Four Fours problem: Any number with One Four

Contributed by Jim Millar
Copyright 2006 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.