This page shows that it is possible to create any number with one four using trigonometric functions. In short:
n = sec(atan(...(sec(atan(4))...)) (n2-42 times), for n>4
n = tan(asec(...(tan(asec(4))...)) (42-n2 times), for n=0, 1, 2, 3
n = 4, for n = 4
Using the trigonometric relation for the tangent and secant functions
tan2(x)+1 = sec2(x)
you can then derive that
sec(atan(x)) = sqrt(x2+1)
(This does not depend on whether you are using degrees or radians!) Therefore
sec(atan(sec(atan(x)))) = sqrt((sqrt(x2+1))2+1) = sqrt(x2+2)
sec(atan(...(sec(atan(x))...)) (k times) = sqrt(x2+k)
Thus repeated application of sec(atan()) will create the next largest square root in a sequence. For example, the table shows a formula that evaluates to 5 using just one 4 and the secant and arctangent functions.
|
x |
sec(atan(x)) |
|
|
4 |
4.123105626 |
sec(atan(4)) |
|
4.123105626 |
4.242640687 |
sec(atan(sec(atan(4)))) |
|
4.242640687 |
4.358898944 |
sec(atan(sec(atan(sec(atan(4)))))) |
|
4.358898944 |
4.472135955 |
sec(atan(sec(atan(sec(atan(sec(atan(4)))))))) |
|
4.472135955 |
4.582575695 |
sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(4)))))))))) |
|
4.582575695 |
4.69041576 |
sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(4)))))))))))) |
|
4.69041576 |
4.795831523 |
sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(4)))))))))))))) |
|
4.795831523 |
4.898979486 |
sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(4)))))))))))))))) |
|
4.898979486 |
5 |
sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(4)))))))))))))))))) |
The last line shows the expression for 5 that uses just one four.
For any integer n greater than 4, there is a k that satisfies the equation above (specifically k = n2-42). This means that the number of times you need to apply sec(atan()) is n2-42 for any n greater than 4. (It takes a lot of operations, but it is possible!)
n = sec(atan(...(sec(atan(4))...)) (n2-42 times), for n>4
To create an expression for the numbers 3, 2, 1, and 0 using one four, the same expression
tan2(x) = sec2(x)-1
yields the formula
tan(asec(x)) = sqrt(x2-1)
So repeated application of tan(asec()) gives the next lowest square root in a sequence.
n = tan(asec(...(tan(asec(4))...)) (42-n2 times), for n=0, 1, 2, 3
The following table shows the expression for 3, 2, 1, and 0.
|
x |
tan(asec(x)) |
|
|
4 |
3.872983346 |
tan(asec(4)) |
|
3.872983346 |
3.741657387 |
tan(asec(tan(asec(4)))) |
|
3.741657387 |
3.605551275 |
tan(asec(tan(asec(tan(asec(4)))))) |
|
3.605551275 |
3.464101615 |
tan(asec(tan(asec(tan(asec(tan(asec(4)))))))) |
|
3.464101615 |
3.31662479 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))) |
|
3.31662479 |
3.16227766 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))))) |
|
3.16227766 |
3 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))))))) |
|
3 |
2.828427125 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))))))))) |
|
2.828427125 |
2.645751311 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))))))))))) |
|
2.645751311 |
2.449489743 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))))))))))))) |
|
2.449489743 |
2.236067977 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))))))))))))))) |
|
2.236067977 |
2 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))))))))))))))))) |
|
2 |
1.732050808 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))))))))))))))))))) |
|
1.732050808 |
1.414213562 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))))))))))))))))))))) |
|
1.414213562 |
1 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))))))))))))))))))))))) |
|
1 |
0 |
tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4)))))))))))))))))))))))))))))))) |
Some people like to do the problem using combinations of different numbers besides four, for example, four zeros. Using the approach above, you can create any number using one of any digit, for example zero. The table below shows the expressions for 0, 1, 2, and 3 using one zero.
|
x |
sec(atan(x)) |
|
|
0 |
1 |
sec(atan(0)) |
|
1 |
1.414213562 |
sec(atan(sec(atan(0)))) |
|
1.414213562 |
1.732050808 |
sec(atan(sec(atan(sec(atan(0)))))) |
|
1.732050808 |
2 |
sec(atan(sec(atan(sec(atan(sec(atan(0)))))))) |
|
2 |
2.236067977 |
sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(0)))))))))) |
|
2.236067977 |
2.449489743 |
sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(0)))))))))))) |
|
2.449489743 |
2.645751311 |
sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(0)))))))))))))) |
|
2.645751311 |
2.828427125 |
sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(0)))))))))))))))) |
|
2.828427125 |
3 |
sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(0)))))))))))))))))) |
The number of operations sec(atan()) you need to create an expression that evaluates to n is n2.
n = sec(atan(...(sec(atan(0))...)) (n2 times), for n>0
The same approach is possible using the cosecant and cotangent functions.
cot2(x)+1 = csc2(x)
Therefore
csc(acot(x)) = sqrt(x2+1)
cot(acsc(x)) = sqrt(x2-1)
So repeated applications of csc(acot()) give the next largest square root in a sequence, similar to sec(atan()) above.