two questions @TedShifrin : 1) I optimized the volume of the box in a sphere, but your question asked for the hemisphere. All my dimensions are $\frac{r}{\sqrt{3}}$. I know the hemisphere is half, but how do I capture the idea in the expression $x^2 + y^2 + z^2 = r^2$?
2) I'm working on your P set question that asks to show a point $a$ exists in the ball $B(\mathbf{0}, 1)$ with $\nabla f(a) = \mathbf{0}$. Based on $f: B(\mathbf{0}, \frac{5}{4}) \to \mathbb{R}$ is $\mathcal{C}^1$, $\nabla f(x) \cdot x > 0$ for all $x$ with $\|x\| = 1$.
I think I got the solution. In essence our function is continuous over the smaller ball contained in the larger ball. Now all the points of norm $1$ make this smaller ball a compact set, so I can invoke the maximum value theorem to claim an extremum exists. but I'm not sure how $\nabla f(x) \cdot x > 0$ fits into the conclusion.....
@dc3rd Your answer is wrong. So how is the function different for the hemisphere and the sphere? Your extremum on the closed ball doesn’t necessarily give vanishing gradient. Why?
@TedShifrin WHich portion of calc/algebra should I go review? cause clearly there is a gap there for something which is most likely obvious and I will eat a shoe over after reviewing the said material...
$\forall \epsilon>0$ $\exists \delta >0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$
I have been thinking for a while if my reasoning is correct or not.
$\forall \epsilon>0$ $\exists \delta >0$ such that $0<|x-a|<\delta$ whenever $|f(x)-L|<\epsilon$
I have been thinking what will happen if I reverse the implication here. So one counter example I came up with is $\sin(x)$. I think reverse implication definition implies that as epsilon gets smaller and smaller then there will be a lot of points it intersect that will get smaller and smaller I mean deltas. Which doesn't work. So…
Intuitively speaking, if $u= x^a, v'= a^x$ then I'll have a term of the form $a/log a$ somewhere and that does not $\to 0$ as $a\to \infty$.
Using series, I get $x^a a^x=x^a (1+\sum_{i=1}^\infty \frac {(x\log a)^i}{i!})$. And integrating both sides from 0 to 1, I get: the integral $I(a)=1/(a+1)+\sum_{i=1}^\infty \frac{1}{(a+i+1)i!}(\log a)^i$
the plan of attack is to follow the proof of an approximate identity
using the series, you can show that for any c<1, the integral from 0 to c goes to 0
then you need to show that the integral from c to 1 tends to 1. for this i think you can add and subtract something nice, maybe (a+1)x^a because this integrates to 1
yeah with a JD after a PhD in math...I'm only on the Bachelors.........and I have the "ambition" to do a masters.....ha....can't even frame an objective function right......pardon this decending into projection....
my people are very eclectic with education. my dad got an M Ed. on the gi bill, my mom had no college. my sister has 'some college' but not the degree. my best friend has almost every associates degree you can get. anything capable of being conferred online, she has. and a BA in womens studies.
akiva: that's why, then. you slacked off. i hope that semester was worth it
@CalvinKhor yes, that's what I meant. I'll elaborate: f is continuous at 1, so given any $\epsilon\gt 0$, there is a c such that $|f(x)-f(1)|<\epsilon$ for all x in (c,1).
so I'll need to add and subtract f(1), where f is integrand.
which is controlled by $$ (1-\epsilon)^{a+1}\frac1{a+1}\sum_{k\ge0} \frac{((1-\epsilon) \log a)^k}{k!} = (1-\epsilon)^{a+1 } \frac{a^{1-\epsilon}}{a+1} $$
OK..... i am seeing an issue.
@Koro idk, try refreshing?
or use a different page. guess there is a \renewcommand{\le}{d} somewhere on that page.
Or install latex
ah, the earlier decomposition is not quite right
ok let me come back when ive worked it out properly lmao
@Semiclassical Thanks. I read that Benford's law is not reliable at some point. For context, check the current state of the Philippine elections. Thanks!
Does the following graph modification have an established name in graph theory?
Consider two edges in some graph G={V,E}, say {u,v} and {v,w} in E. Delete {u,v} and {v,w} from E and add a new edge {u,w} to E. The vertex set is left unmodified.
I am reading through Kajiya - Ray Tracing Volume Densities paper. And I've already got stuck into section 2. I wonder if there's a mistake in that equation.
I'll quote the relevant bit
The quantity to be calculated in a scattering problem is the energy per unit solid angle per unit area
$$
dE = ...
@CalvinKhor I have evaluated the limit of the integral, but I don't want to spoil your fun. I am walking the dog. If you wish, I can post it when I get back.
@robjohn what devil came up with that integral! Haven't thought about it much more since. I might give up eventually but you can share it with koro first (presumably its not a one-liner so i won't understand it without effort anyway)
@robjohn There's a fast algorithm for calculating arcsin. It's one of a family of algorithms for inverse circular & hyperbolic functions, and the natural logarithm. They use a modified AGM with Richardson extrapolation. Carlson (1972). An Algorithm for Computing Logarithms and Arctangents doi.org/10.1090/S0025-5718-1972-0307438-2
I guess those algos aren't great for hand calculation, due to the square roots. OTOH, you don't need many square roots to get good precision, and most of the operations are just simple additions & subtractions, and bit shifting, with a few multiplications and one long division at the final step. I just used it to calculate pi to 10 decimals (using a calculator). I only needed 5 square roots.
