User:Twoxili/sandbox

List of representations of e

As a binomial series

Consider the sequence:

$e_{n}=\left(1+{\frac {1}{n}}\right)^{n}$

By the binomial theorem:

$e_{n}=\sum _{k=0}^{n}{n \choose k}{\frac {1}{n^{k}}}=\sum _{k=0}^{n}{\frac {n^{\underline {k}}}{k!}}{\frac {1}{n^{k}}}$

which converges to $e$ as $n$ increases. The term $n^{\underline {k}}$ is the $k$th falling factorial power of $n$, which behaves like $n^{k}$ when $n$ is large. For fixed $k$ and as $\textstyle n\to \infty$:

${\frac {n^{\underline {k}}}{n^{k}}}\approx 1-{\frac {k(k-1)}{2n}}$

As a product of integrals

The definite integral of $\textstyle f(x)$ over the interval $\textstyle [0,1]$ can be approximated by the limit of a Riemann sum:

$\int _{0}^{1}f(x)\,dx=\lim _{k\to \infty }\sum _{i=1}^{k}f(x_{i})\Delta x$

For the function $\textstyle f(x)=2x$ over the interval $\textstyle [0,1]$, the Riemann sum with partition width $\textstyle \Delta x={\frac {1}{k}}$ is given as:

$U_{f,P_{k}}=\sum _{i=1}^{k}\left(2\cdot {\frac {i}{k}}\cdot {\frac {1}{k}}\right)={\frac {2}{k^{2}}}\sum _{i=1}^{k}i={\frac {k(k+1)}{k^{2}}}=1+{\frac {1}{k}}$

As $k$ approaches infinity, the term $\textstyle {\frac {1}{k}}$ approaches zero:

$1+{\frac {1}{k}}\to 1$

However, the sum $\textstyle 1+{\frac {1}{k}}$ is slightly greater than 1 for any finite $k$ before taking the limit. In the following limit:

$\lim _{k\to \infty }\left(\int _{0}^{1}2x\,dx\right)^{k}$

if we directly evaluate the integral as 1:

$\int _{0}^{1}2x\,dx=\left[x^{2}\right]_{0}^{1}=1^{2}-0^{2}=1$

and raise it to the power $k$, we obtain the precise result:

$\lim _{k\to \infty }1^{k}=e^{0}$

Instead, if we consider the Riemann sum approximation $\textstyle U_{f,P_{k}}=1+{\frac {1}{k}}$ and raise it to the power $k$, we have the product:

$U_{f}^{k}=\lim _{k\to \infty }\left(1+{\frac {1}{k}}\right)^{k}=e$

or $\textstyle L_{f}^{k}=e^{-1}$ for $\textstyle L_{f,P_{k}}=1+{\frac {-1}{k}}$, with 1 as the mean. The Riemann sum approximation, when raised to $k$ (the number of partitions), converges to powers of $e$ in the limit due to the effects of continuous compounding—where small increments precipitate exponential growth over an increasing number of partitions. The result also reflects the balancing interaction between the infinitesimal deviation $\textstyle {\frac {1}{k}}$ (which represents the difference between the theoretical value of the integral and the approximation) and the exponentiation by $k$, which prevents this difference from vanishing in the limit.

Euler's Constant

Properties

Relation to the zeta function

The constant $\gamma$ can also be expressed in terms of the sum of the reciprocals of non-trivial zeros $\rho$ of the zeta function:

$\gamma =\log 4\pi +\sum _{\rho }{\frac {2}{\rho }}-2$

Relation to triangular numbers

Numerous formulations have been derived that express $\gamma$ in terms of sums and logarithms of triangular numbers. One of the earliest of these is a formula for the $n$th harmonic number attributed to Srinivasa Ramanujan where $\gamma$ is related to $\textstyle \ln 2T_{k}$ in a series that considers the powers of $\textstyle {\frac {1}{T_{k}}}$ (an earlier, less-generalizable proof by Ernesto Cesàro gives the first two terms of the series, with an error term):

