Natural logarithm

The natural logarithm of a number is its logarithm to the base of the mathematical constant , which is an irrational and transcendental number approximately equal to . The natural logarithm of is generally written as , , or sometimes, if the base is implicit, simply . Parentheses are sometimes added for clarity, giving , , or . This is done particularly when the argument to the logarithm is not a single symbol, so as to prevent ambiguity.

The natural logarithm of is the power to which would have to be raised to equal . For example, is , because . The natural logarithm of itself, , is , because , while the natural logarithm of is , since .

The natural logarithm can be defined for any positive real number as the area under the curve from to (with the area being negative when ). The simplicity of this definition, which is matched in many other formulas involving the natural logarithm, leads to the term "natural". The definition of the natural logarithm can then be extended to give logarithm values for negative numbers and for all non-zero complex numbers, although this leads to a multi-valued function: see complex logarithm for more.

The natural logarithm function, if considered as a real-valued function of a positive real variable, is the inverse function of the exponential function, leading to the identities:

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

e^{\ln x} &= x \qquad \text{ if } x \in \R_{+}\\

\ln e^x &= x \qquad \text{ if } x \in \R

\end{align}</math>

Like all logarithms, the natural logarithm maps multiplication of positive numbers into addition:

Logarithms can be defined for any positive base other than 1, not only . However, logarithms in other bases differ only by a constant multiplier from the natural logarithm, and can be defined in terms of the latter, <math>\log_b x = \ln x / \ln b</math>.

Logarithms are useful for solving equations in which the unknown appears as the exponent of some other quantity. For example, logarithms are used to solve for the half-life, decay constant, or unknown time in exponential decay problems. They are important in many branches of mathematics and scientific disciplines, and are used to solve problems involving compound interest.

History

The concept of the natural logarithm was worked out by Gregoire de Saint-Vincent and Alphonse Antonio de Sarasa before 1649. Their work involved quadrature of the hyperbola with equation , by determination of the area of hyperbolic sectors. Their solution generated the requisite "hyperbolic logarithm" function, which had the properties now associated with the natural logarithm.

An early mention of the natural logarithm was by Nicholas Mercator in his work Logarithmotechnia, published in 1668, although the mathematics teacher John Speidell had already compiled a table of what in fact were effectively natural logarithms in 1619. It has been said that Speidell's logarithms were to the base ', but this is not entirely true due to complications with the values being expressed as integers.]]

The natural logarithm can be defined as the area under the graph of a rectangular hyperbola with equation <math>y = 1/x</math> between <math>x = 1</math> and <math>x = a</math>. That is,<math display="block">\ln a = \int_1^a \frac{1}{x}\,dx.</math>If is in <math>(0,1)</math>, then the region has negative area, and the logarithm is negative. The form is a specific instance of the general notation for the logarithm to base of a number , which is shown as . (For example, the base-2 logarithm of can be written as .) Some authors use without an explicit base to refer to the natural logarithm. An example can be commonly found in prime number theorem. In addition to mathematics, this usage is commonplace in some programming languages. However, in some other contexts such as chemistry, can be used to denote the common (base 10) logarithm.

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

\ln x &= \int_1^x \frac{1}{t} \, dt = \int_0^{x - 1} \frac{1}{1 + u} \, du \\

&= \int_0^{x - 1} (1 - u + u^2 - u^3 + \cdots) \, du \\

&= (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \frac{(x - 1)^4}{4} + \cdots \\

&= \sum_{k=1}^\infty \frac{(-1)^{k-1} (x-1)^k}{k}.

\end{align}</math>

This is the Taylor series for <math>\ln x</math> around 1. A change of variables yields the Mercator series:

<math display="block">\ln(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1{k} x^k = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots,</math>

valid for <math>|x| \leq 1</math> and <math>x\ne -1.</math>

Leonhard Euler, disregarding <math>x\ne -1</math>, nevertheless applied this series to <math>x=-1</math> to show that the harmonic series equals the natural logarithm of <math display="inline">\frac{1}{1-1}</math>; that is, the logarithm of infinity. Nowadays, more formally, one can prove that the harmonic series truncated at is close to the logarithm of , when is large, with the difference converging to the Euler–Mascheroni constant.

The figure is a graph of and some of its Taylor polynomials around 0. These approximations converge to the function only in the region ; outside this region, the higher-degree Taylor polynomials devolve to worse approximations for the function.

