手册

数学

Discovery

Andrius Kulikauskas

  • ms@ms.lt
  • +370 607 27 665
  • My work is in the Public Domain for all to share freely.

Lietuvių kalba

Introduction E9F5FC

Understandable FFFFFF

Questions FFFFC0

Notes EEEEEE

Software

Math exercises

Combinatorial interpretation of {$e^{A} \cdot e^B = e^{A+B}$}

Given Taylor expansions:

{$$e^A = 1 + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots + \frac{A^i}{i!} + \cdots $$}

{$$e^B = 1 + B + \frac{B^2}{2!} + \frac{B^3}{3!} + \cdots + \frac{B^j}{j!} + \cdots $$}

Multiplying them together we get for the initial terms:

{$$1 + (A + B) + (\frac{A^2}{2!} + AB + \frac{B^2}{2!}) + \cdots = $$}

{$$1 + (A + B) + \frac{1}{2!}(A^2 + 2AB + B^2) + \cdots $$}

and, in general, for the {$n$}th term we have:

{$$ \sum_{i=1}^{n}\frac{A^i}{i!}\frac{B^{n-i}}{(n-i)!} = \frac{1}{n!}\sum_{i=1}^{n}\binom{n}{i}A^iB^{n-i} $$}

and that is the {$n$}th term of {$e^{A+B}$}.

Exercise1


Naujausi pakeitimai


Puslapis paskutinį kartą pakeistas 2018 lapkričio 18 d., 17:08
Tweet