${\begin{aligned}\gamma &=H_{u}-{\frac {1}{2}}\ln 2T_{u}-\sum _{k=1}^{v}{\frac {R(k)}{T_{u}^{k}}}-\Theta _{v}\,{\frac {R(v+1)}{T_{u}^{v+1}}}\end{aligned}}$

From Stirling's approximation follows a similar series:

$\gamma =\ln 2\pi -\sum _{k=2}^{n}{\frac {\zeta (k)}{T_{k}}}$

The series of inverse triangular numbers also features in the study of the Basel problem posed by Pietro Mengoli. Mengoli proved that $\textstyle \sum _{k=1}^{\infty }{\frac {1}{2T_{k}}}=1$, a result Jacob Bernoulli later used to estimate the value of $\zeta (2)$, placing it between $1$ and $\textstyle \sum _{k=1}^{\infty }{\frac {2}{2T_{k}}}=\sum _{k=1}^{\infty }{\frac {1}{T_{k}}}=2$. This identity appears in a formula used by Bernhard Riemann to compute roots of the zeta function, where $\gamma$ is expressed in terms of the sum of roots $\rho$ plus the difference between Boya's expansion and the series of exact unit fractions $\textstyle \sum _{k=1}^{n}{\frac {1}{T_{k}}}$:

$\gamma -\ln 2=\ln 2\pi +\sum _{\rho }{\frac {2}{\rho }}-\sum _{k=1}^{n}{\frac {1}{T_{k}}}$

List of logarithmic identities

Calculus identities

Integral definition

$\ln x=\int _{1}^{x}{\frac {1}{t}}\ dt$

To modify the limits of integration to run from $x$ to $1$, we change the order of integration, which changes the sign of the integral:

$-\int _{1}^{x}{\frac {1}{t}}\,dt=\int _{x}^{1}{\frac {1}{t}}\,dt$

Therefore:

$\ln {\frac {1}{x}}=\int _{x}^{1}{\frac {1}{t}}\,dt$

Riemann Sum

$\ln(n+1)=$

$\lim _{k\to \infty }\sum _{i=1}^{k}{\frac {1}{x_{i}}}\Delta x=$

$\lim _{k\to \infty }\sum _{i=1}^{k}{\frac {1}{1+{\frac {i-1}{k}}n}}\cdot {\frac {n}{k}}=$

$\lim _{k\to \infty }\sum _{x=1}^{k\cdot n}{\frac {1}{1+x/k}}\cdot {\frac {1}{k}}=$

$\lim _{k\to \infty }\sum _{x=1}^{k\cdot n}{\frac {1}{k+x}}=\lim _{k\to \infty }\sum _{x=k+1}^{k\cdot n+k}{\frac {1}{x}}=\lim _{k\to \infty }\sum _{x=k+1}^{k(n+1)}{\frac {1}{x}}$

for $\Delta x=n/k$ and $x_{i}$ is a sample point in each interval.

Series representation

The natural logarithm $\ln(1+x)$ has a well-known Taylor series expansion that converges for $x$ in the open-closed interval $(-1,1]$:

$\ln(1+x)=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}x^{n}}{n}}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}-{\frac {x^{6}}{6}}+\cdots .$

Within this interval, for $x=1$, the series is conditionally convergent, and for all other values, it is absolutely convergent. For $x>1$ or $x\leq -1$, the series does not converge to $\ln(1+x)$. In these cases, different representations or methods must be used to evaluate the logarithm.

Harmonic number difference

It is not uncommon in advanced mathematics, particularly in analytic number theory and asymptotic analysis, to encounter expressions involving differences or ratios of harmonic numbers at scaled indices. The identity involving the limiting difference between harmonic numbers at scaled indices and its relationship to the logarithmic function provides an intriguing example of how discrete sequences can asymptotically relate to continuous functions. This identity is expressed as

$\lim _{k\to \infty }(H_{k(n+1)}-H_{k})=\ln(n+1)$

which characterizes the behavior of harmonic numbers as they grow large. This approximation (which precisely equals $\ln(n+1)$ in the limit) reflects how summation over increasing segments of the harmonic series exhibits integral properties, giving insight into the interplay between discrete and continuous analysis. It also illustrates how understanding the behavior of sums and series at large scales can lead to insightful conclusions about their properties. Here $H_{k}$ denotes the $k$-th harmonic number, defined as