Coz I mean, now it doesn't make much sense to do Abbott after having almost finished Spivak...So i will do baby rudin, then apostal and see whatever is out there.
leslie: If my child ever started to throw tantrums like my parents over this stuff, i'll just hand over the kid to my wife and say "That's a you problem. Goodnight."
prithu: yeah. i think sometimes people get sort of paralyzed by the number of choices. it can be harder to find the 'right' book than it can be to read a good-but-not-great book.
I saw a video in which a leopard showed up in front of a dog and the dog got scared and started running away. Now the question is: why did the dog feel the danger? The dog shouldn't have felt the danger as it had never seen a leopard before.
pretty funny. i disagree with some of it. i hate the idea that it's some kind of acid test for whether you are a mathematician or not. and all the implicit praise in 'it's not a book that holds your hand, but it's good for you.' rudin's style is occasionally impenetrable.
it's like when people say a work of literary fiction is good because they can't easily read it, and they think good means difficult.
i would have criticized chapters 9+ more harshly than the reviewer, but they got it. we agree there.
akiva: are we still on that? it was a cool problem, just: still on it?
My teacher gave this definition of greatest common divisor: "Let $a,b \in \mathbb{Z}$ not both zero. Then $d \in \mathbb{Z}_{\ge 1}$ such that (i) $d|a$ and $d|b$; (ii) for each $d \in \mathbb{Z}$ such that $d'|a$ and $d'|b$, $d'|d$; is called greatest common divisor of $a$ and $b$."
My question is: is (ii) equivalent to say $[(d' \in \mathbb{Z}) \wedge (d'|a \wedge d'|b)] \implies d'|d$?
prithu: this isn't directly evident from the darboux definition of riemann integrability, but one thing you want of the riemann integral is that the limit of any family of riemann sums to be the same across any family of partitions whose maximum interval size goes to 0. that will break if you don't require that condition.
e.g. choosing rationals vs. irrationals inside the points of your partitions should not ultimately affect the limiting value.
i dunno, a lot of methods of approximate integration don't directly reflect the darboux definition. i would rather teach the darboux definition. and i would not prove that the darboux definition is equivalent to the riemann definition in class. i think it's a good fact to know, however.
so you can use it to evaluate series. "this is a family of riemann sums corresponding to partition [ ] of interval [ ] so the limit is [ ]." comes in handy.
work is so boring today. i hate when i get stuff i have to respond to before 8am.
In the Spivak course I needed Riemann to justify things like cylindrical shells, so at that point I did a few minutes and proved the equivalence. And for standard calculus classes, I have never messed with Darboux.
darboux takes the point realizing the inf or sup in every subinterval. riemann allows arbitrary choices.
that's the key, for approximate integration. entirely arbitrary choices.
you get the same dumb broken integral in the end, so i don't know why it matters. i am a nihilist. i believe in nothing.
apparently if you have amazon combine multiple shipments into one, it not only increases the shipping time, but increases the likelihood of your package being stolen. i'm at three packages in a row now where i've chosen that option and it vanishes off of a truck before delivery.
i'm never choosing that option again.
and yes, i would like my AA batteries shipped to me in their own cardboard box, with insulation to make sure they aren't damaged.
In terms of pedagogy, I decided years ago not to do $\sup\{L(f,P)}$ versus $\inf\{U(f,P)\}$. The definition I gave (and used in my multivariable book) is that there should be a unique number $I$ satisfying $L(f,P)\le I\le U(f,P)$ for all $P$. I think students process that more easily, and it leads immediately to the "convenient criterion" anyhow.
there are some scenes around fort mason, i think, or maybe elsewhere in the presidio. and in what used to be a sculpture garden in the east bay before you go over to the bay bridge.
i've not heard of waffle.
ted: does getting older ever get more fun? i'm in a text chain with my best friend about a biopsy. i hate this.
a small opossum got stuck in a rat trap in the garage a few nights ago at 2am. after a fashion i managed to release it, but when closing the garage door it came off the rollers and it took another hour just to close the door. no fun. anyway, the garage guy just left so we're good again.
@Nick It depends on how the tax is assessed, but I would presume that one pays tax only on the taxable purchase price, which should not include other taxes.
For example, in the US, you might have both state and local sales taxes. So if you buy a \$1 candy bar, and there are 8% state taxes and 1.5% local taxes, the total cost of the candy bar is \$1.095 (which rounds down to \$1.09), i.e. $\$1(0.09 + 0.015)$.
However, I believe that VAT works a little differently, in that every time a good is sold along the way, a VAT is paid, so the tax on raw materials gets passed along in the purchase price to the factory which makes a finished good, and so on.
@XanderHenderson ok, agreed that the local won't apply over the state. To wrap my head around the other one, when you say "[VAT is] passed along", is it like there is a tax on the raw material which is paid in parts by checkpoints within the supply chain until the remainder is picked up the retail client. Is that close to reality?
I don't think there could be an argument against it honestly. Because if the problem is not set up right with the right language and ideas, then you can't use the bevy of "tools" one has learned....I was taken by a moment of passion. Now I'm working on finding the existence of the vanishing gradient at the point $a$.
