@robjohn Ok, so are you suggesting if my answer was $x=\pm e^cy$, instead of $x=e^cy,$ then my solution is valid? If so, please consider adding your comment as an answer. This because, Narasimham's answer which I accepted, is then a faulty one, where in the comments section of his answer, he asserted my initial form of the answer is valid, which is indeed misleading as it seems. What do you think ? For convenience, I am posting you the link of the thread.
Find the solution of the differential equation $\frac{xdy-ydx}{x^2+y^2}=0.$
I though this was quite an obvious differential equation. Hence, I proceeded to solve like this:
Given, $\frac{xdy-ydx}{x^2+y^2}=0.$ We can write this as, $$\frac{xdy}{x^2+y^2}=\frac{ydx}{x^2+y^2}\implies xdy=ydx\implies...
Find the solution of the differential equation $xdx+ydy+\frac{xdy-ydx}{x^2+y^2}=0.$
My solution goes like this:
Given, $xdx+ydy+\frac{xdy-ydx}{x^2+y^2}=0.$ We assume $M=x$ and $N=y.$ Now, $$\frac{\partial M}{\partial y}=0=\frac{\partial N}{\partial x}=0,$$ and hence,$xdx+ydy=0$ is an exact diffe...
@TedShifrin The solution I presented in this approach is also an erroneous approach of solving the given differential equation, to begin with due to the same reason you mentioned, here
Solve the differential equation $(2x^3y+4x^3-12xy^2+3y^2-xe^y+e^{2x})dy+(12x^2y+2xy^2+4x^3-4y^3+2ye^{2x}-e^y)dx=0.$
I tried solving the problem like this:
We assume $M=(12x^2y+2xy^2+4x^3-4y^3+2ye^{2x}-e^y)$ and $N=(2x^3y+4x^3-12xy^2+3y^2-xe^y+e^{2x}).$ Now, we observe, $\frac{\partial M}{\parti...
Is it literally just a relabelling? As in what we called "$f$" or "an arrow pointing from $X$ to $Y$" is now relabeled to $f^\text{op}$ or "an arrow pointing from $Y$ to $X$"?
I am confused because nLab says that the collections of objects and morphisms are isomorphic not equal whereas the book by Riehl suggests equality
so this situation reminds me more of say the dual correspondance between bras and kets in quantum mechanics where the objects are not equal but they do provide the same information
Why do schools still teach ode? Especially solving some explicit ode problems. Outside ode, people satisfy with the existence and uniqueness of ode problem and even pde people don't seek explicit solutions (physics people do). They abstract everything and do their theories. Whenever they encounter ode, use computer to solve it.
Find the solution of the differential equation $xdx+ydy+\frac{xdy-ydx}{x^2+y^2}=0.$
My solution goes like this:
Given, $xdx+ydy+\frac{xdy-ydx}{x^2+y^2}=0.$ We assume $M=x$ and $N=y.$ Now, $$\frac{\partial M}{\partial y}=0=\frac{\partial N}{\partial x}=0,$$ and hence,$xdx+ydy=0$ is an exact diffe...
mm, 'dual' is already pretty overloaded. it might make more sense to ask why we don't call more things 'opposite' things, to take the load off of 'dual.'
@TedShifrin Ok. That clears it up. They have taught me just the procedure, to identify, "whether a differential equation of 1st degree and 1st order is exact or not?" and the method, for finding the "solution of an exact differential equation of 1st degree and 1st order" and this problem was an excercise to be dealt with these things.
@TedShifrin Ok, if that post, is fine as you claim, then the thing is: If the terms of a differential equation can be divided such that, we have two groups of terms, and equating each group of terms to zero gives us an exact differential equation
( say, the given differential equation is $Mdx+Ndy=0$ and if we can write this as $Pdx+Qdy+Adx+Bdy=0$ such that, $Pdx+Qdy=0$ and $Adx+Bdy=0$ are both exact differential equations) then we can solve the two equations separately to get respective solutions, $p(x,y)=c$ (on solving for $Pdx+Qdy=0$ ) and $g(x,y)=c$ (on solving $Adx+Bdy=0$ ) and claim, that $p(x,y)+g(x,y)=c$ to be a solution of $Mdx+Qdy=0$.
