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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 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. Sep 26 answered A little confusion with $\int_{1}^{∞}\frac1x$ dx = $\ln|x|$ Sep 24 accepted Dot product of two vectors with respect to their sum and difference. 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: $$\frac{(a+b) \cdot (a+b) - (a - b) \cdot (a - b)}{4}$$ $$\frac{M^2 - N^2}{4}$$ Sep 24 revised Dot product of two vectors with respect to their sum and difference. edited title