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I just wanted to write the iterated version of:

$$ a^2 - b^2 = (a+b)(a-b)$$

As:

$$ n^x - 1 = (n^{ \sqrt{x}} +1) \color{red}{(n^{ \sqrt{x}} -1)} = \\ (n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1) \color{red}{(n^{ \sqrt{\sqrt{x}}} -1)} = \\ (n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1)(n^{ \sqrt{\sqrt{\sqrt{x}}}} +1) \color{red}{(n^{ \sqrt{\sqrt{\sqrt{x}}}} - 1)} = \dots $$

And so:

$$n^x - 1 = (n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1)(n^{ \sqrt{\sqrt{\sqrt{x}}}} +1) ... \Rightarrow \\n^x - 1 = \prod_{k=1}^{\infty} (n^{x^{\left (\frac{1}{2k}\right)}} + 1)$$

But when plugging $n=1$ we get:

$$ 1^x - 1 \equiv 0 = \prod_{k=1}^{\infty} (1^{x^{\left (\frac{1}{2k}\right)}} + 1) \equiv \prod_{k=1}^{\infty} (1+1) = \prod_{k=1}^{\infty} 2 = \infty$$

Where did I go wrong? I think that I "omitted" the $n^{x^{(\frac{1}{2k})}} - 1$ part, but isn't it valid when we talk about infinite series? I would appreciate if you could clear things up.

Thank you!

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Should be instead $$n^x-1 = (n^{x/2}-1)(n^{x/2}+1)$$, as in general,

$$n^x-1 \not = (n^{\sqrt{x}}-1)(n^{\sqrt{x}}+1).$$

This was your error.

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  • $\begingroup$ But if it was like this, and I didn't do this (silly) mistake, would it be true then? I feel like it would still lead to a contradiction $\endgroup$ – MathAsker Nov 27 '20 at 1:34
  • $\begingroup$ That was your first error but yes indeed, it was not your only error. In your infinite product you ignored a term that vanishes, and indeed, $(n^{2^{-k}} -1)$ goes to 0 quite rapidly as $k$ gets larger. And is infact 0 already for $n=1$ no matter what $k$ is. $\endgroup$ – Mike Nov 27 '20 at 2:19

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