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2 | corrected property | ||
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You can approximate the real $n$th root of $k$ as the real solution to $x^n-k=0$. Since this is a monotone function on $[0,k]$, you can use a bisection method to approximate it to an arbitrary degree. So finding a root amounts to iterated computation of $x^n$. So if divide-and-conquer makes computing $x^n$ more efficient so it should make computing $k^{1/n}$ more efficient. There are probably much better methods. I'm no expert in numerical approximation. |
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1 | [made Community Wiki] | ||
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You can approximate the real $n$th root of $k$ as the real solution to $x^n-k=0$. Since this is a function you can use a bisection method to approximate it to an arbitrary degree. So finding a root amounts to iterated computation of $x^n$. So if divide-and-conquer makes computing $x^n$ more efficient so it should make computing $k^{1/n}$ more efficient. There are probably much better methods. I'm no expert in numerical approximation. |
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