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23h
accepted Bound on function increment
Jul
24
comment Dimensional reduction of system of ODEs
Not sure if this helps, but recall that $f(t)\to 0$ does not imply $f'(t)\to 0$ (as $t\to \infty$).
Jul
24
revised Bound on function increment
added 74 characters in body
Jul
23
comment Bounded sequence and every convergent subsequence converges to L
Just a wording suggestion: writing $\lim_{n\to\infty}x_n\neq L$ implies that a limit esists. A more rigorous wording could be "$\lim_{n\to\infty}x_n$ does not exist or it is different from $L$". In either case, your proof works.
Jul
23
asked Bound on function increment
Jul
22
comment $\int_a^b f(x)g(x)dx = \sum \int_a^b f_n(x)g(x)dx.$
Let us continue this discussion in chat.
Jul
22
comment $\int_a^b f(x)g(x)dx = \sum \int_a^b f_n(x)g(x)dx.$
"You last voted on this answer Jul 18 at 5:07. Your vote is now locked in unless this answer is edited."
Jul
22
comment $\int_a^b f(x)g(x)dx = \sum \int_a^b f_n(x)g(x)dx.$
Sure, I understand your point. And now I also suspect that that is the case, although it would still be a good exercise to prove it in the Lebesgue setting (even without dominated convergence). Anyhow, it won't let me take the downvote away now, for some reason. It says my vote is locked. Sorry.
Jul
22
comment $\int_a^b f(x)g(x)dx = \sum \int_a^b f_n(x)g(x)dx.$
It did not mention in the question that the integrals were in the Riemann sense. The tag 'real-analysis' made me think about Lebesgue integral. And in that setting, integrable does not imply bounded.
Jul
18
comment $\int_a^b f(x)g(x)dx = \sum \int_a^b f_n(x)g(x)dx.$
Why does integrable imply bounded? $f(x)=(x-a)^{-1/2}$ is integrable and unbounded.
Jul
18
comment $\int_a^b f(x)g(x)dx = \sum \int_a^b f_n(x)g(x)dx.$
Hint: the sum of a series is the limit of partial sums. What could you say if the sum on the right hand side was truncated to N?
Jul
12
comment Super Simple Proof of Cofactor Expansion
When you say to prove that expression, what "definition" of the determinant can you use?
Jul
12
comment Accumulation points of sets
Note: 0 is outside of $(0,1)$...
Jul
10
answered How to re-write the following equation?
Jul
9
comment Finding an explicit formula from a recursive formula.
This does not look like a linear algebra question... By the way, have you tried using properties of the log on the first expressiong of $g$? It looks like you get something similar to what you have in your update, but simpler. I don't think you can get much more, since you are shuffling exponentials and rational functions...
Jul
2
answered Explanation of $\overline{\lim} A_n$ and $\underline{\lim}A_n$
Jun
17
awarded  Yearling
Apr
19
awarded  Notable Question
Mar
3
answered Operator norm with $\inf$
Feb
20
comment Is $\lim_{x\to 0} (x)$ different from $dx$
Joseph, how do you define an infinitesimal? How do you define arithmetic operations on those? Now, do these definitions fit with the usual real numbers definitions? This should answer your question.