A useful special case for positive integers , taking <math>x = \tfrac{1}{n}</math>, is:

<math display="block"> \ln \left(\frac{n + 1}{n}\right) = \sum_{k=1}^\infty \frac{(-1)^{k-1{k n^k} = \frac{1}{n} - \frac{1}{2 n^2} + \frac{1}{3 n^3} - \frac{1}{4 n^4} + \cdots</math>

If <math>\operatorname{Re}(x) \ge 1/2,</math> then

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

\ln (x) &= - \ln \left(\frac{1}{x}\right) = - \sum_{k=1}^\infty \frac{(-1)^{k-1} (\frac{1}{x} - 1)^k}{k} = \sum_{k=1}^\infty \frac{(x - 1)^k}{k x^k} \\

&= \frac{x - 1}{x} + \frac{(x - 1)^2}{2 x^2} + \frac{(x - 1)^3}{3 x^3} + \frac{(x - 1)^4}{4 x^4} + \cdots

\end{align}</math>

Now, taking <math>x=\tfrac{n+1}{n}</math> for positive integers , we get:

<math display="block"> \ln \left(\frac{n + 1}{n}\right) = \sum_{k=1}^\infty \frac{1}{k (n + 1)^k} = \frac{1}{n + 1} + \frac{1}{2 (n + 1)^2} + \frac{1}{3 (n + 1)^3} + \frac{1}{4 (n + 1)^4} + \cdots</math>

If <math>\operatorname{Re}(x) \ge 0 \text{ and } x \neq 0,</math> then

<math display="block"> \ln (x) = \ln \left(\frac{2x}{2}\right) = \ln\left(\frac{1 + \frac{x - 1}{x + 1{1 - \frac{x - 1}{x + 1\right) = \ln \left(1 + \frac{x - 1}{x + 1}\right) - \ln \left(1 - \frac{x - 1}{x + 1}\right). </math>

Since

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

\ln(1+y) - \ln(1-y)&= \sum^\infty_{i=1}\frac{1}{i}\left((-1)^{i-1}y^i - (-1)^{i-1}(-y)^i\right) = \sum^\infty_{i=1}\frac{y^i}{i}\left((-1)^{i-1} +1\right) \\

&= y\sum^\infty_{i=1}\frac{y^{i-1{i}\left((-1)^{i-1} +1\right)\overset{i-1\to 2k}{=}\; 2y\sum^\infty_{k=0}\frac{y^{2k{2k+1},

\end{align}</math>

we arrive at

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

\ln (x) &= \frac{2(x - 1)}{x + 1} \sum_{k = 0}^\infty \frac{1}{2k + 1} {\left(\frac{(x - 1)^2}{(x + 1)^2}\right)}^k \\

&= \frac{2(x - 1)}{x + 1} \left( \frac{1}{1} + \frac{1}{3} \frac{(x - 1)^2}{(x + 1)^2} + \frac{1}{5} {\left(\frac{(x - 1)^2}{(x + 1)^2}\right)}^2 + \cdots \right) .

\end{align}</math>

Using the substitution <math>x=\tfrac{n+1}{n}</math> again for positive integers , we get:

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

\ln \left(\frac{n + 1}{n}\right) &= \frac{2}{2n + 1} \sum_{k=0}^\infty \frac{1}{(2k + 1) ((2n + 1)^2)^k}\\

&= 2 \left(\frac{1}{2n + 1} + \frac{1}{3 (2n + 1)^3} + \frac{1}{5 (2n + 1)^5} + \cdots \right).

\end{align}</math>

This is, by far, the fastest converging of the series described here.

The natural logarithm can also be expressed as an infinite product:

<math display="block">\ln(x)=(x-1) \prod_{k=1}^\infty \left ( \frac{2}{1+\sqrt[2^k]{x \right )</math>

Two examples might be:

<math display="block">\ln(2)=\left ( \frac{2}{1+\sqrt{2 \right )\left ( \frac{2}{1+\sqrt[4]{2 \right )\left ( \frac{2}{1+\sqrt[8]{2 \right )\left ( \frac{2}{1+\sqrt[16]{2 \right )...</math>

<math display="block">\pi=(2i+2)\left ( \frac{2}{1+\sqrt{i \right )\left ( \frac{2}{1+\sqrt[4]{i \right )\left ( \frac{2}{1+\sqrt[8]{i \right )\left ( \frac{2}{1+\sqrt[16]{i \right )...</math>

From this identity, we can easily get that:

