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

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.

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!

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 $$(x^2+x+1-a)$$.

• very neat solution (+1) – G Cab Oct 22 '18 at 23:59
• @GCab thank you – Ankit Kumar Oct 23 '18 at 4:35

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 $$a>1$$.

• Oh, I never realised that! But does it hold for other values of $a$? He solved $\sqrt {5-x}=5-x^2$ in that video and used that identity to find the answer. Was he wrong? – Mohammad Zuhair Khan Oct 22 '18 at 18:46
• @Raptor I added something. See now. – Michael Rozenberg Oct 22 '18 at 18:57
• Essentially, you both got the same answer, except that he used the equation. Could it be that it only holds for $a\gt \frac 34?$ – Mohammad Zuhair Khan Oct 22 '18 at 18:59
• @Raptor Do you say about the equation $\sqrt{a-\sqrt{a+x}}=x$? – Michael Rozenberg Oct 22 '18 at 19:03
• No, I meant $\frac {\sqrt {4a-3}-1}2$ which is essentially what we are trying to prove $x$ equals. – Mohammad Zuhair Khan Oct 22 '18 at 19:05

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.

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.