Find the value of x: \(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\)

MathCuber Jan 4, 2019

#2**+2 **

This computer summation for about 25 terms shows that x converges to sqrt(2):

c=1E100; listforeach(b,reverse(1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), c=b + 1/c)

**x =1.4142135623730950488016887242097**

**You could also solve it algebraically in this way:**

**Solve for x: 1/(1 + x) + 1 = x**

**Write the left hand side as a single fraction. Bring 1/(x + 1) + 1 together using the common denominator x + 1: (x + 2)/(x + 1) = x**

**Multiply both sides by a polynomial to clear fractions. Multiply both sides by x + 1: x + 2 = x (x + 1)**

**Write the quadratic polynomial on the right hand side in standard form. Expand out terms of the right hand side: x + 2 = x^2 + x**

**Move everything to the left hand side. Subtract x^2 + x from both sides: 2 - x^2 = 0**

**Isolate terms with x to the left hand side. Subtract 2 from both sides: -x^2 = -2**

**Multiply both sides by a constant to simplify the equation. Multiply both sides by -1: x^2 = 2**

**Eliminate the exponent on the left hand side. Take the square root of both sides: x = sqrt(2) **

Guest Jan 4, 2019

edited by
Guest
Jan 4, 2019

edited by Guest Jan 4, 2019

edited by Guest Jan 4, 2019

#3**+2 **

Thanks for that great answer guest,

I will just try to show a little clearer what I think you have done.

First add 1 to both sides.

\(x+1 = 2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\)

now

\(x+1 = 2 + \cfrac{1}{x+1}\\ x-1 = \cfrac{1}{x+1}\\ x^2-1 = 1\\ x^2=2\\ x=\pm\sqrt2 \\ \text{But everything is positive on the RHS so } x>-1\\ x=\sqrt2\)

.Melody Jan 4, 2019