What is the value of sqrt [3 - sqrt (3 + sqrt (3 - sqrt (3 + ....)))] ?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The value of sqrt [3 - sqrt (3 + sqrt (3 - sqrt (3 + ....)))] has to be determined.

Let A = sqrt [3 - sqrt (3 + sqrt (3 - sqrt (3 + ....)))]

As the terms are nested infinitely with the sign alternating between positive and negative, we can write the given expression as in terms of A as:

A = sqrt [3 - sqrt (3 + A)]

=> A^2 = 3 - sqrt(3+A)

=> sqrt(3 + A) = 3 - A^2

=> 3 + A = 9 + A^4 - 6A^2

=> A^4 - 6A^2 - A + 6 = 0

The roots of A^4 - 6A^2 - A + 6 = 0 are 1, -2, (1 - sqrt 13)/2 and (1 + sqrt 13)/2

Of these roots A cannot be -2 or (1- sqrt 13)/2 as the square root of a negative number is complex and a complex number cannot be equal to a real number.

A = (1+sqrt 13)/2  = 2.3027 approximately, but A is the square root of a value less than 3, therefore it cannot be greater than 2.

This leaves the root 1 which is the required value.

The value of sqrt [3 - sqrt (3 + sqrt (3 - sqrt (3 + ....)))] = 1

nathanshields's profile pic

nathanshields | High School Teacher | (Level 1) Associate Educator

Posted on

Let the value of your quantity be called x.

Since it is infinitely long, we can replace the inner part (the "...") with x as well, so we get something like this:

x = sqrt(3 - sqrt(3 + x))

Solve this equation by squaring both sides, subtracting 3, multiplying by -1, squaring again, subtracting 3 again...

x^4 - 6x^2 - x + 6 = 0

The function has four real roots (see link), but two are negative (we know this can't be the case, since the square root function has nonnegative range), and one is larger than 3 (we know this can't be the case, since the square root function has nonnegative domain).  We are left with one root:

x=1

Sources:

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