Loading [MathJax]/jax/output/SVG/config.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

Search

  • Filter Results
  • Location
  • Classification
    • Article type
    • Stage
    • Author
    • Cover Page
    • License
    • Show Page TOC
    • Transcluded
    • PrintOptions
    • OER program or Publisher
    • Autonumber Section Headings
    • License Version
    • Print CSS
    • Screen CSS
    • Number of Print Columns
  • Include attachments
Searching in
About 2 results
  • https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Combinatorics_(Morris)/02%3A_Enumeration/09%3A_Some_Important_Recursively-Defined_Sequences/9.03%3A_Bell_Numbers_and_Exponential_Generating_Functions
    Sometimes a recurrence relation involves factorials, or binomial coefficients. When this happens, it becomes difficult if not impossible to use ordinary generating functions to find an explicit formul...Sometimes a recurrence relation involves factorials, or binomial coefficients. When this happens, it becomes difficult if not impossible to use ordinary generating functions to find an explicit formula for the nth term of the sequence. In some cases, a different kind of generating function, the exponential generating function, may succeed where an ordinary generating function fails.
  • https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Combinatorics_and_Graph_Theory_(Guichard)/03%3A_Generating_Functions/3.03%3A_Exponential_Generating_Functions
    Thus \[ e^x + e^{-x} = \sum_{i=0}^\infty {2x^{2i}\over (2i)!}, \nonumber\] so that \[ \sum_{i=0}^\infty {x^{2i}\over (2i)!} = {e^x+e^{-x}\over 2}. \nonumber\] A similar manipulation shows that \[ \sum...Thus \[ e^x + e^{-x} = \sum_{i=0}^\infty {2x^{2i}\over (2i)!}, \nonumber\] so that \[ \sum_{i=0}^\infty {x^{2i}\over (2i)!} = {e^x+e^{-x}\over 2}. \nonumber\] A similar manipulation shows that \[ \sum_{i=0}^\infty {x^{2i+1}\over (2i+1)!} = {e^x-e^{-x}\over 2}. \nonumber\] Thus, the generating function we seek is \[ {e^x-e^{-x}\over 2}{e^x+e^{-x}\over 2} e^x= {1\over 4}(e^x-e^{-x})(e^x+e^{-x})e^x = {1\over 4}(e^{3x}-e^{-x}). \nonumber\] Notice the similarity to Example 3.2.4.

Support Center

How can we help?