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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
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
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
24
comment Dot product of two vectors with respect to their sum and difference.
Does this sound right?
Sep
24
comment Dot product of two vectors with respect to their sum and difference.
I think I see what to do... $(a + b) \cdot (a + b) = M^2$. So similarly for $(a - b) \cdot (a - b)$ and you get $N^2$. Then consider expanding both terms (like I do above with $(a + b) \cdot (a + b)$ and then solving the system of equations I get: $$\frac{(a+b) \cdot (a+b) - (a - b) \cdot (a - b)}{4}$$ $$\frac{M^2 - N^2}{4}$$
Aug
21
comment Intuition behind $(-\frac{1}{2})! = \sqrt{\pi}$
I'm definitely not sure, but this seems strongly related to this question: math.stackexchange.com/questions/3444/…
Aug
21
comment Intuition behind $(-\frac{1}{2})! = \sqrt{\pi}$
@YellowSkies: wikiwand.com/en/Gamma_function
May
27
comment Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$
$-1$ seems to work...
May
12
comment Quadratic solutions puzzle
(-1/2, -1/2) is a solution, but it is also said in the constraints that $a \neq b$.
May
11
comment Confused about the prime elements of a ring.
@BillDubuque: Ok, thanks! It is just hard to wrap my head around. A lot of things in abstract algebra are requiring some serious rewiring of my brain. sigh
May
11
comment Confused about the prime elements of a ring.
@BillDubuque: Ok, so then we can construct any numerator as a unique combination of primes (well known property of primes) and choose a specific bottom element as a particular unit so this would imply that R is a UFD then?
May
11
comment Confused about the prime elements of a ring.
@BillDubuque: When you say $2$ is a unit, are you referring to the $p \times 2^{-1}$ is unnecessary just as $-1$ is unnecessary in prime factorization?
May
10
comment Need help determining what this center is isomorphic to
Oh, whoops my bad, but $Z$ should have order $2$. That was me slipping up there, I didn't mean to write that, just mixed things around sorry about that. Thanks! (Modified original comment.)