You know, you could just have shown me the set $S = (-1,1), (-1/2,1/2), (-1/3,1/3) . . .$ for a counterexample to the "finite intersections" thing, @Balarka. (For when you get back.)
I'm trying to show that the function $$exp(X) = \displaystyle \sum_{k=0}^{\infty} \frac{X^k}{k!}$$ is Frechet differentiable at every $X \in \mathbb{R}^{n \times n}$
I tried to begin by calculating the increment $f(X+v)-f(X)$
This is equal to $\displaystyle \sum_{k=0}^{\infty}\frac{(X+v)^k-X^k}{k!}$
But if I expand the first bracket, the expression I would end up with as the derivative isn't what I would have thought the derivative would be ($exp(X)v$)
Can anyone give me a hint on how to proceed?
Oh no, wait. I only want the terms linear in $v$, so that does work I think.
Which norm would be best to use in this case?
To show that this is in fact the Frechet derivative?
Theorem - General solution of $y''+p(x)y'+q(x)y=0, x \in I (\star)$
Let $y_1, y_2$ be linearly independent solutions of $(\star)$ in an interval $I$.
Then if $y$ is a solution of $(\star)$ in $I$, there are $c_1, c_2 \in \mathbb{R}$ such that $y(x)= c_1 y_1(x)+ c_2 y_2(x), x \in I$.
For the pr...
he forgot something ..... satan knows how to come with good conter-examples and well argumented , such antithesis caused him been chased away heaven's university without any certificate :D
@Rememberme They are constructed like you construct the reals from Cauchy sequences of rationals, but now using this new metric rather than the old one to determine which sequences are Cauchy
We have $f,g:[-4,4]\rightarrow\mathbb{R}$, $f(x)=x^2+2$ and $g(x)=x+4$. We need to find the crowd area between the graphs f and g.
I know that $A_{(\Gamma_f)}=\int_a^b|f(x)|dx$ but in this case how we can find $A_{(\Gamma_f)}$ ? I think we need to find x (commune abscissa) and it involving that...
hello, please: if i have that $\Sigma=\{u\in W^{1,p}_0, ||u||^p=1\}$ can i say that $$\inf_{u\in W^{1,p}_0\setminus\{0\}}\left(\frac{||u||^p}{||u||_{L^{p^*}}}\right)=\inf_{u\in \Sigma}\left(\frac{1}{||u||_{L^{p^*}}}\right)$$
You can also do it mathematically of course. $f-g=x^2-x-4$, it's a second degree polynomial with a positive leading coefficient so it's negative between its two roots
I want to find two linearly independent solutions of the differential equation
$$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$$
Previously I have seen that the following holds for the differential equation $y''+ \frac{1}{x}y'-\frac{1}{x^2}y=0, x>0$:
We are looking for solutions of t...
We have $f,g:[-4,4]\rightarrow\mathbb{R}$, $f(x)=x^2+2$ and $g(x)=x+4$. We need to find the crowd area between the graphs f and g.
I know that $A_{(\Gamma_f)}=\int_a^b|f(x)|dx$ but in this case how we can find $A_{(\Gamma_f)}$ ? I think we need to find x (commune abscissa) and it involving that...
@Lucas The best thing would be that you answer your own question :-) that way if someone has the kind of problem and find your question, he'll get help too
@TedShifrin I think grad L should be your F. the BH is the Jacobian of the gradient of the Lagrangian. So that should correspond to your $\frac{\partial \mathbf{F}}{\partial \mathbf{y}}$ but isn't grad L a function of only $n+1$ variables?
@Lucas Why do you want to delete the post ? @Lucas The best thing would be that you answer your own question :-) that way if someone has the kind of problem and find your question, he'll get help too
Can someone give me an answer: if i have that $\Sigma=\{u\in W^{1,p}_0, ||u||^p=1\}$ can i say that $$\inf_{u\in W^{1,p}_0\setminus\{0\}}\left(\frac{||u||^p}{||u||_{L^{p^*}}}\right)=\inf_{u\in \Sigma}\left(\frac{1}{||u||_{L^{p^*}}}\right)$$
$$(1-x^2)y''-2xy'+p(p+1)y=0, p \in \mathbb{R} \text{ constant } \\ -1 < x<1$$
At the interval $(-1,1)$ the above differential equation can be written equivalently
$$y''+p(x)y'+q(x)y=0, -1<x<1 \text{ where } \\p(x)=\frac{-2x}{1-x^2} \\ q(x)= \frac{p(p+1)}{1-x^2}$$
$p,q$ can be written as power...
@MikeMiller let $C_n(X)$ be the group of formal $\mathbf Z$-linear combinations of maps from manifolds onto $X$. define the boundary maps $C_n(X) \to C_{n-1}(X)$ by mapping $M \to X$ to $\partial M \to X$. we get a chain complex, as boundaries are compact.
can't we realize bordism groups as homology of this chain complex?
I guess we get back standard unoriented bordism if we consider coefficients in $\mathbb F_2$
@PaulPlummer Hmmm... True. I wasn't really sure what constituted as active. There are messages every few minutes so the message was just to be on the safe side...