<math display="block">\frac{1}{\ln(x)}=\frac{x}{x-1}-\sum_{k=1}^\infty\frac{2^{-k}x^{2^{-k}{1+x^{2^{-k}</math>

For example:

<math display="block">\frac{1}{\ln(2)} = 2-\frac{\sqrt{2{2+2\sqrt{2-\frac{\sqrt[4]{2{4+4\sqrt[4]{2-\frac{\sqrt[8]{2{8+8\sqrt[8]{2 \cdots</math>

Integration

The natural logarithm allows simple integration of functions of the form <math>g(x) = \frac{f'(x)}{f(x)}</math>: an antiderivative of is given by <math>\ln (|f(x)|)</math>. This is the case because of the chain rule and the following fact:

<math display="block">\frac{d}{dx}\ln \left| x \right| = \frac{1}{x}, \ \ x \ne 0</math>

In other words, when integrating over an interval of the real line that does not include <math>x=0</math>, then

where is an arbitrary constant of integration.

Likewise, when the integral is over an interval where <math>f(x) \ne 0</math>,

:<math display="block">\int { \frac{f'(x)}{f(x)}\,dx} = \ln|f(x)| + C.</math>

For example, consider the integral of <math>\tan (x)</math> over an interval that does not include points where <math>\tan (x)</math> is infinite:

<math display="block">\int \tan x \,dx = \int \frac{\sin x}{\cos x} \,dx = -\int \frac{\frac{d}{dx} \cos x}{\cos x} \,dx = -\ln \left| \cos x \right| + C = \ln \left| \sec x \right| + C.

</math>

The natural logarithm can be integrated using integration by parts. Let:

<math display="block">u = \ln x \Rightarrow du = \frac{dx}{x}</math>

<math display="block">dv = dx \Rightarrow v = x</math>

then:

\begin{align}

\int \ln x \,dx & = x \ln x - \int \frac{x}{x} \,dx \\

& = x \ln x - x + C

\end{align}

</math>

==Efficient computation==

For <math>\ln (x)</math> where , the closer the value of is to 1, the faster the rate of convergence of its Taylor series centered at 1.

Natural logarithm of 10

The natural logarithm of 10, a transcendental number approximately equal to , plays a role for example in the computation of natural logarithms of numbers represented in scientific notation, as a mantissa multiplied by a power of 10:

This means that one can effectively calculate the logarithms of numbers with very large or very small magnitude using the logarithms of a relatively small set of decimals in the range .

High precision

To compute the natural logarithm with many digits of precision, the Taylor series approach is not efficient since the convergence is slow. Especially if is near 1, a good alternative is to use Halley's method or Newton's method to invert the exponential function, because the series of the exponential function converges more quickly. For finding the value of to give <math>\exp(y)-x=0</math> using Halley's method, or equivalently to give <math>\exp(y/2) -x \exp(-y/2)=0</math> using Newton's method, the iteration simplifies to

which has cubic convergence to <math>\ln (x)</math>.

Another alternative for extremely high precision calculation is the formula

<math display="block">\ln x \approx \frac{\pi}{2 M(1,4/s)} - m \ln 2,</math>

where denotes the arithmetic-geometric mean of 1 and , and

with chosen so that bits of precision is attained. (For most purposes, the value of 8 for is sufficient.) In fact, if this method is used, Newton inversion of the natural logarithm may conversely be used to calculate the exponential function efficiently. (The constants <math>\ln 2</math> and pi| can be pre-computed to the desired precision using any of several known quickly converging series.) Or, the following formula can be used:

<math display="block">\ln x = \frac{\pi}{M\left(\theta_2^2(1/x),\theta_3^2(1/x)\right)},\quad x\in (1,\infty)</math>

where

\theta_2(x) = \sum_{n\in\Z} x^{(n+1/2)^2},

\quad

\theta_3(x) = \sum_{n\in\Z} x^{n^2}

</math>

are the Jacobi theta functions.

Based on a proposal by William Kahan and first implemented in the Hewlett-Packard HP-41C calculator in 1979 (referred to under "LN1" in the display, only), some calculators, operating systems (for example Berkeley UNIX 4.3BSD or log1p

An identity in terms of the inverse hyperbolic tangent,

<math display="block">\mathrm{log1p}(x) = \log(1+x) = 2 ~ \mathrm{artanh}\left(\frac{x}{2+x}\right)\,,</math>

gives a high precision value for small values of on systems that do not implement .

