Please help me prove by induction that
$\endgroup$ 4$\displaystyle\forall n\in {{\mathbb{N}}^{*}}$, $\displaystyle\forall {{a}_{1}},\ldots ,{{a}_{n}}\in {\mathbb{R}}^{*}_{+}$, $\displaystyle \ln \left( \prod\limits_{j=1}^{n}{{{a}_{j}}} \right)=\sum\limits_{j=1}^{n}{\ln \left( {{a}_{j}} \right)}$.
Deduce that $\displaystyle \forall n\in \mathbb{Z},\forall a\in {\mathbb{R}}^{*}_{+}$, $\displaystyle \ln \left( {{a}^{n}} \right)=n\ln a$.
2 Answers
$\begingroup$In certain "transitions" classes I have taught -- i.e., for undergraduate math majors getting used to formal proof and abstraction -- I assign problems like these as (rather easy) exercises in induction, the point being that the usual "binary" form of the identity is assumed, so here
$\log(xy) = \log x + \log y$.
(I am also guessing that the English translation of "par récurrence" is "by induction" and not, for instance, "by recurrence".)
If this is the case, see e.g. $\S 5$ of this handout on induction for some similar examples of such induction proofs. The main idea here is that you have a product like $x_1 \cdots x_n x_{n+1}$ and you "cleverly" regroup it as $(x_1 \cdots x_n) \cdot x_{n+1}$ -- i.e., first a product of $n$ terms, to which your induction hypothesis applies, and then a binary product, to which your basic identity applies.
$\endgroup$ 1 $\begingroup$HINT $\rm\displaystyle\ \ \ \ f(n)\ =\ \prod_{j\ =\ 1}^n\ a_j $
$\rm\quad \iff\ \ \ \:f\:(n)\ =\:\ a_n\ \: * \ \ f\:(n-1),\:\ \ \ f\:(0)\: = 1$
$\rm\quad \iff\ \ F(n)\ =\ A_n + F(n-1),\ \ F(0) = 0\:,\ $ with $\rm\ \ F(n) = \ln\: f(n)\:,\ \ A_n = \ln\: a_n$
$\rm\displaystyle\quad \iff\ \ F(n)\ =\ \sum_{j\ =\ 1}^n\ A_n$
The first and last equivalences are the recursive definitions of $\rm\:\Pi\:$ and $\rm\:\Sigma\:.$
The middle equivalence follows from $\rm\ \ln\ (x\ *\ y)\ =\ \ln\ x\ +\ \ln\ y\:.$
$\endgroup$ 1
