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comment What is the difference between the following $2$ sets?
What is the difference between $[1+\delta, \infty)$ and $(1,\infty)$? Does one of them contain the other?
Jan
9
comment How to say if Eigenvectors of A are orthogonal or not? without computing eigenvectors
@Andy: en.wikipedia.org/wiki/Symmetric_matrix#Real_symmetric_matrices
Jan
9
comment How to say if Eigenvectors of A are orthogonal or not? without computing eigenvectors
Do you know the Spectral theorem?
Jan
7
comment Predicates and Quantifiers in discrete math
What have you tried so far?
Jan
5
comment Generic method to find a matrix whose null space is given
@Dor: I'm not sure I understand your question. We know we can write $A$ in such a form (with only non-zero elements in one row). By multiplying $A$ with the two vectors $v_1, v_2$ and setting it equal to zero I constrain $A$ such that $\ker(A) = \operatorname{span}(v_1,v_2)$. Does that answer your question? If not, could you try to rephrase it?
Jan
4
comment Eigenvalue of the substraction of 2 matrices
If you have to prove or disprove the statement "If $A$ and $B$ have the same eigenvalues, then $A-B$ has eigenvalue $0$", then one counterexample is sufficient to disprove the statement. If you want to prove that the statement is true, you'll need to show it for general matrices $A$ and $B$ and not just using one example.
Jan
3
comment $0$ and $p^n \mathbb{Z}_p$ are the only ideals of $\mathbb{Z}_p$
@fretty: $n$ can be $0$.
Jan
3
comment Diagonalization versus s.d. product for non-commuting Hermitian matrices
@nomadreid: Yes, they are equivalent, which is what I have shown in my post. First, I have shown that $[A, B] \neq 0$ is equivalent to (1) and then, I have given you the theorem that it is also equivalent to (2). Because equivalence of statements is an equivalence relation, (1) and (2) are equivalent to each other. This also applies to your third point: The point of my post was to show that (1) and (2) are not just consequences of $A$ and $B$ being non-commutative, but all three statements are equivalent!
Jan
2
comment Diagonalization versus s.d. product for non-commuting Hermitian matrices
@nomadreid: Which bases do you mean? My point was that both statement (1) and (2) are equivalent to $[A,B] \neq 0$, in case it wasn't clear. I thought you were trying to understand how one comes up with these two results.
Jan
2
comment If A is an open set does A the complement of A have to be closed? I know the opossite is true by definition, is this?
A set is closed if its complement is open.
Jan
2
comment How does $\sum_{k=0}^n (pe^t)^k{n\choose k}(1-p)^{n-k} = (pe^t+1-p)^n$?
en.wikipedia.org/wiki/Binomial_theorem#Statement_of_the_theorem
Jan
2
comment The set of the roots of all polynomials in one variable with integer coefficients
What have you tried so far?
Jan
1
comment Integrate $ \int {1\over x^2(x-1)^3} \, dx $
@user1904218: You don't actually (explicitly) need substitution, you can use the chain rule which gives $((x-1)^{-1})' = -(x-1)^{-2}$ immediately.
Jan
1
comment A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?
By multiplying both sides of the second last inequality with $n(n-1)$, we get $-3(n-1) - n + 1 \leq -(n-1)$ which is equivalent to $-3n + 3 - n + 1 \leq -n + 1$, and by adding $3n-3$ to both sides, I ended up with $-n+1 \leq 2n-2$, if you're wondering. It seems simpler to just add $n-1$ to both sides, to retrieve $n \geq 1$ immediately. I changed the answer accordingly.
Jan
1
comment A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?
Also, your last line implies that $\frac{3}{n+1} \leq 0$, which is wrong.
Jan
1
comment A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?
You have a mistake in your calculations. Note that $$- \frac{n+1}{n(n+1)} - \frac{1}{n(n+1)} = \frac{-n-1-1}{n(n+1)} = -\frac{n+2}{n(n+1)}.$$
Jan
1
comment Function and sequence
How do you get $u_0 > u_1$ in question 5?
Jan
1
comment Conversion of the Gauss law $\nabla \cdot E = \frac{\rho } {\epsilon_0}$ into integral form
Basically, you're going back from Maxwell's first equation $\nabla \cdot E = \frac{\rho}{\epsilon_0}$ to Gauss' law $\Phi_E = \frac{Q}{\epsilon_0}$. As Siminore has pointed out, your notation is a bit problematic.
Jan
1
comment Cauchy product of two different series
What did you try?
Jan
1
comment What is $\rightarrowtail$ used for?
Knowing a map is an injection is often better than only knowing it is a map. See: en.wikipedia.org/wiki/Injective_function