List of logarithmic identities

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Template:Short description In mathematics, many logarithmic identities exist. The following is a compilation of the notable of these, many of which are used for computational purposes.

Trivial identities

Trivial mathematical identities are relatively simple (for an experienced mathematician), though not necessarily unimportant. The trivial logarithmic identities are as follows:

if <math>\log_b(1) = 0 </math> then <math> b^0 = 1</math>
if <math>\log_b(b) = 1 </math> then <math> b^1 = b</math>

Explanations

By definition, we know that:

<math display="block">\log_b(y) = x \iff b^x = y,</math>

where <math>b \neq 0 </math> and <math>b \neq 1</math>.

Setting <math>x = 0</math>, we can see that:

<math display="block"> b^x = y \iff b^{(0)} = y \iff 1 = y \iff y = 1 </math>

So, substituting these values into the formula, we see that:

<math display="block"> \log_b (y) = x \iff \log_b (1) = 0, </math>

which gets us the first property.

Setting <math>x = 1</math>, we can see that:

<math display="block"> b^x = y \iff b^{(1)} = y \iff b = y \iff y = b </math>

So, substituting these values into the formula, we see that:

<math display="block"> \log_b (y) = x \iff \log_b (b) = 1, </math>

which gets us the second property.

Cancelling exponentials

Logarithms and exponentials with the same base cancel each other. This is true because logarithms and exponentials are inverse operationsTemplate:Sndmuch like the same way multiplication and division are inverse operations, and addition and subtraction are inverse operations:<ref>Template:Cite web</ref>

<math display="block">b^{\log_b(x)} = x\text{ because }\mbox{antilog}_b(\log_b(x)) = x</math> <math display="block">\log_b(b^x) = x\text{ because }\log_b(\mbox{antilog}_b(x)) = x</math>

Both of the above are derived from the following two equations that define a logarithm: (note that in this explanation, the variables of <math>x</math> and <math>x</math> may not be referring to the same number)

<math display="block">\log_b (y) = x \iff b^x = y</math>

Looking at the equation <math> b^x = y </math>, and substituting the value for <math>x</math> of <math> \log_b (y) = x </math>, we get the following equation:

<math display="block"> b^x = y \iff b^{\log _b(y)} = y \iff b^{\log_b (y)} = y, </math>

which gets us the first equation.

Another more rough way to think about it is that <math> b^{\text{something}} = y</math>, and that that "<math>\text{something}</math>" is <math> \log_b (y) </math>.

Looking at the equation <math> \log_b (y) = x</math>, and substituting the value for <math> y </math> of <math>b^x = y</math>, we get the following equation:

<math display="block"> \log_b (y) = x \iff \log_b(b^x) = x \iff \log_b(b^x) = x, </math>

which gets us the second equation.

Another more rough way to think about it is that <math> \log_b (\text{something}) = x</math>, and that that something "<math>\text{something}</math>" is <math> b^x</math>.

Using simpler operations

Logarithms can be used to make calculations easier. For example, two numbers can be multiplied just by using a logarithm table and adding. These are often known as logarithmic properties, which are documented in the table below.<ref>Template:Cite web</ref> The first three operations below assume that Template:Math and/or Template:Math, so that Template:Math and Template:Math. Derivations also use the log definitions Template:Math and Template:Math.

<math>\log_b(xy)=\log_b(x)+\log_b(y)</math> because <math>b^c b^d=b^{c+d}</math>
<math>\log_b(\tfrac{x}{y})=\log_b(x)-\log_b(y)</math> because <math>\tfrac{b^c}{b^d}=b^{c-d}</math>
<math>\log_b(x^d)=d\log_b(x)</math> because <math>(b^c)^d=b^{cd}</math>
<math>\log_b\left(\sqrt[y]{x}\right)=\frac{\log_b(x)}{y}</math> because <math>\sqrt[y]{x}=x^{1/y}</math>
<math>x^{\log_b(y)}=y^{\log_b(x)}</math> because <math>x^{\log_b(y)}=b^{\log_b(x)\log_b(y)}=(b^{\log_b(y)})^{\log_b(x)}=y^{\log_b(x)}</math>
<math>c\log_b(x)+d\log_b(y)=\log_b(x^c y^d)</math> because <math>\log_b(x^c y^d)=\log_b(x^c)+\log_b(y^d)</math>

Where <math>b</math>, <math>x</math>, and <math>y</math> are positive real numbers and <math>b \ne 1</math>, and <math>c</math> and <math>d</math> are real numbers.

