@Jakobian I have to ask, though I'm not expecting a reply, why does the author state that the map $E\mapsto E_1$ commutes with unions, intersections and complements and hence $\sigma(E)=n\cdot m(E_1)$ is a Borel measure on $S^{n-1}$? Does it not suffice to simply say that it commutes with unions? What's the relevance of including intersections and complements?
since the map $E\mapsto E_1$ commutes with unions, we have for a pairwise disjoint sequence of sets $\{U_j\}$ that $$\sigma\left(\bigcup_1^\infty U_j\right)=n\cdot m\left(\bigcup_1^\infty (U_j)_1\right).$$Now, since $(U_j)_1=\Phi^{-1}((0,1]\times U_j)$ and preimages preserve disjoint sets, the countable additivity follows from $m$.
Let $(A_n)_n$ be a sequence of subsets of a set $X$. Consider these two results: $\liminf_{n\to +\infty} A_n = \{x \in X \ | \ x \in A_n \ \text{for all but finitely many} \ n \in \mathbb{N}\}$ and $\limsup_{n\to +\infty} A_n = \{x \in X \ | \ x \in A_n \ \text{for infinitely many} n \in \mathbb{N}\}$. I was able to prove the equality for liminf, and I was thinking if I can obtain the second by taking the complements. My work is the following:
I am unsure about the equality $\{x \in X \ | \ x \in A_n^c \ \text{for all but finitely many} \ n \in \mathbb{N}\})^c=\{x \in X \ | \ x \in A_n \ \text{for finitely many} \ n \in \mathbb{N}\})^c$. Is the logic correct?
not particularly. someone may hop all over this remark with some strained analogy, but the "kernel" in the sense it is used for integral transforms like the laplace transform has little relationship with the "kernel" in the linear algebra sense. at the very least, the "linear algebra kernel" of a linear map that is defined by an integral transform having a "kernel" in the integral transform sense is a different object. it's not the same "kernel" in both cases
the "integral operator kernel" is certainly something that goes into the definition of a linear map, and hence affects what the "linear algebra kernel" of that map will be. but i don't think there's an analogy at the level you would be hoping for.
in the sense of meanings of words, i don't actually understand the use of the term 'kernel' for the nullspace of a linear map. i do understand it in the integral transform sense, in the same way that a kernel of corn can give rise to a whole plant of corn, the kernel of an integral operator gives rise to an integral operator determined by the kernel.
the genetic material in a seed determines a plant in a way that a linear map is not determined by its linear algebra kernel. so i don't know why the linear algebra kernel is a "kernel."
but, like so many sheeple before me, i have simply accepted this fact and moved on with my life.
Yeah, but then I am planning to do a minor in Mathematics ( I am majoring currently in Chemical Engineering ) Can you suggest some books to learn those Transforms to solve differential equations?
well, a lot of standard, 'methods' oriented (as opposed to theory oriented) books would do that. you do not need to interrogate the nature of a general integral transform to learn various properties of any given integral transform.
books will tabulate all sorts of formulas specific to the laplace transform. these are unique things, in some sense. there won't be a similarly well developed set of formulas for any integral transform you can think of
> If you believe your question or someone else's question or answer should be converted to a community wiki, you may flag it for moderator attention.
On Physics.SE, resource recommendation questions are normally made community wiki. We have a few fairly extensive ones on Physics.SE, but nothing with the huge scope of your question.
I appreciate what you're trying to do, but IMHO its scope is way too broad. It may be better to split it up a bit. Maybe have a very general broad overview that acts as an index, and then have several separate questions for different fields.
The problem is that I have no idea about anything on Physics so I don't know what to ask or what to search for etc in other words I am too ignorant to even start studying physics
@pie There's a flag button at the bottom of your question. That brings up the flag dialog. Select "In need of moderator intervention", which lets you write a comment explaining the reason for the flag.
But at this stage, I recommend you wait a bit, and discuss it in chat before flagging.
