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seen Dec 16 at 11:18

Dec
12
comment Show that two metric space are equaly
What have you tried?
Nov
23
comment How to mathematically describe a loop over a set with two indexes.
As it stands, $1_A(x) 1_B(y) = 0$ if and only if either $x \notin A$ or $y \notin B$. I couldn't quite read off if that was your criterion.
Nov
18
comment How to mathematically describe a loop over a set with two indexes.
Hm, I'm not sure what you mean by $d1$ and $d2$. But you could write $1_{D_{i,j}}(x)1_{D_{i,j+1}}(y)$. Another thing that confused me about your sum was that from your description, it seems like you did /not/ want $y$ to be in that particular set.
Nov
18
answered How to mathematically describe a loop over a set with two indexes.
Nov
18
comment How to mathematically describe a loop over a set with two indexes.
The index pattern is unclear; it seems that the number of sets with a given first index depends on the index if there are only three sets with first index $0$, but looking at the sum you wrote, that's not the case.
Nov
13
answered Linear transformation between matrices of different dimensions
Nov
5
comment The sum of finite exponential series with a quadratic phase
Do you know the residue theorem? Super roughly: If so, you can construct a function whose poles include the first $K$ integers and whose residue at each of the poles is the corresponding summand. Then you can get the left hand side by integrating over a rectangular contour around those integers.
Oct
30
comment Is this function one-to-one and onto?
$f(1) = f(2)$ but $1 \not= 2$.
Oct
29
comment For a differentiable map $f: \mathbb{R^n}\to \mathbb{R^n}$, Show that $f^*({dy_1 \wedge\cdots \wedge dy_n})=\det(df)dx_1\wedge \cdots\wedge dx_n$
I see, perhaps someone else can point out a proof that is closer in spirit to that of the first chapter of that book, then. If you like, you can just think of the $\frac{\partial}{\partial x_j}$ as the dual of $dx_j$. That is, $dx_i \left( \frac{\partial}{\partial x_j} \right) = 1$ if $i = j$ and it equals $0$ if $i \not= j$. Perhaps that's sufficient for the above to make sense.
Oct
29
comment For a differentiable map $f: \mathbb{R^n}\to \mathbb{R^n}$, Show that $f^*({dy_1 \wedge\cdots \wedge dy_n})=\det(df)dx_1\wedge \cdots\wedge dx_n$
The $\frac{\partial}{\partial x_i}$ are the standard basis vectors of the tangent space at a point of interest. This is fairly standard notation, but perhaps you are used to a different one?
Oct
29
comment For a differentiable map $f: \mathbb{R^n}\to \mathbb{R^n}$, Show that $f^*({dy_1 \wedge\cdots \wedge dy_n})=\det(df)dx_1\wedge \cdots\wedge dx_n$
You have a wedge product of vectors equaling a number; depending on what setup OP is used to, this might require a bit of explanation.
Oct
29
comment For a differentiable map $f: \mathbb{R^n}\to \mathbb{R^n}$, Show that $f^*({dy_1 \wedge\cdots \wedge dy_n})=\det(df)dx_1\wedge \cdots\wedge dx_n$
My point is that it's a bit easier to only do it at a single point as by definition, $$f^* (dy_1 \wedge \cdots \wedge dy_n) \left(\frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial x_n}\right) = (dy_1 \wedge \cdots \wedge dy_n) \left(f_* \frac{\partial}{\partial x_1}, \dots, f_* \frac{\partial}{\partial x_n}\right),$$ so by the formula of the comment, this is the determinant of the matrix whose $(i,j)$'th entry is $dy_i \left( \sum_k \frac{\partial f_k}{\partial x_j} \frac{\partial}{\partial y_k}\right) = \frac{\partial f_i}{\partial x_j}$.
Oct
29
answered For a differentiable map $f: \mathbb{R^n}\to \mathbb{R^n}$, Show that $f^*({dy_1 \wedge\cdots \wedge dy_n})=\det(df)dx_1\wedge \cdots\wedge dx_n$
Oct
29
comment For a differentiable map $f: \mathbb{R^n}\to \mathbb{R^n}$, Show that $f^*({dy_1 \wedge\cdots \wedge dy_n})=\det(df)dx_1\wedge \cdots\wedge dx_n$
Are you aware that in general, for covectors $u^i$ and vectors $v_i$, one has $$u^1 \wedge \cdots \wedge u^n(v_1,\dots,v_n) = \det ((u^i(v_j))_{i,j})?$$
Oct
23
comment Euclid algorithm greatest common factor
There's an example of Wikipedia which might be helpful.
Oct
17
comment How to determine the derivative of $ f $ at $ x=2$ by looking at the graph only?
Then $f(3) = 21$, which certainly does not appear to be the case. :)
Oct
17
comment How to determine the derivative of $ f $ at $ x=2$ by looking at the graph only?
Then $f(3) = 21$, which certainly does not appear to be the case. :)
Oct
17
comment How to determine the derivative of $ f $ at $ x=2$ by looking at the graph only?
The other answers are also reasonable, as they tell you that by making further assumptions, you can give more concrete answers.
Oct
17
comment A function $f$ such that the limit of $f(x^2)$ exists but not $f(x)$.
You're welcome.
Oct
17
answered How to determine the derivative of $ f $ at $ x=2$ by looking at the graph only?