The laws result from canceling exponentials and the appropriate law of indices. Starting with the first law:

<math display="block">xy = b^{\log_b(x)} b^{\log_b(y)} = b^{\log_b(x) + \log_b(y)} \Rightarrow \log_b(xy) = \log_b(b^{\log_b(x) + \log_b(y)}) = \log_b(x) + \log_b(y)</math>

The law for powers exploits another of the laws of indices:

<math display="block">x^y = (b^{\log_b(x)})^y = b^{y \log_b(x)} \Rightarrow \log_b(x^y) = y \log_b(x)</math>

The law relating to quotients then follows:

<math display="block">\log_b \bigg(\frac{x}{y}\bigg) = \log_b(x y^{-1}) = \log_b(x) + \log_b(y^{-1}) = \log_b(x) - \log_b(y)</math> <math display="block">\log_b \bigg(\frac{1}{y}\bigg) = \log_b(y^{-1}) = - \log_b(y)</math>

Similarly, the root law is derived by rewriting the root as a reciprocal power:

<math display="block">\log_b(\sqrt[y]x) = \log_b(x^{\frac{1}{y}}) = \frac{1}{y}\log_b(x)</math>

Derivations of product, quotient, and power rules

These are the three main logarithm laws, rules, or principles,<ref> Template:Cite web </ref> from which the other properties listed above can be proven. Each of these logarithm properties correspond to their respective exponent law, and their derivations and proofs will hinge on those facts. There are multiple ways to derive or prove each logarithm lawTemplate:Sndthis is just one possible method.

Logarithm of a product

To state the logarithm of a product law formally:

<math display="block">\forall b \in \mathbb{R}_+, b \neq 1, \forall x, y, \in \mathbb{R}_+, \log_b(xy) = \log_b(x) + \log_b(y)</math>

Derivation:

Let <math>b \in \mathbb{R}_+</math>, where <math>b \neq 1</math>, and let <math>x, y \in \mathbb{R}_+</math>. We want to relate the expressions <math>\log_b(x)</math> and <math>\log_b(y)</math>. This can be done more easily by rewriting in terms of exponentials, whose properties we already know. Additionally, since we are going to refer to <math>\log_b(x)</math> and <math>\log_b(y)</math> quite often, we will give them some variable names to make working with them easier: Let <math>m = \log_b(x)</math>, and let <math>n = \log_b(y)</math>.

Rewriting these as exponentials, we see that

<math display="block">\begin{align} m &= \log_b(x) \iff b^m = x, \\ n &= \log_b(y) \iff b^n = y. \end{align}</math>

From here, we can relate <math>b^m</math> (i.e. <math>x</math>) and <math>b^n</math> (i.e. <math>y</math>) using exponent laws as

<math display="block">xy = (b^m)(b^n) = b^m \cdot b^n = b^{m + n}</math>

To recover the logarithms, we apply <math>\log_b</math> to both sides of the equality.

<math display="block">\log_b(xy) = \log_b(b^{m + n})</math>

The right side may be simplified using one of the logarithm properties from before: we know that <math>\log_b(b^{m + n}) = m + n</math>, giving

<math display="block">\log_b(xy) = m + n</math>

We now resubstitute the values for <math>m</math> and <math>n</math> into our equation, so our final expression is only in terms of <math>x</math>, <math>y</math>, and <math>b</math>.

<math display="block">\log_b(xy) = \log_b(x) + \log_b(y)</math>

This completes the derivation.

Logarithm of a quotient

To state the logarithm of a quotient law formally:

<math display="block">\forall b \in \mathbb{R}_+, b \neq 1, \forall x, y, \in \mathbb{R}_+, \log_b \left( \frac{x}{y} \right) = \log_b(x) - \log_b(y)</math>

Derivation:

Let <math>b \in \mathbb{R}_+</math>, where <math>b \neq 1</math>, and let <math>x, y \in \mathbb{R}_+</math>.

We want to relate the expressions <math>\log_b(x)</math> and <math>\log_b(y)</math>. This can be done more easily by rewriting in terms of exponentials, whose properties we already know. Additionally, since we are going to refer to <math>\log_b(x)</math> and <math>\log_b(y)</math> quite often, we will give them some variable names to make working with them easier: Let <math>m = \log_b(x)</math>, and let <math>n = \log_b(y)</math>.

Rewriting these as exponentials, we see that:

<math display="block">\begin{align} m &= \log_b(x) \iff b^m = x, \\ n &= \log_b(y) \iff b^n = y. \end{align}</math>

From here, we can relate <math>b^m</math> (i.e. <math>x</math>) and <math>b^n</math> (i.e. <math>y</math>) using exponent laws as

<math display="block">\frac{x}{y} = \frac{(b^m)}{(b^n)} = \frac{b^m}{b^n} = b^{m - n}</math>

To recover the logarithms, we apply <math>\log_b</math> to both sides of the equality.

