Is there a way to reduce the expression $P+\sqrt{P^2+\sqrt{P^4+\sqrt{ P^8+\cdots)}}}$?

  • $\begingroup$ What is $P$? Any real number, a real number from some specific interval, something else? $\endgroup$ – Daniel Fischer Jun 15 '15 at 11:35
  • $\begingroup$ P is any real number....question was just to simplify the given expression.. $\endgroup$ – Sumit Jha Jun 15 '15 at 11:36
  • 1
    $\begingroup$ Assuming existence / convergence, $P + \sqrt{P^2+\sqrt{P^4+\cdots}} = P(1+\sqrt{1+\sqrt{1+\cdots}}) = PS$, say. Can you find $S$? $\endgroup$ – Macavity Jun 15 '15 at 11:39
  • $\begingroup$ I'm not sure but I think $P$ is prime, however, primes are usually denoted with a $p$. $\endgroup$ – Aleksandar Jun 15 '15 at 11:39
  • $\begingroup$ You may also need mathworld.wolfram.com/HerschfeldsConvergenceTheorem.html $\endgroup$ – Macavity Jun 15 '15 at 12:01

So our target expression is:

$$ P + \sqrt{P^2 + \sqrt{P^4...}}$$

An indirect approach that is quite simple is to consider

$$ P \left( 1 + \sqrt{1 + \sqrt{1 ...}} \right)$$

It follows that of we distribute the P we will have

$$ P +P \sqrt{1 + \sqrt{1 ...}} $$

Which yields

$$ P + \sqrt{P^2 + P^2\sqrt{1 + ...}}$$

And again we can distribute under the next root for $P^4$ then $P^8$ etc...

So now the trick is to evaluate:

$$S = 1+\sqrt{1+\sqrt{...}}$$

It follows that

$$(S-1)^2 = 1+\sqrt{1+\sqrt{...}}$$

And therefore

$$ (S-1)^2 = S$$

Using this we can solve and find

$$S^2-3S + 1 = 0$$

Which has solutions given by the quadratic formula:

$$S = \frac{3 \pm \sqrt{5}}{2}$$

A simple test of evaluating the first few terms of

$$S = 1+\sqrt{1+\sqrt{...}}$$

Suggests that solution closer to $2$ is in order this we have

$$S = \frac{3 + \sqrt{5}}{2}$$

And this to evaluate:

$$ P + \sqrt{P^2 + P^2\sqrt{1 + ...}}$$

Is just

$$P \left( \frac{3 + \sqrt{5}}{2} \right)$$


If the series converges to $S$ then $S= P + \sqrt{P^2+\sqrt{P^4+\cdots}} = P\left(1+\sqrt{1+\sqrt{1+\cdots}}\right) = PA$,

where $A$ is $1 + \sqrt{1+\sqrt{1+\cdots}}$.

Now, $(A-1)^{2}=A$ so

$A$= $\frac{3+\sqrt 5}{2}$ or $\frac{3-\sqrt 5}{2}$, but we discard the latter root, because if you look at the series $A$ it is obvious that $A>2$.

Therefore, $S=\frac{3+\sqrt 5}{2}P$.


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