# Understanding what equations mean

##### Full Member
For a start, let's agree that all symbols {i, j, n} represent Natural numbers (i.e., {1, 2, 3, …}).

Symbols i and j are variables (i.e., they each take on more than one value) and symbol n is a parameter (i.e., a constant; however, its value may change when starting a new application of the equation).

The constant n limits how large j can be. Symbol j starts at 1 in value and increments up, to take on the value n as its final value. For example, if n is set at 10, then j takes on the values {1,2,3,4,5,6,7,8,9,10} -- one at a time.

That is what the line 1 ≤ j ≤ n tells us.

r() and v() are called "Function Notation". Are you familiar with function notation?
thank you very much!
yes I'm familiar with, but what'd like to say why I need to take all possibilities of j?! it didn't clear that I need to take all possibilities .. it just said that j can move from 1 to n .. so what I would like to say is the logic confirms to take all possibilities of j? or actually logic confirms whenever there're possibilities of something, then I must take them all to solve the problem properly ..

#### Otis

##### Senior Member
We don't have a definition for any function r(x). We don't have a definition for function v(j), either.

Also, for the purposes of discussion, let's limit i and say: 1 ≤ i ≤ n

The equation assigns a value to r(i), for each i input. The value for each r(i) is the smallest value found in a set of related numbers. Each set is defined by:

{ 1 + r(i - v(j)) }, for j from 1 through n.

So we see that each value r(i) is defined in terms of another function r output: r(i-v(j)). That is, the values of r(i-v(j)) must be listed first (for each j), before r(i) can be determined.

Each value in the related list is increased by 1, and the smallest number appearing in the resulting set is assigned to r(i).

Let r(x) = 2x

In other words, function r outputs its input doubled.

Let v(j) = 1 - j/10

In other words, function v outputs the difference between 1 and one-tenth its input.

Let n = 3

In other words, variables i and j will each take on values {1,2,3}. Variable j will loop through its values three times (once for each i), while variable i loops through its values only once.

Code:
[FONT=courier new]
i   j   v(j)   i - v(j)   r(i - v(j))   {1 + r(i-v(j))}   r(i) = MIN{}
----------------------------------------------------------------------
1   1   0.9      0.1          0.2            {1.2,        r(1) = 1.2
1   2   0.8      0.2          0.4             1.4,
1   3   0.7      0.3          0.6             1.6}

2   1   0.9      1.1          2.2            {3.2,        r(2) = 3.2
2   2   0.8      1.2          2.4             3.4,
2   3   0.7      1.3          2.6             3.6}

3   1   0.9      2.1          4.2            {5.2,        r(3) = 5.2
3   2   0.8      2.2          4.4             5.4,
3   3   0.7      2.3          4.6             5.6}[/FONT]
Again, that example is contrived; it's meant only to show how the pieces of the equation go together. :cool:

Last edited:

#### Ryan\$

##### Full Member
We don't have a definition for any function r(x). We don't have a definition for function v(j), either.

Also, for the purposes of discussion, let's limit i and say: 1 ≤ i ≤ n

The equation assigns a value to r(i), for each i input. The value for r(i) is the smallest value in a set of numbers. This set is defined by:

{ 1 + r(i - v(j)) }, for each j from 1 through n.

So we see that each value r(i) is defined in terms of r(i-v(j)). That is, the values of r(i-v(j)) must be listed first (for each j), before r(i) can be determined.

Each value in that list is increased by 1, and the smallest number appearing in the resulting set is assigned to r(i).

Let r(x) = 2x

In other words, function r outputs its input doubled.

Let v(j) = 1 - j/10

In other words, function v outputs the difference between 1 and one-tenth its input.

Let n = 3

In other words, variables i and j will each take on values {1,2,3}. Variable j will loop through its values three times (once for each i); variable i will loop through its values only once.

Code:
[FONT=courier new]
i   j   v(j)   i - v(j)   r(i - v(j))   {1 + r(i-v(j))}   r(i) = MIN{}
----------------------------------------------------------------------
1   1   0.9      0.1          0.2            {1.2         r(1) = 1.2
1   2   0.8      0.2          0.4             1.4
1   3   0.7      0.3          0.6             1.6}

2   1   0.9      1.1          2.2            {3.2         r(2) = 3.2
2   2   0.8      1.2          2.4             3.4
2   3   0.7      1.3          2.6             3.6}

3   1   0.9      2.1          4.2            {5.2         r(3) = 5.2
3   2   0.8      2.2          4.4             5.4
3   3   0.7      2.3          4.6             5.6}[/FONT]
Again, that example is contrived; it's meant only to show how the pieces of the equation go together. :cool:

thanks very much!!