It's probably not the right time of day for most of the mods, though.
it's $T_1$ because points are compact, but a $T_1$ normal space is a Hausdorff space, but the Alexandroff extension is not Hausdorff if $X$ is not locally compact
oh ok I was under the impression that you were already proficient in your field (which I don't know, that's why I can't really judge to begin with hahaha) so I thought you were doing research already
While studying about PDEs I found that we can calculate the general solution of a PDE from it's complete solution.
The method goes like this:
If say, a PDE $F(x,y,z,p,q)=0$ where $p=z_x,q=z_y$ is given, and $f(x,y,z,a,b)=0$ is a complete solution of $F,$ then assume $b$ to be an unknown function of $a$ say, $b=\phi(a).$
So, the complete solution $f$ becomes, $f(x,y,z,a,\phi(a))=0\tag 1$
Now, differentiating the above equation wrt $a$ we get, $f_a(x,y,z,a,\phi(a))+f_b(x,y,z,a,\phi(a))\phi'(a)=0\tag 2$
I have a doubt regarding $$f : \mathbb{R}^n \to \mathbb{R}, f = x^TAx+b^Tx+c,A \mathbf{b} \in \mathbb{R}^n,c \in \mathbb{R} $$. Moreover, $f \in C^2(\mathbb{R}^n)$. Therefore I know that, in order for f to be convex, the Hessian of f must be must be positive semi-definite $\forall x \in R^n$. Due to $Hf(x) = 2A$ , I know that A must be a positive semidef matrix.
now, how do I show that for f to be strictly convex, $A$ must be positive definite? I feel like it could just be a consequence of the previous result, but then the exercise wouldn't ask to prove it :p
You also need to assume $A$ is symmetric, otherwise hessian is $A + A^\top$ If $A$ is not positive definite, it has a zero eigenvalue, that is, a nontrivial kernel $f$ restricted to the kernel of $A$ will be an affine linear function, which is not strictly convex
@VladimirLysikov If I wanted to show it without writing that affine linear functions are not strictly convex, I can just use your hint and say, for instance: let $u,v \in Ker(A)$ and since $\hat{f} :=f\rvert_{Ker(A)} = b^Tx+c$, then we have $$ \hat{f}(\lambda u +(1-\lambda) v)=\lambda b^Tu+(1-\lambda)b^Tv+c +\lambda c-\lambda c = \lambda f(u)+(1-\lambda)f(v)$$ which is not strictly convex
@VladimirLysikov First of all thanks for the hint, restricting to the kernel of A is a great idea :p I'm checking the solution to see If I'm on the right path, but yeah my definition for strict convexity is $f(tx+(1-t)y)<tf(x)+(1-t)f(y), \forall x,y \text{ and } \forall t \in (0,1)$ so it should be ok
Unfortunately, the solution uses the other definition, which involves showing that the function is always above the tangent plane (it uses Taylor expansion with Lagrange remainder)
But I'm sure your solution is perfectly valid and more elegant in my opinion @VladimirLysikov, I'll ask the professor tomorrow just in case.
This is a follow-up question to this question. Consider Theorem 2.49 in Folland's real analysis text:
We denote $S^{n-1}$ the unit sphere $\{x\in\mathbb{R}^n:|x|=1\}$. If $x\in\mathbb{R}^n\setminus\{0\}$, the polar coordinates of $x$ are
$$r=|x| \in (0,\infty),\quad x'=x/|x| \in S^{n-1}$$
The ma...
Choose three marked curves on the flat torus in $\Bbb R^4.$ Define an immersion $f:T^2 \to \Bbb R^3$ that maps these marked curves into a topological octahedral graph (where the intersections of the marked curves form the vertices of the graph).
Hi everyone, I have a question about expansion when considering terms of different scale
If we have $|\vec x - \vec R_{\vec n}|$ where $\vec x$ is some arbitrary vector far away from the origin and $\vec R_{\vec n}$ is just a fixed vector from the origin. If we consider that $\frec{|\vec R_{\vec n}|}{x}<<1$. How does one expand $|\vec x - \vec R_{\vec n}|$ ?
