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| bio | website | |
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| age | ||
| visits | member for | 1 year, 5 months |
| seen | May 13 at 7:00 | |
| stats | profile views | 49 |
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May 7 |
comment |
Compute the square root of $2$ in $R=\mathbb{Z}/n\mathbb{Z}$ where $n=2^s+1$ and $s=2^k$ The reason I am so curious is because @Justin's formula and the entire discussion here doesn't care that $n$ is a Fermat number, but only that $n$ is divisible by 4. |
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May 7 |
comment |
Compute the square root of $2$ in $R=\mathbb{Z}/n\mathbb{Z}$ where $n=2^s+1$ and $s=2^k$ @Jyrki : Is there a danger that $ux \equiv -x$, i.e. that there might be repeats among $x,-x,ux,-ux$? |
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May 7 |
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Compute the square root of $2$ in $R=\mathbb{Z}/n\mathbb{Z}$ where $n=2^s+1$ and $s=2^k$ Sorry, what I meant was that there are for sure 2 distinct solutions given by $x_1=2^{2^{k-2}}(2^{2^{k-1}}-1)$ and $x_2=2^{2^k}+1-x_1$. I was wondering if there are other roots besides these 2, I can't seem to find an argument against such a possibility. |
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May 7 |
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Compute the square root of $2$ in $R=\mathbb{Z}/n\mathbb{Z}$ where $n=2^s+1$ and $s=2^k$ EDIT: Sorry for the double comment. |
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May 7 |
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Compute the square root of $2$ in $R=\mathbb{Z}/n\mathbb{Z}$ where $n=2^s+1$ and $s=2^k$ Is there another square root besides the $\pm$ of the solution above? |
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May 6 |
accepted | Non-uniqueness of solutions of the heat equation |
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May 6 |
accepted | Generalized addition function |
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May 6 |
comment |
Generalized addition function Your solution is correct, I made a mistake in my last edit, there is no problem with the last statement in its current form. |
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May 6 |
revised |
Generalized addition function deleted 393 characters in body |
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May 5 |
revised |
Generalized addition function added 397 characters in body |
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May 5 |
comment |
Generalized addition function $(-x)\oplus 0= x\oplus 0$ is a desired property |
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May 5 |
comment |
Generalized addition function About the $0$, $0\oplus 0=(0+0)0\oplus 0=0$. |
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May 4 |
comment |
Generalized addition function Thanks, I have changed the fourth condition to avoid this inconsistency. |
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May 4 |
revised |
Generalized addition function added 105 characters in body |
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May 4 |
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Generalized addition function Thanks, I added it in the statement of the problem. |
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May 4 |
revised |
Generalized addition function added 73 characters in body |
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May 4 |
asked | Generalized addition function |
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Apr 27 |
comment |
Deriving the exponential distribution from a shift property of its expectation (equivalent to memorylessness). I don't think $f$ is uniquely determined. It cannot be derived from the differential equation $F(a)+\mu F'(a)=1$ since this equation is not valid on a set of measure zero. At least I don't know how to solve such equations. I agree that the exponential distribution is a solution to the above problem. I have a feeling that it might not be unique and I'd like to have an argument for or against my feeling... |
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Apr 26 |
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Deriving the exponential distribution from a shift property of its expectation (equivalent to memorylessness). Thank you for replying, I'd love a clarification for my own self. Correct me if I'm wrong, but I thought that the FTOC can only guarantee that the derivative exists almost everywhere. This implies that the first order differential equation is only almost everywhere satisfied by $F(x)$, and therefore is not uniquely given by the exponential solution any more. |
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Apr 26 |
answered | A Question on Outer Measure |
