Let $\xi_1, \xi_2, \cdots, \xi_n$ be indeterminates. Define the following indeterminates: $$s_k := \sum\limits_{i=1}^n\xi_i^k, 1\le k <\infty ,$$ $$\sigma_k := \sum\limits_{1\le i_1 How to show $$ \prod\limits_{i=1}^n(1-\xi_it)=1-\sigma_1t+\sigma_2t^2-\cdots+(-1)^n\sigma_nt^n=\exp\left(-\sum\limits_{j=1}^\infty s_j\frac{t^j}{j}\right)?$$ Thanks.
A Newton-like identity
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$\begingroup$
calculus
combinatorics
sequences-and-series
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0The left side is simple foiling... The right side... hmm what is $s_j$? – 2011-11-06
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0@user9176: It is defined in line two of the post. – 2011-11-06
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3Ups, I probably need new glasses. For the right side, plug in $s_j$ and use the Taylor series of $\ln$ ;) – 2011-11-06
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1@user9176: why not post an answer? :-) – 2011-11-06
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5@user9176: I think my English is quite okay for a non-native speaker but until a few days ago I wouldn't have had the slightest clue what "the left side is simple *preventing something undesirable to happen*" could possibly mean. Anyone outside the US/Schaum-system would be similarly stumped. – 2011-11-06
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1You'll want to see page 427 of [Charalambides's book](http://books.google.com/books?id=PDMGA-v5G54C&pg=PA427). – 2011-11-06
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0@t.b. `Foiling' is an affectionate term for distributing. The name comes from expansion of two binomials: in the case of $(a+b)(c+d)$, the expanded product is the sum $ac+ad+bc+bd$, i.e. First, Inner, Outer, Last. – 2011-11-06
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0@AWalker: Thanks! As I said, the first time I saw that was [a few days ago](http://math.stackexchange.com/questions/77606/) and [I found out a few minutes later](http://chat.stackexchange.com/transcript/36?m=2309330#2309330) :) – 2011-11-06
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3@t.b.: If it's any consolation, I find the acronyms most of my students use that are related to math just as confusing (I'm also not a native speaker either). And I positively dislike "FOIL" and its many derivations. As far as I can tell, it actually makes students *unable* to expand products unless they happen to be binomial times binomial... – 2011-11-06
1 Answers
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The power series of the logarithm gives
$$\log (1-\xi t) = -\sum_{k=1}^{\infty} \xi^k \frac{t^k}{k}.$$
Summing this identity for the different values of $\xi$ and then taking the exponential gives the desired identity.