Well under most definitions of eigenvalues, they are only defined for nonzero eigenvectors but in the quotient space $V/V$ only contains the zero vector. Note that the same problem would occure even in just $V$ if we considered "eigenvalues" of the 0 vector @imu96
(and normally eigenvectors are defined to be non zero too)
I went to schools using the British curriculum, and we did it there. (Although we didn't call it that)
Okay, so in the same pdf on page 12, in problem 35, they ask us to show that if $\lambda$ is an eigenvalue of $T/U$ then it is an eigenvalue of $T$ where $U$ is a subspace of a finite dimensional vectorspace, $V$
What I have so far is: if $\lambda$ is an eigenvalue then there exists $v \in V$ such that $(T/U) (v+U) = Tv + U = \lambda v + U \biconditional Tv - \lambda v \in U$
$\Rightarrow (T - \lambda I)u \in U$
Since I want to show that $ \lambda$ is an eigenvalue, I need to show that $T - \lambda I)v = 0$ for some nonzero $v$.
Problem 36 hints at the fact that it has to do with $V$ being of finite dimension, but I don't really see what this would have to do with anything.
I was trying to do this problem with induction on the dimension of V, but didn't really make any progress with that.
"In terms of prerequisites, the present book assumes the read er has some familiar- ity with the content of the standard undergraduate courses i n algebra and point-set topology. In particular, the reader should know about quoti ent spaces, or identifi- cation spaces as they are sometimes called, which are quite i mportant for algebraic topology. Good sources for this concept are the textbooks [A rmstrong 1983] and [J ̈ anich 1984] listed in the Bibliography."
@LeGrandDODOM $f^{-1} \colon f(\mathbb{R}^n) \to \mathbb{R}^n$ is a continuous bijection. If $f$ were surjective, we'd have a homeomorphism between $\mathbb{R}$ and $\mathbb{R}^n$.
@DanielFischer Oops. I got the wrong statement (how did you get the continuity of $f^{-1}$ anyway ?) The correct one is " let $f:\mathbb R^n\to \mathbb R$ with $n\geq 2$ be a continuous closed map. Prove that $f$ is not surjective". Since $f$ need not be injective, one cannot resort to some homeo-argument
@MikeMiller Don't ask me about PDEs, I have fortunately forgotten almost all of the little I once knew. But yes, there are some regularity theorems for parabolic PDEs. You could ask Slim (Yes, Behaviour, ...), he is into PDEs.
@LeGrandDODOM The closedness of an injective map gives the continuity of the inverse, since $g\colon X \to Y$ is continuous if and only if $g^{-1}(F)$ is closed for all closed $F\subset Y$. Now $(f^{-1})^{-1}(F) = f(F)$ is closed in $\mathbb{R}$ by assumption, hence a fortiori in the subspace $f(\mathbb{R}^n)$. Okay, the new one is harder, let me think about it a bit. But I have some moderation to do first.
@DanielFischer please if i have that $\Sigma=\{u\in W^{1,p}_0, ||u||^p=1\}$ can i sauy that $$\inf_{u\in W^{1,p}_0\setminus\{0\}}\left(\frac{||u||^p}{||u||_{L^{p^*}}}\right)=\inf_{u\in \Sigma}\left(\frac{1}{||u||_{L^{p^*}}}\right)$$
@Vrouvrou If $\lVert u\rVert^p$ is the norm of $u$ then yes, but if it's the $p$-th power of the norm, then only for $p = 1$. Since the infimum would not be interesting if the latter was the case and $p \neq 1$, it's probably the former. Then you have the equality by the homogeneity of norms.
anyway, if you draw one copy of S^1 above S^1 for each z->z^p, then copy this diagram infinitely, then take the inverse limit you get R x Z^hat mod Z, equivalently A_Q mod Q
@anon not sure if you've ever though about it : standard p-adic solenoids inherit a natural action of $\mathbf Z_p$ from monodromy. can you do the same with something like a solenoidal algebraic variety to get it to inherit a natural action of Gal(\bar Q/Q)?
@anon interesting : no idea.
@anon erm, no, that doesn't work. $\varprojlim S^1$ with bonding maps being $z \mapsto z^p$ at each level spits a Z_p-bundle over S^1 at you.
Theorem - General solution of $y''+p(x)y'+q(x)y=0, x \in I (\star)$
Let $y_1, y_2$ be linearly independent solutions of $(\star)$ in an interval $I$.
Then if $y$ is a solution of $(\star)$ in $I$, there are $c_1, c_2 \in \mathbb{R}$ such that $y(x)= c_1 y_1(x)+ c_2 y_2(x), x \in I$.
For the pr...
We have $f,g:[-4,4]\rightarrow\mathbb{R}$, $f(x)=x^2+2$ and $g(x)=x+4$. We need to find the crowd area between the graphs f and g.
I know that $A_{(\Gamma_f)}=\int_a^b|f(x)|dx$ but in this case how we can find $A_{(\Gamma_f)}$ ? I think we need to find x (commune abscissa) and it involving that...