<math display="block">\log_b \left( \frac{x}{y} \right) = \log_b \left( b^{m -n} \right)</math>

The right side may be simplified using one of the logarithm properties from before: we know that <math>\log_b(b^{m - n}) = m - n</math>, giving

<math display="block">\log_b \left( \frac{x}{y} \right) = m -n</math>

We now resubstitute the values for <math>m</math> and <math>n</math> into our equation, so our final expression is only in terms of <math>x</math>, <math>y</math>, and <math>b</math>.

<math display="block">\log_b \left( \frac{x}{y} \right) = \log_b(x) - \log_b(y)</math>

This completes the derivation.

Logarithm of a power

To state the logarithm of a power law formally:

<math display="block">\forall b \in \mathbb{R}_+, b \neq 1, \forall x \in \mathbb{R}_+, \forall r \in \mathbb{R}, \log_b(x^r) = r\log_b(x)</math>

Derivation:

Let <math>b \in \mathbb{R}_+</math>, where <math>b \neq 1</math>, let <math>x\in \mathbb{R}_+</math>, and let <math>r \in \mathbb{R}</math>. For this derivation, we want to simplify the expression <math>\log_b(x^r)</math>. To do this, we begin with the simpler expression <math>\log_b(x)</math>. Since we will be using <math>\log_b(x)</math> often, we will define it as a new variable: Let <math>m = \log_b(x)</math>.

To more easily manipulate the expression, we rewrite it as an exponential. By definition, <math>m = \log_b(x) \iff b^m = x</math>, so we have

<math display="block">b^m = x</math>

Similar to the derivations above, we take advantage of another exponent law. In order to have <math>x^r</math> in our final expression, we raise both sides of the equality to the power of <math>r</math>:

<math display="block"> \begin{align} (b^m)^r &= (x)^r \\ b^{mr} &= x^r \end{align} </math>

where we used the exponent law <math>(b^m)^r = b^{mr}</math>.

To recover the logarithms, we apply <math>\log_b</math> to both sides of the equality.

<math display="block">\log_b(b^{mr}) = \log_b(x^r)</math>

The left side of the equality can be simplified using a logarithm law, which states that <math>\log_b(b^{mr}) = mr</math>.

<math display="block">mr = \log_b(x^r)</math>

Substituting in the original value for <math>m</math>, rearranging, and simplifying gives

<math display="block"> \begin{align} \left( \log_b(x) \right)r &= \log_b(x^r) \\ r\log_b(x) &= \log_b(x^r) \\ \log_b(x^r) &= r\log_b(x) \end{align} </math>

This completes the derivation.

Changing the base

Most calculators have buttons for natural logarithms (Template:Fake link) and common logarithms (log or Template:Fake link), but not all calculators have buttons for the logarithm of an arbitrary base. Accordingly, it is sometimes useful to change the base of a logarithm.Template:PbAs briefly mentioned in the Template:Fake link section, the definition of a logarithm is: <math>\log_b(y) = x \iff b^x = y.</math> Solving for Template:Mvar in the exponential equation Template:Nobr could be done on most calculators by using common or natural logarithms: <math>b^x = y \iff x = \log_b(y) = \frac{\log(y)}{\log(b)} = \frac{\ln(y)}{\ln(b)}.</math> For example, the binary logarithm (Template:Fake link), which is widely used in Template:Fake link, could be calculated on most calculators as: <math>2^x = y \iff x = \log_2(y) = \frac{\log(y)}{\log(2)} = \frac{\ln(y)}{\ln(2)}</math>Template:PbMore generally, the change of base formula can be formally defined as:<math>\forall a, b \in \mathbb{R}_+, a, b \neq 1, \forall x \in \mathbb{R}_+, \log_b(x) = \frac{\log_a(x)}{\log_a(b)}</math>

Proof and derivation

Let <math>a, b \in \mathbb{R}_+</math>, where <math>a, b \neq 1</math> Let <math>x \in \mathbb{R}_+</math>. Here, <math>a</math> and <math>b</math> are the two bases we will be using for the logarithms. They cannot be 1, because the logarithm function is not well defined for the base of 1.Template:Citation needed The number <math>x</math> will be what the logarithm is evaluating, so it must be a positive number. Since we will be dealing with the term <math>\log_b(x)</math> quite frequently, we define it as a new variable: Let <math>m = \log_b(x)</math>.

