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11h
accepted Proving trefoil group is isomorphic to a fundamental group.
11h
comment Proving trefoil group is isomorphic to a fundamental group.
It think so, it can now be shown that $x_1 = b^2a^-1$ and that $x_2 = ab^{-1}$, so $a$, $b$ also generates the trefoil group, therefore we have an alternative presentation, and the two groups must be isomorphic.
12h
asked Proving trefoil group is isomorphic to a fundamental group.
Jan
13
revised integral problem - what is the quickest solution?
Placed equation in center of the page.
Jan
12
suggested approved edit on integral problem - what is the quickest solution?
Oct
31
comment Confused about basics of subsequences
You do realize this is a sequence and not a series? I once confused the two and I sense that maybe you are confusing the two...
Oct
29
accepted How can I manipulate this substitution of variables?
Oct
29
answered How to find $\lim_{x \to \infty} [x]/x$?
Oct
28
asked How can I manipulate this substitution of variables?
Oct
26
comment Mathematical texts: white background or tan
I don't really have a problem with white colored background. I would guess, that it's just easier to get a hold of white colored paper since it is "more standard." Not so much math but I feel like at least some of the paper writing formats (MLA, APA, Chicago etc.) require white as the color of choice. There would be less contrast between the black font and tan which may make it hard for people with poor vision to read. Just ask the professor. (This is ALL ENTIRELY speculative and a very unsupported guess, more of something to consider).
Oct
25
comment Give a regular expression that generates C.
This is off topic but I have never seen a language use (in particular) the #. Usually it is something like: /* and */
Oct
12
comment Neighbors of Irrational Numbers on Real Number Line
Maybe this can help convince you otherwise: wikiwand.com/en/Archimedean_property Instead of arguing about what infinitesimal neighbor 0 has, the Archimedean property indicates that over $\mathbb{R}$ there exists no infinitesimal objects in the first place. Hence, the "next neighbor" is neither rational nor irrational as it doesn't exist.
Oct
9
comment Relationship between perfect squares and infinite series (zeta function)
I don't believe there is any nice formula: wikiwand.com/en/Ap%C3%A9ry%27s_constant
Oct
1
awarded  Critic
Sep
29
comment Mathematical Symbols
Possible duplicate: xkcd.com/927
Sep
29
comment Help with verifying integral inequality.
@mercio: Plugging in the linear transformation $t$ gives an equality for the first statement. So then looking at the comment above you can use this fact to conclude: $\int_x^1 f(t) dt \geq \int_x^1 t dt$. Therefore: $$\left|\int_0^1 f(x)dx \right|^2 \geq \int_0^1 x dx \cdot \int_0^1 x dx$$ and similarly from the Cauchy-Schwarz inequality: $$\int_0^1 |f(x)|^2 dx = \int_0^1 x f(x) dx$$
Sep
29
comment Help with verifying integral inequality.
@Aravind: Sorry, but do you mind clarifying why we cannot conclude thusly?
Sep
29
comment Help with verifying integral inequality.
Ok, so if I understand this correctly, then: $$\int_x^1 [f(t) - t] dt \geq 0$$ $$\int_x^1 f(t) dt - \int_x^1 t dt \geq 0$$ $$\int_x^1 f(t) dt \geq \int_x^1 t dt$$ So from this: $$\left| \int_0^1 f(x) \right|^2 \geq \int_0^1 xf(x) dx$$ and from the above Cauchy-Schwarz inequality one can conclude: $$\int_0^1 |f(x)|^2 dx \geq \int_0^1 x f(x) dx$$
Sep
29
comment Help with verifying integral inequality.
Oh, I see silly me. The negative sign should flip the inequality. I still believe it results in the same final step though unless I am misunderstanding?
Sep
29
asked Help with verifying integral inequality.