1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Large Operator for Composite Functions

  1. Oct 13, 2011 #1
    Sometimes I like to find patterns in certain functions, for example, repeated Sigma (Summation) notation.
    But what if I wanted to do an arbitrary number of nested summations? Or something similar with other functions? Is there a compressed way of writing this? For example:

    [tex] \sum_{k_5=1}^{1} \left (k_5 \sum_{k_4=2}^{4} \left (k_4 \sum_{k_3=4}^{16} \left (k_3 \sum_{k_2=8}^{64} \left (k_2 \sum_{k_1=16}^{256}\left ( k_1f(x) \right ) \right ) \right ) \right ) \right ) [/tex]

    I have come up with a large operator for this, very similar to the Sigma notation, but using a pair of extra square brackets and Omega for the operator.

    [tex] \Omega_{l=1}^{5}\left [ \sum_{k_l=2^{5-l}}^{4^{5-l}} (k_lx) \right ]_x\left ( f(x) \right ) [/tex]

    It looks similar to this, but the Omega would be larger and the subscript/superscript parts would be on the bottom/top of the Omega respectively.

    This notation means:
    • Whatever is in the round brackets is the base.
    • The subscript pronumeral next to the square brackets denotes what is being replaced inside the square brackets by the base.
    • The function inside the square brackets is executed like any other large operator, then the value of l rises from 1 to 2 and the x in the square brackets is then replaced by what you have now.
    • This is then repeated until you have executed the 5th function.
    This could be used with any function, even a single Sigma or Pi large operation or just 1+2!
    Keep in mind that this is a sample of what it could do, even though there may be a much simpler way of writing it, but there are heaps of other functions that this may be useful for (compression-wise, at least, or for finding patterns).

    [tex] \Omega_{l=1}^{n} \left [ f_l(x) \right ]_x \left ( x \right ) = f_n(f_{n-1}(\cdots f_2(f_1(x))\cdots))[/tex]
    Or...
    [tex] \Omega_{l=1}^{1} \left [ x+2 \right ]_x \left ( 1 \right ) = 3[/tex]
     
    Last edited: Oct 13, 2011
  2. jcsd
  3. Oct 13, 2011 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    I looks like your expression is equal to
    [tex] \prod_{i = 1}^5 \left( \sum_{2^{a_i}}^{2^{4a_i}} k_i \right) f(x)[/tex]
    where [itex]a_i[/itex] is some expression that I don't really feel like thinking about right now.
     
  4. Oct 13, 2011 #3

    disregardthat

    User Avatar
    Science Advisor

    Isn't that just (256-16)*(64-8)*(16-4)*(4-2)? ( oh, and *f(x) for some reason)

    EDIT: I mean (1+256-16)*(1+64-8)*(1+16-4)*(1+4-2).
     
    Last edited: Oct 13, 2011
  5. Oct 14, 2011 #4
    [tex]\underset{n}{\underbrace {\int \cdots \int }}\left ( \sum_{i=1}^{\infty}\left ( a_ix^{b_i} \right ) \right ) dx^n

    = \sum_{i=1}^{\infty}\left ( \frac{a_ix^{b_i-n}}{\prod_{j=b_i}^{b_i+n-1}\left ( j \right )} \right )

    + \sum_{i=1}^{n}\left ( \frac{c_ix^{n-i}}{(n-i)!} \right )[/tex]


    This is somewhat abuse of notation, but my Omega operator makes it "feel" correct. This is what it would look like:

    [tex]\Omega_{l=1}^{n}\left [ \int z\, dx \right ]_z \left ( \sum_{i=1}^{\infty}\left ( a_ix^{b_i} \right ) \right )

    = \sum_{i=1}^{\infty}\left ( \frac{a_ix^{b_i-n}}{\prod_{j=b_i}^{b_i+n-1}\left ( j \right )} \right )

    + \sum_{i=1}^{n}\left ( \frac{c_ix^{n-i}}{(n-i)!} \right )[/tex]
     
    Last edited: Oct 14, 2011
  6. Oct 20, 2011 #5
    This is incorrect; made correction:

    [tex]\underset{n}{\underbrace {\int \cdots \int }}\left ( \sum_{i=1}^{\infty}\left ( a_ix^{b_i} \right ) \right ) dx^n

    = \sum_{i=1}^{\infty}\left ( \frac{a_ix^{b_i+n}}{\prod_{j=b_i}^{b_i+n-1}\left ( j \right )} \right )

    + \sum_{i=1}^{n}\left ( \frac{c_ix^{n-i}}{(n-i)!} \right )[/tex]


    [tex]\Omega_{l=1}^{n}\left [ \int z\, dx \right ]_z \left ( \sum_{i=1}^{\infty}\left ( a_ix^{b_i} \right ) \right )

    = \sum_{i=1}^{\infty}\left ( \frac{a_ix^{b_i+n}}{\prod_{j=b_i}^{b_i+n-1}\left ( j \right )} \right )

    + \sum_{i=1}^{n}\left ( \frac{c_ix^{n-i}}{(n-i)!} \right )[/tex]
     
    Last edited: Oct 20, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Large Operator for Composite Functions
  1. Composite functions (Replies: 3)

  2. Composite function (Replies: 18)

Loading...