To more easily manipulate the expression, it can be rewritten as an exponential. <math display="block">b^m = x </math>

Applying <math>\log_a</math> to both sides of the equality,

<math display="block">\log_a(b^m) = \log_a(x) </math>

Now, using the logarithm of a power property, which states that <math>\log_a(b^m) = m\log_a(b)</math>,

<math display="block">m\log_a(b) = \log_a(x)</math>

Isolating <math>m</math>, we get the following:

<math display="block">m = \frac{\log_a(x)}{\log_a(b)}</math>

Resubstituting <math>m = \log_b(x)</math> back into the equation,

<math display="block">\log_b(x) = \frac{\log_a(x)}{\log_a(b)}</math>

This completes the proof that <math>\log_b(x) = \frac{\log_a(x)}{\log_a(b)}</math>.

This formula has several consequences:

<math display="block"> \log_b a = \frac 1 {\log_a b} </math> <math display="block"> \log_{b^n} a = {\log_b a \over n} </math> <math display="block"> \log_{b} a = \log_b e \cdot \log_e a = \log_b e \cdot \ln a </math> <math display="block"> b^{\log_a d} = d^{\log_a b} </math> <math display="block"> -\log_b a = \log_b \left({1 \over a}\right) = \log_{1/b} a</math>


<math display="block"> \log_{b_1}a_1 \,\cdots\, \log_{b_n}a_n = \log_{b_{\pi(1)}}a_1\, \cdots\, \log_{b_{\pi(n)}}a_n, </math>

where <math display="inline">\pi</math> is any permutation of the subscripts Template:Math. For example

<math display="block"> \log_b w\cdot \log_a x \cdot \log_d c \cdot \log_d z = \log_d w \cdot \log_b x \cdot \log_a c \cdot \log_d z. </math>

Summation and subtraction

The following summation and subtraction rule is especially useful in probability theory when one is dealing with a sum of log-probabilities:

<math>\log_b (a+c) = \log_b a + \log_b \left(1 + \frac{c}{a}\right)</math> because <math>\left(a + c \right) = a \times \left(1 + \frac{c}{a} \right)</math>
<math>\log_b (a-c) = \log_b a + \log_b \left(1 - \frac{c}{a}\right)</math> because <math>\left(a - c \right) = a \times \left(1 - \frac{c}{a} \right)</math>

Note that the subtraction identity is not defined if <math>a=c</math>, since the logarithm of zero is not defined. Also note that, when programming, <math>a</math> and <math>c</math> may have to be switched on the right hand side of the equations if <math>c \gg a</math> to avoid losing the "1 +" due to rounding errors. Many programming languages have a specific log1p(x) function that calculates <math>\log_e (1+x)</math> without underflow (when <math>x</math> is small).

More generally:

<math display="block">\log _b \sum_{i=0}^N a_i = \log_b a_0 + \log_b \left( 1+\sum_{i=1}^N \frac{a_i}{a_0} \right) = \log _b a_0 + \log_b \left( 1+\sum_{i=1}^N b^{\left( \log_b a_i - \log _b a_0 \right)} \right)</math>

Exponents

A useful identity involving exponents:

<math display="block"> x^{\frac{\log(\log(x))}{\log(x)}} = \log(x)</math>

or more universally:

<math display="block"> x^{\frac{\log(a)}{\log(x)}} = a</math>

Other or resulting identities

<math display="block"> \frac{1}{\frac{1}{\log_x(a)} + \frac{1}{\log_y(a)}} = \log_{xy}(a)</math> <math display="block"> \frac{1}{\frac{1}{\log_x(a)}-\frac{1}{\log_y(a)}} = \log_{\frac{x}{y}}(a)</math>

Inequalities

Based on,<ref>Template:Citation</ref>

<math display="block">\frac{x}{1+x} \leq \ln(1+x) \leq \frac{x(6+x)}{6+4x} \leq x \mbox{ for all } {-1} < x</math> <math display="block">\begin{align} \frac{2x}{2+x}&\leq3-\sqrt{\frac{27}{3+2x}}\leq\frac{x}{\sqrt{1+x+x^2/12}} \\[4pt] &\leq \ln(1+x)\leq \frac{x}{\sqrt{1+x}}\leq \frac{x}{2}\frac{2+x}{1+x} \\[4pt] &\text{ for } 0 \le x \text{, reverse for } {-1} < x \le 0 \end{align}</math>

All are accurate around <math>x=0</math>, but not for large numbers.

Tropical identities

The following identity relates log semiring to the min-plus semiring.

