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seen Aug 26 at 9:09

Jul
2
awarded  Curious
Jun
6
accepted Basic question about wedge product
May
11
awarded  Popular Question
Apr
30
comment Regarding usage of $\pm$ sign
@TuckerRapu These are completely different questions, and have completely different answers.
Feb
13
awarded  Popular Question
Dec
4
comment Proving that matrix is positive definite
I think that in your answer "if and only if all $a_i$ are non-zero" should be changed to "if and only if not all $a_i$ are zero".
Dec
3
comment Question about continuous differentiability
Thank you. This helped me to understand as well: freemathhelp.com/forum/threads/…
Dec
3
accepted Question about continuous differentiability
Dec
3
comment Question about continuous differentiability
So, can we say that $f(x,y)$ is continuously differentiable with respect to $x$?
Dec
2
revised Question about continuous differentiability
deleted 17 characters in body
Dec
2
comment Question about continuous differentiability
@user99680 In the numerator you have the term $1+\frac{y^2}{\sqrt{x^2+y^2}}$. If you substitute x and y by zeros, u get $1 + \frac{0}{0}$ which is undefined.
Dec
1
comment Question about continuous differentiability
@user99680 But there is division by zero in the numerator.
Dec
1
comment Question about continuous differentiability
@user99680 Sorry I wrote the wrong function. I have corrected.
Dec
1
revised Question about continuous differentiability
fixed the function
Dec
1
asked Question about continuous differentiability
Nov
26
awarded  Critic
Nov
26
comment Solving equation in Clifford algebra
@ShuchangZhang $G_{p,q}$ is a Clifford algebra over $R^{p+q}$ where $p$ orthonormal basis vectors of $R^{p+q}$ square to $1$ and $q$ orthonormal basis vectors of $R^{p+q}$ square to $-1$. For example $G_{1,1}$ is defined over $R^2$ which has the orthonormal basis of $e_1 = [1,0]^T$ and $e_2=[0,1]^T$ for which we define that $e_1^2=1$ and $e_2^2=-1$. Then the basis of our clifford algebra is given by $g_1=1$, $g_2=e_1$, $g_3=e_2$, and $g_4=e_1 e_2$. Then we have that $g_1^2=1^2=1$, $g_2^2=e_1^2=1$, $g_3^2=e_2^2=-1$, and $g_4^2=(e_1 e_2)^2=e_1 e_2 e_1 e_2 = -e_1 e_1 e_2 e_2 = -e_1^2 e_2^2 = 1$.
Nov
26
comment Solving equation in Clifford algebra
@rschwieb The actual equation uses the operator from my previous question: math.stackexchange.com/questions/269335/… With this operator it looks like $x^*(-b)x=a^*ba$.
Nov
26
comment Solving equation in Clifford algebra
@ShuchangZhang Consider the case of $G_{0,1}$, then this clifford algebra is isomorphic to complex numbers, so $g_1=1$ and $g_2=i$. Then $g_1^3=1$ and $g_2^3=i*i*i=-i$.
Nov
26
revised Solving equation in Clifford algebra
added x definition