A recurrence relation for the odd order moments of the Fabius function

S{\o}ren G. Have

arXiv:1703.01904·math.CA·March 7, 2017

A recurrence relation for the odd order moments of the Fabius function

S{\o}ren G. Have

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TL;DR

This paper derives recurrence relations for the odd order moments of the Fabius function, linking them to even moments and providing a closed-form for the coefficients, advancing understanding of its moment structure.

Contribution

It introduces a recurrence relation for the odd order moments of the Fabius function and a closed-form expression for the coefficients involved.

Findings

01

Recurrence relation for odd moments in terms of even moments

02

Matrix formulation of the recurrence relations

03

Closed-form expression for the coefficients

Abstract

A simple recurrence relation for the even order moments of the Fabius function is proven. Also, a very similar formula for the odd order moments in terms of the even order moments is proved. The matrices corresponding to these formulas (and their inverses) are multiplied so as to obtain a matrix that correspond to a recurrence relation for the odd order moments in terms of themselves. The theorem at the end gives a closed-form for the coefficients.

Equations76

F (0) = 0, F^{'} (x) = 2 F (2 x), \forall x \in [0, 1] : F (1 + x) = 1 - F (1 - x)

F (0) = 0, F^{'} (x) = 2 F (2 x), \forall x \in [0, 1] : F (1 + x) = 1 - F (1 - x)

P_{i} (x) = F (2^{- i} (1 + x)) - (- 1)^{i} F (2^{- i} (1 - x))

P_{i} (x) = F (2^{- i} (1 + x)) - (- 1)^{i} F (2^{- i} (1 - x))

P_{i}^{'} (x) =

P_{i}^{'} (x) =

2^{1 - i} (F (2^{1 - i} (1 + x)) - (- 1)^{i - 1} F (2^{1 - i} (1 - x)) = 2^{1 - i} P_{i - 1} (x)

P_{i + 1} (x) = 2^{- i} \int_{0}^{x} P_{i} (u) d u + P_{i + 1} (0) = 2^{- i} \int_{0}^{x} P_{i} (u) d u + {0 2 F (2^{- i - 1}) i is odd else

P_{i + 1} (x) = 2^{- i} \int_{0}^{x} P_{i} (u) d u + P_{i + 1} (0) = 2^{- i} \int_{0}^{x} P_{i} (u) d u + {0 2 F (2^{- i - 1}) i is odd else

c_{i} (0) = {0 2 F (2^{- i - 1}) i is odd else, c_{i} (j) = {2^{1 - i} c (i - 1, j - 1) / j 0 j < i else

c_{i} (0) = {0 2 F (2^{- i - 1}) i is odd else, c_{i} (j) = {2^{1 - i} c (i - 1, j - 1) / j 0 j < i else

c_{i} (j) = {\frac{2 ^{(1 - i) j} F ( 2 ^{j - i} )}{2 ^{(1 - j) j /2 - 1} j !} 0 0 \leq j < i and j - i is odd else

c_{i} (j) = {\frac{2 ^{(1 - i) j} F ( 2 ^{j - i} )}{2 ^{(1 - j) j /2 - 1} j !} 0 0 \leq j < i and j - i is odd else

j = 0 \sum i + 1 c_{i + 2} (j) = P_{i + 2} (1) = F (2^{- i - 1}) - (- 1)^{i + 2} F (0) = F (2^{- i - 1})

j = 0 \sum i + 1 c_{i + 2} (j) = P_{i + 2} (1) = F (2^{- i - 1}) - (- 1)^{i + 2} F (0) = F (2^{- i - 1})

F (2^{- i - 1}) =

F (2^{- i - 1}) =

2^{- i} F (2^{- i - 1}) + j = 1 \sum i /2 \frac{4 ^{j (j - i)} F ( 2 ^{2 j - i - 1} )}{2 ^{i + j} ( 2 j + 1 )!} ⟹ F (2^{- i - 1}) = j = 1 \sum i /2 \frac{4 ^{j (j - i)} F ( 2 ^{2 j - i - 1} )}{( 2 ^{i + j} - 2 ^{j} ) ( 2 j + 1 )!}

F (2^{- i}) = j = 0 \sum i c_{i + 1} (j) = j = 0 \sum i /2 c_{i + 1} (2 j) = j = 0 \sum i /2 \frac{2 ^{(1 - (i + 1)) 2 j} F ( 2 ^{2 j - (i + 1)} )}{2 ^{(1 - 2 j) (2 j) /2 - 1} ( 2 j )!} = j = 0 \sum i /2 \frac{4 ^{j (j - i)} F ( 2 ^{2 j - i - 1} )}{2 ^{j - 1} ( 2 j )!}

