Even little comments will be of great help to me. And I am very grateful for that. You can at least tell under the question whether I made a METHODOLOGY error or not. I understand everyone. Your time is precious. So at least you can make a methodological opinion.
Problem says:
Let $a,b>0$ and $2(a^2+b^2)-(a+b)=2ab$
Find the minimum of
$$\frac{a^3+2020}{b}+\frac{b^3+2020}{a}$$
The things I have done:
$$\begin{align}\frac{2ab+(a+b)}{2}≥2ab \end{align}$$
$$\begin{align} &\iff 2ab+(a+b)≥4ab \\
&\iff a+b≥2ab\end{align}$$
$$\begin{align}&2(a^2+b^2)-(a+b)=2a...
If a comment is made under the question, it is most likely that I will receive a notification. Usually there is no notification from here. I haven't figured out how this place works yet.
@leslietownes Thanks. I know this is not correct. But I need deeply understand this fact. I need "meat" of the error. Why this doesn't work..I need rigorous mathematical arguments.
google seems only recently to have begun answering wolfram-like requests to graph something. google.com/…
at least on my version of chrome.
you can fiddle with rotating the graph and the x, y, and z ranges. pretty cool although i wonder why they did it.
lonestudent, i don't have a good sense for constrained optimization problems of the sort in your question. the presence of 2020 suggests it is a contest problem, and while i bear no ill will toward contest problems, they sometimes prioritize being clever over using generally applicable techniques. which is not my vibe.
i realize that there are forums full of people who have worked these techniques down to something resembling a science, but i am not among those people. :)
i don't think it's irrelevant. the fact that the constraint is symmetric is probably the beginning of something you need for a solution to a constrained optimization problem involving a symmetric function to be guaranteed to have symmetry.
but nor am i going to work on it. that's a lot of math there and contest math is not my thing. other people may be more likely to dive into the deep end of that.
in a comment, someone's written the constraint in an interesting way. if you write g(a,b) = (a-b)^2 + (a - 1/2)^2 + (b - 1/2)^2 then the constraint is that g(a,b) = g(1,1). which is interesting because not only is g symmetric, but (1,1) is too, and that happens to be where f achieves an extremum.
there is probably a general result along these lines but i'm guessing that isn't what the contest would expect.
i still don't imagine that's how the problem is 'expected' to be done. you're supposed to apply AM-GM to something pulled out of thin air. that's usually how it works.
ha, it even has my example with citation. buniakovsky, 1854.
well, he traveled forward in time to steal from my intellect, and then back in time to cover his tracks.
the russians have always been sneaky.
i wonder if that example has just been passed down from book to book because he chose it. i certainly didn't come up with that idea on my own, although once you see it you remember it.
@TedShifrin it took me even longer (nearly all week) still to use the substitution method to solve $\int f'(x)\frac1{f(x)} dx$ despite having been given the answer. But I got there!
copper from what i've been able to gather, contest inequalities / optimization problems might not be as clean as "(1,1)" in terms of numbers, but will be clean in terms of, not drawing upon things that you couldn't get from a small number of standard inequalities
"the smallest positive root of [irreducible quartic]" is unlikely to be the answer to a contest math problem
I am asking is that can we say for a function f(z) to be 'entire function' if it is an analytic function where domain of analytic function is z which belongs to Complex numbers.
up above when you deduced x < a from x being positive and |x| < a, you're also assuming that x is positive. so x > - a also holds in that instance because x > 0 and 0 > -a. it's not just something that you can only deduce in the case that x is negative.
whereas if x > a, you can't deduce x < -a from that. that's maybe part of the vibe of why you get an and in one case and an or in the other.
also helps to draw a picture. i don't know where you are in all of this, whether tihs is being done from order axioms or something else.
i can see how that would be confusing. i don't think it's helpful to think of a function as the same thing as an implication (or set of implications) that might characterize the rule of the function.
even the idea of a 'piecewise' function as a separate class of function, not too helpful. useful in constructing examples, but not really a property of a function.
$\sqrt{x^2}$ probably isn't a piecewise function. $\frac{2}{\pi} \int_0^{\infty} \frac{x^2}{x^2 + t^2} \, dt$ isn't certainly isn't a piecewise function and doesn't even have square roots in it. but $|x|$ is. oh wait they're all the same.
@user1591719 The number of interesting questions that are "acceptable" these days is very thin.
There are some interesting questions, but since they are asked in a "context-free" manner, one cannot answer.
PSQs (problem statement questions) are discouraged, and if someone has no idea what to do, it is hard to include much context.
So helping people who are in the most need of help is discouraged because there are many who have abused the system to get homework questions answered here.
I did come across a nice way to compute $\int_0^{\pi/2}\log^a(\sin(x))\log^b(\cos(x))\,\mathrm{d}x$
@robjohn It should deserve many upvotes (I'm sure that many creative ways - speaking in general - can be obtained by reducing everything to simple differential equations).
Maybe did you get the inspiration from some work on differential equations?
If f is continuous only [a,b] then its continuous extension g exists that $g(x)=f(x)$ for $x\in [a,b]$ and that g is continuous everywhere on $\mathbb R$
That’s fine
But if we have $(a,b)$ is there instead of $ [a,b]$ then this need not be true
I disagree with this totally because if $f$ is continuous on $(a,b)$ then clearly we can define its continuous extension as $g(x)=f(x)$ for $x\in (a,b)$ and $g(x)=f(a+)$ for $x\le a$ and $g(x)=f(b-)$ for $x\ge b$.
