lfdahl said:Prove the following identity ($n = 1,2,3,...$):
\[\sqrt{n - \sqrt{n+\sqrt{n-\sqrt{n +...}}}} = \sqrt{(n-1)-\sqrt{(n-1)-\sqrt{(n-1)-...}}}\]
Prove It said:$\displaystyle \begin{align*} x &= \sqrt{n - \sqrt{n + \sqrt{n - \sqrt{n + \sqrt{ \dots } }}}} \\ x^2 &= n - \sqrt{ n + \sqrt{ n - \sqrt{ n + \sqrt{\dots }}}} \\ x^2 - n &= - \sqrt{ n + \sqrt{ n - \sqrt{ n + \sqrt{ \dots } } }} \\ \left( x^2 - n \right) ^2 &= n + \sqrt{ n - \sqrt{ n + \sqrt{ n - \sqrt{ n + \sqrt{ \dots } }}}} \\ \left( x^2 - n \right) ^2 - n &= \sqrt{ n -
\sqrt{n + \sqrt{ n - \sqrt{ n + \sqrt{ \dots }}}}} \\ \left( x^2 - n \right) ^2 - n &= x \end{align*}$
Next:
$\displaystyle \begin{align*} y &= \sqrt{ n - 1 - \sqrt{ n - 1 - \sqrt{ n - 1 - \sqrt{ n - 1 - \sqrt{ \dots }}}}} \\ y^2 &= n - 1 - \sqrt{ n - 1 - \sqrt{ n - 1 - \sqrt{ n - 1 - \sqrt{ \dots }}}} \\ y^2 &= n - 1 - y \\ y^2 + y &= n - 1 \\ y^2 + y + \left( \frac{1}{2} \right) ^2 &= n - 1 + \left( \frac{1}{2} \right) ^2 \\ \left( y + \frac{1}{2} \right) ^2 &= n - \frac{3}{4} \\ y + \frac{1}{2} &= \frac{\pm \sqrt{ 4\,n - 3 }}{2} \\ y &= \frac{-1 \pm \sqrt{ 4\,n - 3 }}{2} \end{align*}$
As $\displaystyle \begin{align*} y > 0 \end{align*}$ that means $\displaystyle \begin{align*} y = \frac{-1 + \sqrt{ 4\,n -3}}{2} \end{align*}$, now we need to check if $\displaystyle \begin{align*} x = y \end{align*}$...
$\displaystyle \begin{align*} \left( y^2 - n \right) ^2 - n &= \left[ \left( \frac{-1 + \sqrt{4\,n - 3}}{2} \right) ^2 - n \right] ^2 - n \\ &= \left( \frac{1 - 2\,\sqrt{4\,n - 3} + 4\,n - 3}{4} - n \right) ^2 - n \\ &= \left( \frac{4\, n - 2 - 2\,\sqrt{ 4\,n - 3 }}{4} - n \right) ^2 - n \\ &= \left( \frac{ 2\,n - 1 - \sqrt{ 4\,n - 3}}{2} - n \right) ^2 - n \\ &= \left( \frac{-1 - \sqrt{4\,n - 3}}{2} \right) ^2 - n \\ &= \frac{1 + 2\,\sqrt{4\,n - 3} + 4\,n - 3}{4} - n \\ &= \frac{4\, n- 2 + 2\,\sqrt{4\,n - 3} }{4} - n \\ &= \frac{2\,n - 1 + \sqrt{4\,n - 3 }}{2} - n \\ &= \frac{-1 + \sqrt{4\,n - 3}}{2} \\ &= y \end{align*}$
which satisfies $\displaystyle \begin{align*} \left( x^2 - n \right) ^2 - n = x \end{align*}$. Thus the identity holds.
kaliprasad said:let LHS = b
so we get $b= \sqrt{n-\sqrt{n+b}}$ or $b^2 = n- \sqrt{n+b}$
or $(b^2-n)^2 = n+ b\cdots(1)$
let RHS be a
we have $a^2 = n- 1-a$
or $a^2 -n = -(a+1)\cdots(2)$
we need to show that a satisfies (1)
square (2) to get
$(a^2-n)^2 = (a+1)^2 = a^2 + 2a + 1 = n- (a+1) + 2a + 1 = n + a$
so a and b satisfy (1)
and there is only one value of a which is positive hence they are same
The nested radicals identity is a mathematical expression that involves an infinite number of nested square roots. It can be represented as √(n−√(n+√(n−√(n+...), where n is a positive integer. This expression has a unique solution that can be expressed in terms of the positive solution to the equation x^2 = n.
Proving the nested radicals identity is important in mathematics because it helps establish the validity of this expression and provides a solution for it. It also has applications in various fields, such as in calculating the area of a regular n-sided polygon.
The nested radicals identity can be proved using mathematical induction, which involves showing that the identity holds true for a base case (usually n = 1) and then showing that if it holds true for any given value of n, it also holds true for n+1. This process is repeated infinitely, proving the identity for all positive integers n.
The nested radicals identity is significant in mathematics because it demonstrates the concept of infinite nested radicals and their solutions. It also has connections to other mathematical concepts, such as continued fractions and the golden ratio.
Yes, the nested radicals identity has real-life applications in fields such as geometry, physics, and engineering. For example, it can be used to calculate the side lengths of regular polygons, the trajectories of objects in motion, and the resistance in electrical circuits.