UPDATE: Some more recent results can be found here.

### Equation 1

This equation can be "derived" by assuming that the expression on the right converges, setting it equal to its limit L, replacing the 2nd radical and all its contents with L, and solving for L. Once this is done, all that remains is to prove that the expression on the right does indeed converge.

General Equation
```         -----------------------------------------------------------------------
|                      -------------------------------------------------
|                     |                      ---------------------------
|        n      n-1   |        n      n-1   |        n      n-1    -----
x =    | (1-q) x  + q x      | (1-q) x  + q x      | (1-q) x  + q x      | ...
|                     |                     |                     |
n |                   n |                   n |                   n |
\/                    \/                    \/                    \/
```

Conditions Resulting Special Case
n = 2
a = (1-q) x²
b = q x
```                       -----------------------------
________      |        ---------------------
b + \/b^2 + 4a      |       |        -------------
---------------- =   |       |       |        -----
2             |a + b  | a + b | a + b | ...
|       |       |       |
\/      \/      \/      \/
```
q = 1
```        -----------------------------------
|          -------------------------
|         |          ---------------
|   n-1   |   n-1   |   n-1    -----
x =    |  x      |  x      |  x      | ...
|         |         |         |
n |       n |       n |       n |
\/        \/        \/        \/
```
q = 1
n = 2
```           ---------------------------------
|       --------------------------
|      |       -------------------
|      |      |       ------------
|      |      |      |       -----
x =   |  x   |  x   |  x   |  x   | ...
|      |      |      |      |
|      |      |      |      |
\/     \/     \/     \/     \/
```

### Equation 2

This equation can be derived from equation 1 by taking each term multiplying a radical and pushing it through the radical, continuing from left to right for all the radicals. It is also possible to derive equation 1 from equation 2 by performing the inverse operation: pulling out appropriate terms from each radical, from right to left.

Another way to derive equation 2 involves taking the expression on the left and expanding it into the expression on the right, one radical at a time.

Notice how some of the special cases are more elegant and inspiring ("beautiful") than the general equation.

General Equation
```                 --------------------------------------------------------------------------
|                        --------------------------------------------------
|                       |                        --------------------------
k            |   k+1                 |   k+2                 |   k+3
n -1          |  n   -n               |  n   -n               |  n   -n
------   j     | --------        j+1   | --------        j+2   | --------        j+3
n-1    n      |   n-1          n      |   n-1          n      |   n-1          n
q       x   =   |q        (1-q) x    +  |q        (1-q) x    +  |q        (1-q) x    + ...
|                       |                       |
n |                     n |                     n |
\/                      \/                      \/
```

Conditions Resulting Special Case
q = 1/2
```                -------------------------------------------------------
|                  -------------------------------------
|                 |                  -------------------
k         |     k+1         |     k+2         |     k+3
1-n          |  1-n            |  1-n            |  1-n
-----   j     | -----     j+1   | -----     j+2   | -----     j+3
n-1   n      |  n-1     n      |  n-1     n      |  n-1     n
2      x   =   |2        x    +  |2        x    +  |2        x    + ...
|                 |                 |
n |               n |               n |
\/                \/                \/
```
q = 1/2
n = 2
```               ----------------------------------------------------
|                 -----------------------------------
|                |                 ------------------
k   j     |    k+1   j+1   |    k+2   j+2   |    k+3   j+3
1-2   2      | 1-2     2      | 1-2     2      | 1-2     2
2     x   =   |2       x    +  |2       x    +  |2       x    + ...
|                |                |
|                |                |
\/               \/               \/
```
q = 1/2
n = 2
x = 1
```           ---------------------------------------------------------
|           ----------------------------------------------
|          |           -----------------------------------
|          |          |           ------------------------
|          |          |          |           -------------
k     |     k+1  |     k+2  |     k+3  |     k+4  |     k+5
1-2      |  1-2     |  1-2     |  1-2     |  1-2     |  1-2
2     =   | 2      + | 2      + | 2      + | 2      + | 2      + ...
|          |          |          |          |
|          |          |          |          |
\/         \/         \/         \/         \/
```
q = 1/2
n = 2
x = 1
k = -1
```          -----------------------------------------------------
|          -------------------------------------------
|         |          ---------------------------------
|         |         |          -----------------------
|         |         |         |          -------------
_      |   2     |   2     |   2     |   2     |   2
\/2 =    | ----- + | ----- + | ----- + | ----- + | ----- + ...
|    0    |    1    |    2    |    3    |    4
|   2     |   2     |   2     |   2     |   2
|  2      |  2      |  2      |  2      |  2
\/        \/        \/        \/        \/
```
n = 2
k = j = 1
a = (1-q)/q²
b = (q x)²
```                      --------------------------------
|         -----------------------
|        |         --------------
b         ____      |        |        |         -----
--- (1 + \/1+4a) =   |   2    |   4    |   8    |
2                   | ab  +  | ab  +  | ab  +  | ...
|        |        |        |
\/       \/       \/       \/
```

### Equation 3

This equation appears rather weird at first, but can be readily understood as the limit of equation 2 when x = 1 and k goes to negative infinity. So equation 3 is really a limit of a limit, which explains the presence of two ellipses.