<math display="block">\lim_{T \rightarrow 0} -T\log(e^{-\frac{s}{T}} + e^{-\frac{t}{T}}) = \mathrm{min}\{s,t\}</math>

Calculus identities

Limits

<math display="block">\lim_{x\to 0^+}\log_a(x)=-\infty\quad \mbox{if } a > 1</math> <math display="block">\lim_{x\to 0^+}\log_a(x)=\infty\quad \mbox{if } 0 < a < 1</math> <math display="block">\lim_{x\to\infty}\log_a(x)=\infty\quad \mbox{if } a > 1</math> <math display="block">\lim_{x\to\infty}\log_a(x)=-\infty\quad \mbox{if } 0 < a < 1</math> <math display="block">\lim_{x\to \infty}x^b\log_a(x)=\infty\quad \mbox{if } b > 0</math> <math display="block">\lim_{x\to\infty}\frac{\log_a(x)}{x^b}=0\quad \mbox{if } b > 0</math>

The last limit is often summarized as "logarithms grow more slowly than any power or root of x".

Derivatives of logarithmic functions

<math display="block">{d \over dx} \ln x = {1 \over x }, x > 0</math> <math display="block">{d \over dx} \ln |x| = {1 \over x }, x \neq 0</math> <math display="block">{d \over dx} \log_a x = {1 \over x \ln a}, x > 0, a > 0, \text{ and } a\neq 1</math>

Integral definition

<math display="block">\ln x = \int_1^x \frac {1}{t}\ dt </math>

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

<math display="block">-\int_1^x \frac{1}{t} \, dt = \int_x^1 \frac{1}{t} \, dt</math>

Therefore:

<math display="block">\ln \frac{1}{x} = \int_x^1 \frac{1}{t} \, dt</math>

Riemann Sum

<math display="block">\ln(n + 1) = </math> <math display="block">\lim_{k \to \infty} \sum_{i=1}^{k} \frac{1}{x_i} \Delta x = </math> <math display="block">\lim_{k \to \infty} \sum_{i=1}^{k} \frac{1}{1 + \frac{i-1}{k}n} \cdot \frac{n}{k} =</math> <math display="block">\lim_{k \to \infty} \sum_{x=1}^{k \cdot n} \frac{1}{1 + \frac{x}{k}} \cdot \frac{1}{k} =</math> <math display="block">\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}</math>

for <math>\textstyle \Delta x = \frac{n}{k}</math> and <math>x_{i}</math> is a sample point in each interval.

Series representation

The natural logarithm <math>\ln(1 + x)</math> has a well-known Taylor series<ref>Template:Cite web</ref> expansion that converges for <math>x</math> in the open-closed interval Template:Open-closed:

<math display="block">\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.</math>

Within this interval, for <math>x = 1</math>, the series is conditionally convergent, and for all other values, it is absolutely convergent. For <math>x > 1</math> or <math>x \leq -1</math>, the series does not converge to <math>\ln(1 + x)</math>. In these cases, different representations<ref>To extend the utility of the Mercator series beyond its conventional bounds one can calculate <math display="inline"> \ln(1 + x)</math> for <math display="inline"> x = -\frac{n}{n+1}</math> and <math display="inline"> n \geq 0</math> and then negate the result, <math display="inline"> \ln\left(\frac{1}{n+1}\right)</math>, to derive <math display="inline"> \ln(n + 1)</math>. For example, setting <math display="inline"> x = -\frac{1}{2}</math> yields <math display="inline"> \ln 2 = \sum_{n=1}^{\infty} \frac{1}{n 2^n}</math>.</ref> 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.<ref name="Flajolet2009AnalyticCombinatorics">Template:Cite book See page 117, and VI.8 definition of shifted harmonic numbers on page 389 </ref> 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<ref name="Deveci2022DoubleSeries">Template:Cite arXiv See Theorem 5.2. on pages 22 - 23</ref>

<math display="block">\lim_Template:K \to \infty (H_Template:K(n+1) - H_k) = \ln(n+1)</math> which characterizes the behavior of harmonic numbers as they grow large. This approximation (which precisely equals <math>\ln(n+1)</math> 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 <math>H_k</math> denotes the <math>k</math>-th harmonic number, defined as

<math display="block">H_k = \sum_Template:J=1^k \frac{1}{j}</math>

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.<ref>Template:Cite book </ref><ref name="Flajolet2009AnalyticCombinatorics"/><ref>Template:Cite web See formula 13. </ref> As <math>k</math> tends towards infinity, the difference between the harmonic numbers <math>H_{k(n+1)}</math> and <math>H_k</math> converges to a non-zero value. This persistent non-zero difference, <math>\ln(n+1)</math>, precludes the possibility of the harmonic series approaching a finite limit, thus providing a clear mathematical articulation of its divergence.<ref>Template:Cite report See Proofs 23 and 24 for details on the relationship between harmonic numbers and logarithmic functions. </ref><ref>Template:Cite journal </ref> 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:

<math display="block">\sum_Template:J=k+1^{k(n+1)} \frac{1}{j} \approx \int_k^{k(n+1)} \frac{dx}{x}</math>

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 <math>k+1</math> to <math>k(n+1)</math>:

<math display="block">H_Template:K(n+1) - H_k = \sum_Template:J=k+1^{k(n+1)} \frac{1}{j}</math>

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

<math display="block">\lim_Template:K \to \infty \sum_Template: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)</math>

As the window's lower bound begins at <math>k+1</math> and the upper bound extends to <math>k(n+1)</math>, both of which tend toward infinity as <math>k \to \infty</math>, 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 <math>1</math> to <math>n+1</math> where the onset <math>k</math> implies this minimally discrete region.

