Search results for query: *

Here is a depiction of the problem a) The potential at any point P due to a charge q is given by ##\frac{kq}{r}=\frac{kq}{\lvert \vec{r}_s-\vec{r}_P \rvert}##, where ##r## is the distance from the charge to point P, which is the length of the vector difference between ##\vec{r}_s##, the...

Let's think now about the energy to create a KF molecule from neutral atoms. First we need to ionize both K and F: K loses an electron and F gains an electron. Then we have to bring the ions together. The first ionization energy of K is 418 kJ/mol and for F is 1681 kJ/mol. The electron...

The answer at the end of the book says ##-\sqrt{2}##. Is this correct or is my solution correct? Here is a depiction of the path where we are integrating

Indeed, thanks. So the integral itself is correct? It's just a question of solving it.

Here is figure 2.16.6 Here is the picture I drew to set up the problem My first question is if the reasoning and integrals are correct. I used Maple to compute the three integrals. The first two result in 0, which makes sense by symmetry. Maple can't seem to solve the last integral.

Here is a picture of the problem It is not clear to me how to really prove that the equation for ##\theta(t)## is simple harmonic motion, and what the period of this motion is.

Looks like the answer is a silly oversight. Since the induction is over ##n##, this includes the part about the function ##R## being ##(n+1)##-times differentiable. After the inductive hypothesis, we want to prove the result for some ##k+1##. To do this we assume that the function ##R## is...

In Chapter 20 of Spivak's Calculus is the lemma shown below (used afterward to prove Taylor's Theorem). My question is about a step in the proof of this lemma. Here is the proof as it appears in the book My question is: how do we know that ##(R')^{n+1}## is defined in ##(2)##? Let me try to...

Is it the case that someone has looked into it and there is no easy solution, or is it just an issue on the backburner? If the latter, is it possible for a regular user to look into it and possibly create a pull request?

My posts all include many equations written in Latex. It seems to me like there are a few bugs related to usage of Latex specifically when just starting a new thread. I've made a short 2 minute screen recording to show the issue I face every single time I want to start a new thread, and the way...

Perhaps not obvious but also not not obvious. I am only using concepts from the current and previous chapter I am on in Spivak, so no bijections yet. Also not inverse functions, which is what I think might be needed to consider the case of replacing ##x^3## with ##\sin{x}##. However, using...

c) Why is the assertion ##\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} f(x^3)## obvious? First of all I don't think it is obvious but here is an explanation of why the limits are the same. ##\lim\limits_{x\to0} f(x^3)=l_2## means we are looking at points with ##x## close to zero and...

It's the proof of this, which was a step in the main proof. It's required for the latter.

The proposition should actually be with a ##\leq##, ie ##a-\delta \leq f(x)##. Is this what you mean?

Proving that ##a-\delta<f(x)## seems like a side-proof in the main proof, not the main thing to be proved. Here's an attempt at a proof: Proposition: ##b<x \implies f(b) \leq f(x)## Proof: Assume ##0<b<x<1##. If ##x## contains no digit 7 in its decimal expansion then ##f(x)=x>b \geq f(b)...

It's Spivak actually, and he's not expecting an epsilon-delta proof in this case; he's simply asking where the limit exists, not a proof. This epsilon-delta madness is the product of my curiosity.

Let me try and fill in the parts that you left implicit for the case of an ##a## with a decimal expansion with no digit ##7## in it, ending in zeros, e.g. ##a=0.6##. If ##x>a## then ##x\geq f(x)>a=f(a)##. This means that ##f(x)## is always between ##x## and ##a##, so ##f(x)-a## is between 0...

If ##0<a<1##, then ##\forall \epsilon>0## choose ##\delta = \sqrt{a}\epsilon##. Then ##|x-a|<min(a,\delta) \implies |\sqrt{x}-\sqrt{a}|<\epsilon## My god, so simple.

