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| bio | website | |
|---|---|---|
| location | ||
| age | 41 | |
| visits | member for | 10 months |
| seen | Oct 26 '12 at 6:50 | |
| stats | profile views | 260 |
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Sep 22 |
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How to divide currency? The standard monetary series $1,2,5,10,20,50,\dots$ is a compromise between several goals: (1) it should contain the powers of 10, (2) there should be an approximately constant factor between the different values, (3) that approximate constant shouldn't be too large. |
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Sep 22 |
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Show that for vectors $\bf u$ and $\bf v$ in $ℝ^3$, $\bf u \times v = (-v) \times u$ Easier to implement than the trivial component-wise formula (which I guess would even be more efficient because there are no unnecessary multiplications with $1$)? |
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Sep 22 |
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What is $dx$ in integration? The advantage of writing $\mathrm dx$ at the beginning is that for nested integrals with limits, it's more easily seen which limits belong to which variable, compare $\int_1^2\mathrm dx\int_3^4\mathrm dy\,f(x^2+g(x,y))h(x+y-3)$ with $\int_1^2\int_3^4 f(x^2+g(x,y))h(x+y-3)\,\mathrm dy\,\mathrm dx$ |
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Sep 22 |
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Show that for vectors $\bf u$ and $\bf v$ in $ℝ^3$, $\bf u \times v = (-v) \times u$ Which raises the question: Why? |
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Sep 22 |
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Show that for vectors $\bf u$ and $\bf v$ in $ℝ^3$, $\bf u \times v = (-v) \times u$ Note that the question was explicitly about using the properties of the determinant for proving it (otherwise the most obvious proof would be to just calculate both and compare). |
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Sep 22 |
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What is the remainder when $4^{100}$ is divided by 6? @GeoffRobinson: You surely meant $\frac{2^{199}-2}{3}$. |
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Sep 22 |
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What is the remainder when $4^{100}$ is divided by 6? But in integer division, $2^{200}/6$ is not the same as $2^{199}/3$. You can easily see this with lower exponentials: $2^4/6 = 16/6 = 2,\text{ remainder }4$ but $2^3/3 = 8/3 = 3,\text{ remainder }2$. |
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Sep 22 |
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Is $[0,1)\times[0,1]$ a linear continuum? Maybe a way to make the order-isomorphy work would be to use intervals starting at the points of Cantor's discontinuum. |
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Sep 21 |
awarded | Custodian |
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Sep 21 |
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Why are measures real-valued? @NateEldredge: But then, that condition could be put specifically on probability measures instead of measures in general. |
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Sep 21 |
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Is there a bijective map from $(0,1)$ to $\mathbb{R}$? And the inverse is $f(x)=\ln\left(\frac{1}{x}-1\right)$ |
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Sep 21 |
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Is there a bijective map from $(0,1)$ to $\mathbb{R}$? Actually you don't even have to generalize the argument: If you have the bijection between $(0,1)$ and $(0,1)^2$, you get a bijection from $(0,1)$ to $(0,1)^3$ by just applying the same bijection to one of the two factors of $(0,1)^2$. Of course the same way you get to $(0,1)^n$. |
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Sep 21 |
awarded | Nice Answer |
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Sep 20 |
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$Y^3$ congruent to $1 \pmod {p}$ But for $p=3$, the only non-zero elements are $1$ and $2$, so there's no way you get 3 different roots. |
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Sep 20 |
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$Y^3$ congruent to $1 \pmod {p}$ If it is a finite field, then $p$ is a prime. Except $3$ itself, no prime is a power of $3$. |
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Sep 20 |
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Proof of derivative of $e^x$ is $e^x$ without using chain rule Note that my corrected version still needs no interchange of limits. |
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Sep 20 |
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If A and C are independent, is P(A,C) = P(A)*P(C) always true? They really multiply P(not B) with P(C)? Since C implies not B (they are mutually exclusive!), P(C)=P(C and not B). "and" only means "multiply probabilities" if the events are independent. In short, their solution is wrong. |
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Sep 20 |
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Let $a$ be any nonzero vector, If $v=a-\alpha e_1$, where … @Siminore: No, $v$ is a column. And therefore $v^T$ is the transpose of a column, which is a row. Ah, I now notice that I mixed up the two products in my previous comments. Now fixed. |
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Sep 20 |
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Let $a$ be any nonzero vector, If $v=a-\alpha e_1$, where … @Siminore: No, $vv^T$ is a matrix (unlike $v^Tv$ which indeed is a number). And $\|\cdot\|_2$ is the standard way to write the Euclidean norm. |
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Sep 20 |
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Let $a$ be any nonzero vector, If $v=a-\alpha e_1$, where … Hint: $a = v + \alpha e_1$ |