F (2^{- i}) = j = 0 \sum i c_{i + 1} (j) = j = 0 \sum i /2 c_{i + 1} (2 j) = j = 0 \sum i /2 \frac{2 ^{(1 - (i + 1)) 2 j} F ( 2 ^{2 j - (i + 1)} )}{2 ^{(1 - 2 j) (2 j) /2 - 1} ( 2 j )!} = j = 0 \sum i /2 \frac{4 ^{j (j - i)} F ( 2 ^{2 j - i - 1} )}{2 ^{j - 1} ( 2 j )!}

d_{i} = 2^{i (i + 1) /2} i! F (2^{- i - 1});

d_{i} = 2^{i (i + 1) /2} i! F (2^{- i - 1});

d_{i} =

d_{i} =

j = 0 \sum (i + 1) /2 \frac{2 ^{i (i + 1) /2} i !}{2 ^{j - 1} ( 2 j )!} \frac{4 ^{j (j - i - 1)} d _{i - 2 j + 1}}{2 ^{(i - 2 j + 1) ((i - 2 j + 1) + 1) /2} ( i - 2 j + 1 )!} = j = 0 \sum (i + 1) /2 (2 j i + 1) \frac{d _{2 j} / 2 ^{i}}{i + 1}

d_{i} =

d_{i} =

j = 1 \sum i /2 \frac{2 ^{i (i + 1) /2} i !}{( 2 ^{i + j} - 2 ^{j} ) ( 2 j + 1 )!} \frac{4 ^{j (j - i)} d _{i - 2 j}}{2 ^{(i - 2 j) ((i - 2 j) + 1) /2} ( i - 2 j )!} = j = 0 \sum i /2 - 1 (2 j i) \frac{d _{2 j} / ( 2 ^{i} - 1 )}{i - 2 j + 1}

μ_{n} = \int_{0}^{1} F (x) x^{n} d x = \frac{1}{n + 1} - d_{n}

μ_{n} = \int_{0}^{1} F (x) x^{n} d x = \frac{1}{n + 1} - d_{n}

\displaystyle\left(\begin{array}[]{c}d_{1}\\ d_{3}\\ \vdots\\ d_{2i-1}\end{array}\right)=M_{i}\left(\begin{array}[]{c}d_{0}\\ d_{2}\\ \vdots\\ d_{2i}\end{array}\right)\ \Longrightarrow\ \left(\begin{array}[]{c}2d_{0}\\ d_{1}\\ d_{3}\\ \vdots\\ d_{2i-1}\end{array}\right)=\left(\begin{array}[]{c}2e_{1}^{i+1}\\ M_{i}\\ \end{array}\right)\left(\begin{array}[]{c}d_{0}\\ d_{2}\\ \vdots\\ d_{2i}\end{array}\right)\hskip 113.81102pt

\displaystyle\left(\begin{array}[]{c}d_{1}\\ d_{3}\\ \vdots\\ d_{2i-1}\end{array}\right)=M_{i}\left(\begin{array}[]{c}d_{0}\\ d_{2}\\ \vdots\\ d_{2i}\end{array}\right)\ \Longrightarrow\ \left(\begin{array}[]{c}2d_{0}\\ d_{1}\\ d_{3}\\ \vdots\\ d_{2i-1}\end{array}\right)=\left(\begin{array}[]{c}2e_{1}^{i+1}\\ M_{i}\\ \end{array}\right)\left(\begin{array}[]{c}d_{0}\\ d_{2}\\ \vdots\\ d_{2i}\end{array}\right)\hskip 113.81102pt

\displaystyle\left(\begin{array}[]{c}d_{0}\\ d_{2}\\ \vdots\\ d_{2i+2}\end{array}\right)=\left(\begin{array}[]{c}I_{i+1}\\ R_{i+1}\\ \end{array}\right)\left(\begin{array}[]{c}2e_{1}^{i+1}\\ M_{i}\\ \end{array}\right)^{-1}\left(\begin{array}[]{c}2d_{0}\\ d_{1}\\ d_{3}\\ \vdots\\ d_{2i-1}\end{array}\right)\ \Longrightarrow\hskip 85.35826pt

\displaystyle\hskip 85.35826pt\left(\begin{array}[]{c}d_{1}\\ d_{3}\\ \vdots\\ d_{2i+1}\end{array}\right)=M_{i+1}\left(\begin{array}[]{c}I_{i+1}\\ R_{i+1}\\ \end{array}\right)\left(\begin{array}[]{c}2e_{1}^{i+1}\\ M_{i}\\ \end{array}\right)^{-1}\left(\begin{array}[]{c}2d_{0}\\ d_{1}\\ d_{3}\\ \vdots\\ d_{2i-1}\end{array}\right)\

