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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 2 years, 5 months |
| seen | May 17 at 23:25 | |
| stats | profile views | 83 |
Hello.
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May 13 |
awarded | Caucus |
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May 11 |
awarded | Critic |
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Apr 18 |
answered | Finding a position opposite to a rotated point |
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Jun 8 |
accepted | Solving an inequality modulo 1 |
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May 2 |
suggested | suggested edit on Adjacent sequences example |
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May 2 |
comment |
Solving an inequality modulo 1 Ah, that makes sense. I'll definitely try to see what I can do with this. |
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May 2 |
comment |
Solving an inequality modulo 1 Thank you for the detailed answer, I really appreciate it. Is there any reason you chose $3$ and $2$ for the bounds in this expression: $\frac{n}3<i<\frac{n}2$? The reason why I'm looking for an analytic solution for $i$ is because I'm trying to factor a large number, just to see how efficiently I can do it. Each of the possible $i$'s corresponds to a possible factor of a number $n$. I've found graphically that these $i$'s grow very predictably for arbitrary $n$, as you can see in this graph I tried to make (each peaks correspond to an $i$). |
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May 2 |
comment |
Solving an inequality modulo 1 I guess you're right. Thanks for the help! |
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May 2 |
comment |
Solving an inequality modulo 1 I'm basically trying to factor a number with only two prime factors. If you look at the value of the function for $i$ ranging from $0$ to $\sqrt{n}$, you can see that the distance between the peaks of the function is growing predictably. Instead of trying to factor the number by trying every possible factor, I thought it might be a bit faster to try only the factors that correspond to the peaks, as the set of all prime factors of the number will exist within the solution set of the inequality I posted in my question. Hopefully that makes at least some sense. |
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May 2 |
comment |
Solving an inequality modulo 1 The code that I have works exactly the same was as you have described, but I'm trying to find a way to analytically solve for $i$ or at least make the solution easier to calculate. |
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May 2 |
comment |
Solving an inequality modulo 1 @Kaz: Right, I'm only worrying about the fractional part of the real number. $i$ will always be smaller than $\sqrt{n}$. |
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May 2 |
revised |
Solving an inequality modulo 1 added 7 characters in body |
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May 2 |
comment |
Solving an inequality modulo 1 @BrettFrankel: Ah, my bad. Thanks for the clarification. |
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May 2 |
revised |
Solving an inequality modulo 1 added 17 characters in body |
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May 2 |
comment |
Solving an inequality modulo 1 @BrettFrankel: I'm not entirely sure about notation for inequalities involving the modulo operator, but I think this makes a bit more sense: $\frac{n}{i}\text{ (mod 1) } < \frac{n}{i + 1}\text{ (mod 1) for } n, i\in\mathbb{N}$ |
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May 2 |
comment |
Solving an inequality modulo 1 @BrianM.Scott: Yes, that's exactly what I'm trying to do. |
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May 2 |
asked | Solving an inequality modulo 1 |
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Dec 5 |
awarded | Yearling |
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Sep 20 |
comment |
How to find the function $f$ given $f(f(x)) = 2x$? I've updated the answer. @Asaf, would saying that $f(f(-x)) = -f(f(x))$ be enough to prove that $f(x)$ is invertible? |
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Sep 20 |
revised |
How to find the function $f$ given $f(f(x)) = 2x$? added 19 characters in body |
