It's odd when an OP says "I didn't realize that the upper semicircle would indeed produce a greater surface area than the lower one until I read your comment" but no voting ensues.
@copper.hat I'm not tall (only 5'10") but my leg-torso ratio is such that when I sit I am as tall as my dad who was 6'1". That does interfere with getting into cars sometimes.
Bro, my mind had a literal eruption just now. Clicked on this random bprp video in my recs and said, "area under the tangent line", and it clicked in my mind. We can use $\frac{d^2}{d^2 x} f(x)$ to compute the integral of $f(x)$.
Much easier to work with. We just need to know how the area changes over time and then the conjunction of the area under two tangents over time, and then we can take the limit as those tangents approach another tangent for every point on $f'(x)$ to then get an integral, $f(x)$.
Yeah, I realized that after the fact. Nevertheless, yes, I'm aware any given $f(x)$ has nothing to do with that particular triangle. I'm attempting to generalize to any triangle defined by $f'(x)$ and $f''(x)$.
Where the three vertices are the origin, and the two intersection points of the tangent line.
Besides, I got a better idea anyways
We create a recursive formula that takes the area of the previous triangle, then computes the area of the current triangle, then subtracts their conjunction. If the formula has a known explicit formula, then we have something that at least resembles $f(x)$.
Ideally we have for the recursive formula $a(n)$ that $a(0)$ is the actual area under the curve of $f'(x)$ between two values of $x$ if not an approximation so that we can take a limit and have the true $f(x)$.
So, e.g. we have already computed in closed form the integral $a(0) = \int_0^{\epsilon} f'(t) dt$ or an approximation. We then construct the triangle as detailed above for some $x > \epsilon$ to get an area; compute the conjunction of $a(0)$ and this newly constructed triangle and subtract it from the sum of the two areas to get an approximation of $\int_0^{x} f'(t) dt$. That's what I meant to say.
Having any sort of weighting difference where practice mutually affects overall grade whatsoever undermines the purpose of tests and examinations in the first place which exist... to test and examine a person's competence in the subject according to some ideally objective and well-designed metric.
At that point, you may as well just equate the two and say that tests and exams are just homework that weighs more.
My stats class has weekly hw due on Friday, and I feel it's frustrating to have to rush through it without a chance to go through the textbook chapter and do the problems there first.
I try to do it simultaneously, but there just isn't enough time. I end up having to finish the chapter over the weekend.
I see... so at least over here, elementary, middle, and high school spoonfeed us information, and then university slaps you with the responsibility of learning to find information by throwing you a textbook heavy enough to kill a man.
There is a well-known inequality relating the arithmetic and geometric mean. It follows from this that if the arithmetic mean is less than a certain number, then the geometric mean is also less than this number.
My question is: is it also possible, with some sum, to estimate a lower bound for the geometric mean?
I mean $(abc)^{\frac{1}{n}}≥?$. And this question should also be the sum of the elements.
I determined that this could be the following: $(abc)^{\frac{1}{n}}≥\frac{a+b+c}{n+1}$ but this is only on condition that $a>>1,b>>1,c>>1$
@dtn this cannot scale properly. increase $a,b,c$ to $\lambda a,\lambda b,\lambda c$ then $(abc)^{1/n}$ increases to $\lambda^{3/n}(abc)^{1/n}$ while $\frac{a+b+c}{n+1}$ increases to $\lambda\frac{a+b+c}{n+1}$
The geometric mean can be bounded below by the harmonic mean
Hello, if I have two real power series $\sum_{n=0}^\infty a_n x^n$ and $\sum_{n=0}^\infty b_n x^n$ that both converge and agree in the open interval $(1,2)$, which does not contain their center $0$, does it follow that $a_n=b_n$ for all $n$?
But the identity theorem requires that I have a sequence $(x_n)_n$ with $x_n \neq x_0$ and $x_n \to x_0$ where both series agree, where $x_0$ is the center, doesn't it?
there is a theorem which says you can center a series at any point of convergence and it will converge to the nearest end of the interval of convergence of the original series
Suppose that $A=\{a_1,a_2,\cdots, a_n\}$. Then by definition of anti-symmetric relation(let's call it R), if $a_i R a_j$ and $a_j R a_i$ then $a_i R a_i$.
Given any $(a_i,a_j)\in A\times A$ and if R is anti-symmetric relation, we have three cases: 1. $a_i R a_j$ and $a_j Ra_i$ so that $a_i Ra_i$ 2. $a_i R a_j$ and $(a_j, a_i)\notin R$. 3. $a_j Ra_i$ and $(a_i, a_j)\notin R$ 4. $(a_i,a_j)\notin R$ and $(a_j,a_i)\notin R$
you don't have to learn all the tools of general topology from the start, a lot of universities make you learn about metric spaces first, so you get more intuition for how things work, at least in theory
Let the function $K:\mathbb{R}^d \to \mathbb{R}$. Let $\rho$ be a probability density on $\mathbb{R}^d$ with finite second moment. What conditions can I possibly put on $K$ to get the following to hold ?
@robjohn The MathJax issue has been fixed in most places. It's still present in a few lesser important places, like when you view the list of all your actions.