General Equation
```                ------------------------------------------------------------------
|          --------------------------------------------------------
|         |                  --------------------------------------
|         |                 |                  --------------------
|         |   -1            |   0             |   1
1       |         |  n  -n          |  n - n          |  n - n
-----     |         | -------         | -------         | -------
n-1      |         |   n-1           |   n-1           |   n-1
(1/q)      =   | ... +   |q       (1-q) +  |q       (1-q) +  |q       (1-q) + ...
|         |                 |                 |
n |       n |               n |               n |
\/        \/                \/                \/
```

Conditions Resulting Special Case
q = 1/2
```            ---------------------------------------------------
|          -----------------------------------------
|         |              ---------------------------
|         |             |             --------------
|         |      -1     |      0     |      1
1       |         |   1-n       |   1-n      |   1-n
-----     |         |  -----      |  -----     |  -----
n-1      |         |   n-1       |   n-1      |   n-1
2      =   | ... +   | 2       +   | 2      +   | 2      + ...
|         |             |            |
n |       n |           n |          n |
\/        \/            \/           \/
```
q = 1/2
n = 2
```        --------------------------------------------------------------
|         -----------------------------------------------------
|        |          -------------------------------------------
|        |         |          ---------------------------------
|        |         |         |          -----------------------
|        |         |         |         |          -------------
|        |   2     |   2     |   2     |   2     |   2
2 =    | ... +  | ----- + | ----- + | ----- + | ----- + | ----- + ...
|        |    -2   |    -1   |    0    |    1    |    2
|        |   2     |   2     |   2     |   2     |   2
|        |  2      |  2      |  2      |  2      |  2
\/       \/        \/        \/        \/        \/
```

### Equation 4

This equation is due to Ramanujan. It can be derived by expanding the expression on the left into the expression on the right, one radical at a time. In doing so, a recurrence relationship will become apparent, allowing expansion to an arbitrary number of radicals. Once this is achieved, all that remains to be proven is that the infinite expansion converges, and converges to the expression on the left.

Parameter Ranges General Equation
x >= 0
n >= 0
a >= 0
```           ---------------------------------------------------------------------------------------
|              -------------------------------------------------------------------------
|             |                      ---------------------------------------------------
|             |                     |                        ---------------------------
|        2    |            2        |             2         |             2          ---
x+n+a =   |ax+(n+a) +x  |a(x+n)+(n+a) +(x+n)  |a(x+2n)+(n+a) +(x+2n)  |a(x+3n)+(n+a) +(x+3n)  |...
|             |                     |                       |                       |
|             |                     |                       |                       |
\/            \/                    \/                      \/                      \/
```

Conditions Resulting Special Case
a = 0
```           ---------------------------------------------------------
|          -----------------------------------------------
|         |              ---------------------------------
|         |             |               ------------------
|  2      |  2          |  2           |  2            ---
x + n =   | n  + x  | n  + (x+n)  | n  + (x+2n)  | n  + (x+3n)  |...
|         |             |              |              |
|         |             |              |              |
\/        \/            \/             \/             \/
```
a = 0
n = 1
```           ----------------------------------------------------
|         -------------------------------------------
|        |             ------------------------------
|        |            |             -----------------
|        |            |            |             ----
x + 1 =   | 1 + x  | 1 + (x+1)  | 1 + (x+2)  | 1 + (x+3)  | ...
|        |            |            |            |
|        |            |            |            |
\/       \/           \/           \/           \/
```
a = 0
n = 1
x = 2
```       ---------------------------------------------
|          -----------------------------------
|         |          -------------------------
|         |         |          ---------------
|         |         |         |          -----
3 =   | 1 + 2   | 1 + 3   | 1 + 4   | 1 + 5   | ...
|         |         |         |         |
|         |         |         |         |
\/        \/        \/        \/        \/
```
a = 0
n = 0
```           ---------------------------------
|       --------------------------
|      |       -------------------
|      |      |       ------------
|      |      |      |       -----
x =   |  x   |  x   |  x   |  x   | ...
|      |      |      |      |
|      |      |      |      |
\/     \/     \/     \/     \/
```

### Equation 5

This section is unfinished. I was hoping I could find a generalization of the below "special case", but I haven't had any luck yet.

Parameter Ranges General Equation
...
```...
```

Conditions Resulting Special Case
...
```   ---------------------------
|        -------------------
|       |        -----------                                     ________
|       |       |        ---             a                 b + \/b^2 + 4a
|a + b  |a + b  |a + b  |...  =  b + ------------------ = ----------------
|       |       |       |                    a                     2
\/      \/      \/      \/              b + ------------
a
b + ------
...

Setting a = b = 1, we get the golden ratio.
```
...
```...
```

Contributors: Michael McGuffin and Brian Wong
Related Sites
Nested Radicals (Eric's Treasure Trove of Math)
Continued Square Roots
Herschfeld's Convergence Theorem
Simplifying Finitely Nested Radicals (Susan Landau)
Continued Fractions