\displaystyle G_{i+1}=M_{i+1}\left(\begin{array}[]{c}I_{i+1}\\ R_{i+1}\\ \end{array}\right)\left(\begin{array}[]{c}2e_{1}^{i+1}\\ M_{i}\\ \end{array}\right)^{-1}

\displaystyle G_{i+1}=M_{i+1}\left(\begin{array}[]{c}I_{i+1}\\ R_{i+1}\\ \end{array}\right)\left(\begin{array}[]{c}2e_{1}^{i+1}\\ M_{i}\\ \end{array}\right)^{-1}

(M_{i})_{k, j} = (2 j - 2 2 k) \frac{1}{4 ^{k} k}, (R_{i})_{j} = (2 j - 2 2 i) \frac{1/ ( 4 ^{i} - 1 )}{2 i - 2 j + 3}

(M_{i})_{k, j} = (2 j - 2 2 k) \frac{1}{4 ^{k} k}, (R_{i})_{j} = (2 j - 2 2 i) \frac{1/ ( 4 ^{i} - 1 )}{2 i - 2 j + 3}

(G_{i})_{i, j} = (\frac{E _{2 (i - j + 1)}}{4 ^{2 - j}} + \frac{ζ ( 2 j - 2 i - 3 )}{2 j - 2 i - 3} (1 - 4^{i - j + 2})) \frac{4}{1 - 4 ^{i}} {(2 j - 3 2 i - 1) 1/ (2 i) j \neq = 1 else

(G_{i})_{i, j} = (\frac{E _{2 (i - j + 1)}}{4 ^{2 - j}} + \frac{ζ ( 2 j - 2 i - 3 )}{2 j - 2 i - 3} (1 - 4^{i - j + 2})) \frac{4}{1 - 4 ^{i}} {(2 j - 3 2 i - 1) 1/ (2 i) j \neq = 1 else

d_{i} = j \in 2 N_{0} - 2 j < i - 1 \sum (\frac{ζ ( j - i )}{j - i} - \frac{2 ^{j} E _{i - j - 1}}{2 ^{i - j + 1} - 1}) \frac{2 ^{i - j + 3} - 4}{2 ^{i + 1} - 1} {(j + 1 i) d_{j + 1} 1/ (i + 1) j \neq = - 2 else

d_{i} = j \in 2 N_{0} - 2 j < i - 1 \sum (\frac{ζ ( j - i )}{j - i} - \frac{2 ^{j} E _{i - j - 1}}{2 ^{i - j + 1} - 1}) \frac{2 ^{i - j + 3} - 4}{2 ^{i + 1} - 1} {(j + 1 i) d_{j + 1} 1/ (i + 1) j \neq = - 2 else

\displaystyle\left(M_{i}\left(\begin{array}[]{c}I_{i}\\ R_{i}\\ \end{array}\right)\right)_{i,j}=(M_{i})_{i,j}+(M_{i})_{i,i+1}(R_{i})_{j}=\

\displaystyle\left(M_{i}\left(\begin{array}[]{c}I_{i}\\ R_{i}\\ \end{array}\right)\right)_{i,j}=(M_{i})_{i,j}+(M_{i})_{i,i+1}(R_{i})_{j}=\

(2 j - 2 2 i - 1) \frac{2 ^{1 - 2 i} ( j - i - 1 ) + 2 i - 2 j + 3}{( 1 - 4 ^{i} ) ( j - i - 1 ) ( 2 i - 2 j + 3 )}

\displaystyle\left(G_{i}\left(\begin{array}[]{c}2e_{1}^{i}\\ M_{i-1}\\ \end{array}\right)\right)_{i,j}=\left(\sum_{m=2}^{i}(G_{i})_{i,m}(M_{i-1})_{m-1,j}\right)+\begin{cases}2(G_{i})_{i,1}&j=1\\ 0&\text{else}\end{cases}=

\displaystyle\left(G_{i}\left(\begin{array}[]{c}2e_{1}^{i}\\ M_{i-1}\\ \end{array}\right)\right)_{i,j}=\left(\sum_{m=2}^{i}(G_{i})_{i,m}(M_{i-1})_{m-1,j}\right)+\begin{cases}2(G_{i})_{i,1}&j=1\\ 0&\text{else}\end{cases}=

m = 2 \sum i (\frac{E _{2 (i - m + 1)}}{4 ^{2 - m}} + \frac{ζ ( 2 m - 2 i - 3 )}{2 m - 2 i - 3} (1 - 4^{i - m + 2})) \frac{4}{1 - 4 ^{i}} (2 m - 3 2 i - 1) (2 j - 2 2 m - 2) \frac{4 ^{1 - m}}{m - 1} +

