ProofOfAbelsLemmaByInduction 2003-05-14 17:30:59 2004-04-21 11:21:55 Proof Abel's lemma 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \PMlinkescapeword{limit} \PMlinkescapeword{normal} \emph{Proof.} The proof is by induction. However, let us first recall that sum on the right side is a piece-wise defined function of the upper limit $N-1$. In other words, if the upper limit is below the lower limit $0$, the sum is identically set to zero. Otherwise, it is an ordinary sum. We therefore need to manually check the first two cases. For the trivial case $N=0$, both sides equal to $a_0 b_0$. Also, for $N=1$ (when the sum is a normal sum), it is easy to verify that both sides simplify to $a_0 b_0 + a_1 b_1$. Then, for the induction step, suppose that the claim holds for some $N\ge 1$. For $N+1$, we then have \begin{eqnarray*} \sum_{i=0}^{N+1} a_i b_i &=& \sum_{i=0}^{N} a_i b_i + a_{N+1} b_{N+1} \\ &=& \sum_{i=0}^{N-1}A_i(b_i-b_{i+1})+A_N b_N + a_{N+1} b_{N+1} \\ &=& \sum_{i=0}^{N}A_i(b_i-b_{i+1})-A_{N}(b_{N}-b_{N+1})+ A_N b_N + a_{N+1} b_{N+1}. \end{eqnarray*} Since $-A_{N}(b_{N}-b_{N+1})+ A_N b_N + a_{N+1} b_{N+1} = A_{N+1} b_{N+1}$, the claim follows. $\Box$.