nyuuzyou/wikis huggingface dataset:
Locus of Midpoints of Parallel Chords of Central Conic passes through Center

== Theorem ==
Let $\KK$ be a central conic.
Let $\DD$ be the locus of the midpoints of a system of parallel chords of $\KK$.
Then $\DD$ passes through the center of $\KK$.
== Proof ==
We note from Circle is Ellipse with Equal Major and Minor Axes that a circle is a special case of the ellipse.
Hence there is no need to investigate the circle separately.
It remains to demonstrate the result for the ellipse and the hyperbola.
Let the central conic be expressed in reduced form.
By definition of reduced form, the center of $\KK$ lies at the origin.
Hence it is sufficient to demonstrate that the locus of the midpoints of an arbitrary system of parallel chords of $\KK$ passes through the origin.
From Equation of Ellipse in Reduced Form and Equation of Hyperbola in Reduced Form, $\KK$ can be expressed as:
:$\dfrac {x^2} {a^2} \pm \dfrac {y^2} {b^2} = 1$
Let $\family \CC_{\alpha \mathop \in I}$ be an indexed family of parallel chords of $\KK$ for a given indexing set $I$.
By Condition for Straight Lines in Plane to be Parallel, each of the elements of $\family \CC$ have the same slope.
Let $\LL$ be an arbitrary element of $\family \CC$.
By Equation of Straight Line in Plane: Slope-Intercept Form, $\LL$ therefore be expressed as:
:$y = m x + c$
where $m$ is the slope of $\LL$.
Because $\LL$ is a chord of $\KK$, it intersects $\KK$ in $2$ points.
Let these $2$ points be $\tuple {x_1, y_1}$ and $\tuple {x_2, y_2}$.
Let its midpoint be $\tuple {h, k}$.
We have:
:$h = \dfrac {x_1 + x_2} 2$
:$k = \dfrac {y_1 + y_2} 2$
and so:
:$\tuple {x_2, y_2} = \tuple {2 h - x_1, 2 y - y_1}$
By construction, either $h \ne x_1$ or $y \ne y_1$ or both.
, suppose $h \ne x_1$.
We have that $\tuple {x_1, y_1}$ and$\tuple {x_2, y_2}$ both lie on $\KK$ and so:
:$\dfrac { {x_1}^2} {a^2} \pm \dfrac { {y_1}^2} {b^2} = 1$
:$\dfrac { {x_2}^2} {a^2} \pm \dfrac { {y_2}^2} {b^2} = 1$
Substituting for $\tuple {x_2, y_2}$:
By definition, the slope $m$ of $\LL$ is given by:
Substituting for $k - y_1$ in $(1)$:
We have that $\tuple {h, k}$ is the midpoint of $\LL$.
We have that $\LL$ is an element of a family of chords such that $m$, $a$ and $b$ are constant, it follows that the locus of their midpoints is the set of points $\tuple {h, k}$ such that $\dfrac {k m} {b^2} = \pm \dfrac h {a^2}$.
Hence $\tuple {h, k}$ is an arbitrary point on that locus.
It follows that this locus can be expressed by the equation:
:$\dfrac m {b^2} y = \pm \dfrac 1 {a^2} x$
which is in the form of Equation of Straight Line in Plane: Slope-Intercept Form:
:$y = m' x + c$
where $m' = \pm \dfrac {b^2} {m a^2}$ and $c = 0$.
But $c$ is the $y$-intercept.
Because $c = 0$, it follows that $\dfrac m {b^2} y = \pm \dfrac 1 {a^2} x$ is the equation of a straight line passing through the origin.
Hence the result.
== Sources ==
* 
* 
Category:Diameters of Conic Sections
Category:Central Conics
Log of Gamma Function is Convex on Positive Reals

== Theorem ==
Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers.
Let $\ln$ denote the natural logarithm function.
Then the composite mapping $\ln \circ \operatorname \Gamma$ is a convex function.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $11.14 \ \text{(iv)}$
Category:Gamma Function
Category:Log of Gamma Function is Convex on Positive Reals
Log of Gamma Function is Convex on Positive Reals/Proof 1

== Theorem ==
== Proof ==
By definition, the Gamma function $\Gamma: \R_{> 0} \to \R$ is defined as:
:$\ds \map \Gamma z = \int_0^{\infty} t^{z - 1} e^{-t} \rd t$
:$\forall z > 0: \map \Gamma z > 0$, as an integral of a strictly positive function in $t$.
 