Is this implication, that I concluded above the right conclusion, @TedShifrin ? This is what I wanted to know to be very elaborate.
@FdstZfsy In principle every ODE $P\,dx+Q\,dy=0$ with either $P$ or $Q$ nonzero everywhere has at least a local integrating factor in the plane, but generally impossible to find.
Well, it may be a stupid solution. Remember my $dx-dy=0$ example. Work it out. No, your sum is not in general a valid solution. Just the intersection of the two solutions.
@TedShifrin Well, I was asking about math.stackexchange.com/questions/4689974/…. Are you talking about the one, you answered. If so, then I admit, there was a sign error.
( say, the given differential equation is $Mdx+Ndy=0$ and if we can write this as $Pdx+Qdy+Adx+Bdy=0$ such that, $Pdx+Qdy=0$ and $Adx+Bdy=0$ are both exact differential equations) then we can solve the two equations separately to get respective solutions, $p(x,y)=c$ (on solving for $Pdx+Qdy=0$ ) and $g(x,y)=c$ (on solving $Adx+Bdy=0$ ) and claim, that $p(x,y)+g(x,y)=c$ to be a solution of $Mdx+Qdy=0$.
This line of comments was in reference to the post,
Find the solution of the differential equation $xdx+ydy+\frac{xdy-ydx}{x^2+y^2}=0.$
My solution goes like this:
Given, $xdx+ydy+\frac{xdy-ydx}{x^2+y^2}=0.$ We assume $M=x$ and $N=y.$ Now, $$\frac{\partial M}{\partial y}=0=\frac{\partial N}{\partial x}=0,$$ and hence,$xdx+ydy=0$ is an exact diffe...
The sum of solutions is not a solution in general, no. Did you work out the simple case for $dx-dy=0$. You would get the sum of $x=c_1$ and $y=c_2$, and $x+y=c$ is not a solution,
yeah, why would it be a typo? its a basic appeal to continuity of the adjoint in the norm topology (and presumably P_n is self adjoint as ted suggests)
the result is also true for more general banach spaces (where the banach space adjoint plays the role of T*) although you cannot use the implication 'compact' => 'is a norm limit of finite rank operators' in general. in fact, i think it's an if and only if even if the banach space is not reflexive.
there's stuff very close to 'is a norm limit of finite rank operators' that holds in general, but its definitely a weaker condition than that, in general.
onepotato: if you want a more general approach (with an if and only if, for banach spaces) see arxiv.org/abs/1010.1298
the 'lemma 1' there is playing the banach role of being a norm limit of finite rank operators.
ted there was actually a bit of movement in what people wanted compact operators to be, for a while. of course it's all blended up together in the hilbert space case.
it was "maps weakly convergent sequences to norm convergent ones" for a while. that's equivalent to compactness / norm approximability via finite rank operators in the hilbert space case, but not in general.
any book or article before 1960 or so is going to use different words, or use familiar words given an unfamiliar meaning.
more seriously i realize it is something of a point of annoyance for you, so i'll avoid it. any material in the wrong hands can be annoying, and it sounds like your program is full of the wrong hands.
If I am teaching for example Banach Alaouglo theorem: I'll put emphasize on "compactness of closed unit ball in a normed space". closed unit ball in a normed space is compact iff the space is finite dimensional.
But in dual of X, the unit ball is compact in weak* topo.
Instead of just writing: Theorem: [...] Proof: [...]
I understood its importance through self study and not from class. In the class, it was just written on the board in the above format, that's it.
Richard Hell and the Voidoids were an American punk rock band, formed in New York City in 1976 and fronted by Richard Hell, a former member of the Neon Boys, Television and the Heartbreakers.