$H_{k}=\sum _{j=1}^{k}{\frac {1}{j}}$

The harmonic numbers are a fundamental sequence in number theory and analysis, known for their logarithmic growth. This result leverages the fact that the sum of the inverses of integers (i.e., harmonic numbers) can be closely approximated by the natural logarithm function, plus a constant, especially when extended over large intervals. As $k$ tends towards infinity, the difference between the harmonic numbers $H_{k(n+1)}$ and $H_{k}$ converges to a non-zero value. This persistent non-zero difference, $\ln(n+1)$, precludes the possibility of the harmonic series approaching a finite limit, thus providing a clear mathematical articulation of its divergence. The technique of approximating sums by integrals (specifically using the integral test or by direct integral approximation) is fundamental in deriving such results. This specific identity can be a consequence of these approximations, considering:

$\sum _{j=k+1}^{k(n+1)}{\frac {1}{j}}\approx \int _{k}^{k(n+1)}{\frac {dx}{x}}$

Harmonic limit derivation

The limit explores the growth of the harmonic numbers when indices are multiplied by a scaling factor and then differenced. It specifically captures the sum from $k+1$ to $k(n+1)$:

$H_{k(n+1)}-H_{k}=\sum _{j=k+1}^{k(n+1)}{\frac {1}{j}}$

This can be estimated using the integral test for convergence, or more directly by comparing it to the integral of $1/x$ from $k$ to $k(n+1)$:

$\lim _{k\to \infty }\sum _{j=k+1}^{k(n+1)}{\frac {1}{j}}=\int _{k}^{k(n+1)}{\frac {dx}{x}}=\ln(k(n+1))-\ln(k)=\ln \left({\frac {k(n+1)}{k}}\right)=\ln(n+1)$

As the window's lower bound begins at $k+1$ and the upper bound extends to $k(n+1)$, both of which tend toward infinity as $k\to \infty$, the summation window encompasses an increasingly vast portion of the smallest possible terms of the harmonic series (those with astronomically large denominators), creating a discrete sum that stretches towards infinity, which mirrors how continuous integrals accumulate value across an infinitesimally fine partitioning of the domain. In the limit, the interval is effectively from $1$ to $n+1$ where the onset $k$ implies this minimally discrete region.

Double series formula

The harmonic number difference formula for $\ln(m)$ is an extension of the classic, alternating identity of $\ln(2)$:

$\ln(2)=\lim _{k\to \infty }\sum _{n=1}^{k}\left({\frac {1}{2n-1}}-{\frac {1}{2n}}\right)$

which can be generalized as the double series over the residues of $m$:

$\ln(m)=\sum _{x\in \langle m\rangle \cap \mathbb {N} }\sum _{r\in \mathbb {Z} _{m}\cap \mathbb {N} }\left({\frac {1}{x-r}}-{\frac {1}{x}}\right)=\sum _{x\in \langle m\rangle \cap \mathbb {N} }\sum _{r\in \mathbb {Z} _{m}\cap \mathbb {N} }{\frac {r}{x(x-r)}}$

where $\langle m\rangle$ is the principle ideal generated by $m$. Subtracting $\textstyle {\frac {1}{x}}$ from each term $\textstyle {\frac {1}{x-r}}$ (i.e., balancing each term with the modulus) reduces the magnitude of each term's contribution, ensuring convergence by controlling the series' tendency toward divergence as $m$ increases. For example:

$\ln(4)=\lim _{k\to \infty }\sum _{n=1}^{k}\left({\frac {1}{4n-3}}-{\frac {1}{4n}}\right)+\left({\frac {1}{4n-2}}-{\frac {1}{4n}}\right)+\left({\frac {1}{4n-1}}-{\frac {1}{4n}}\right)$

This method leverages the fine differences between closely related terms to stabilize the series. The sum over all residues $r\in \mathbb {N}$ ensures that adjustments are uniformly applied across all possible offsets within each block of $m$ terms. This uniform distribution of the "correction" across different intervals defined by $x-r$ functions similarly to telescoping over a very large sequence. It helps to flatten out the discrepancies that might otherwise lead to divergent behavior in a straightforward harmonic series.

