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seen Apr 12 at 5:51

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
Nov
26
comment Solving equation in Clifford algebra
@ShuchangZhang $$x=\sum_{i=1}^{2^{p+q}}(x_ig_{i})$$
Nov
25
revised Solving equation in Clifford algebra
edited tags
Nov
25
revised Solving equation in Clifford algebra
deleted 62 characters in body
Nov
25
asked Solving equation in Clifford algebra