== History ==
Kentucky-born Richard Meyers moved to New York City after dropping out of high school in 1966, aspiring to become a poet. He and his best friend from high school, Tom Miller, founded the rock band the Neon Boys which became Television in 1973. The pair adopted stage names; Miller called himself Verlaine after Paul Verlaine, a French poet he admired, and Meyers became Richard Hell because, as...
Suppose $f:U\subset\Bbb C\to\Bbb C$ is a holomorphic function with $f(z_0) = z_0$ for some $z_0\in U$. If $f'(z_0) = \mu$, $0\leq |\mu|<1$ then I can find a small nbd $V$ of $z_0$ such that $f$ is contraction on $V$: $|f(z)-f(z_0)|\leq C|z-z_0|$ for $C<1$ on $z\in V$.
In numerical analysis, the Lagrange interpolating polynomial is the unique polynomial of lowest degree that interpolates a given set of data.
Given a data set of coordinate pairs
(
x
j
,
y
j
)
{\displaystyle (x_{j},y_{j})}
with
0
≤
j
≤
k
,
{\displaystyle 0\leq j\leq k,}
the
x...
The question was: E is Banach space, F is proper subspace of E. Dual of E, E* is strictly convex then given any bounded linear f in F*, it extends uniquely to a norm preserving map in E*.
Simple Lagrange polys are fine for interpolation, but they usually go crazy if you need to extrapolate outside the region of the points you used to create the poly. Admittedly, Chebyshev & Remez minimax polys aren't well-behaved outside the target interval, either.
No worries. It can be fun looking for good approximations, but it can also be a bit frustrating. ;) Analysis can guide the search, but there's also a bit of luck, guesswork, and experience involved.
@Jakobian My attempt: Given a completely regular space X, if X has unique compactification, its Stone-Čech compactification coincides with its one-point compactification. Since the Stone-Čech compactification is Hausdorff, the one-point compactification is Hausdorff also. We conclude X was locally compact Hausdorff to begin with.
I am trying to understand the proof that if $f$ is continuous on $[a,b]$, then it is bounded on $[a,b]$. The following is a proof by Spivak in Calculus:
> Choose $x_n$ with $f(x_n)>n$. There is a subsequence $x_{n_j}$ which converges to a point $x$, which is in $[a,b]$. Thus for every $\epsilon>0$ there are infinitely many $x_{n_j}$with $|x-x_{n_j}|<\epsilon$, and consequently $f$ is unbounded on $[x-\epsilon,x+\epsilon]$, contradicting the fact that $f$ is continuous at $x$.
I do not understand the part "...and consequently $f$ is unbounded on $[x-\epsilon,x+\epsilon]$...". If someone could clarify, I'd be very grateful.
I'm trying to prove that, for a differentiable function $f:(a,b)\to\mathbb{R}$ such that $f'(x) \ne 0$ in $(a,b)$ and such that $\lim_{x \to a^+} f(x)=+\infty$, there exists a right neighborhood of $a$ in which $f$ is decreasing. I tried this: since $f'(x) \ne 0$, for each $x \in (a,b)$ it is $f'(x)>0$ or $f'(x)<0$. Assuming $f'(x)>0$ in $(a,a+r)$ for some $r>0$, we have that $f$ is increasing in $(a,a+r)$ and so $f(x)<f(a+r)$ for each $x\in (a,a+r)$.
But $f(x) \to +\infty$ as $x \to a^+$, so there exists $R>0$ such that $f(x)>f(a+r)$ for each $x \in (a,a+R)$. So, for each $x \in \left(a,a+\min\{r,R\}\right)$ it is $f(x)>f(a+r)$ and $f(x)<f(a+r)$, a contradiction. So, it must be $f'(x)<0$ in $(a,a+r)$, that is $f$ is decreasing in a right neighborhood of $a$.
Is this proof valid?