The function is smooth according to Gamma Function is Smooth on Positive Reals, and
:$\ds \forall k \in \N: \map {\Gamma^{\paren k} } z = \int_0^{\infty} \map \ln t^k t^{z - 1} e^{-t} \rd t$
Let $\map f z := \map \ln {\map \Gamma z}$.
:$f$ is smooth because $\Gamma$ is smooth and positive.
Then:
:$\map {f'} z = \dfrac {\map {\Gamma'} z} {\map \Gamma z}$
:$\map {f^{\paren 2} } z = \dfrac {\map {\Gamma^{\paren 2} } z \map \Gamma z - \map {\Gamma'} z^2} {\map \Gamma z^2} > 0$
The numerator is positive due to the Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces applied to the scalar products:
:$\ds \innerprod g h = \int_0^\infty \map g t \map h t t^{z - 1} e^{-t} \rd t \quad \forall z \gt 0$
applied to $g = \ln$ and $h = 1$.
:$\forall z \in \R_{>0}: \map {f^{\paren 2} } z > 0 \implies f$ is convex.
Category:Log of Gamma Function is Convex on Positive Reals
Log of Gamma Function is Convex on Positive Reals/Proof 2

== Theorem ==
== Proof ==
The strategy is to show that:
:$\map \ln {\map \Gamma {\dfrac x 2 + \dfrac y 2} } \le \dfrac 1 2 \map \ln {\map \Gamma x} + \dfrac 1 2 \map \ln {\map \Gamma y}$
Let $0 < \delta < \Delta$.
Then:
Letting $\delta \to 0$ and $\Delta \to \infty$, $(1)$ becomes equivalent to:
:$\paren {\map \Gamma {\dfrac {x + y} 2} }^2 \le \paren {\map \Gamma x} \paren {\map \Gamma y}$
The result follows by definition of convex function.
== Sources ==
* : $\S 17.4 \ (6)$
Category:Log of Gamma Function is Convex on Positive Reals
Log of Gamma Function is Convex on Positive Reals/Proof 3

== Theorem ==
== Proof ==
The strategy is to use the Euler Form of the Gamma function and directly calculate the second derivative of $\ln \map \Gamma z$.
The limit interchange is justified because Gamma Function is Smooth on Positive Reals.
Logarithmic convexity then follows from Real Function with Strictly Positive Second Derivative is Strictly Convex.
Category:Log of Gamma Function is Convex on Positive Reals
Logarithm Base 10 of 2 is Irrational

== Theorem ==
The common logarithm of $2$:
:$\log_{10} 2 \approx 0.30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$
is irrational.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $10$
Category:Common Logarithms
Category:2
Category:10
Category:Logarithm Base 10 of 2 is Irrational
Logarithm Base 10 of 2 is Irrational/Proof 1

== Theorem ==
== Proof ==
 $\log_{10} 2$ is rational.
Then:
Both $10^p$ and $2^q$ are integers, by construction.
But $10^p$ is divisible by $5$, while $2^p$, which has only $2$ as a prime factor, is not.
So $10^p \ne 2^q$.
So, by Proof by Contradiction, it follows that $\log_{10} 2$ is irrational.
Category:Logarithm Base 10 of 2 is Irrational
Logarithm Base 10 of 2 is Irrational/Proof 2

== Theorem ==
== Proof ==
Because $5$ is a divisor of $10$ but not $2$, it cannot be the case that $2^a = 10^b$ for $a, b \in \Z_{>0}$.
Hence this is a special case of Irrationality of Logarithm.
Category:Logarithm Base 10 of 2 is Irrational
Logarithm Tends to Infinity

== Theorem ==
Let $x \in \R_{>0}$ be a strictly positive real number.
Let $\ln x$ be the natural logarithm of $x$.
Then:
:$\ln x \to +\infty$ as $x \to +\infty$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Logarithm Tends to Negative Infinity
== Sources ==
* : Appendix $A$: Properties of the Natural Logarithmic Function
Category:Logarithms
Category:Logarithm Tends to Infinity
Logarithm Tends to Infinity/Proof 1