Double series formula

The harmonic number difference formula for <math>\ln(m)</math> is an extension<ref name="Deveci2022DoubleSeries"/> of the classic, alternating identity of <math>\ln(2)</math>:

<math display="block">\ln(2) = \lim_{k \to \infty} \sum_{n=1}^{k} \left( \frac{1}{2n-1} - \frac{1}{2n} \right)</math>

which can be generalized as the double series over the residues of <math>m</math>:

<math display="block">\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)} </math>

where <math>\langle m \rangle</math> is the principal ideal generated by <math>m</math>. Subtracting <math>\textstyle \frac{1}{x}</math> from each term <math>\textstyle \frac{1}{x-r}</math> (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 <math>m</math> increases. For example:

<math display="block">\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)</math>

This method leverages the fine differences between closely related terms to stabilize the series. The sum over all residues <math>r \in \N</math> ensures that adjustments are uniformly applied across all possible offsets within each block of <math>m</math> terms. This uniform distribution of the "correction" across different intervals defined by <math>x-r</math> 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. Note that the structure of the summands of this formula matches those of the interpolated harmonic number <math>H_x</math> when both the domain and range are negated (i.e., <math>-H_{-x}</math>). However, the interpretation and roles of the variables differ.

Deveci's Proof

A fundamental feature of the proof is the accumulation of the subtrahends <math display="inline"> \frac{1}{x}</math> into a unit fraction, that is, <math display="inline"> \frac{m}{x} = \frac{1}{n}</math> for <math>m \mid x</math>, thus <math>m = \omega + 1</math> rather than <math>m = |\mathbb{Z}_{m} \cap \mathbb{N}|</math>, where the extrema of <math>\mathbb{Z}_{m} \cap \mathbb{N}</math> are <math>[0, \omega]</math> if <math>\mathbb{N} = \mathbb{N}_{0}</math> and <math>[1, \omega]</math> otherwise, with the minimum of <math>0</math> being implicit in the latter case due to the structural requirements of the proof. Since the cardinality of <math>\mathbb{Z}_{m} \cap \mathbb{N}</math> depends on the selection of one of two possible minima, the integral <math>\textstyle \int \frac{1}{t} dt</math>, as a set-theoretic procedure, is a function of the maximum <math>\omega</math> (which remains consistent across both interpretations) plus <math>1</math>, not the cardinality (which is ambiguous<ref>Template:Cite arXiv A synopsis on the nature of 0 which frames the choice of minimum as the dichotomy between ordinals and cardinals.</ref><ref>Template:Cite journal See section 3.1</ref> 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 <math>m</math>-tuple—over the harmonic series, advancing the window by <math>m</math> positions to select the next <math>m</math>-tuple, and offsetting each element of each tuple by <math display="inline"> \frac{1}{m}</math> relative to the window's absolute position. The sum <math display="inline">\sum_{n=1}^{k} \sum \frac{1}{x - r}</math> corresponds to <math>H_{km}</math> which scales <math>H_{m}</math> without bound. The sum <math display="inline"> \sum_{n=1}^{k} -\frac{1}{n}</math> corresponds to the prefix <math>H_{k}</math> trimmed from the series to establish the window's moving lower bound <math>k+1</math>, and <math>\ln(m)</math> is the limit of the sliding window (the scaled, truncated<ref>The <math>k+1</math> shift is characteristic of the right Riemann sum employed to prevent the integral from degenerating into the harmonic series, thereby averting divergence. Here, <math display="inline"> -\frac{1}{n}</math> functions analogously, serving to regulate the series. The successor operation <math>m = \omega + 1</math> signals the implicit inclusion of the modulus <math>m</math> (the region omitted from <math>\mathbb{N}_{1}</math>). The importance of this, from an axiomatic perspective, becomes evident when the residues of <math>m</math> are formulated as <math>e^{\ln(\omega + 1)}</math>, where <math>\omega + 1</math> is bootstrapped by <math>\omega = 0</math> to produce the residues of modulus <math>m = \omega = \omega_{0} + 1 = 1</math>. Consequently, <math>\omega</math> represents a limiting value in this context.</ref> series):

