# Large Operator for Composite Functions

1. Oct 13, 2011

### bjshnog

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:

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

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.

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

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).

$$\Omega_{l=1}^{n} \left [ f_l(x) \right ]_x \left ( x \right ) = f_n(f_{n-1}(\cdots f_2(f_1(x))\cdots))$$
Or...
$$\Omega_{l=1}^{1} \left [ x+2 \right ]_x \left ( 1 \right ) = 3$$

Last edited: Oct 13, 2011
2. Oct 13, 2011

### CompuChip

I looks like your expression is equal to
$$\prod_{i = 1}^5 \left( \sum_{2^{a_i}}^{2^{4a_i}} k_i \right) f(x)$$
where $a_i$ is some expression that I don't really feel like thinking about right now.

3. Oct 13, 2011

### disregardthat

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
4. Oct 14, 2011

### bjshnog

$$\underset{n}{\underbrace {\int \cdots \int }}\left ( \sum_{i=1}^{\infty}\left ( a_ix^{b_i} \right ) \right ) dx^n = \sum_{i=1}^{\infty}\left ( \frac{a_ix^{b_i-n}}{\prod_{j=b_i}^{b_i+n-1}\left ( j \right )} \right ) + \sum_{i=1}^{n}\left ( \frac{c_ix^{n-i}}{(n-i)!} \right )$$

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

$$\Omega_{l=1}^{n}\left [ \int z\, dx \right ]_z \left ( \sum_{i=1}^{\infty}\left ( a_ix^{b_i} \right ) \right ) = \sum_{i=1}^{\infty}\left ( \frac{a_ix^{b_i-n}}{\prod_{j=b_i}^{b_i+n-1}\left ( j \right )} \right ) + \sum_{i=1}^{n}\left ( \frac{c_ix^{n-i}}{(n-i)!} \right )$$

Last edited: Oct 14, 2011
5. Oct 20, 2011

### bjshnog

$$\underset{n}{\underbrace {\int \cdots \int }}\left ( \sum_{i=1}^{\infty}\left ( a_ix^{b_i} \right ) \right ) dx^n = \sum_{i=1}^{\infty}\left ( \frac{a_ix^{b_i+n}}{\prod_{j=b_i}^{b_i+n-1}\left ( j \right )} \right ) + \sum_{i=1}^{n}\left ( \frac{c_ix^{n-i}}{(n-i)!} \right )$$
$$\Omega_{l=1}^{n}\left [ \int z\, dx \right ]_z \left ( \sum_{i=1}^{\infty}\left ( a_ix^{b_i} \right ) \right ) = \sum_{i=1}^{\infty}\left ( \frac{a_ix^{b_i+n}}{\prod_{j=b_i}^{b_i+n-1}\left ( j \right )} \right ) + \sum_{i=1}^{n}\left ( \frac{c_ix^{n-i}}{(n-i)!} \right )$$