== Theorem ==
== Proof ==
From Natural Logarithm of 2 is Greater than One Half:
:$\ln 2 \ge \dfrac 1 2$
From the definition of infinite limit at infinity, our assertion is:
:$\forall M \in \R_{>0} : \exists N > 0 : x > N \implies \ln x > M$.
As $x \to +\infty$, we will restrict our attention to sufficiently large $M$.
From Logarithm is Strictly Increasing:
:$\ln x$ is strictly increasing.
So, for sufficiently large $M$:
:$x > 2^{2 M} \implies \ln x > \ln 2^{2 M}$
From the Laws of Logarithms:
Choosing $N = \ln 2^{2 M}$:
:$\forall M \ge a: \exists N > 0: x > N \implies \ln x > M$
for some $a \in \R$.
Hence the result, by the definition of infinite limit at infinity.
== Sources ==
* : Appendix $A$: Properties of the Natural Logarithmic Function
Category:Logarithm Tends to Infinity
Logarithm Tends to Negative Infinity

== Theorem ==
Let $x \in \R$ be a real number such that $x > 0$.
Let $\ln x$ be the natural logarithm of $x$.
Then:
:$\ln x \to -\infty$ as $x \to 0^+$
== Proof ==
From the definition of natural logarithm:
The result follows from Integral of Reciprocal is Divergent.
== Also see ==
* Logarithm Tends to Infinity
Category:Logarithms
Logarithm is Strictly Concave

== Theorem ==
Let $x \in \R$ be a real number such that $x > 0$.
Let $\ln x$ be the natural logarithm of $x$.
Then:
:$\ln x: x > 0$ strictly concave.
== Proof ==
From Logarithm is Strictly Increasing, $\ln x$ is strictly increasing on $x > 0$.
From Second Derivative of Natural Logarithm Function:
:$D^2 \ln x = -\dfrac 1 {x^2}$
Thus $D^2 \ln x$ is strictly negative on $x > 0$ (in fact is strictly negative for all $x \ne 0$).
Thus from Derivative of Monotone Function, $D \dfrac 1 x$ is strictly decreasing on $x > 0$.
So from Real Function is Strictly Concave iff Derivative is Strictly Decreasing, $\ln x$ is strictly concave on $x > 0$.
== Sources ==
* : $\S 14.1$
Category:Logarithms
Logarithm is Strictly Increasing

== Theorem ==
Let $x \in \R$ be a real number such that $x > 0$.
Let $\ln x$ be the natural logarithm of $x$.
Then:
:$\ln x: x > 0$ is strictly increasing.
=== Corollary ===
== Proof ==
From Derivative of Natural Logarithm Function $D \ln x = \dfrac 1 x$, which is strictly positive on $x > 0$.
From Derivative of Monotone Function it follows that $\ln x$ is strictly increasing on $x > 0$.
== Sources ==
* : $\S 14.1$
Category:Logarithms
Category:Examples of Strictly Increasing Real Functions
Logarithm is Strictly Increasing/Corollary

== Corollary to Logarithm is Strictly Increasing ==
Let $\ln$ be the natural logarithm.
Then $\ln$ is injective on $\R_{>0}$.
== Proof ==
From Logarithm is Strictly Increasing, $\ln$ is strictly increasing on $\R_{> 0}$.
From Ordering on Real Numbers is Total Ordering, $\struct {\R_{> 0}, \le}$ is totally ordered.
From Strictly Monotone Mapping with Totally Ordered Domain is Injective, $\ln$ is injective.
Category:Logarithms
Logarithm of Absolute Value of 2 times Sine of pi x is Replicative Function

== Theorem ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \log \, \size {2 \sin \pi x}$
Then $f$ is a replicative function.
== Proof ==
We have that:
Thus to demonstrate that $f$ is replicative, it is sufficient to demonstrate that:
:$\ds \prod_{k \mathop = 0}^{n - 1} \paren {2 \sin \pi \paren {x + \frac k n} } = 2 \sin \pi n x$
By Product Formula for Sine:
Hence the result.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $39 \ \text{f)}$
Category:Replicative Functions
Logarithm of Base

