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Sep
22
comment Heat Equation & Fundamental Theorem of Calculus
Note that the whole expression on the left of (1) has two free variables (and none of them is $x$).
Sep
22
comment Can I keep adding more dimensions to complex numbers?
Well, it does break the total ordering property: Unlike for real numbers, there's no total ordering of the complex numbers which is compatible with its algebraic structure.
Sep
22
comment 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.
Sep
22
comment 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 $1$)?
Sep
22
comment 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 $\int_1^2\int_3^4 f(x^2+g(x,y))h(x+y-3)\,\mathrm dy\,\mathrm dx$
Sep
22
comment Show that for vectors $\bf u$ and $\bf v$ in $ℝ^3$, $\bf u \times v = (-v) \times u$
Which raises the question: Why?
Sep
22
comment 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).
Sep
22
comment What is the remainder when $4^{100}$ is divided by 6?
@GeoffRobinson: You surely meant $\frac{2^{199}-2}{3}$.
Sep
22
comment 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 $2^3/3 = 8/3 = 3,\text{ remainder }2$.
Sep
22
comment 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.
Sep
21
awarded  Custodian
Sep
21
comment Why are measures real-valued?
@NateEldredge: But then, that condition could be put specifically on probability measures instead of measures in general.
Sep
21
comment Is there a bijective map from $(0,1)$ to $\mathbb{R}$?
And the inverse is $f(x)=\ln\left(\frac{1}{x}-1\right)$
Sep
21
comment 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 $(0,1)^n$.
Sep
21
awarded  Nice Answer
Sep
20
comment $Y^3$ 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.
Sep
20
comment $Y^3$ 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 $3$.
Sep
20
comment Proof of derivative of $e^x$ is $e^x$ without using chain rule
Note that my corrected version still needs no interchange of limits.
Sep
20
comment 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.
Sep
20
comment 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.