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If the function $f(x)$ is such that $$f^2(x)=x+f(x+1),$$ find a closed-form expression for $f$.

I found $$f(x)=\sqrt{x+\sqrt{x+1+\sqrt{x+2+\sqrt{x+3+\cdots}}}}$$ is such an $f$. Does anyone have other solutions? Thank you.

TBrendle
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math110
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    As far as I can tell you can define $f(x)$ arbitrarily in the interval $[0,1)$ and extend it from there according to the functional equation. This should produce infinitely many solutions. – user1337 Dec 09 '13 at 17:24
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    You need to be careful here. Do you mean $f(f(x))$ or $f(x) \cdot f(x)$? Usually $f^2(x) \equiv f(f(x))$ and $f(x) \cdot f(x)$ is written as $f(x)^2$. – Carl Dec 09 '13 at 18:07

2 Answers2

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You can always take either square-root, so $$ f(x)=\pm\sqrt{x\pm\sqrt{x+1\pm\sqrt{x+2\pm\sqrt{x+3\pm\cdots}}}} $$ Gives you uncountably many solutions...

GEdgar
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Notice that $f(0)$ is nothing else than the square root of the Nested Radical Constant, which is yet unknown to possess a closed form. Obviously, if f were to possess such a form, then so would this constant, meaning your question is still open.

Lucian
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