== Theorem ==
Let $b \in \R_{>0}$ such that $b \ne 1$.
Then:
:$\log_b b = 1$
where $\log_b$ denotes the logarithm to base $b$.
== Proof ==
Let $a \in \R_{>0}$ such that $a \ne 1$.
Then:
Category:Logarithms
Logarithm of Beta Function is Convex on Positive Reals

== Theorem ==
Let $x, y \in \R$ be real numbers.
Let $\map \Beta {x, y}$ be the Beta function:
:$\ds \map \Beta {x, y} := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \paren {1 - t}^{y - 1} \rd t$
Let $y \in \R_{>0}$ be given.
Then $\map \ln {\map \Beta {x, y} }$ is a convex function of $x$ on $\R_{>0}$.
== Proof ==
== Sources ==
* : $\S 17.7 \ (4)$
Category:Beta Function
Logarithm of Convergent Product of Real Numbers

== Theorem ==
Let $\sequence {a_n}$ be a sequence of strictly positive real numbers.
:$(1): \quad$ The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ converges to $a \in \R_{\ne 0}$.
:$(2): \quad$ The series $\ds \sum_{n \mathop = 1}^\infty \ln a_n$ converges to $\ln a$.
== Proof ==
Let $p_n$ denote the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty a_n$.
Let $s_n$ denote the $n$th partial sum of $\ds \sum_{n \mathop = 1}^\infty \ln a_n$.
By Sum of Logarithms, $s_n = \map \ln {p_n}$.
=== 1 implies 2 ===
Let $\ds \prod_{n \mathop = 1}^\infty a_n$ converge to $a>0$.
Then $p_n \to a$.
By Real Natural Logarithm Function is Continuous:
:$s_n \to \ln a$
=== 2 implies 1 ===
Let $\ds \sum_{n \mathop = 1}^\infty \ln a_n$ converge to $\ln a$.
Then $s_n \to \ln a$.
By Exponential Function is Continuous, $p_n \to a$.
Because $a \ne 0$, $\ds \prod_{n \mathop = 1}^\infty a_n$ converges to $a$.
== Also see ==
* Logarithm of Infinite Product of Real Numbers, for similar results
* Logarithm of Infinite Product of Complex Numbers
Category:Infinite Products
Category:Convergence
Logarithm of Dirichlet L-Functions

== Theorem ==
Let $\chi$ be a Dirichlet character modulo $q$.
The Dirichlet series:
:$\map f s = \ds \sum_{n \mathop \ge 1} \sum_p \frac {\map \chi p^n} {n p^{n s} }$
converges absolutely to an analytic function, where $p$ ranges over the primes.
Moreover, $\map f s$ defines a branch of $\ln \map L {s, \chi}$.
== Proof ==
By Convergence of Dirichlet Series with Bounded Coefficients, $\map f s$ converges absolutely on $\map \Re s > 1$ to an analytic function.
For fixed $s \in \set {\map \Re s > 1}$:
Hence $\map f s$ is an analytic branch of $\ln$ on $\map \Re s > 1$.
Category:Analytic Number Theory
Logarithm of Divergent Product of Real Numbers

== Theorem ==
Let $\left\langle{a_n}\right\rangle$ be a sequence of strictly positive real numbers.
=== Divergence to zero ===
=== Divergence to infinity ===
== Also see ==
* Logarithm of Infinite Product of Real Numbers, for similar results
Category:Infinite Products
Category:Convergence
Category:Logarithms
Category:Logarithm of Divergent Product of Real Numbers
Logarithm of Divergent Product of Real Numbers/Infinity

== Theorem ==
Let $\sequence {a_n}$ be a sequence of strictly positive real numbers.
:
:$(1): \quad$ The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ diverges to $+\infty$.
:$(2): \quad$ The series $\ds \sum_{n \mathop = 1}^\infty \log a_n$ diverges to $+\infty$.
== Proof ==
Let $p_n$ denote the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty a_n$.
Let $s_n$ denote the $n$th partial sum of $\ds \sum_{n \mathop = 1}^\infty\log a_n$.
By Sum of Logarithms:
:$s_n = \map \log {p_n}$
== Also see ==
* Logarithm of Infinite Product of Real Numbers, for similar results
Category:Logarithm of Divergent Product of Real Numbers
Logarithm of Divergent Product of Real Numbers/Zero