Deveci's Proof

A fundamental feature of the proof is the accumulation of the subtrahends $\textstyle {\frac {1}{x}}$ into a unit fraction, that is, $\textstyle {\frac {m}{x}}={\frac {1}{n}}$ for $m\mid x$, thus $m=\omega +1$ rather than $m=|\mathbb {Z} _{m}\cap \mathbb {N} |$, where the extrema of $\mathbb {Z} _{m}\cap \mathbb {N}$ are $[0,\omega ]$ if $\mathbb {N} =\mathbb {N} _{0}$ and $[1,\omega ]$ otherwise, with the minimum of $0$ being implicit in the latter case due to the structural requirements of the proof. Since the cardinality of $\mathbb {Z} _{m}\cap \mathbb {N}$ depends on the selection of one of two possible minima, the integral $\textstyle \int {\frac {1}{t}}dt$, as a set-theoretic procedure, is a function of the maximum $\omega$ (which remains consistent across both interpretations) plus $1$, not the cardinality (which is ambiguous due to varying definitions of the minimum). Whereas the harmonic number difference computes the integral in a global sliding window, the double series, in parallel, computes the sum in a local sliding window—a shifting $m$-tuple—over the harmonic series, advancing the window by $m$ positions to select the next $m$-tuple, and offsetting each element of each tuple by $\textstyle {\frac {1}{m}}$ relative to the window's absolute position. The sum $\textstyle \sum _{n=1}^{k}\sum {\frac {1}{x-r}}$ corresponds to $H_{km}$ which scales $H_{m}$ without bound. The sum $\textstyle \sum _{n=1}^{k}-{\frac {1}{n}}$ corresponds to the prefix $H_{k}$ trimmed from the series to establish the window's moving lower bound $k+1$, and $\ln(m)$ is the limit of the sliding window (the scaled, truncated series):

$\sum _{n=1}^{k}\sum _{r=1}^{\omega }\left({\frac {1}{mn-r}}-{\frac {1}{mn}}\right)=\sum _{n=1}^{k}\sum _{r=0}^{\omega }\left({\frac {1}{mn-r}}-{\frac {1}{mn}}\right)$

$=\sum _{n=1}^{k}\left(-{\frac {1}{n}}+\sum _{r=0}^{\omega }{\frac {1}{mn-r}}\right)$

$=-H_{k}+\sum _{n=1}^{k}\sum _{r=0}^{\omega }{\frac {1}{mn-r}}$

$=-H_{k}+\sum _{n=1}^{k}\sum _{r=0}^{\omega }{\frac {1}{(n-1)m+m-r}}$

$=-H_{k}+\sum _{n=1}^{k}\sum _{j=1}^{m}{\frac {1}{(n-1)m+j}}$

$=-H_{k}+\sum _{n=1}^{k}\left(H_{nm}-H_{m(n-1)}\right)$

$=-H_{k}+H_{mk}$

$\lim _{k\to \infty }=H_{km}-H_{k}=\sum _{x\in \langle m\rangle \cap \mathbb {N} }\sum _{r\in \mathbb {Z} _{m}\cap \mathbb {N} }\left({\frac {1}{x-r}}-{\frac {1}{x}}\right)=\ln(\omega +1)=\ln(m)$

Asymptotic definitions

As a consequence of the harmonic number difference, the natural logarithm is asymptotically approximated by a finite series difference, representing a truncation of the infinite series at $k=n$:

$H_{2T[n]}-H_{n}\sim \ln(n+1)$

where $T[n]$ is the nth is triangular number, and $2T[n]$ is the sum of the first n even integers. Since the nth pronic number is asymptotically equivalent to the nth perfect square, it follows that:

$H_{n^{2}}-H_{n}\sim \ln(n+1)$

This bound converges more slowly than the bound involving the pronic number.

The prime number theorem provides the following asymptotic equivalence:

${\frac {n}{\pi (n)}}\sim \ln n$

where $\pi (n)$ is the prime counting function. This relationship is equal to:

${\frac {n}{H(1,2,\ldots ,x_{n})}}\sim \ln n$

where $H(x_{1},x_{2},\ldots ,x_{n})$ is the harmonic mean of $x_{1},x_{2},\ldots ,x_{n}$.

