I could find $||S||=||T||=1$ I could Show that $S\circ T \neq T\circ S$. I could also show that $||S\circ T ||\leq 1$ and $||T\circ S||\leq 1$ I tried a lot to find a function $x(t): ||ST(x(t))||_{\infty}=1$. How do I find $||S\circ T ||$ and $||T\circ S ||$?
proof:Let c$\in$C. Note that since f is surjective, $\forall$ b$\in$B $\exists$a$\in$A such that f(a)=b. Similarily, since g is surjective $\exists$b $\in$ B : g(b)=c. So setting a$\in$A to be such that f(a)=b we have (g$\circ$f)(a)=g(f(a))=g(b)=c.
Next year I have the option of taking either Calc 3 at a local university (assuming I do well on the AP exam) or doing an independent study for math. I'm not sure which is more worthwhile for when I go to college. I know if I take calc 3, I'll be able to possibly get to knock out two math credits (semester classes, so I could take linear algebra or something as well next year). On the other hand, if I do an independent study, I could just learn calc 3 through your or Spivak's book.
Basically, I don't know what will help me more in the long run since I kind of have a head start.
the method by which Galois theory studies root of polynomials is to consider these permutations of the roots that respect the relation between the roots (that's how Galois thought about Galois theory)
Calc III is badly taught lots of places. Spivak won't get you Calc III. That's rigorous Calc I and II (and beginning analysis). My book is Calc III and beyond.
If you ultimately go to a place that has a really good math program, you might have options of doing something like my book/course. But you might not. If you do, it's worth waiting to get credit for doing it in your college.
@TedShifrin Thanks so much, I am trying to adopt my own style, and your videos are helping me with that. Another question I have is with regards to proofs of the form (if you don't mind me asking) $\forall$ $\exists$, for instance in epsilon delta proofs
my professor told me to always start with let $\epsilon>0$, but in some text books, like the james stewart (if i rememeber correctly), they start with some sort of note prior to the proof, at first it made sense to me that both are correct, but then I was told that I shouldn't start that way, but I haven't understood why
so if you take for example $x^2+1$ over $\Bbb R$, then you have the two roots $i$ and $-i$. The Galois group has two elements, the identity and the permuation which switches $i$ and $-i$. In the modern formulation, these correspond to the identity automorphism and the complex conjugation of $\Bbb C$
@TedShifrin is that because thats how the definition is contructed? or is there is some other argument as to why thats how the official proof should be?
with the field-automorphism approach is easily explained how these are symmetries: if $L$ is a field containing a subfield $K$, for example $\Bbb C \supset \Bbb R$, then we can look at the automorphism group of all ring isomorphisms $L \to L$ that restrict to the identity on $K$
these are symmetries because these are those bijections that preserve the structure that we have in this situation: addition, multiplication and the subfield $K$
in a generalized sense, structure-preserving bijections from an object to itself constitute symmetries
if $F$ is a field and $f$ is a polynomial over $f$, then one can construct a field that contains $F$ and all roots of $f$ (and is minimal wrt those properties), called the splitting field, then one can consider symmetries of the splitting field of $f$ over $F$
@SharathZotis There might be, but it's very tricky---lots of sums of different numbers might be the same. There's an easy upper bound of |A||B| though, and for small lists just writing it all out might be preferable.
(Or programming a thing to do that for you)
Like, {1,2}, {4,7} has 4 distinct sums, but {1,2}, {3,4} has three, since 2 + 3 = 1 + 4.
@Jacksoja so suppose we have $\Bbb Q$ and we have the polynomial $x^2-2$, then one can consider the subfield of $\Bbb C$ consisting of numbers of the form $\{a+b\sqrt{2} \mid a,b \in \Bbb Q\}$
the reason perhaps is that $\Bbb C$ always works: any polynomial with coefficients in $\Bbb Q$ (or even $\Bbb C$) has all its roots inside $\Bbb C$, that's the fundamental theorem of algebra
@Jacksoja right, but things can get more complicated
but I think this picture is good enough for now , I shall return with more questions to you guys once I start the lectures ! @MatheinBoulomenos @TedShifrin
so to add one more thing, the reason that x^2-2 is so simple is that one root automatically gives us the other one: $\sqrt{2}$ is one root and if we want to have a field, we also need to have the additive inverse $-\sqrt{2}$ in there, which is the other root
@TedShifrin there are still open conjectures about quadratic number fields
Let $k>1$ be an integer.
Consider equations of type
$$f_k(x+y) = f_k(x) + f_k(y) + f_k^k(x) f_k(y) + f_k^k(y) f_k(x) , f(-z) = f(z) $$
Valid for all real $x,y,z$ and $f$ is real-analytic.
Consider the nonconstant solutions.
When considering $f_2$ I came to the conclusion that $f_2(z) = 0 $ ...
A person decides to jump from a plank that is 1 meter high which is inclined at 30 degrees, to get to the other side he has to cross 50 meters, what should be his initial velocity
so like I seperated them into their x and y velocity vectors, I tried finding the initial velocity in the y first hoping I would be able find time using one of the four main equations, I tried finding the velocity using this formula $ Vf^2 = Vi^2 + 2*a*(xf-xi)$ the vf has to be zero when it touches the ground and the the change in x is just 1 so I thought I could be able to find the initial y velocity component but for some reason this didn't work, I just wanna know why this wouldn't work
Let $k>1$ be an integer.
Consider equations of type
$$f_k(x+y) = f_k(x) + f_k(y) + f_k^k(x) f_k(y) + f_k^k(y) f_k(x) , f(-z) = f(z) $$
Where $f_k$ is the $k$ th function and $f_k^k $ is the $k$ th power of $f_k$.
Valid for all real $x,y,z$ and $f$ is real-analytic.
Consider the nonconstant ...
I know about the probability of a sequence of graphs being contained in an Erdös-Renyi graph as subgraph when the index goes to infinity, but that's all I know about them.
Say is it legit to ask your help to get a gold badge? I mean, I believe I'm missing 1 up vote on my answer to this question math.stackexchange.com/questions/3039363/… in order to get the populist (gold) tag (Highest scoring answer that outscored an accepted answer with score of more than 10 by more than 2x). But you must not upvote the accepted answer.
Anonymous
@Yanko It says "outscored an accepted answer with score of more than 10 by more than 2x".
Anonymous
By the way, asking others to vote on your answer, in chat, is generally frowned upon. So, no, it's not "legit". However, if it's a one-time thing then it may be fine. Just don't make it a habit. :P
@Blue sorry I didn't realize more than means strictly more than. Also since it's frowned upon (and won't work anyway), do you know if it's possible to delete the message then?
Anonymous
@Yanko You (and other non-moderators) can only delete your own messages within a 2 minute window. Since that time is over, I could delete it on your behalf, if you wish. I don't think it's not necessary to delete it in this case though. Just don't repeat it in the future.
Let $k>1$ be an integer.
Consider equations of type
$$f_k(x+y) = f_k(x) + f_k(y) + f_k^k(x) f_k(y) + f_k^k(y) f_k(x) , f(-z) = f(z) $$
Valid for all real $x,y,z$ and $f$ is real-analytic.
Consider the nonconstant solutions.
When considering $f_2$ I came to the conclusion that $f_2(z) = 0 $ ...
