MHB Prove the nested radicals identity √(n−√(n+√(n−√(n+....

[lfdahl]

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]

Prove the following identity ($n = 1,2,3,...$):

\[\sqrt{n - \sqrt{n+\sqrt{n-\sqrt{n +...}}}} = \sqrt{(n-1)-\sqrt{(n-1)-\sqrt{(n-1)-...}}}\]

Spoiler

$\displaystyle \begin{align*} x &= \sqrt{n - \sqrt{n + \sqrt{n - \sqrt{n + \sqrt{ \dots } }}}} \\ x^2 &= n - \sqrt{ n + \sqrt{ n - \sqrt{ n + \sqrt{\dots }}}} \\ x^2 - n &= - \sqrt{ n + \sqrt{ n - \sqrt{ n + \sqrt{ \dots } } }} \\ \left( x^2 - n \right) ^2 &= n + \sqrt{ n - \sqrt{ n + \sqrt{ n - \sqrt{ n + \sqrt{ \dots } }}}} \\ \left( x^2 - n \right) ^2 - n &= \sqrt{ n -
\sqrt{n + \sqrt{ n - \sqrt{ n + \sqrt{ \dots }}}}} \\ \left( x^2 - n \right) ^2 - n &= x \end{align*}$

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]

Spoiler

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

 

[lfdahl]

Spoiler

$\displaystyle \begin{align*} x &= \sqrt{n - \sqrt{n + \sqrt{n - \sqrt{n + \sqrt{ \dots } }}}} \\ x^2 &= n - \sqrt{ n + \sqrt{ n - \sqrt{ n + \sqrt{\dots }}}} \\ x^2 - n &= - \sqrt{ n + \sqrt{ n - \sqrt{ n + \sqrt{ \dots } } }} \\ \left( x^2 - n \right) ^2 &= n + \sqrt{ n - \sqrt{ n + \sqrt{ n - \sqrt{ n + \sqrt{ \dots } }}}} \\ \left( x^2 - n \right) ^2 - n &= \sqrt{ n -
\sqrt{n + \sqrt{ n - \sqrt{ n + \sqrt{ \dots }}}}} \\ \left( x^2 - n \right) ^2 - n &= x \end{align*}$

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.

Good job, Prove It! Thankyou for your participation. (Yes)

 

[lfdahl]

Spoiler

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

What a smart solution!(Cool) Thankyou, kaliprasad for your participation!