== Theorem ==
Let $\sequence {a_n}$ be a sequence of strictly positive real numbers.
The following are equivalent:
:The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$.
:The series $\ds \sum_{n \mathop = 1}^\infty \log a_n$ diverges to $-\infty$.
== Proof ==
Let $p_n$ denote the $n$th partial product of $\ds \prod_{n \mathop = 1}^\infty a_n$.
Let $s_n$ denote the $n$th partial sum of $\ds \sum_{n \mathop = 1}^\infty \log a_n$.
By Sum of Logarithms:
:$s_n = \map \log {p_n}$
== Also see ==
* Logarithm of Infinite Product of Real Numbers, for similar results
Category:Logarithm of Divergent Product of Real Numbers
Logarithm of Factorial

== Theorem ==
where:
:$\ds a_m = \dfrac {\paren {-1}^m} {m!} \sum_{k \mathop \ge 1} \paren {-1}^k \binom {m - 1} {k - 1} \ln k$
== Proof ==
== Historical Note ==
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: $(12)$
Category:Factorials
Category:Falling Factorials
Category:Logarithm of Factorial
Logarithm of Factorial/Historical Note

== Historical Note on Logarithm of Factorial ==
The sequence of $a_1, a_2, \ldots$ was established by during the course of his attempt extend the factorial to the real numbers.
However, although he established that $\paren {\dfrac 1 2}! = \dfrac {\sqrt \pi} 2$, he was not able to prove that this sum defined $n!$ for a general non-integer $n$.
It was who finally proved in $1900$ that this formula does indeed define $n!$, by demonstrating that it is identical to the Euler form of the Gamma function.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials
Category:Historical Notes
Logarithm of Infinite Product of Complex Functions

== Theorem ==
Let $X$ be a weakly locally compact topological space.
Let $\sequence {f_n}$ be a sequence of everywhere nonzero continuous mappings $f_n: X \to \C$.
Then the following are equivalent:
:$(1): \quad$ The product $\ds \prod_{n \mathop = 1}^\infty f_n$ converges locally uniformly to $f$.
:$(2): \quad$ The series $\ds \sum_{n \mathop = 1}^\infty \ln f_n$ converges locally uniformly to $\ln f + 2k\pi i$ for some mapping $k:K\to\Z$.
=== Corollary ===
== Proof ==
=== 1 implies 2 ===
It suffices to show that:
:$(1): \quad \ds \sum_{n \mathop = 1}^\infty \map \Re {\ln f_n} = \map \Re {\ln f}$ locally uniformly
:$(2): \quad \ds \sum_{n \mathop = 1}^\infty \map \Im {\ln f_n} = \map \Im {\ln f} + 2 k \pi$ locally uniformly for some $k: K \to \Z$
==== Real part ====
By Absolute Value of Uniformly Convergent Product, $\ds \prod_{n \mathop = 1}^\infty \cmod {f_n} = \cmod f$ locally uniformly.
By Logarithm of Infinite Product of Real Functions $\ds \sum_{n \mathop = 1}^\infty \ln \cmod {f_n} = \ln \cmod f$ locally uniformly.
Hence the result.
==== Imaginary Part ====
Let $K\subset X$ be compact.
Let $P_n$ denote the $n$th partial product.
Then $P_n \to f$ uniformly on $K$.
Let $\theta$ be the argument of $f$.
Let $\theta_n = \arg P_n$ be the argument of $P_n$ in the half-open interval $\hointl {\theta - \pi} {\theta + \pi}$.
By the corollary to Uniform Convergence of Complex Functions in Polar Form:
:$\theta_n \to \theta$ uniformly on $K$.
Let $k_n: K \to \Z$ be such that:
:$\ds \sum_{j \mathop = 1}^n \map \Im {\ln \map {f_n} x} = \map {\theta_n} x + 2 \map {k_n} x \pi$
We show that $k_n$ is eventually equal to a fixed function $k: K \to \Z$.
By the Triangle Inequality for Complex Numbers:
:$2 \pi \sup_{x \mathop \in K} \size {\map {k_{n + 1} } x - \map {k_n} x} \le \sup_{x \mathop \in K} \cmod {\map {\theta_{n + 1} } x - \map {\theta_n} x} + \sup_{x \mathop \in K} \size {\map \Im {\ln \map {f_{n + 1} } x } }$
By Factors in Uniformly Convergent Product Converge Uniformly to One, $f_n \to 1$ uniformly on $K$.
Let $n_0 \in \N$ be such that $\cmod {f_n - 1} \le \dfrac 1 2$ for $n \ge n_0$.
By Complex Logarithm is Continuous Outside Branch and Heine-Cantor Theorem, $\ln$ is uniformly continuous on $\map {\overline B} {1, \dfrac 1 2}$.
By Uniformly Continuous Function Preserves Uniform Convergence, $\ln f_n \to 0$ uniformly on $K$.
Thus $\ds 2 \pi \sup_{x \mathop \in K} \cmod {\map {k_{n + 1} } x - \map {k_n} x} \to 0$.
So the sequence $k_n$ is eventually constant, say equal to $k: K \to \Z$.
Then:
:$\ds \sum_{j \mathop = 1}^n \map \Im {\ln f_n} \to \theta + 2 k \pi$ uniformly on $K$.
=== 2 implies 1 ===
Follows from Complex Exponential is Uniformly Continuous on Half-Planes.
== Also see ==
* Logarithm of Infinite Product of Complex Numbers, of which this proof is no more than an adaptation
* Logarithmic Derivative of Infinite Product of Analytic Functions
Category:Infinite Products
Logarithm of Infinite Product of Complex Functions/Corollary

