@blue I must be missing how to do it since it wasn't spelled out for me; surely it's trivial if $(U\vee V)\wedge(U\vee W)$, but I wasn't sure how it applied when $(U\vee V)\wedge(\neg U\vee W)$
after one expansion, I now have for the left side $\forall x, ((x\in A\wedge x \notin B) \vee (x\in B))\wedge ((x\in A\wedge x \notin B)\vee (x \notin A))$
@Hello any idea why it's called the "trapezoid" rule? do you know how to compute the area of a trapezoid using the left and right side lengths A and B, and the base length C?
riemann sums are about accumulating areas, after all
@blue oh I see you can now make an argument that each of those statements in parenthesis is going to end up being equal to their right hand side counterparts
if you look at the graph, usually this doesn't make sense (there may be no derivative at 1, so you can't evaluate), but left and right derivatives make sense so we would technically need to use those if we want to be rigorous.
when you first play an area in a video game, you don't know the map or where you need to go necessarily, in which case one must go exploring. students must apply this same idea to everything - we must explore if we don't already know where to go.
I don't see how any set of distributive expansions of $((P\wedge \neg Q)\vee Q)\wedge ((P\wedge \neg Q)\vee \neg P)$ could produce a set that can be made like the set of distributive expansions on the right
we want $f$ to be continuous at $1$; $n1^3-1$ is the limit of $f$ as $x$ approaches $1$ from the left, and $m1^2+5$ is the limit of $f$ as $x$ approaches $1$ from the right, and we need these to be equal if $f$ is to be continuous at $1$
math problems are sometimes like sudokus. you see a chance to fill out new information, you take that chance instead of wasting it. here, we have a ripe opportunity to solve for m. which turns out to give us the solution, because we can plug the value for m in to get n
don't waste opportunities to deduce new information
@rehband Well, I try to do some generalizations of some old limits and then to continue the work on some of the Ramanujan's results that I want to generalize.
@blue I am most tempted to give a Galois theoretic interpretation of the manipulations in polynomial factorizations/root finidings when people claim that these tricks develop by a lot of practice. For example, some guy reduced a general-looking quartic to a quartic with no linear and cubic term via a Tschrinhausen transformation and some other guy claims that this is just an accident.
They just don't realize that it happens iff the extension has no cubic element, i.e., order of the galois group is < 24 but not divisible by 3. i.e., either $8$ or $4$.
@skullpatrol I cannot afford that, I need to work hard on my results, I plan to publish an exceptional book, so I try to work as much as I can, less rest, less party. You know, to do such things it's hard since when you're about to create something big, you lose the support of the people you think would support you in such situations.
a chatroom might be good enough to figure out how to do some problems, but it won't help you understand what you're doing or why it works (and that also makes it harder to remember the methods)
the sum over $j$ is tends to about $n\log(n/2)^2$ and the sum over $i$ to $\frac{n}{\log(n/2)}$ so the whole thing tends to $n^2\log(n/2)\sim n^2\log(n)$
@Chris'ssis Yes, but the stupid thing is I am having trouble working out $\mathrm{Li}_3\left(\frac12\right)$. I could just copy it from a table, but I want to work it out.
@Alizter I'm creating some series. Did you see this one? $$\lim_{n\to\infty}\frac{1}{n^2 \log(n)}\sum_{i=2}^{n-2}\sum_{j=2}^{n-2} \frac{\log(j)\log(n-j)}{\sqrt{\log(i) \log(n-i)}}$$
@Alizter hmmm, maybe not now, but a bit later after getting some mor experience. However, it's worth looking at it a bit. Such limits I'll have in my book.
haha, okay. it just came up in a solution to a question i'm working on, and i'd like to include it in my notes. i guess i can include it in my exponents section... are there other similar formulas that i should know, in this "family"?
@topper: If you look at the expression on the left hand side as a function $f(x) = 1-x^4$, sometimes you can immediately see some of its zeroes (in this case $f(1) = 0$ and thus $x=1$ is one if its zeroes). Then, you can divide by $(x-1)$ using polynomial division and get the right hand side.
@Huy thanks, to be honest it's a bit over my head right now. i'm more at the level of learning common formulas, not proving them - that's for real mathematicians. but thanks for the info :)
@topper: It really doesn't have anything with a proof or something but it is a concept that can be used in a lot of situations. Instead of trying to learn formulas by heart, you should try to understand when and how you can apply this concept. Where I live, it is being taught to pupils in high school at age 14. A lot of people will be overwhelmed at first but if you actually spend some time trying to figure out what's happening, you'll grasp it eventually, I promise.
@Huy Noted, seriously. I wholeheartedly admit I'm trying to do the minimum possible - not to bore you but I'm about to do a second degree in business, and because my first degree in computer science was more than a certain time ago, I'm forced to do prerequisite courses in maths, statistics, and economics. So there isn't really time to get too much into the first principles in this case (as much as I'm aware how bad that may sound)
@topper: I have a lot of friends who do not care about any more mathematical background than they really need to know and I respect that and then exactly explain to them how much they really need to know to pass their tests. In your case, I would tell you to study polynomial division - if you think you will encounter more similar problems. It doesn't require any further insight.
@Huy Thanks, it's a bit touch and go this time - real life got in the way of my preparations, but there's always another chance, it's just that if I don't pass this exam, I can't take certain modules in the MBA until I do pass it. Not the end of the world.
The nice part is that $$\lim_{n\to\infty}\frac{1}{n^2 \log(n)}\sum_{i=2}^{n-2}\sum_{j=2}^{n-2} \frac{\log(j)\log(n-j)}{\sqrt{\log(i) \log(n-i)}} $$ can be entirely computed with the high school tools, in a very elegant way.
you're talking about hatcher's book, and about homotopic maps from (X,Y) to, say, (A,B). yes, in this context, every $f_t$ must still send Y to B; that is, $Y \times [0,1]$ must still be sent to B
@MikeMiller you are right , but i find an other prove where the homotopy P is just between i and i' where as f and g are homotopic there is a homotopy h such that $f\circ h=i $ and $g=h\circ i'$
so $P$ is defined from C_n(X)\rightarrow C_n(X\times [0,1])
getting stuck into second derivatives and their implications. i understand that whether the s.d. is greater than or less than zero determines whether we have a local maximum or minimum, but this has to be checked with the table method to determine whether the point is an inflection point? but how can a maximum or minimum not be an inflection? is this when the point we are checking has not already been determined to be a maximum or minimum (by equating the first derivative to zero)?
ah, i think i can answer my own confusion (or make it worse). in $y=x^2)$, we have a minimum. but it's not an inflection i.e. the graph is always "smiley"
sorry, i'll try not to think out loud in here again
@VibhavPant I didn't start writing the book, but I have all the material needed (some thousands of questions). Many of them are still on paper, they are pretty old ...
Let $S$ be a nonempty set of positive nonzero integers such that if $x$ is in $S$ then so is $4x$ and $\lfloor \sqrt{x} \rfloor$. Prove that $S = \Bbb N \setminus \{0\}$