A correction: in the last part, I meant: "$f(x)>f\left(a+\min\{r,R\}\right)$ and $f(x)<f\left(a+\min\{r,R\}\right)$, a contradiction"
consider the group $\mathrm{SO}(3)$ of all rotations of a sphere $S^2.$ Let $x$ be the north pole $(0,0,1).$ Then a rotation which does not change $x$ must turn about the usual axis, leaving the north pole and the south pole fixed. These rotations correspond to the action of the circle group $S^1$ on the equator. So are $x$ and its antipodal point $(0,0,-1)$ stabilizers?
@TedShifrin you're just gonna have to wait until I actually become familiar with it to a reasonable degree. I even made a comment here that "unlearning" the concept of angle is a bit tricky
There's a story I remember hearing/reading, but I'm not sure where. Essentially the idea is that a professor gave a task to some of his graduate students. To make a program that, in general, when fend an image, could identify whether or not a book was in the image. Then three of the graduate students got their PhDs just trying to pose the problem. Does this ring a bell for anyone?
Re: $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$ - You can reduce to the case of $\sqrt{3}\notin\mathbb{Z}[\sqrt{2}]$ by saying that $\sqrt{3}=\frac{a}{b}+\frac{c\sqrt{2}}{d}+\frac{e}{f\sqrt{2}}$, and then clear denominators to get an equality that uses only integers, $\sqrt{2}$ and $\sqrt{3}$
Is it possible to approximate $frac{\cos(n)}{\log(n)+\cos(n)}\sim \frac{\cos(n)}{\log(n)}-\frac{\cos^2(n)}{\log^2(n)}$ to conclude the series $\sum\limits_{n\geq 1} \dfrac{\cos(n)}{\ln(n)+\cos(n)}$ diverges, or is this approximation not good enough?
There are 5 holes on a wall and a total of 17 pigeons have to be kept in them. Which of the following options(choices) are correct and can be the best suitable arrangement? a) Every filled/occupied hole has a minimum of 3 pigeons b) Every filled /occupied hole has at least 2 pigeons but not more than 4 pigeons c) Every filled/occupied hole has not less than 2 pigeons d) All of the above e) The filled/occupied holes have not less than 4 pigeons
The answer key says it's the second option (b) but I really cant wrap my head around why the first option (a) is wrong
@TedShifrin I think "best suitable" tries to imply that one should attempt to make it least crowded crowded for the pigeons but indeed the language confusing
My thought was to use that $frac{\cos(n)}{\log(n)+\cos(n)}\sim \frac{\cos(n)}{\log(n)}-\frac{\cos^2(n)}{\log^2(n)}$, since $\lim_{n\to\infty} \frac{frac{\cos(n)}{\log(n)+\cos(n)}}{\frac{\cos(n)}{\log(n)}-\frac{\cos^2(n)}{\log^2(n)}}=1$. So we can compare and since $\sum\limits_{n\geq 2} \dfrac{\cos(n)}{\log(n)}$ converges and $\sum\limits_{n\geq 2} \dfrac{\cos^2(n)}{\log^2(n)}$ diverges, the original series diverges also.
I was just asking, if my conclusion is right, if I can prove this above. For convergence for the series see for example there: math.stackexchange.com/questions/744125/… The limit I checked with WA. If the conclusion is alright, I will try to prove the limit "by hand".
Ah, funny that I even answered that one. You just packed way too much in here for those of us not thinking about it. Why the divergence? Removing $\cos n$ from your limit, it certainly looks wrong. @DavidP
Now I understand why $Q(2^{1/4})$ is field isom. to $Q(i2^{1/4})$
In one of my exams, I wrote since the former has no element x with the property that $x^2<0$, they are not isom.
They are isom. because of the following: If $\sigma: K\to L$ is a field isom. $f$ is irred. over K. Suppose it has a root a. Suppose $\sigma (f)$ has a root b. Then, $K(a)$ is isom. to $L(b)$.
In the above, take $K=L=Q,\sigma= id, f(x)= x^4-2$
I was not removing the $\cos(n)$, I wanted compare with (asymptotic equivalent sequence) and write (formally) $\sum\limits_{n\geq 1} \frac{\cos(n)}{\log(n)+\cos(n)} \approx \sum\limits_{n\geq 2} \frac{\cos(n)}{\log(n)}-\sum\limits_{n\geq 2} \frac{\cos^2(n)}{\log^2(n)}$ and since on the right hand side the first series converges and the second diverges, so does the left hand side. Is that a valid conclusion?