I think there's no way to find the shape of base of a polyhedron.
except guessing and drawing a lot of pictures.
For example: for a dodecahedron, one will start with a triangle and try make 12 faces and see if they get polyhedron with that or not. If not, then start will be with square and then from a pentagon and so on until that polyhedron is found.
Sure. or either. or neither. And there are other correct definitions, as well. Which definition is most "correct" will depend on what you want to do with those definitions. Without further context, it is impossible know which definition is appropriate.
@CroCo It may also be worth pointing out that a definition cannot be "correct" or "incorrect". A definition just tells us how we are going to use a word or notation. Definitions can be more or less useful, but we can define words to mean whatever we want them to mean.
@Koro Well, there are only 5 regular polyhedra. That's pretty easy to show. Of course, there are lots of semi-regular ones, eg en.wikipedia.org/wiki/Archimedean_solid
@XanderHenderson but it seems to me the second definition may break the radical concept but the first has no problem because this is the way we define it. I may be wrong.
Also, defining $i$ by the property that $i^2 = -1$ could cause problems of its own, as that equation has two solutions. One then has to make a choice about which of those solutions is actually $i$ (though the choice is irrelevant... something something automorphisms something something field extension something something).
@Koro Every edge is shared by 2 faces. And in a regular polyhedron, k faces meet at a vertex for some k>=3. And the angle sum at a vertex must be <360°, so there aren't a lot of options. 3, 4, 5 triangles, 3 squares, 3 pentagons.
@Koro Sort of. You can expand that equation V-E+F=2. Let n be the number of edges on a face, and k as above. Then E = Fn/2 and V=Fn/k. You can simplify that to give an equation of n in terms of F and k, and then try various k until you get an integer for n.
Here's a semiregular solid called the small stellated dodecahedron. You can think of it as a dodecahedron with a pentagonal pyramid stuck to each face. Or as 12 pentagonal stars stuck together.
In a perfectly regular solid , all faces are congruent regular polygons. And each vertex looks like any other vertex. In contrast, the solid in my avatar image, the rhombic dodecahedron, the faces are rhombuses, and all the vertices aren't the same. Some have 3 edges, some have 4.
There's an entire section in my algebra book on symmetry groups of the regular polyhedra. (And in another section the classification based on group actions.)
@Croco $\sqrt{-1}$ is not well-defined.
So you have to say $\pm i = \sqrt{-1}$, and then we don't know which is which.
@Odestheory12 Exactly. And, quoting Asaf's very find answer: "The question whether something has a root or not must include a setting. Definitions do not appear magically, they require some preexisting framework."
What is with people these days? I get it that once the OP corrected his post, it looked like I was insisting that $\setminus$ be $-$. I don't give a flip. Is it true that only those of us over 60 write $A-B$ instead of $A\setminus B$?
I don't object to the notation, @Jakobian. I'm just pointing out that $\sqrt$ need not be a function, and is not working over any field other than $\Bbb R$. The field extension doesn't care which root you adjoin, of course.
Lads, I'm trying to prove the continuous mapping theorem - $X_n \rightarrow{\text{p}} X \implies h(X_n) \rightarrow{\text{p}} h(X)$ for a continuous function $h$. My idea was to just use the fact that $h$ is continuous to bound $|X_n - X| < \delta$ and then use $|h(X_n) - h(X)| < \epsilon$ to complete the proof and show convergence in probability. Is this right?
I wonder if $\setminus$ really is an ageist thing. In my mathematical upbringing, I believe I only saw $A\setminus B$ in the context of quotient objects although $B/A$ is more common for that; it depends on who's acting on which side.
@TedShifrin No, the comments have been flagged as "no longer necessary". It looks like your correspondent deleted their comments, then flagged your responses as no longer necessary, as the inciting comments are gone.
For what it is worth, in the area I work in $A - B$ is the Minkowski difference, i.e. $$ A-B = \{ a-b : a\in A, b\in B\}. $$ Hence I use $A \setminus B$ for the relative complement.
But I have zero problem with $A - B$ as long as there is no chance of ambiguity (and there is none in the question being asked).
Munkres's Topology book still uses $-$, not $\setminus$.
Yes, I have used left and right quotients throughout my career. Because of Chern, my principal bundles seem to have the group action on the opposite side from almost everyone else. :)
So I do have to write $H\backslash G$ instead of $G/H$. Sigh.
But the spacing is all wrong with setminus.
So I write \backslash.
Anyhow, I agree that that algebraist is pissing me off and calls me rude for saying he was "ridiculous."
So $X_n$ and $X$ are functions on our probability space, mapping to $\Bbb R$? And $h$ is a function on the space of random variables. Surely your course/textbook has made these things precise.
Why is it that if I take an integral like $$\int_{0}^l sin(\frac{n\pi x}{l})sin(\frac{m\pi x}{l})\mathrm dx$$ and I calculate the integral, I get a formula which is undefined at $n=m$ for $n,m\in \Bbb N$, but if I set $n=m$ to be equal integers in the integral itself I get a finite value?
I feel like in this case, since the integral is equal to whatever it evaluates when you work it out, you end up with two equivalent $f(n,m)$, one of which is undefined at $n=m$ and another which is not
I feel like this relates to the Dirac delta function