(\frac{E _{2 i}}{4} + \frac{ζ ( - 2 i - 1 )}{- 2 i - 1} (1 - 4^{i + 1})) \frac{4}{1 - 4 ^{i}} \frac{1}{i} {10 j = 1 else

\displaystyle\frac{1}{Q}\left(M_{i}\left(\begin{array}[]{c}i_{i}\\ R_{i}\\ \end{array}\right)\right)_{i,j}=\frac{2^{1-2i}(j-i-1)+2i-2j+3}{4(j-i-1)(2i-2j+3)}

\displaystyle\frac{1}{Q}\left(M_{i}\left(\begin{array}[]{c}i_{i}\\ R_{i}\\ \end{array}\right)\right)_{i,j}=\frac{2^{1-2i}(j-i-1)+2i-2j+3}{4(j-i-1)(2i-2j+3)}

\displaystyle\frac{1}{Q}\left(G_{i}\left(\begin{array}[]{c}2e_{1}^{i}\\ M_{i-1}\\ \end{array}\right)\right)_{i,j}=\left(\frac{E_{2i}}{4}+\frac{\zeta(-2i-1)}{-2i-1}(1-4^{i+1})\right)\frac{1}{i}\begin{cases}1&j=1\\ 0&\text{else}\end{cases}\ +\

\displaystyle\frac{1}{Q}\left(G_{i}\left(\begin{array}[]{c}2e_{1}^{i}\\ M_{i-1}\\ \end{array}\right)\right)_{i,j}=\left(\frac{E_{2i}}{4}+\frac{\zeta(-2i-1)}{-2i-1}(1-4^{i+1})\right)\frac{1}{i}\begin{cases}1&j=1\\ 0&\text{else}\end{cases}\ +\

m = 2 \sum i (\frac{E _{2 (i - m + 1)}}{4 ^{2 - m}} + \frac{ζ ( 2 m - 2 i - 3 )}{2 m - 2 i - 3} (1 - 4^{i - m + 2})) (2 m - 3 2 i - 1) (2 j - 2 2 m - 2) \frac{4 ^{1 - m}}{m - 1} / (2 j - 2 2 i - 1) =

m = 1 \sum i (\frac{E _{2 (i - m + 1)}}{4 ^{2 - m}} + \frac{ζ ( 2 m - 2 i - 3 )}{2 m - 2 i - 3} (1 - 4^{i - m + 2})) \frac{4 ^{1 - m}}{i - m + 1} (2 m - 2 j 2 i - 2 j + 1)

\displaystyle\frac{1}{Q}\left(M_{i}\left(\begin{array}[]{c}i_{i}\\ R_{i}\\ \end{array}\right)\right)_{i,j}=\frac{2^{-2i-1}}{2j+1}-\frac{1}{4j}

\displaystyle\frac{1}{Q}\left(M_{i}\left(\begin{array}[]{c}i_{i}\\ R_{i}\\ \end{array}\right)\right)_{i,j}=\frac{2^{-2i-1}}{2j+1}-\frac{1}{4j}

\displaystyle\frac{1}{Q}\left(G_{i}\left(\begin{array}[]{c}2e_{1}^{i}\\ M_{i-1}\\ \end{array}\right)\right)_{i,j}=\sum_{m=1}^{i}\left(\frac{E_{2m}}{4^{m-i+1}}+\frac{\zeta(-1-2m)}{-1-2m}(1-4^{m+1})\right)\frac{4^{m-i}}{m}\binom{2j-1}{2j-2m}

\frac{1}{4 j} - \frac{2 ^{- 2 i - 1}}{2 j + 1} + m = 1 \sum i (\frac{E _{2 m}}{4 ^{m - i + 1}} + \frac{B _{2 (m + 1)}}{2 ( m + 1 ) ( 2 m + 1 )} (1 - 4^{m + 1})) \frac{4 ^{m - i}}{m} (2 j - 2 m 2 j - 1) =

\frac{1}{4 j} - \frac{2 ^{- 2 i - 1}}{2 j + 1} + m = 1 \sum i (\frac{E _{2 m}}{4 ^{m - i + 1}} + \frac{B _{2 (m + 1)}}{2 ( m + 1 ) ( 2 m + 1 )} (1 - 4^{m + 1})) \frac{4 ^{m - i}}{m} (2 j - 2 m 2 j - 1) =

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TopicsMatrix Theory and Algorithms · Advanced Mathematical Theories and Applications · Scientific Research and Discoveries

Full text

A recurrence relation for the odd order moments of the Fabius function

Søren G. Have

A simple recurrence relation for the even order moments of the Fabius function is proven. Also, a very similar formula for the odd order moments in terms of the even order moments is proved. The matrices corresponding to these formulas (and their inverses) are multiplied so as to obtain a matrix that correspond to a recurrence relation for the odd order moments in terms of themselves. The theorem at the end gives a closed-form for the coefficients.