The proof I presented was for the specific case of a number ##a## ending in ##6\bar{9}## (originally I specified this as ending in ##7\bar{0}##). These are points at the left end of one of the infinite intervals composing the graph of ##f##. ##a=0.800005## isn't part of the available values in...

I didn't succeed with this suggestion. Here is an alternative way I found: Let (i) ##0<a<1## (ii) ##\delta<min(a,1-a) \leq 0.5##. $$|x-a|<\delta \implies a-\delta<x<a+\delta$$ $$0<a\leq 0.5 \implies \delta < a \implies 0<x<2a<1$$ $$0.5 \leq a < 1 \implies \delta < 1-a \implies 0<2a-1<x<1$$...

It seems to be tricky to show that ##\lim\limits_{x \to a} \sqrt{x}=\sqrt{a}## for a general ##a##. More specifically, I think I can prove it for ##a \geq 1##, but having trouble with ##0 < a < 1##. The proof I gave for ##a=1## is valid for ##a \geq 1## with slight modifications: Let...

Not sure if you saw this, but this was my initial sketch and attempt above. It is convoluted. I'm actually going to come back to this problem at a later time, I've spent too long on it today! Proposition: ##\lim\limits_{x\to a}f(x)## exists for any number ##a## that is not on the right (closed)...

The first part yes, and this is why the intervals are closed on the right. About the second part (##f(0.abc...hij8)=0.abc...hij8##) I am not sure. This number ends in a string of zeros, so it is equal to another number that ends in ##7\bar{9}##. On the line ##y=x##, at ##x=0.7\bar{9}##, there...

Not quite. See my updated version above. The limit exists except where ##x## is of the form ##0.abc...hij7\bar{9}##, ie it ends in ##7\bar{9}##

Indeed, I've spent a few hours on this problem and have not yet understood exactly what is meant by the passage I quoted. I understand that ##0.7\bar{9}## and ##0.8\bar{0}## are equal. It's more murky what it means that "we will always use the one ending in 9's". Does this mean that the notation...

Missed this message earlier :) Here is the exact wording:

I mean actually formally prove using an epsilon-delta proof. From simply looking at the graph and the function definition, I would say the limit exists for every ##x## except at places such as ##0.8##. The reason is that the limit from below exists and is equal to 0.7, but the limit from above...

Just to be extra precise about the absolute value issue. Isn't it always implicit that there is the ##0<## portion? Absolute value is always non-negative, by definition. To make sure ##x \geq 0## it is only necessary to write $$|x-1|<\delta\leq 1$$

I've been thinking that maybe I should read a book or take a course on mathematical logic, to really learn to write mathematics correctly (is that the best way to learn the actual "grammar" of math, if I can call it that?). I actually understand what you mean with the points you brought up...

Isn't it rather ##\forall x: 0.7 \leq x \leq 0.7\bar{9}, f(x)=0.7##? (Note that I made the interval closed not open). The solution manual graph that I posted actually shows ##\forall x: 0.7 < x \leq 0.7\bar{9}, f(x)=0.7##. This is actually one question I have: why is the interval open on the...

In general, if we write ##|x-1|<\delta## this means that ##1-\delta<x<1+\delta##. If ##\delta>1## then ##x## can be negative. If we write ##0<\delta<min(1,\epsilon)## this implies that ##x\geq 0## (as I showed in the previous post). Indeed I forgot to put in the ##0<## part. If we write...

$$f(0.73)=0.7<0.73$$

I believe the x-axis is vertical here. The graph is composed of i) an infinite number of intervals that start on the ##y=x## line and finish at some ##x## with a decimal expansion ending in ##7\bar{9}##. E.g., from ##0.67## to ##0.67\bar{9}## which is considered ##0.68##. Other examples of...

##\forall \epsilon>0##, we are looking for ##\delta>0:|x-1|<\delta \implies|\sqrt{x}-1|<\epsilon## The domain of ##f## is ##[0,+\infty)## so ##x\geq 0##. Consider ##0<\delta \leq 1 ## $$|x-1|<\delta \implies 1-\delta<x<1+\delta \implies x \geq 0$$...

