You have to use the limit definition. The question is this: If $f$ is continuous on $[a,a+\epsilon)$ and $\lim_{x\to a^+} f'(x)$ exists, is $f$ differentiable at $a$?
I've commented on this before. I sat in on calculus lectures by two instructors with PhDs (admittedly in applied math) who taught their beginning calculus students exactly what dc3rd said. I had to reprimand them afterward!
The point of the material in beginning calculus is to show the left-hand and right-hand difference quotient limits exist and are equal.
Assuming continuity, it is correct, but it's a theorem beyond where the students are at the beginning of Calc I. Certainly not what you should be putting on the board as the correct approach.
Well, you can compute left- and right-hand derivatives here with your two functions. Or you can use what I'm leading you to, which is easier because of implicit differentiation.
glad you said that cause you were about to wrack my head, with this condition violation
so I know that this point $\phi'(1)$ exists due to the MVT. THe only way I can envision calculating it is by taking the limits of one of my other functions.....I don't see how I can calculate it using implicit differentiation
MVT doesn't get used enough....always drops out of sight until it unexpectadly shows up....
@Ted okay that makes sense. So I guess you're saying that one can apply MVT to the pieces of a piecewise-defined function to get the existence of the derivative at the patching point?
@TedShifrin If one comes across an arbitrary multifunction $f(\pm x)$ in the course of a computation, can this be reduced to an arbitrary function $g(x)$ as one would reduce $f(2x)$ to $g(x)$?
@user10478 if $g:A \to B$ and we assume that $-A \subset A$, then you can consider $A \to B^2, a \mapsto (g(a),g(-a))$. That's a function in some sense, but probably not quite what you want
@Semiclassical I am trying to use the similarity solution method to solve a PDE. I have a relation between scaling coefficients $p^2 r = q^2$, thus $q = \pm p \sqrt r$, and ultimately I have a solution form which has $\pm$ in its remaining argument. I wasn't sure if I could just drop the $\pm$ by calling it a different arbitrary function. I might have to finish the solution process with the positive and negative versions separately.
maybe you can get rid of that ambiguity with boundary conditions?
though that's more typically in the case of deciding between two trial solutions, e.g., $e^{x}$ is not a valid solution if we want to be bounded over $[0,\infty)$
Maybe, I just wanted to try the method on a nonlinear PDE so I made up an example that has a nice solution in Mathematica. I don't have specific boundary conditions, or I could make up my own.
If $S$ is an embedded submanifold in $M$ then for each $p\in S$, there exists a chart $U$ in $M$ on which $U\cap S$ is defined by the vanishing of some coordinates.
My question is, does this then imply that $U\cap S$ is an embedded submanifold of $S$ (because then it is defined by the vanishing o...
I can compute the derivatives, but taking the limits of each of them is leading to some zeroes in the denominators.....is there a trick I should use?......I'm going to put this one aside after this....
cause it is............interesting.
mainly actually it is the $\arccos(x)$ term that is leading me into problems once I differentiate it.
the other one as I work on it looks like it is going to be just a large algebraic expression and cancellation should occur.
Legend has it Peter Lax was asked once during an interview how he had developed such a fruitful career as a researcher. He replied: "I integrated by parts."
lax's functional analysis text is very good. it has deceptively simple proofs of things that people still include as lemmas in papers (with longer, more complicated proofs).
see this is what sometimes fcks me up. My approach should have worked as well. Because I still can t see what I did that is actually mathematically incorrect. going from $R_2 - V_{out}R_2$ to $R_2(1-V_{out})$ looks correct to me
I might have chosen a more "difficult" path but it somehow should have given the same result
@robjohn Assume that in addition to $f$, $g$, and $h$ being functions in $L^{2} (\mathbb{R})$, $h$ is also in $L^{1} (\mathbb{R})$. Then would it always be true that $$\int_{\mathbb{R}^{3}} f(x) g(t) e^{-ixt} h(u) e^{-ixu} \, \mathrm du \, \mathrm dt \, \mathrm dx =\int_{\mathbb{R}^{3}} f(x) g(t) e^{-ixt} h(u) e^{-ixu} \, \mathrm dx \, \mathrm du \, \mathrm dt?$$ If not, then I'm misunderstanding the issue you described earlier.
So individually these are both implicit equations and as such level sets.
And as a system of equations I'm solving for the intersection of these level surfaces. So would I be calling my equation $F(\mathbf{x})$ and implicit equation as well? I was under the impression implicit equations had to be level sets.
I know how to do the question, but want to get the right idea behind it.
well rather I should say I'll take the derivative and do all the necessary work to determine if it is a manifold.
@RandomVariable (as before), I don't have context, but essentially you're asking whether Fubini holds. And looks like the answer is no, since $f, g$, are not $L^1$
