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Graduate student trying to learn something.


1h
reviewed Approve suggested edit on Zero divisor and Field
1h
comment Convergence of series (putnam training)
Here you go math.stackexchange.com/questions/264980/…
1h
revised Is there an injective continuous map $\mathbb{R}^2 \rightarrow \mathbb{R}$?
edited body
1h
revised Is there an injective continuous map $\mathbb{R}^2 \rightarrow \mathbb{R}$?
added 2 characters in body
1h
revised Is there an injective continuous map $\mathbb{R}^2 \rightarrow \mathbb{R}$?
added 172 characters in body
1h
answered Is there an injective continuous map $\mathbb{R}^2 \rightarrow \mathbb{R}$?
1h
answered Univalent function and one-to-one function
2h
reviewed Approve suggested edit on Problem on Linear Regression
8h
reviewed Approve suggested edit on I just don't see what I do wrong - number of surjections seems higher than number of functions.
14h
accepted Why is $[\widetilde{v},\widetilde{w}]_p(f)=0$ when $f$ has a critical point at $p$?
15h
comment Why is $[\widetilde{v},\widetilde{w}]_p(f)=0$ when $f$ has a critical point at $p$?
Ah, I see. So I guess this should have been obvious - basically we are saying that differentiating $f$ in the direction of $[\widetilde{v},\widetilde{w}]_p$ at $p$ is zero because the total derivative of $f$ is zero. This actually crossed by mind but then I got confused about why $\widetilde{w}(f)$ wouldn't just be zero and hence $\widetilde{v}_p\widetilde{w}(f)$. But I guess this is because $f$ has a critical point only at $p$ so $\widetilde{w}(f)$ is not actually the zero function, it is only zero at $p$. So differentiating gives something nonzero in general. Thanks!
17h
revised Why is $[\widetilde{v},\widetilde{w}]_p(f)=0$ when $f$ has a critical point at $p$?
edited title
17h
asked Why is $[\widetilde{v},\widetilde{w}]_p(f)=0$ when $f$ has a critical point at $p$?
17h
reviewed Approve suggested edit on finding maximum likelihood estimators
17h
reviewed Approve suggested edit on A doubt in the rigorous definition of limits.
1d
reviewed Approve suggested edit on Let $f(x)=x^2+bx+4$ in $\mathbb{R}[x]$. For each $b \in \mathbb{R}$, factor $f(x)$ into a product of irreducible polynomials in $\mathbb{R}[x]$.
1d
answered What's the relation between prime spectrum and affine space?
1d
comment Show that in Z/2Z[x] two polynomials are associates if and only if they are equal.
Yes, thats what I was thinking.
1d
comment Show that in Z/2Z[x] two polynomials are associates if and only if they are equal.
Does associate just mean differ by multiplication by a unit? If so, just think about what the units are in Z/2Z[x]
1d
comment Proof that the coefficients of a polynomial are real
@gniourf_gniourf exactly... I thought of the complex conjugation trick right away and then I realized I wasn't sure why it would solve the problem. I wondered if evaluation at all real numbers are real implied the coefficients were real (since that solves the problem) and then saw this answer. Hence my comments above.