== Corollary to Logarithm of Infinite Product of Complex Functions ==
Let $X$ be a locally compact and locally connected metric space.
Let $\sequence {f_n}$ be a sequence of continuous mappings $f_n: X \to \C$.
Let the product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge locally uniformly to $f$.
Let $x_0 \in X$.
Then there exist $n_0 \in \N$, $k \in \Z$ and a neighborhood $U$ of $x_0$ such that:
:$(1): \quad \map {f_n} x \ne 0$ for $n \ge n_0$ and $x \in U$
:$(2): \quad$ The series $\ds \sum_{n \mathop = n_0}^\infty \ln f_n$ converges uniformly on $U$ to $\ln g + 2 k \pi i$, where $g = \ds \prod_{n \mathop = n_0}^\infty f_n$.
== Outline of Proof ==
We construct a neighborhood of $x_0$ on which $\ds \sum_{n \mathop = n_0}^\infty \ln f_n$ and $\ln g$ are continuous, so that $k$ is continuous and thus constant.
== Proof ==
Let $K$ be a compact neighborhood of $x_0$.
By Tail of Uniformly Convergent Product Converges Uniformly to One, there exists $N \in \N$ such that $\ds \prod_{n \mathop = N}^\infty \map {f_n} x \notin \R^-$ for all $x \in K$.
By Factors in Uniformly Convergent Product Converge Uniformly to One, there exists $M \in \N$ such that $\size {\map {f_n} x - 1} \le \dfrac 1 2$ for $n \ge M$ and $x \in K$.
Let $n_0 = \map \max {N, M}$.
Let $g = \ds \prod_{n \mathop = n_0}^\infty f_n$.
By Logarithm of Infinite Product of Complex Functions, there exists $k: K \to \Z$ such that $\ds \sum_{n \mathop = n_0}^\infty \ln f_n = \ln g + 2 k \pi i$ uniformly on $K$.
We show that $k$ is constant on some neighborhood $U \subset K$.
By Uniform Product of Continuous Functions is Continuous and Complex Logarithm is Continuous Outside Branch, $\ln g$ is continuous on $K$.
By the Heine-Cantor Theorem, $\ln$ is uniformly continuous on $\map {\overline B} {1, \dfrac 1 2}$.
By Uniformly Continuous Function Preserves Uniform Convergence, $\ds \sum_{n \mathop = n_0}^\infty \ln f_n$ converges uniformly on $K$.
By the Uniform Limit Theorem, $\ds \sum_{n \mathop = n_0}^\infty \ln f_n$ is continuous.
Because $\ln g$ and $\ds \sum_{n \mathop = n_0}^\infty \ln f_n$ are continuous, so is $k$.
Let $U \subset K$ be a connected neighborhood of $x_0$.
By Continuous Mapping from Connected to Discrete Space is Constant $k$ is constant on $U$.
Category:Infinite Products
Logarithm of Infinite Product of Complex Numbers

