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Please help me prove by induction that

$\displaystyle\forall n\in {{\mathbb{N}}^{*}}$, $\displaystyle\forall {{a}_{1}},\ldots ,{{a}_{n}}\in {\mathbb{R}}^{*}_{+}$, $\displaystyle \ln \left( \prod\limits_{j=1}^{n}{{{a}_{j}}} \right)=\sum\limits_{j=1}^{n}{\ln \left( {{a}_{j}} \right)}$.

Deduce that $\displaystyle \forall n\in \mathbb{Z},\forall a\in {\mathbb{R}}^{*}_{+}$, $\displaystyle \ln \left( {{a}^{n}} \right)=n\ln a$.

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    thx u Arturo very mux for editing...it is now appreciated....2011-03-10

2 Answers 2

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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.

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    oui~ merci pour ça...2011-03-11
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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 $\rm\ \ln\ (x\ *\ y)\ =\ \ln\ x\ +\ \ln\ y\:.$

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    merci beaucoup~2011-03-11