Pascal's triangle

Extensions

To arbitrary bases

Isaac Newton once observed that the first five rows of Pascal's Triangle, considered as strings, are the corresponding powers of eleven. He claimed without proof that subsequent rows also generate powers of eleven. In 1964, Dr. Robert L. Morton presented the more generalized argument that each row $n$ can be read as a radix $a$ numeral, where $\lim _{n\to \infty }11_{a}^{n}$ is the hypothetical terminal row, or limit, of the triangle, and the rows are its partial products. He proved the entries of row $n$, when interpreted directly as a place-value numeral, correspond to the binomial expansion of $(a+1)^{n}=11_{a}^{n}$. More rigorous proofs have since been developed. To better understand the principle behind this interpretation, here are some things to recall about binomials:

• A radix $a$ numeral in positional notation (e.g. $14641_{a}$) is a univariate polynomial in the variable $a$, where the degree of the variable of the $i$th term (starting with $i=0$) is $i$. For example, $14641_{a}=1\cdot a^{4}+4\cdot a^{3}+6\cdot a^{2}+4\cdot a^{1}+1\cdot a^{0}$.
• A row corresponds to the binomial expansion of $(a+b)^{n}$. The variable $b$ can be eliminated from the expansion by setting $b=1$. The expansion now typifies the expanded form of a radix $a$ numeral, as demonstrated above. Thus, when the entries of the row are concatenated and read in radix $a$ they form the numerical equivalent of $(a+1)^{n}=11_{a}^{n}$. If $c=a+1$ for $c<0$, then the theorem holds for $a=\{c-1,-(c+1)\}\;\mathrm {mod} \;2c$ with odd values of $n$ yielding negative row products.

By setting the row's radix (the variable $a$) equal to one and ten, row $n$ becomes the product $11_{1}^{n}=2^{n}$ and $11_{10}^{n}=11^{n}$, respectively. To illustrate, consider $a=n$, which yields the row product $n^{n}\left(1+{\frac {1}{n}}\right)^{n}=11_{n}^{n}$. The numeric representation of $11_{n}^{n}$ is formed by concatenating the entries of row $n$. The twelfth row denotes the product:

$11_{12}^{12}=1:10:56:164:353:560:650:560:353:164:56:10:1_{12}=27433a9699701_{12}$

with compound digits (delimited by ":") in radix twelve. The digits from $k=n-1$ through $k=1$ are compound because these row entries compute to values greater than or equal to twelve. To normalize the numeral, simply carry the first compound entry's prefix, that is, remove the prefix of the coefficient ${n \choose n-1}$ from its leftmost digit up to, but excluding, its rightmost digit, and use radix-twelve arithmetic to sum the removed prefix with the entry on its immediate left, then repeat this process, proceeding leftward, until the leftmost entry is reached. In this particular example, the normalized string ends with $01$ for all $n$. The leftmost digit is $2$ for $n>2$, which is obtained by carrying the $1$ of $10_{n}$ at entry $k=1$. It follows that the length of the normalized value of $11_{n}^{n}$ is equal to the row length, $n+1$. The integral part of $1.1_{n}^{n}$ contains exactly one digit because $n$ (the number of places to the left the decimal has moved) is one less than the row length. Below is the normalized value of $1.1_{1234}^{1234}$. Compound digits remain in the value because they are radix $1234$ residues represented in radix ten:

$1.1_{1234}^{1234}=2.885:2:35:977:696:\overbrace {\ldots } ^{\text{1227 digits}}:0:1_{1234}=2.717181235\ldots _{10}$

Other proposals for this edit

Note: add this citation for "to integers" section, for second approach to extension, borrowing from Hilton and Pedersen's

The Value of a Row subsection under Rows will be replaced with the following:

The $n$th row reads as the numeral $11_{a}^{n}$ for all $a$. See Extension to arbitrary bases.

The comment to this edit (the "Edit Summary") will be:

Replaced bullet point on powers of 11 with a more robust description. A discussion of this edit can be found on the Talk page.