# Solution to puzzle 88: Nested radicals

Solve the equation = x.

(All square roots are to be taken as positive.)

Consider f(x) = .

Then f(f(x)) = = x.

A solution to f(x) = x, if it exists, will also be a solution to f(f(x)) = x.

## Solving f(x) = x

Consider, then, f(x) = = x.

Let y = . Then y^{2} = 4 − x.

We also have x = , from which x^{2} = 4 + y.

Subtracting, we have x^{2} − y^{2} = x + y.

Hence (x + y)(x − y − 1) = 0.

Since x 0 and y 0, x + y = 0 x = 0, which does not satisfy f(x) = x.

Therefore we take x − y − 1 = 0, or y = x − 1.

Substituting into x^{2} = 4 + y, we obtain x^{2} = x + 3, or x^{2} − x − 3 = 0.

Rejecting the negative root, we have x =

## Proving uniqueness

We must now show that this is the *only* solution to f(f(x)) = x. This is necessary as f(f(x)) = x does not necessarily imply f(x) = x. (Consider f(x) = 4 − x.) That is, there may be other solutions for which f(x) x.

For any solution, since each square root is positive, we must have 0 x 4.

Considering each nested radical in turn, from the innermost outwards, we see also that f(f(x)) is strictly increasing over this range.

We have also: f(f(0)) = 2.29 and f(f(4)) = 2.33, correct to two decimal places.

We conclude that the graph of y = f(f(x)), for 0 x 4, is almost a straight line, with average slope approximately 0.01.

Since f(f(x)) is a continuous function, it follows that y = f(f(x)) intersects the line y = x, for 0 x 4.

Furthermore, y = f(f(x) will intersect y = x precisely once, *provided that* no section of y = f(f(x)) has a slope greater than 1.

This is clearly the case, and may be verified, if necessary, by differentiating f(f(x)) with respect to x.

Therefore the only solution to = x, is x =

## Remarks

If f is a *strictly increasing* function over a given range, then we can show that f(f(x)) = x f(x) = x.

Suppose x_{0} is a solution to f(f(x)) = x.

If f(x_{0}) < x_{0}, then x_{0} = f(f(x_{0})) < f(x_{0}) < x_{0}, which is a contradiction.

Similarly, if f(x_{0}) > x_{0}, then x_{0} = f(f(x_{0})) > f(x_{0}) > x_{0}, which again is a contradiction.

We conclude that f(x_{0}) = x_{0}.

Alas, we cannot use this approach above, as f(x) = is a decreasing function.

## Further reading

- Nested Radical

Source: Original

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