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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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Questions in the imperative mode are generally not appreciated. As far as I understand French, these infinitives are something similar to an imperative mode? Contentwise, in this sort of question, it's usually a good idea to indicate what it is you want to or are allowed to use to show this, since it's a relatively elementary fact and it's hard to tell which facts you (or your teacher/course/book) are considering as less elementary than this one (and thus a suitable basis for proving it). –  joriki Mar 10 '11 at 15:35
    
quelle definition du logarithm est-ce-que vous utilisez? the logarithm is the unique continuous function with the property that $f(ab)=f(a)+f(b)$ –  yoyo Mar 10 '11 at 15:40
    
sorry then...I just took it out from my exercises papers... it's in imperatif originally... I just wonder what's wrong with this exercise coz none did it~ –  Jesse Mar 10 '11 at 16:02
    
thx u Arturo very mux for editing...it is now appreciated.... –  Jesse Mar 10 '11 at 22:55

2 Answers 2

up vote 4 down vote accepted

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... –  Jesse Mar 11 '11 at 4:23

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~ –  Jesse Mar 11 '11 at 4:22

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