# Wolfram Alpha simplifying $\sqrt{1+x^\sqrt{1+x^\sqrt{1+...}}}$ incorrectly [closed]

I am pretty new to Wolfram Alpha. I want Wolfram Alpha to find the derivative of $$\sqrt{1+x^\sqrt{1+x^\sqrt{1+...}}}$$. At the moment, I am having trouble with Wolfram Alpha simplifying it incorrectly.

I had thought that Fold[sqrt(1+{x}^#,2)&, x, Range[∞]]]

would work but Wolfram says it equals $$\sqrt{x^x+1}$$ which I don't think is actually equivalent.

For example, if I plug in 0 to the original equation:

$$f(x)=\sqrt{1+x^\sqrt{1+x^\sqrt{1+...}}}$$

$$f(0)=\sqrt{1+0^\sqrt{1+0^\sqrt{1+...}}}$$

$$f(0)=\sqrt{1+0^{f(0)}}$$

$$f(0)$$ must be greater than $$0$$ because $$\sqrt{1}>0$$ so we know $$0^{f(0)}=0$$.

$$f(0)=\sqrt{1+0}$$

$$f(0)=1$$

Now, let's plug it into the equation Wolfram Alpha gave me:

$$g(x)=\sqrt{x^x+1}$$

$$g(0)=\sqrt{0^0+1}$$

$$g(0)=\sqrt{1+1}$$

$$g(0)=\sqrt{2}$$

And we all know: $$1\ne\sqrt{2}$$

• In Fold[sqrt(1+{x}^#,2)&, x, Range[∞]]] your brackets don't match and the ,2 seems oddly placed. If I change that to Fold[sqrt(1+x^#)&, x, Range[4]] then WA seems to understand that. WA does have a limited buffer size for input and calculations. I have seen error and omissions when it runs out of space, but I can't tell if that is what you are seeing. WA is probably isn't the tool for this task.
– Bill
Mar 8, 2023 at 16:09
• Note that you function can be written by an implicit function like: y= Sqrt[1+x^y]  Mar 8, 2023 at 16:16
• I'm afraid that Questions about Wolfram Alpha are off-topic in this site. Mar 8, 2023 at 16:41
• Welcome to Mathematica StackExchange. First, as mentioned by @rhemans, this site is only about Mathematica and Wolfram Language. Second, what you observe is – in my opinion – a buggy behaviour of WolframAlpha. Writing Range[∞] is not a valid expression in Wolfram Language, and if you evaluate your Fold expression in Mathematica, you get an error. However, you also get a result where only one iteration was performed, and it seems that Wolfram|Alpha blatantly discards this error and uses the output, which is clearly wrong, as a result. Mar 8, 2023 at 19:09
• If you have access to Wolfram Mathematica, please use it, and appropriately rephrase your question, otherwise it will probably get closed ... Mar 8, 2023 at 19:18

$$A = \sqrt{1 + x^{\sqrt{1 + x^{\sqrt{1+x...}}}}}$$
So $$A^2 = 1 + x^A$$
Take derivative with respect to $$x$$:
$$2 A(x) A^\prime (x) = x^{A(x)} \left( A(x)/x + \ln (x) A^\prime(x)\right)$$
This has not direct analytic solution for $$A^\prime (x)$$ (as far as I can tell).
• In 13.1, you can calculate the implicit derivative with ImplicitD[A==Sqrt[1+x^A], A, x] to get the "result" -((A x^(-1 + A))/(-2 Sqrt[1 + x^A] + x^A Log[x])) Mar 8, 2023 at 20:26