<math display="block">\begin{align} \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} \end{align}</math> <math display="block">\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)</math>

Integrals of logarithmic functions

<math display="block">\int \ln x \, dx = x \ln x - x + C = x(\ln x - 1) + C</math> <math display="block">\int \log_a x \, dx = x \log_a x - \frac{x}{\ln a} + C = \frac{x (\ln x - 1)}{\ln a} + C</math>

To remember higher integrals, it is convenient to define:

<math display="block">x^{\left [n \right]} = x^{n}(\log(x) - H_n)</math>

where <math>H_n</math> is the nth harmonic number:

<math display="block">x^{\left [ 0 \right ]} = \log x</math> <math display="block">x^{\left [ 1 \right ]} = x \log(x) - x</math> <math display="block">x^{\left [ 2 \right ]} = x^2 \log(x) - \begin{matrix} \frac{3}{2} \end{matrix}x^2</math> <math display="block">x^{\left [ 3 \right ]} = x^3 \log(x) - \begin{matrix} \frac{11}{6} \end{matrix}x^3</math>

Then:

<math display="block">\frac{d}{dx}\, x^{\left[ n \right]} = nx^{\left[ n-1 \right]}</math> <math display="block">\int x^{\left[ n \right]}\,dx = \frac{x^{\left [ n+1 \right ]}}{n+1} + C</math>

Approximating large numbers

The identities of logarithms can be used to approximate large numbers. Note that Template:Math, where a, b, and c are arbitrary constants. Suppose that one wants to approximate the 44th Mersenne prime, Template:Math. To get the base-10 logarithm, we would multiply 32,582,657 by Template:Math, getting Template:Math. We can then get Template:Math.

Similarly, factorials can be approximated by summing the logarithms of the terms.

Complex logarithm identities

The complex logarithm is the complex number analogue of the logarithm function. No single valued function on the complex plane can satisfy the normal rules for logarithms. However, a multivalued function can be defined which satisfies most of the identities. It is usual to consider this as a function defined on a Riemann surface. A single valued version, called the principal value of the logarithm, can be defined which is discontinuous on the negative x axis, and is equal to the multivalued version on a single branch cut.

Definitions

In what follows, a capital first letter is used for the principal value of functions, and the lower case version is used for the multivalued function. The single valued version of definitions and identities is always given first, followed by a separate section for the multiple valued versions.

The multiple valued version of Template:Math is a set, but it is easier to write it without braces and using it in formulas follows obvious rules.

When k is any integer:

<math display="block">\log(z) = \ln(|z|) + i \arg(z)</math> <math display="block">\log(z) = \operatorname{Log}(z) + 2 \pi i k</math> <math display="block">e^{\log(z)} = z</math>

Constants

Principal value forms:

<math display="block">\operatorname{Log}(1) = 0</math> <math display="block">\operatorname{Log}(e) = 1</math>

Multiple value forms, for any k an integer:

<math display="block">\log(1) = 0 + 2 \pi i k</math> <math display="block">\log(e) = 1 + 2 \pi i k</math>

Summation

Principal value forms:<ref name=":0">Template:Cite book</ref>

<math display="block">\operatorname{Log}(z_1) + \operatorname{Log}(z_2) = \operatorname{Log}(z_1 z_2) \pmod {2 \pi i}</math> <math display="block">\operatorname{Log}(z_1) + \operatorname{Log}(z_2) = \operatorname{Log}(z_1 z_2)\quad (-\pi <\operatorname{Arg}(z_1)+\operatorname{Arg}(z_2)\leq \pi; \text{ e.g., } \operatorname{Re}z_1\geq 0 \text{ and } \operatorname{Re}z_2 > 0)</math> <math display="block">\operatorname{Log}(z_1) - \operatorname{Log}(z_2) = \operatorname{Log}(z_1 / z_2) \pmod {2 \pi i}</math> <math display="block">\operatorname{Log}(z_1) - \operatorname{Log}(z_2) = \operatorname{Log}(z_1 / z_2) \quad (-\pi <\operatorname{Arg}(z_1)-\operatorname{Arg}(z_2)\leq \pi; \text{ e.g., } \operatorname{Re}z_1\geq 0 \text{ and } \operatorname{Re}z_2 > 0)</math>

Multiple value forms:

<math display="block">\log(z_1) + \log(z_2) = \log(z_1 z_2)</math> <math display="block">\log(z_1) - \log(z_2) = \log(z_1 / z_2)</math>

Powers

A complex power of a complex number can have many possible values.