Going through Spivak's Calculus, every chapter, every problem

Consider item ##vii##, which specifies the function ##f(x)=\sqrt{|x|}## with ##a=0## Case 1: ##\forall \epsilon: 0<\epsilon<1## $$\implies \epsilon^2<\epsilon<1$$ $$|x|<\epsilon^2\implies \sqrt{|x|}<\epsilon$$ Case 2: ##\forall \epsilon: 1\leq \epsilon < \infty## $$\epsilon\leq\epsilon^2...

A thin shell in reality doesn't have zero thickness. Consider the image below, showing a cross-section of a small portion of the shell: Here we are considering a more general case in which we have electric fields of magnitude ##E_1## and ##E_2## on each side of the shell. Gauss's Law...

Interesting, and makes sense now. I see a lot of geometric arguments used so far to discuss fields - electric and gravitational so far. There are a lot of corollaries to the inverse ##r^2## relationship of distance and field. I am finding it eery how many of these passages there are that aren't...

Ok, looks like I did miss the point. I was trying to use Coulomb's law to calculate the electric field at an arbitrary point P, to then use Gauss's law. But for what? The point of GL is to have a way to calculate the electric field at any point using a surface integral. If you already have an...

I tried to do it just now. Seems pretty intractable for me from the point of view of calculating the integrals, as far as I did it. My strategy was: - Use Coulomb's law to find the electric field generated by the ring at any point ##P##. If I have this then I have an expression to use in a...

And I have stated my conclusion: yes, I can find E by applying it to all of the objects. I've actually done all calculations except for the ring and sphere. Infinite Line $$\vec{E}=\frac{\lambda}{2\pi r\epsilon_0}\hat{r}$$ Infinite Cylinder, uniform charge density Inside the cylinder...

So just to confirm: we can apply Gauss's Law to all of the charges in the pictures in my original post? If so, then the course is actually asking something slightly different; but I think it doesn't matter because in that case the question is simply formulated in a way that is not clear enough.

What am I missing? I also don't get the title of the section: "Charge distributions with enough symmetry for Gauss's Law". I thought Gauss's Law was valid for any closed surface enclosing a charge. I don't understand what "enough symmetry" means in the title above. I get that with symmetry...

Using Gauss's Law By using a symmetry argument, we expect the magnitude of the electric field to be constant on planes parallel to the non-conducting plane. We need to choose a Gaussian surface. A straightforward one is a cylinder, ie a "Gaussian pillbox". The charge enclosed is...

They definitely do cancel, that is clear by simple inspection. You are totally right about adding ##\varphi##, my question is more about where in my original calculations I missed adding that in. And here's what I came up with. There was a step where I wrote...

I thought I didn't have to worry about ##\varphi## because I wrote the integral in terms of infinitesimal rings. ##dq## is the charge on an infinitesimal ring When I integrate only over ##r##, aren't I summing up all the rings already? I can see the intuition that I do need to somehow integrate...

I am interested in particular in the second integral, in the ##\hat{r}## direction. Here is my depiction of the problem: As far as I can tell, due to the symmetry of the problem, this integral should be zero. $$\int_0^R \frac{r^2}{(x^2+r^2)^{3/2}}dr\hat{r}$$ I don't believe I need to...

I can't believe I missed the integral limits! My god. I am aware of the symmetry of the problem. At this stage I am doing the full calculations to get practice doing the calculus. But I will try to just use symmetry to also get practice with identifying such shortcuts from now on as well.

The strategy will be to figure out what ##dq##, ##\hat{r}_{dq,p}##, and ##r_{dq,p}## are, plug them into the expression for ##d\vec{E}_{p_r}##, then integrate over ##d\vec{E}_{p_r}## to obtain ##\vec{E}_{p_r}##, the electric field at ##P## due to the arc on the right. Then I will repeat the...