Sum of Logarithms/Natural Logarithm/Proof 3

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Theorem

Let $x, y \in \R$ be strictly positive real numbers.


Then:

$\ln x + \ln y = \map \ln {x y}$

where $\ln$ denotes the natural logarithm.


Proof

Let $\sequence {f_n}$ be the sequence of mappings $f_n : \R_{>0} \to \R$ defined as:

$\map {f_n} x = n \paren {\sqrt [n] x - 1}$

Let $\map M t = \max \set {\size {t - 1}, \size {\dfrac {t - 1} t} }$

From Bounds of Natural Logarithm:

$\forall t \in \R_{>0}: \size {\map {f_n} t} \le \map M t$

Fix $x, y \in \R_{>0}$.

Then:

\(\ds \size {\map {f_n} {x y} - \paren {\map {f_n} x + \map {f_n} x} }\) \(=\) \(\ds \size {n \paren {\sqrt [n] {x y} - 1 - \sqrt [n] x + 1 - \sqrt [n] y + 1} }\) Definition of $\map {f_n} x$
\(\ds \) \(=\) \(\ds n \size {\sqrt [n] {x y} - \sqrt [n] x - \sqrt [n] y + 1}\) Absolute Value Function is Completely Multiplicative
\(\ds \) \(=\) \(\ds n \size {\sqrt [n] x - 1} \size {\sqrt [n] y - 1}\) Absolute Value Function is Completely Multiplicative
\(\ds \) \(=\) \(\ds n \frac {\size {\map {f_n} x} } n \frac {\size {\map {f_n} y} } n\) Definition of $\map {f_n} x$
\(\ds \) \(=\) \(\ds \frac 1 n \size {\map {f_n} x} \size {\map {f_n} y}\)
\(\ds \) \(\le\) \(\ds \frac {\map M x \map M y} n\) Inequality of Product of Unequal Numbers
\(\ds \leadsto \ \ \) \(\ds \size {\map \ln {x y} - \paren {\map \ln x + \map \ln y} }\) \(=\) \(\ds \size {\lim_{n \mathop \to \infty} \map {f_n} {x y} - \paren {\lim_{n \mathop \to \infty} \map {f_n} x + \lim_{n \mathop \to \infty} \map {f_n} y} }\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\map {f_n} {x y} - \paren {\map {f_n} x + \map {f_n} y} }\) Modulus of Limit
\(\ds \) \(\le\) \(\ds \lim_{n \mathop \to \infty} \frac {\map M x \map M y} n\) Limit of Bounded Convergent Sequence is Bounded
\(\ds \) \(=\) \(\ds \map M x \map M y \lim_{n \mathop \to \infty} \frac 1 n\) Multiple Rule for Real Sequences
\(\ds \) \(=\) \(\ds 0\)

Thus:

$\ds \lim_{n \mathop \to \infty} \map {f_n} {x y} = \lim_{n \mathop \to \infty} \paren {\map {f_n} x + \map {f_n} y}$

Hence the result, from the definition of $\ln$.

$\blacksquare$


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