# Four Fours problem: Any number with One Four

Contributed by Jim Millar

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

# Numbers larger than four

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

# Numbers less than 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))))))))))))))))))))))))))))))))

# Any number using one zero

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

# Other trigonometric functions

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.