@TedShifrin Well, $\mathbb{Q}(\sqrt{2})$ is a field (parentheses rather than square brackets), meaning that it contains $\frac{1}{\sqrt{2}}$, which isn't in $\mathbb{Q}[\sqrt{2}]$, so yeah, I think you do need it. As for how to clear denominators, just multiply the whole thing by $bdf\sqrt{2}$. You get $n\sqrt{6}=k+m\sqrt{2}$ with $k,m,n\in\mathbb{Z}$, after the correct replacements.
So, it's more $n\sqrt{6}\notin\mathbb{Z}[\sqrt{2}]$, so not the exact same statement, but you can apply the same approaches
@DavidP I cannot keep up with all the bits and pieces. Is limit comparison valid for non-positive series? I still don’t see the limit comparison as valid, anyhow.
Well, what is a rectangle in 3 dimensions? Do you mean a 2D rectangle embedded into 3-space? Is this meant to be the boundary of a cuboid? The edge-frame of a cuboid?
@冥王Hades Well, again, depends on your definition. A cuboid is generally the object and all points interior to it. Do you mean the edges that define the cuboid? (Also, for that matter, a rectangle also generally includes the interior, and I kinda assumed you meant the edges of the rectangle, only)
Also, question: What about the unit square $Bd([0,1]^2)$ with the line being $x=0$? Wouldn't that have infinitely many intersections?
I got what you were saying initially, with the line only intersecting a quadrilateral at two points (after all, I corrected you with convexity), but, thinking about it, you need the line to be through a point interior to the quadrilateral
@Rithaniel I mean, you aren't wrong, but I'm quite sure the question doesn't care about those "edge cases" no pun intended. It probably just means a plane that intersects a cuboid to form another 2D shape when those points are joined
@Rithaniel sure but that "interior point" won't define the shape that forms
Pick two opposite vertices on the cube. Consider the two collections of three points that are a singular edge away from one of those two points. Construct the two planes that are through those three points, respectively. These are distinct planes, and any plane fully between those two planes will intersect six edges of the cube
What I would look into is whether a tesseract-frame can be intersected by a 3D subspace at 20 points (because then that might imply that you're looking at something of the form $\binom{2(d-1)}{d-1}$, where $d$ is the dimension of the parent space)
Actually, you might be able to get 24, which would imply that the formula is just $d!$
is there a mathematical way to express a bound on the number of terms of a polynomial? say I am looking for the best polynomial approximation to a transcendental function s.t. the polynomial has at most 3 terms
but say, is there an analytic/algebraic expression of this space?
or an analytic/algebraic expression of the space of such polynomials, polynomials made of at most 3 monomials, and we're looking inside of it to find the best approximation
it's clearly something smaller than the polynomials in a ring of polynomials
for any fixed trio of monomials you at least have a vector space (its span), and you can use that structure in approximation. depending on what 'best approximation' means it might not be that easy to compute, but at least e.g. hilbert space methods would work. i don't know how you'd minimize over all trios of monomials. that's not a vector space, anyway.
it's also not particularly simple to single out from the point of view of an abstract vector space, where the monomials are just one of many possible bases. they diagonalize various operators that act on polynomials, but i don't know how you distiguish "three mononomials" from "three things in the span of an eigenspace of something" without more.
theres something kind of related, which is finding the best approximation to an operator (not a vector) in terms of operators of finite rank. "what's the closest thing to [this] that has rank at most 3" is a question that sometimes has an answer, for some notions of "closer." but i don't see an immediate relation of that theory to this, for the reason that "rank 3" or "spanned by 3 vectors" is not capturing "spanned by 3 mononomials."
its pretty easy to come up with optimization problems that don't have theoretically simple solutions, and this might be one of them.