== Theorem ==
Let $\sequence {z_n}$ be a sequence of nonzero complex numbers.
:$(1): \quad$ The infinite product $\ds \prod_{n \mathop = 1}^\infty z_n$ converges to $z \in \C_{\ne 0}$.
:$(2): \quad$ The series $\ds \sum_{n \mathop = 1}^\infty \log z_n$ converges to $\log z + 2 k \pi i$ for some integer $k \in \Z$. 
== Proof ==
=== $(1)$ implies $(2)$ ===
Suppose $\ds \prod_{n \mathop = 1}^\infty z_n$ converges to $z$.
By Convergence of Series of Complex Numbers by Real and Imaginary Part, it suffices to show that:
:$\ds \sum_{n \mathop = 1}^\infty \Re \log z_n = \Re \log z$
:$\ds \sum_{n \mathop = 1}^\infty \Im \log z_n = \Im \log z + 2 k \pi$
==== Real Part ====
Let $P_n$ denote the $n$th partial product.
Then $P_n \to z$.
By Modulus of Limit:
:$\cmod {P_n} \to \cmod z$
By Real Natural Logarithm Function is Continuous;
:$\log \cmod {P_n} \to \log \cmod z$
By definition of complex natural logarithm and Real Part of Sum of Complex Numbers:
:$\ds \log \cmod {P_n} = \sum_{j \mathop = 1}^n \Re \log z_j$
==== Imaginary Part ====
It remains to show that:
:$\ds \sum_{j \mathop = 1}^n \Im \log z_j \to \arg z + 2 k \pi$
for some $k \in \Z$.
Let $\theta = \arg z$ be the argument of $z$.
Let $\theta_n = \arg P_n$ be the argument of $P_n$ in the half-open interval $\hointl {\theta - \pi} {\theta + \pi}$.
By the corollary to Convergence of Complex Sequence in Polar Form:
:$\theta_n \to \theta$
Let $k_n \in \Z$ be such that:
:$\ds \sum_{j \mathop = 1}^n \Im \log z_j = \theta_n + 2 k_n \pi$
We show that $k_n$ is eventually constant.
By the Triangle Inequality:
:$2 \pi \cmod {k_{n + 1} - k_n} \le \cmod {\theta_{n + 1} - \theta_n} + \cmod {\Im \log z_{n + 1} }$
By Factors in Convergent Product Converge to One, $z_n \to 1$.
By Complex Logarithm is Continuous Outside Branch, $\log z_n \to 0$.
Thus $2 \pi \cmod {k_{n + 1} - k_n} \to 0$.
So $k_n$ is eventually constant, say equal to $k$.
Then:
:$\ds \sum_{j \mathop = 1}^n \Im \log z_j \to \theta + 2 k \pi$
=== $(2)$ implies $(1)$ ===
Suppose $\ds \sum_{n \mathop = 1}^\infty \log z_n = \log z + 2 k \pi i$.
By Exponential of Series Equals Infinite Product:
:$\ds \prod_{n \mathop = 1}^\infty z_n = z$
Hence the result.
== Also see ==
* Logarithm of Infinite Product of Real Numbers
* Logarithm of Infinite Product of Complex Functions
== Sources ==
* $\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S 5$: Weierstrass Factorization Theorem: Proposition $5.2$
Category:Infinite Products
Logarithm of Logarithm in terms of Natural Logarithms

== Theorem ==
Let $b, x \in \R_{>0}$ be (strictly) positive real numbers.
Then:
:$\map {\log_b} {\log_b x} = \dfrac {\map \ln {\ln x} - \map \ln {\ln b} } {\ln b}$
where $\ln x$ denotes the natural logarithm of $x$.
== Proof ==
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $21$
Category:Logarithms
Category:Natural Logarithms