1

Here's my exercise: EDIT: $v=nu$, not $\nu$ (same Latex code but one is without )

$\forall n \in \mathbb{N}$ $\forall$ ${i \in \{1,...n\} }$ $ |a_{i}|\leq u $ and $nu<0.01$ and $u=2^{-t-1}$ prove that there exists $\eta$ such that $\prod_{i=0}^{n}(1+a_{i})$=1+$\eta$ and $\eta \leq 1.01\nu$ .

In my classroom we have proven the following theorem:

$\forall n \in \mathbb{N}$ if $\forall$ ${i \in \{1,...n\} } $ $ |a_{i}|\leq u, p_{i} \in \{1,-1\} $ and $\nu<1$ where $u=2^{-t-1}$ for some integer $t$

then

there exists $\theta$ such that $\prod_{i=0}^{n}(1+a_{i})^{p_{i}}$=1+$\theta$ and $|\theta| \leq \frac{\nu}{1-\nu}$

At first I thought that my exercise demanded nothing more that using the theorem we have proven, for the case where all the $p$'s were $1$'s. Using that I put $\eta=\theta$ , but then I got an estimation $|\eta|\leq \frac{\nu}{1-\nu} < 1.(01)\nu$, which is slightly more than I wanted, and I see no other way to approach this exercise... Can anyone help?

Arek Krawczyk
  • 2,114
  • 1
    I can see no difference between what you should get and what you got. Btw is $\nu=nu$? – Algebraic Pavel Oct 13 '13 at 15:57
  • Yes, $v=nu$, for some reason Bill Liu changed it. What I got is that $\theta <1.(01)nu$, and what I wanted is $\theta<=1.01nu$. I have no way of knowing, that $\theta$ isn't for example $\theta=1.0101nu$, which would go against what I'm trying to prove, and be completely right with the second theorem. – Arek Krawczyk Oct 13 '13 at 16:04
  • 1
    OK sorry I misread the notation. The proof is a bit more complicated than I thought :) – Algebraic Pavel Oct 13 '13 at 18:34

1 Answers1

2

While the "class theorem" can be proved simply by induction, this one can be a bit "tricky". Let $\prod_{i=1}^n (1+a_i)=1+\theta$, so that $$ \theta=\prod_{i=1}^n(1+a_i)-1. $$ Since $|a_i|\leq u$, we have $$ |\theta|=\left|\prod_{i=1}^n(1+a_i)-1\right|\leq(1+u)^n-1. $$ Use the binomial theorem to get $$ \begin{split} |\theta|&\leq\sum_{k=1}^n\begin{pmatrix}n \\k\end{pmatrix}u^k \leq\sum_{k=1}^n\frac{(nu)^k}{k!}\leq nu\sum_{k=0}^n\left(\frac{nu}{2}\right)^k \\&\leq nu\sum_{k=0}^{\infty}\left(\frac{nu}{2}\right)^k\leq\frac{nu}{1-nu/2}\leq\frac{nu}{0.995}<1.01nu. \end{split} $$

Algebraic Pavel
  • 22,928
  • Thank you so much! You're a life saver :) – Arek Krawczyk Oct 13 '13 at 18:52
  • 1
    You're welcome! :) Btw the proof based on $(1+u)\leq\exp(u)$ can be found here: http://books.google.fr/books?id=epilvM5MMxwC&lpg=PP1&hl=cs&pg=PA68#v=onepage&q&f=false I thought that using the binomial expansion might be more elementary. Both require to know that $\begin{pmatrix}n\k\end{pmatrix}\leq\frac{n^k}{k!}$ and $k!\geq 2^k$ anyway. – Algebraic Pavel Oct 13 '13 at 18:59
  • I'm probably asking too much, but I get all of the inequalities, except this one = $\leq\sum_{k=1}^n\frac{(nu)^k}{k!}\leq nu\sum_{k=0}^n\left(\frac{nu}{2}\right)^k$? How did you come to that? EDIT: Nevermind, I figured it myself :P – Arek Krawczyk Oct 13 '13 at 19:06