Computational complexity

The computational complexity of computing the natural logarithm using the arithmetic-geometric mean (for both of the above methods) is <math>\text{O}\bigl(M(n) \ln n \bigr)</math>. Here, is the number of digits of precision at which the natural logarithm is to be evaluated, and is the computational complexity of multiplying two -digit numbers.

Continued fractions

While no simple continued fractions are available, several generalized continued fractions exist, including:

\begin{align}

\ln(1+x) & =\frac{x^1}{1}-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots \\[5pt]

& = \cfrac{x}{1-0x+\cfrac{1^2x}{2-1x+\cfrac{2^2x}{3-2x+\cfrac{3^2x}{4-3x+\cfrac{4^2x}{5-4x+\ddots}

\end{align}

</math>

\begin{align}

\ln\left(1+\frac{x}{y}\right) & = \cfrac{x} {y+\cfrac{1x} {2+\cfrac{1x} {3y+\cfrac{2x} {2+\cfrac{2x} {5y+\cfrac{3x} {2+\ddots \\[5pt]

& = \cfrac{2x} {2y+x-\cfrac{(1x)^2} {3(2y+x)-\cfrac{(2x)^2} {5(2y+x)-\cfrac{(3x)^2} {7(2y+x)-\ddots

\end{align}

</math>

These continued fractions—particularly the last—converge rapidly for values close to 1. However, the natural logarithms of much larger numbers can easily be computed, by repeatedly adding those of smaller numbers, with similarly rapid convergence.

For example, since 2 = 1.253 × 1.024, the natural logarithm of 2 can be computed as:

\begin{align}

\ln 2 & = 3 \ln\left(1+\frac{1}{4}\right) + \ln\left(1+\frac{3}{125}\right) \\[8pt]

& = \cfrac{6} {9-\cfrac{1^2} {27-\cfrac{2^2} {45-\cfrac{3^2} {63-\ddots

+ \cfrac{6} {253-\cfrac{3^2} {759-\cfrac{6^2} {1265-\cfrac{9^2} {1771-\ddots.

\end{align}

</math>

Furthermore, since 10 = 1.2510 × 1.0243, even the natural logarithm of 10 can be computed similarly as:

\begin{align}

\ln 10 & = 10 \ln\left(1+\frac{1}{4}\right) + 3\ln\left(1+\frac{3}{125}\right) \\[10pt]

& = \cfrac{20} {9-\cfrac{1^2} {27-\cfrac{2^2} {45-\cfrac{3^2} {63-\ddots

+ \cfrac{18} {253-\cfrac{3^2} {759-\cfrac{6^2} {1265-\cfrac{9^2} {1771-\ddots.

\end{align}

</math>

The reciprocal of the natural logarithm can be also written in this way:

<math display="block">\frac {1}{\ln(x)} = \frac {2x}{x^2-1}\sqrt{\frac {1}{2}+\frac {x^2+1}{4x\sqrt{\frac {1}{2}+\frac {1}{2}\sqrt{\frac {1}{2}+\frac {x^2+1}{4x}\ldots</math>

For example:

<math display="block">\frac {1}{\ln(2)} = \frac {4}{3}\sqrt{\frac {1}{2} + \frac {5}{8 \sqrt{\frac {1}{2} + \frac {1}{2} \sqrt{\frac {1}{2} +\frac {5}{8} \ldots</math>

Complex logarithms

The exponential function can be extended to a function which gives a complex number as for any arbitrary complex number ; simply use the infinite series with =z complex. This exponential function can be inverted to form a complex logarithm that exhibits most of the properties of the ordinary logarithm. There are two difficulties involved: no has ; and it turns out that . Since the multiplicative property still works for the complex exponential function, , for all complex and integers .

So the logarithm cannot be defined for the whole complex plane, and even then it is multi-valued—any complex logarithm can be changed into an "equivalent" logarithm by adding any integer multiple of at will. The complex logarithm can only be single-valued on the cut plane. For example, or or , etc.; and although can be defined as , or or , and so on.

Image:NaturalLogarithmRe.png|

Image:NaturalLogarithmImAbs.png|

Image:NaturalLogarithmAbs.png|

Image:NaturalLogarithmAll.png| Superposition of the previous three graphs

</gallery>

Notes

References

de:Logarithmus#Natürlicher Logarithmus

History

Integration

Natural logarithm of 10

High precision

Computational complexity

Continued fractions

Complex logarithms

See also

Notes

References