Quoting Henry Baker <hbaker1@pipeline.com>:
Is there a simple/elegant/one-line proof that the coefficients of (1+x)^n approximate a Gaussian curve? I know that it is the convolution of n boxes, and that a large convolution of non-impulses approaches a Gaussian, but is there a quick/obvious proof of this that also gives you the appropriate mean & std deviation parameters?
Why talk about convolutions? - The expansion of (1+x)^n has the same coefficients as (a+B)^n and you get them by figuring out how many ways you can get so many a's and the rest b's, having applied the distributive law to the product. But that is the number of (average plus epsilon)'s relative to the number of (average minus epsilons)'s you get while deriving the Gaussian distribution. QED! - hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos