I have three sets A, B, C. h: A -> C and g: B -> C. what necessary and sufficient condition on h and g guarantees that there's a function f: A -> B such that h(a) = g(f(a)) for all a in A?
so I'm tinkering with the peano axioms at the moment
here's my problem: prove proposition 2.2.5 (Hint: fix two of the variables and induct on the third.) Proposition 2.2.5: (Addition is associative). For any natural numbers a, b, c, we have (a + b) + c = a + (b + c).
I'm way out of practice with little proofs like this.
what does it mean to fix two of the variables and induct on the third?
something like, fix a and b, and use the induction principle to prove "for all c in N, (a+b)+c = a + (b+c)." you could fix another two but i chose a and b.
I can think of this way though: Let's define a sequence $c_n$ such that $c_{2k-1}=a_{2k-1}$ and $c_{2k}=b_{2k}$ then limit the only only limit of $c_n$ is $\infty$
professor Robjohn, can you please guide me to a source where I can find error estimate on trapezoidal law? I ask because using that, one can prove Stirling's formula.
Suppose we want to show that $$ n! \sim \sqrt{2 \pi} n^{n+(1/2)}e^{-n}$$
Instead we could show that $$\lim_{n \to \infty} \frac{n!}{n^{n+(1/2)}e^{-n}} = C$$ where $C$ is a constant. Maybe $C = \sqrt{2 \pi}$.
What is a good way of doing this? Could we use L'Hopital's Rule? Or maybe take the log...
If $g(a)=g(b)=0$, then integrating by parts twice gives $$ \begin{align} \int_a^bg(x)\,\mathrm{d}x &=-\int_a^b\left(x-\frac{a+b}2\right)g'(x)\,\mathrm{d}x\\ &=\int_a^b\frac{(x-a)(x-b)}2g''(x)\,\mathrm{d}x\\ &=g''(\xi)\int_a^b\frac{(x-a)(x-b)}2\,\mathrm{d}x\\ &=-\frac{(b-a)^3}{12}g''(\xi)\tag1 \end{align} $$ Apply $(1)$ to $g(x)=f(x)-\frac{f(a)(b-x)+f(b)(x-a)}{b-a}$ $$ \int_a^b\left(f(x)-\frac{f(a)(b-x)+f(b)(x-a)}{b-a}\right)\mathrm{d}x=\int_a^bf(x)\,\mathrm{d}x-\frac{f(a)+f(b)}{2}(b-a)\tag2 $$
the last part of $(1)$ is the intermediate value theorem
:) By Rolle's, I was referring to how $f$ was introduced (Geometrically speaking $g$ is difference between graph of f and straight line connecting $(a,f(a))$ and $(b,f(b))$) and this is how Lagrange's theorem is also proved :) For MVT, yes I understood that $(1)$ is consequence of MVT for integrals and continuity of $g''(x)$ is in hypothesis for that to be applicable. :)
@Semiclassical I just changed it to x= 2^(-x+3) and noticed x=2 has to be the only solution as if x>2, RHS is always <2, LHS is always >2, vice versa for x<2
I don't know if that's cheating but inspection was what I could think of
"It is the view of the Ministry that a theoretical knowledge will be sufficient to get you through your examinations, which after all, is what school is all about." -Dolores Umbridge
We were the first or second round of 'new' math, I believe. I think it was rather good and enjoyed it more than the classical geometry (we did that too, but less emphasis).
@copper.hat Interesting. I always loved math for the artistry and the aesthetics of proofs. I like "black and white" computations, but I certainly do not love them.
@Koro This is one of the exercises I actually added to Spivak in the third edition. Did you find it?
@TedShifrin professor Ted, yes I found it. The exercise problem is in parts-One is in I think Integration chapter or in sequence chapter wherein $\frac{n!^\frac 1n} n \to \frac 1e$ is to be proven. And this is followed by a note which says to look up chapter 27 (on complex power series) and last three problems in chapter develop Stirling in a series of exercises. I'm glad that those exercise have been incorporated there. I found them yesterday. I've not yet solved that exercise though. :)
It's in integration in elementary terms chapter. I knew those methods when I studied numerical methods (a course I took at my college) but that didn't have any error term, which I discussed today with professor Rob. I'll try that problem now. Thanks a lot professor Ted :)
What makes Simpson's rule amazing is the following neat fact: If $f$ is a cubic polynomial on $[a,b]$ and you take the quadratic $Q$ agreeing with it at $a, (a+b)/2, b$, then $\int_a^b Q = \int_a^b f$, so you get an extra order of accuracy for free.
@copper I actually taught myself calculus in high school and then, after I took the AP test, my teacher (who didn't know what he was doing) had me teach the class for the remaining month and a half or so. That was interesting.
The principal came and observed me one day. I think he was serious when he said to let him know if I needed a job :P
So Ted sorry to bother you again, if my knot is a circumference what I'm thinking in my example is to remove the same thing constructed with a cilinder? Is it an ok analogy?
