Sep
20
comment Help with finding the real zeros of a polynomial
sorry, now we have wasted the saved time and space anyway.
Sep
20
comment Help with finding the real zeros of a polynomial
$(x+1),(x-1),(x+2),(x-2),(x+4),(x-4)$ are not the possible zeros of the polynomial, they are possible divisors or linear factors of the polynomial. Th possible zeros are $+1,-1,+2,-2,+4,-4$.
Sep
18
reviewed Approve Jacobian matrix of the inverse of a bijective function
Sep
17
comment How to calculate running time of code?
-1 One way to solve the problem is only mentioned in two sentences. It doesn't answer the question becuase it does not show how to calculate the formulas. I don't think that one can derive a 4th degree polynomial by looking at your patterns
Sep
16
comment Proving a limit with epsilon delta definition
I don't understand what you mean by "f(x)=L+- ε 2=(1/2+-ε)x + 1 x= 3+- 2/ε". can you state in your post using the $\varepsilon$-$\delta$ notation what you have to prove?
Sep
16
revised Negation of a proposition of the form “not(p) & q”
format
Sep
16
reviewed Approve Finding example of a special type of continuous differentiable function
Sep
16
reviewed Approve Whether the sequence following is convergent?
Sep
15
reviewed Approve Finding which base number given operations
Sep
14
reviewed Reject Question about a passage in the Bicommutant Theorem's proof.
Sep
14
reviewed Approve Radius of curvature polar
Sep
14
reviewed Approve Counting Enumeration Problem
Sep
12
reviewed Edit condition for transitivity
Sep
12
revised condition for transitivity
removed an unneeded package; latexing
Sep
12
reviewed Approve Prove that $2|ab| \leq a^2 + b^2$ and $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$
Sep
12
revised Proof of special case of Fermat's Last Theorem
changing title, adding category. I hope the title is not too pompous. If so, then change it
Sep
12
answered Negation of a proposition of the form “not(p) & q”
Sep
12
comment Negation of a proposition of the form “not(p) & q”
yes, I think this is correct
Sep
12
comment Negation of a proposition of the form “not(p) & q”
If that what you expressed in English is "p & not(q)" then it is wrong because you have to express "p | not(q)". I think you have actually expressed "p & not(q)", so it is wrong.
Sep
12
comment Negation of a proposition of the form “not(p) & q”
not( not(p) & q) <=> (p | not(q)) or better $$\neg(\neg p \land q) \Leftrightarrow ( p \lor \neg q) $$