I'm a bit tangled up as to whether the distributivity rule of limits takes precedent over apply the delta function
I have a product that can be factored out to have a delta function as one of the factors
The other factors permit their individual limits, and I'm confused as to whether I can split the products or not
So for example I have $\prod_i f_i g_i h_i$ with all smooth functions and variables, everything is real and analytic, yadda yadda. Normally I would allow them to be split as $\prod_i f_i \prod_i g_i \prod_i h_i$ but then one of the products is just the trivial product (?!)
How would I interpret that?
So to make it clear, if I take my limit, I can take the limit over each factor individually. One of the products resolves to be $\exp[A*\delta(x-\lambda_i)]$ where $\lambda_i$ and an indeterminate residuel parameter of the compound product above
Actually let's take $\lambda$ defined in the reals to make it "easy"
Okay, so I have a product of factors over which I'm taking a limit
If I write $\prod_i f_i g_i h_i$ as $\prod_i f_i \prod_i g_i \prod_i h_i$, then take the limit over each of the individual products, one of the products becomes a Dirac delta
But I believe that the limit transition has been made at that point, and the delta function will not restrict the form of the other products
@1010011010 if $X - a \mid P(X)$ then it's obvious. Suppose $a$ is a root of $P$ but $X-a \nmid P(X)$. Then $P(X) = (X-a)Q(X) + r(X)$ with $\deg r < \deg(X-a) = 1$ so $r(X)$ is constant, say $k$. Then since $P(a) = 0$ you have $P(a) = (a - a)Q(a) + k$ so that $k = 0$, contradiction.
We have the laplac problem in $(0,1)^2$ with the boundary conditions $$u(x,0)=f(x) , \ u(x,1)=0, \ \ x\in [0,1] \\ u(0,y)=u_x(1,y)=0, \ \ y\in [0,1]$$ Using the separtaion of variables we get the following two problems $$X''(x)=\lambda X(x)=0, X(0)=X'(1)=0 \\ Y''(y)-\lambda Y(y)=0, \ \ Y(1)=0$$ F...
@schn Start by using $|x|\le \sqrt{x^2+y^2}$ and the same for $|y|$.
@TheEastWind This is implicit differentiation. You need to write down the chain rule for differentiating $F(x,y) = (f(x,y,u(x,y),v(x,y)),g(x,y,u(x,y),v(x,y))$.
The simplest case of implicit differentiation is to consider $f(x,y)=0$ defining $y=y(x)$. Then $F(x)=f(x,y(x))=0$ for all $x$. Take the derivative carefully. You get $F'(x)=0=\partial f/\partial x + \partial f/\partial y \cdot dy/dx$.
This tells you that $dy/dx = -\dfrac{\partial f/\partial x}{\partial f/\partial y}$.
You want to do the same thing with more variables. Using matrices is easiest, as I said.
Basically, you do the same argument, treating $x$ and $y$ in that last example as vectors (and replacing the division with multiplication (on the left) by the inverse matrix).
But the problem arises when extra variables that we included are themselves functions of the already present variables, like in the above case. This confuses me.
The grand bargain we made with engineering departments --- get teaching in exchange for writing the curriculum in an order suited to them --- is a cancer
There is absolutely no pedagogical reason to teach lin alg after multivar differentiation. It's abominable
Yes, that's right, but you have to do $f$ and $g$ both, as well as the $y$ partials and then work with determinants. This is why I suggested writing vectors and the chain rule using matrices.
The point, @TheEastWind, is that you can think of the derivative matrix $D\vec F(\vec x,\vec u)$ as a block matrix $\begin{bmatrix} \frac{\partial\vec F}{\partial\vec x} & \frac{\partial\vec F}{\partial\vec u}\end{bmatrix}$.
This is a $2\times 4$ matrix that I'm visualizing in terms of two $2\times 2$ blocks. Play with that until you understand it.
More interestingly, if you look at it in $\Bbb P^3$ it's a basic example of something that is not a complete intersection. Try as hard as you want, you cannot give it as the intersection of just two surfaces.
You need $3$ equations to define it. $2$ will always give you something extra.
Yes, complete intersections are very nice and hard to come by, in general. All sorts of nice things happen for them (if you think back to diff top, you get things like the normal bundle is the sum of the normal bundles of the hypersurfaces, etc.). Of course, I omitted to say that you have transverse intersections of the hypersurfaces.
Demonark: You should be able to write down equations in $\Bbb P^3$ and see you need 3 of them. As you learn AG you can mull over how you'd prove it's impossible to do it with just 2.
A couple of us independently asked for one and when we met up and decided on elliptic curves (assuming he thought it was a good idea) one of us emailed him
Okay. But if the squeeze theorem shows that the absolute value of the original expression converges to 0, how does that prove it for without the absolute value of it?
@TedShifrin $f(x,y)\leq|f(x,y)|\leq0$ when $(x,y)\to0$. If there’s a lower bound for $f(x,y)$ that is 0, then I see how we could conclude that $f(x,y)$ also converges to 0.
Probably stole it. He's obviously a very different flavor of mathematician from me, I've noticed. I wonder if he even covers things like multiple integrals and differential forms "properly."
Yeah, Spivak has a question like that in his book, and I don't think I ever assigned it. I mean, understanding order of quantifiers is important, but ...
I was wondering whether there is any good justification for giving a circle map rotation number $1$ as opposed to $0$ (the rotation number is taken mod $1$). There are cases in which it seems to make sense to do so. For instance, considering a continuously varying family of circle maps starting with $\rho(f_0) = 0$ and ending with $\rho(f_1) = 1$. Then every rotation number in between is attained, by continuity of $\rho$.
Consider the following multivariable function: $\frac{e^{|\textbf{u}|^2}-1}{|\textbf{u}|^2+x^2y+y^2z}$ where $\textbf{u}=(x,y,z)$ and $|\textbf{u}|=\sqrt{x^2+y^2+z^2}$. How would one find its limit as $\textbf{u}\to0$?
If one approaches the origin from where $y=0$, the terms $x^2y+y^2z$ in the denominator equal 0. Substituting $|\textbf{u}|^2$ for $t$ yields the expression $\frac{e^t-1}{t}$, which has limit 1 as $\textbf{u}\to\textbf{0}$ and thus $t\to0$. So the limit should be 1 if it exists.
Right. Now how do you make the last step rigorous?
Well, of course, you take the limit as $x\to 0$. Note where the $x^2$ and $3x$ came from. They came from factoring out the $x$. Can you approach the multivariable problem in a similar fashion?
Right, by the properties of limits I would justify the last step by the sum rule and the fact (which also is sometimes stated as a "rule") that $\lim\limits_{x\to 0} x=0$.
Regarding the multivariable problem, is there something that one can factor out in the numerator?
No, but you already know a basic limit fact that you used. You already know $\lim\limits_{\mathbf u\to 0} \frac{e^{\|\mathbf u\|^2}-1}{\|\mathbf u\|^2}$. So how should you proceed?
Since these are continuous functions, one could simply plug in $x=0,y=0,z=0$ in the expression and together with the fact that $\lim\limits_{\mathbf u\to 0} \frac{e^{\|\mathbf u\|^2}-1}{\|\mathbf u\|^2}=1$ evaluate the limit. Yet in $\dfrac{x^2y+y^2z}{\|\mathbf u\|^2}$ it remains to determine whether the numerator or denominator is bigger...
Alright, so that together with the triangle inequality (to separate the sum in the numerator) and the squeeze theorem should show that $\dfrac{x^2y+y^2z}{\|\mathbf u\|^2}$ equals 0 as $x,y,z\to0$?