Ahh I confused continuity of f on (a,b) with “ end points limits should exists”
Also I have one more question: if f is continuous in one set S with metric function $d_1$, then is it true that f is continuous on S under metric function $d_2$ also ?
Background of this question is like this :-
Suppose that E is compact then $f$ is continuous on $E$ if and only if $\{(x,f(x)): x\in E\}$ is compact.
@Koro Consider on any set $X$ the metric given by $d(x,y)=1$ if $x\neq y$ and $d(x,y)=0$ if $x=y$. Can you tell me which functions on $X$ are continuous with respect to that metric and do you see how that answers your question?
I observe that this set on Right side is subset of $E\times f(E)$ and that $f(E)$ is compact by continuity of $f$. There may be metric function on $E\times f(E)$ which makes it a metric space.
And then I plan to prove the function $g(x)=(x,f(x))$ on E continuous proving 1st half of the theorem. @Thorgott I’ll respond to you also.
:57996021: Thorgott, I think that limit at any point in this metric will not exist and that’s because for every $\epsilon \gt 0$ I should have a $\delta \gt 0$ such that $d(x,a)\lt \delta$ such that ….
That answers my question on continuity on a set under different metrics. The example of this metric which is inherent to every set shows that continuity need not remain as we change metric. Thanks a lot @robjohn @Thorgott
For a function to be continuous at any $a\in X$: for $\epsilon \gt 0.02$ (in particular) there exists a $\delta =1.2$ such that $d(x,a)\lt \delta\implies d(f(x),f(a))\lt0.02$ which is possible only when $d(f(x),f(a))=0$ that is $f(x)=f(a)$.
@Thorgott
True for every $x$ in X because of the way we chose delta
@Koro This delta doesn't work, but continuity says that there exist some delta, so you cannot pick one delta that doesn't work and conclude that the function is discontinuous
Every point in X is an interior point why? Because for every x in X, I can find an r (by taking it less or equal to 1) such that $B(x,r)=\{t\in X: d(t,x)\lt r\}\subset X$
Sir @robjohn: My apologies if I caused confusion. I responded to Thorgott question wherein he introduced X as a metric space which separates unlike points by 1
I use this definition for disconnectedness: set S is connected if it can not be written as union of two non empty separated sets A and B. By separated I mean: closure of A is disjoint with B and closure of B is disjoint with A.
Continuous image of connected is connected, but that does not imply that continuous image of a disconnected space is disconnected. In fact every compact metric space is a continuous image of the totally disconnected Cantor set
@robjohn: I referred to the link of totally disconnected space (I never knew it before) but I have read separated sets and not connected sets (disconnected word was not used). So since there is no connected subset, f would vacuously…
I have a conundrum. If $a,b,c\in\{0,1\}\subset\mathbb{Z}$, can you find a closed expression using addition, subtraction, max, and min such that if $a=1$, the expression returns $min(b,c)$ and if $a=0$, the expression returns $max(b,c)$? I feel like I've done something similar to this, before, but it is evading me
Let S be any set in X then for any $s\in S$ take $r\le 1$ and then consider $B(s,r)=\{t\in X: d(t,s)\lt r\} =\{s\}$ is open in X and $S=\cup_{s \in S \} \{s \}$ and hence S is open in X because union of open sets in open @Thorgott
(Not sure if I used the term "closed expression," correctly, tbh, but hopefully what I mean is clear: A single mathematical expression using a finite number of $+$, $-$, $max$, and $min$)
Got it, I think: $min(max(1-a,b),max(1-a,c),max(b,c))$
Yeah, that'll work (I translated them into booleans and then worked out a truth table, and then converted them back to max/mins (so I wasn't entirely sure at a first glance))
@Thorgott: That would mean that every function pulls back an open set that is inverse image of every function is open
thereby making every function continuous
Am I correct?
For example: we take the set $S=\{a,b\ }\subset X$ and define $f(a)=c, f(b)=d$ so that $f(S)=\{c,d\ }$, every inverse image of $f$ here is open and that makes $f$ continuous. :)
On Pg. 33 here, why is it enough to show that vol(\partial\tilde K) \le 2n vol(\tilde K)^{(n-1)/n}? The author hasn't provided any argument for it. library.msri.org/books/Book31/files/ball.pdf
Even in the first sentence after the statement of Theorem 6.2, the author says it is enough to prove vol(K) \le 2^n (which comes from the previous fact)
Quick question: Assume a function h(.,.) defined on R^n*R^m, and for any given y, h(.,y) is a convex function of x. Define f(x)=max_{y\in C} h(x,y) where C is a convex set. is the max function here nondecreasing? meaning f(x) is convex?
draw a bunch of random lines, and then color in the region that lies beneath all of them. it's important to color. i learned this from my daughter, who likes crayons.
that's tomorrow's activity. i'm keeping my hip healthy by only doing that once a week.
i also carry my daughter to and from the mailbox and hold her while she opens it (she takes a while because it has a key). upper body strength is important too.
Problem says:
Let $a,b>0$ and $2(a^2+b^2)-(a+b)=2ab$
Find the minimum of
$$\frac{a^3+2020}{b}+\frac{b^3+2020}{a}$$
The things I have done:
$$\begin{align}\frac{2ab+(a+b)}{2}≥2ab \end{align}$$
$$\begin{align} &\iff 2ab+(a+b)≥4ab \\
&\iff a+b≥2ab\end{align}$$
$$\begin{align}&2(a^2+b^2)-(a+b)=2a...
