How to prove $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$

Question

So, I was watching this video by blackpenredpen where he mentions that $$\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$$ so I wanted to try and prove it myself.

Let $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=x$
But$\sqrt {a-\sqrt {a+\underbrace{\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}_x}}=x$
$\therefore \sqrt {a-\sqrt {a+x}}=x$
$a-\sqrt {a+x}=x^2$
$x^2-a=-\sqrt {a+x}$
$x^4-2ax^2+a^2=a+x$
$x^4-2ax^2-x+a^2-a=0$
Note that this is of the form $y^4+py^2+qy+r=0$ so we can use Ferrari-Cardano.

We need to find a $z$ such that $(2z-p)y^2-qy+(z^2-r)$ has a discriminant of $0$. The discriminant of $(2z-p)y^2-qy+(z^2-r)$ is equal to $q^2 - 4(2z - p)(z^2 - r),$ which simplifies to $8z^3 - 4pz^2 - 8rz + (4pr - q^2) = 0$

Substituting values from $x^4-2ax^2-x+a^2-a=0$ into $8z^3 - 4pz^2 - 8rz + (4pr - q^2) = 0$ gives us
$8z^3-4\cdot(-2a)\cdot z^2-8\cdot(-a)\cdot z+\left(4\cdot (-2a) \cdot (-a) - (-1)^2 \right)=0 \implies 8z^3+8az^2+8az+(8a^2-1)=0$

Using Cardano's formula, or in my case Wolfram Alpha, we get that $$z_1 = \frac {\sqrt [3]{-16 a^3 - 144 a^2 + 3 \sqrt 3 \sqrt {256 a^5 + 512 a^4 + 224 a^3 - 288 a^2 + 27} + 27}}{6\sqrt[3]2} - \frac {192 a - 64 a^2}{48\cdot 2^{\frac 23} \sqrt [3]{-16 a^3 - 144 a^2 + 3\sqrt 3 \sqrt {256 a^5 + 512 a^4 + 224 a^3 - 288 a^2 + 27} + 27}} - \frac a3$$

I simply can not solve that quintic and continue as it is already too cluttered. Was there a mistake in my problem or is there any other way to do it? Also, I am sincerely sorry but I am not sure how to tag this question.

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Edit $1:$

After Ross Millikan's answer, I snooped around in the comment section of the video and found someone who found that it is true using alternating root series. Was his proof correct as $\frac {\sqrt {4a-3}-1}2$ does not seem to have real values for $a \lt \frac 34$? Thank you!

Answer 1

It is CORRECT!

$x^4-2ax^2-x+a^2-a=0$ can be factorised as $(x^2-x-1)*(x^2+x+1-a)$. You'll get the desired answer from the right factor $(x^2+x+1-a)$.

Answer 2

It's wrong! You can try $a=0$.

Also, try $a=1$.

If your sequence converges then we need to solve the following equation $$\sqrt{a-\sqrt{a+x}}=x.$$ Let $a+x=y^2,$ where $y\geq0$.

Hence, $$y^2-x=a$$ and $$x^2+y=a,$$ where $x\geq0$ and $a\geq0.$

Thus, $$y^2-x^2-x-y=0$$ or $$(x+y)(y-x-1)=0.$$

If $x+y=0$ so $x=y=a=0$ and your formula is still wrong.

If $y=x+1$ then $x^2+x+1-a=0,$ which gives $$x=\frac{-1+\sqrt{4a-3}}{2}.$$ Now, since $x\geq0$, we have $$\frac{-1+\sqrt{4a-3}}{2}\geq0,$$ which gives $a\geq1$.

But for $a=1$ our sequence divergences and it should be $a>1$.

Answer 3

If the nested radical converges to $x$, $$\sqrt{a-\sqrt{a+x}}=x,$$

$$\sqrt{a+x}=a-x^2,$$

$$a+x=(a-x^2)^2.$$

This should be verified by $x=\dfrac{\sqrt{4a-3}-1}2$. Indeed,

$$(a-x^2)^2=\left(a-\frac{4a-2-2\sqrt{4a-3}}4\right)^2=\frac{4a-2+2\sqrt{4a-3}}4=a+x.$$

We can show that

$$\sqrt{a-\sqrt{a+x}}-x$$ is monotonic, so that there is at most one solution.

Anyway, this still doesn't prove that the nested radical converges.

Answer 4

Here's an easier solution that doesn't require you going out to a fourth degree polynomial. Let $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=x$, $-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\ldots}}}}}}=y$.

Then $x^2 - y = a$, and $y^2-x = a$. Subtracting these two equations, we have that $x^2 - y^2 + x-y =0 \implies (x-y)(x+y+1) = 0$. Since $x \neq y$ unless both are equal to zero, $x+y = -1$. Adding these equations, we have $x^2 +y^2 - (x+y) = 2a$, and substituting everything in, we have $x^2 + x = a-1$, which we can evaluate via the quadratic formula to $x=\frac{-1+\sqrt{4a-3}}{2}$, where we have ignored the other solution since it is always negative.