Principal value form:

<math display="block">{z_1}^{z_2} = e^{z_2 \operatorname{Log}(z_1)} </math> <math display="block">\operatorname{Log}{\left({z_1}^{z_2}\right)} = z_2 \operatorname{Log}(z_1) \pmod {2 \pi i}</math>

Multiple value forms:

<math display="block">{z_1}^{z_2} = e^{z_2 \log(z_1)}</math>

Where Template:Math, Template:Math are any integers:

<math display="block">\log{\left({z_1}^{z_2}\right)} = z_2 \log(z_1) + 2 \pi i k_2</math> <math display="block">\log{\left({z_1}^{z_2}\right)} = z_2 \operatorname{Log}(z_1) + z_2 2 \pi i k_1 + 2 \pi i k_2</math>

Asymptotic identities

Pronic numbers

As a consequence of the harmonic number difference, the natural logarithm is asymptotically approximated by a finite series difference,<ref name="Deveci2022DoubleSeries"/> representing a truncation of the integral at <math>k = n</math>:

<math display="block">H_{{2T[n]}} - H_n \sim \ln(n+1)</math>

where <math>T[n]</math> is the Template:Mvarth triangular number, and <math>2T[n]</math> is the sum of the [[Pronic number#As figurate numbers|first Template:Mvar even integers]]. Since the Template:Mvarth pronic number is asymptotically equivalent to the Template:Mvarth perfect square, it follows that:

<math display="block">H_Template:N^2 - H_n \sim \ln(n+1)</math>

Prime number theorem

The prime number theorem provides the following asymptotic equivalence:

<math display="block">\frac{n}{\pi(n)} \sim \ln n</math>

where <math>\pi(n)</math> is the prime counting function. This relationship is equal to:<ref name="Deveci2022DoubleSeries"/>Template:Rp

<math display="block">\frac{n}{H(1, 2, \ldots, x_n)} \sim \ln n</math>

where <math>H(x_1, x_2, \ldots, x_n)</math> is the harmonic mean of <math>x_1, x_2, \ldots, x_n</math>. This is derived from the fact that the difference between the <math>n</math>th harmonic number and <math>\ln n</math> asymptotically approaches a small constant, resulting in <math>H_{n^2} - H_n \sim H_n</math>. This behavior can also be derived from the properties of logarithms: <math>\ln n</math> is half of <math>\ln n^2</math>, and this "first half" is the natural log of the root of <math>n^2</math>, which corresponds roughly to the first <math>\textstyle \frac{1}{n}</math>th of the sum <math>H_{n^2}</math>, or <math>H_n</math>. The asymptotic equivalence of the first <math>\textstyle \frac{1}{n}</math>th of <math>H_{n^2}</math> to the latter <math>\textstyle \frac{n-1}{n}</math>th of the series is expressed as follows:

<math display="block">\frac{H_n}{H_{n^2}} \sim \frac{\ln \sqrt{n}}{\ln n} = \frac{1}{2}</math>

which generalizes to:

<math display="block">\frac{H_n}{H_{n^k}} \sim \frac{\ln \sqrt[k]{n}}{\ln n} = \frac{1}{k}</math> <math display="block">k H_n \sim H_{n^k}</math>

and:

<math display="block">k H_n - H_n \sim (k - 1) \ln(n+1)</math> <math display="block">H_Template:N^k - H_n \sim (k - 1) \ln(n+1)</math> <math display="block">k H_n - H_Template:N^k \sim (k - 1) \gamma</math>

for fixed <math>k</math>. The correspondence sets <math>H_n</math> as a unit magnitude that partitions <math>H_{n^k}</math> across powers, where each interval <math>\textstyle \frac{1}{n}</math> to <math>\textstyle \frac{1}{n^2}</math>, <math>\textstyle \frac{1}{n^2}</math> to <math>\textstyle \frac{1}{n^3}</math>, etc., corresponds to one <math>H_n</math> unit, illustrating that <math>H_{n^k}</math> forms a divergent series as <math>k \to \infty</math>.

Real Arguments

These approximations extend to the real-valued domain through the interpolated harmonic number. For example, where <math>x \in \mathbb{R}</math>:

<math display="block">H_Template:X^2 - H_x \sim \ln x</math>

Stirling numbers

The natural logarithm is asymptotically related to the harmonic numbers by the Stirling numbers<ref>Template:Cite arXiv</ref> and the Gregory coefficients.<ref>Template:Cite book</ref> By representing <math>H_n</math> in terms of Stirling numbers of the first kind, the harmonic number difference is alternatively expressed as follows, for fixed <math>k</math>:

<math display="block">\frac{s(n^k+1, 2)}{(n^k)!} - \frac{s(n+1, 2)}{n!} \sim (k-1) \ln(n+1)</math>

See also

References

Template:Reflist

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