infty} \frac{n^2}{\ln n} \Big(\frac{r(n)}{r(n+1)}-1\Big) = 1}.$

If the above conjecture is true than we can have explicit bounds on $p_n ^{1/n}$ in terms of $p_{n+1}$ .

Corollary 1: For every $\epsilon, 0<\epsilon<1$ , there exists a sufficiently large natural number $N_{\epsilon}$ , which depends only on $\epsilon$ , such that for all $n>N_{\epsilon}$ ,