The Fabius function $F:\mathbb{R}_{0}^{+}\mapsto\mathbb{R}$ satisfies

[TABLE]

We define $P_{i}:[0,1]\mapsto\mathbb{R}$ by

[TABLE]

The 3rd property in (1) implies $P_{0}(x)=0$ . Moreover

[TABLE]

Hence

[TABLE]

Let $c_{i}(j)$ denote the coefficient of $x^{j}$ in $P_{i}(x)$ . It follows from (2) that

[TABLE]

Which has the solution

[TABLE]

It holds that

[TABLE]

Equating the RHS and LHS of (3) we get for $i\in 2\mathbb{N}$ the following formula for $F(2^{-i-1})$ :

[TABLE]

From the same equation we get for $i\in 2\mathbb{N}$ the following formula for $F(2^{-i})$ :

[TABLE]

The formulas depends on $F(1/2)$ which equals $1/2$ cf. the 3rd property in (1).

We define $d_{i}$ by

[TABLE]

Clearly $d_{0}=1/2$ , and if $i\in 2\mathbb{N}-1$

[TABLE]

If $i\in 2\mathbb{N}$ we get

[TABLE]

The following holds c.f. theorem 6 of [1].

[TABLE]

Hence the substitution $d_{i}=1/(i+1)-\mu_{i}$ turns any recurrence relation for $d_{i}$ into a recurrence relation for the moments of the Fabius function.

For all $i\in\mathbb{N}_{0}$ the recurrence relations above imply the following matrix forms, where $M_{i}$ is an $i\times(i+1)$ matrix, $R_{i}$ is a length $i$ row, $e_{1}^{i}$ is the length $i$ unit vector $(1,0,...,0)$ , and $I_{i}$ is the $i\times i$ identity matrix

[TABLE]

Invertibility is no issue, because a pseudo-inverse suffices. The last matrix form expresses $d_{2i+1}$ as a linear combination of previous odd-indexed $d_{i}$ s and $2d_{0}=1$ . The coefficients are the last row of the matrix:

[TABLE]

The $(k,j)$ ’th entry of $M_{i}$ , and the $j$ ’th entry of $R_{i}$ are known from the recurrence relations:

[TABLE]

Theorem:

For all $i\in\mathbb{N}$ the $j$ ’th column entry of the last row of $G_{i}$ is

[TABLE]

And so for all $i\in 2\mathbb{N}-1$ the following recurrence relation holds:

[TABLE]

Proof:

It suffices to show that the last row of $G_{i}\left(\begin{array}[]{c}2e_{1}^{i}\\ M_{i-1}\\ \end{array}\right)$ and $M_{i}\left(\begin{array}[]{c}I_{i}\\ R_{i}\\ \end{array}\right)$ are the same for $i\in\mathbb{N}$ .

Each entry in the last row for the second matrix product consists of just 2 terms:

[TABLE]

The last row for the first matrix product:

[TABLE]

At this point the last row for both matrix products are divided by $Q=\frac{4}{1-4^{i}}\binom{2i-1}{2j-2}$ :

[TABLE]

Next $j$ is substituted by $i-j+1$ and $m$ is substituted by $i-m+1$ :

[TABLE]

The rows are subtracted from one another and the $\zeta$ -function is expressed with Bernoulli numbers:

[TABLE]

We show that both parentheses equal 0. Simplifying the first parenthesis yields the summation:

[TABLE]

The sum is rewritten as the coefficient of a generating function:

[TABLE]

Hence from linear independence of the monomials we have that the first parenthesis is 0 for $j\in\mathbb{N}$ .

Rewriting the second parenthesis makes it

[TABLE]

the sum term of which is used as the coefficient of a generating function:

[TABLE]

Hence from linear independence of the monomials we have that

[TABLE]

∎

Bibliography1

The reference list from the paper itself. Each links out to its DOI / PubMed record.

1[1] Juan Arias de Reyna, An infinitely differentiable function with compact support: Definition and properties, Rev. Real Acad. Ciencias Madrid, 76 (1982) 21-38, English version: 1702.05442 [math.CA] .