At least this is the idea that comes to me drawing
In $\mathbb R^+$, $s=x+\frac{1}{2}(y-x)$ implies $x<s<y$ if $x<y$ and $y<s<x$ if $y<x$. Is there some sort of algebraic field where $s=x+\frac{1}{2}(y-x)$ implies the above for $\mathbb R$?
@TedShifrin I'll do that exercise. I think I'd skipped Integration chapter's exercises in Spivak as I thought I knew integration. But now that I see the exercises of the chapter in Spivak, I see lot of new things like -Wallis formula etc. I'll do exercise problems of this chapter also now. :)
So what I'm doing it's just considering this? I can use the red line for the Seifert circle?
At least they intersect once I guess..
Or meridional slope is just the whole circle? I'm not sure about the wording "slope"
english is not my mother tangue and I've always encountered slope as in a line or a mountain, I would not know what use I should make of meridional slope
is the set of real coordinate pairs with $y<-x<x<-y$ any sort of well-known structure?
it makes a whole class of real analysis proofs easier
it has the property that $s=x+(1/2)(y-x)$ implies $y<s<x$, so it conserves "in-betweness", which is not the case for an arbitrary choice of positive x and negative y
shin, your s is going to be between x and y for any unequal x and y. if y > x, this would be expressed by y > s > x instead of y < s < x, but that's still an example of what most people would allow as 'betweenness'
the signs of x and y don't really impact this conclusion. you could also use ps + (1-p)y for any p satisfying 0 < p < 1 (not just 1/2) and get the same result
i'm looking for the property that reduces two-case proofs to one-case proofs with a single chosen in-between number $s$. the $y<-x<x<-y$ property guarantees that you can't choose, for instance, $x=3, y=-1$, which does not make $s$ be in-between $x,y$ (but does make another choice of number be in-between, just not the same)
i'm checking your ps + (1-p)y
oh, is this a property of convexity in general, then?
hm, you're right that requiring instead $y>x$ is still in-betweeness
neat formula by the way, seems to work for any choice of x,y with 0<p<1
i'm not immediately seeing where this would come up enough to simplify anything, to be honest. but one reason people like 'open set' proofs in elementary R^1 analysis is that they avoid explicit reference to order and hence any case analysis involving order
even though the underlying topology is of course an order topology
you can write inequality-based proofs that avoid case analysis too, but sometimes it's not the first thing that comes to mind
yeah, the proof that there's a cauchy sequence of rational numbers between any two cauchy sequence of rational numbers, in-betweeness defined as, there exists some $N$ after which order is preserved
@leslietownes hm, but interleaving two separate equivalence classes of cauchy sequences doesn't make a converging sequence, no?
interleaving equivalent cauchy sequences would. it's just a vague memory i had of something someone else was talking about, i don't know if it would address whatever issues you are addressing
Then if $(x_n)<(y_n)$ you can find $N$ so that $x_k<y_k$ for $k>N$ and define $(z_n)$ to be whatever up to $N$ and $1/2(x_n+y_n)$ afterward to get a sequence in between $(x_n)$ and $(y_n)$
@AlessandroCodenotti you gotta use absolute values in that case, no? what was bothering me was precisely the fact that negatives and positives force at least a second case (i might be wrong though)
@AlessandroCodenotti i was trying to identify precisely what exactly would make the argument the same up to renaming the sequences, without needing to worry about the behaviour of crossing $0$ on order relations
yeah, the analytic translation, i.e., exact definition of the middle number is what makes it difficult, i think. consider defining the middle number as $s_n = x_n + \frac{1}{2}(y_n - x_n)$
that definition fails if we do not have $y < -x < x < y$, or x,y with same signs
LOL ... well, it sorta explains why Screech was starting to act a little differently ... presumably getting ready to go into heat soon. Just what I don't need. Olivia and Munchkin are evidently cut from the same cloth.
Can we compute time complexities for computations done in non-abelian groups ?
Suppose $H \rtimes_{\phi} K$ is a semidirect product of two finite groups, say $H$ and $K$. For an element $(h,k) \in H \rtimes_{\phi} K$, the inverse is $(\phi_k^{-1}(h^{-1}, k^{-1})$. If a person chooses any element $(h,k) \in H \rtimes_{\phi} K$ and compute its inverse, how can I explain about the time complexity of that computation?
@TedShifrin Thanks for the reply. I thought that because the question says "find the set of points", so I wasn't sure if geometric descriptions counted.
Would this work as a solution? $\{z\in \mathbb{C}: z \in \mathrm{Circle\ centered\ at\ (-2,2i)\ with\ radius\ \sqrt8}\ \cap z \notin \mathrm{Circle\ centered\ at\ (2,2i)\ with\ radius\ \sqrt8}\}$
I think you need to have a conversation with your adviser or a topologist about this, @SMC. It really isn't amenable to chatroom help beyond what I've said.
Suppose we want to show that $$ n! \sim \sqrt{2 \pi} n^{n+(1/2)}e^{-n}$$
Instead we could show that $$\lim_{n \to \infty} \frac{n!}{n^{n+(1/2)}e^{-n}} = C$$ where $C$ is a constant. Maybe $C = \sqrt{2 \pi}$.
What is a good way of doing this? Could we use L'Hopital's Rule? Or maybe take the log...