I have never expanded such a thing, which contains vectors
it is not a good idea to ask for necessary and sufficient conditions in analysis. that is an algebra thing.
on [a,b], i would hesitate to make sense of that kind of "expansion" outside of L^2[a,b], but L^2[a,b] is a set of functions that includes all continuous functions on [a,b].
one issue which isn't maybe clear in that image is that the use of the L^2 norm is implicit in whatever the dot dot dot is, or "formally expand." or something like that.
i would argue that you are implicitly incorporating the L^2 norm into whatever you are doing, if you are announcing some stuff like that.
there's no such thing as an infinite sum without a notion of convergence, which usually means a notion of topology, which can be supplied by a norm (such as the L^2 norm). so "dot dot dot" means choices if you are talking infinite sums.
so yeah, one difference between a "finite" closed interval [a,b] and the whole real line R is that the set of continuous functions on [a,b] is a dense subset of L^2[a,b]. while the set of continuous functions on R is not a subset (let alone a dense subset) of L^2(R)
well that's the whole thing. what is "they're" in "they're orthogonal to 1,"? it's not a choice, those functions are orthogonal to 1, which means that whole sum involving any a_n's for n >= 1 goes away.
So, I study physics, but they idea is that I am considering the following function $f(x)=\sqrt{1 - \frac{2R}{x}}$ (where R<<<x). So the fraction inside the root is 0<fraction<1
Now, If I were to taylor expand the expression given, I would taylor expand around this point x, which is non-zero
But, you can argue that one can write the expression as
If you consider $\sqrt{1-\epsilon}$ and expand around 0 (for some reason), you get an expansion of the form $\sqrt{1-\epsilon}=1 -\frac{\epsilon}{2}\pm..$. Now if you consider $f(x)=\sqrt{1 - \frac{2xR}{x^2}}$ where $R<<x$,you can say that this expression is similar to $\sqrt{1-\epsilon}$ and this we can expand $f(x)=\sqrt{1 - \frac{2Rx}{x^2}}=1 - \frac{xR}{x^2}$ in similar fashion , where $\epsilon$ is $\frac{2Rx}{x^2}$ in this case. I want the same result, but directly by working with $f(x)=\sqrt{1 - \frac{2Rx}{x^2}}$ and expanding around some point $x'$
the point is you can approximate in various ways, expand using $g$ and substitute ${2 R \over x}$, or expand $f$ about $x'$. But if you want these expressions to match in some way, you need to expand the $h$ as well.
Yeah I want that my main expression has the final expression and not to reach to that by the help of considering some other function with is what I did
@imbAF Suppose $F$ is the linear approximation of $f$ based at $x'$, $G$ is the linear approximation of $g$ at $0$ and $H$ the linear approximation of $h$ at $x'$. Then $F = G \circ H$. But you can (depending on circumstances) just use $G\circ h$, since this is slightly more accurate. But $G \circ H$ and $G \circ h$ will, in general, be different.
I simply performed Tailor expansion of $f(x)=\sqrt{1 - \frac{2xR}{x^2}}$ and I got: $f(x)=\sqrt{1 - \frac{2x'R}{x'^2}}+\frac{R}{x^2 \sqrt{1 - \frac{2x'R}{x'^2}}}(x-x')$. If you compare it to the result I got for the same function but with the help of how $\sqrt{1-\epsilon}$ behaves you will see that these are different, $f(x)=1 - \frac{xR}{x^2}$
Let $\Gamma$ be a finite connected undirected $4$-regular graph embedded on $S^2$. Enrich $\Gamma$ with the Ihara zeta function, $Z_{\Gamma}(u)$. Is there a direct relationship between the fixed points of the holonomy group generated by parallel transport along closed paths in $\Gamma$ and the poles of $Z_{\Gamma}(u)$?
Can you consider a taylor expansion to some arbitrary point, without a specific value? like x, and you need to expand around it. So you'd take the "other variable" $\bar x$, so in your expansion you would have smth like $....(\bar x - x)^i$, where i=0,1,2...
well, since the point is far away and occurs in the denominator the resulting expression will be small and so approximations about that value (meaning $0$ as in ${2 R \over \infty}$ as such) may be meaningful.
