TITLE: Finding x using logarithms
QUESTION [0 upvotes]: $$x^{1/3} - 4x^{-1/3}= 3$$
I dont quite understand what to do with the 4 in the term 4x so that I can rewrite the two terms with the same base.
Also since there are two terms how do i solve this? I'm supposed to find x

REPLY [2 votes]: If you multiply by $x^{1/3}$ you get
$$x^{2/3}-4 = 3x^{1/3}$$
Then
$$x^{2/3}-3x^{1/3} -4 = (x^{1/3} -4)(x^{1/3}+1) = 0$$
which gives you two solutions.  Logarithms aren't needed.

REPLY [0 votes]: You can solve it in the following way: 1) Rewrite the equation by substituting $y=x^\frac{1}{3}$. Note that then $\frac{1}{y}=x^{-\frac{1}{3}}$. 2) Multiply the equation by $y$. 3) Now you have a quadratic equation in $y$. Solve it using the usual formula. 4) When you have found the solutions, substitute back (ie. write $x^\frac{1}{3}$ instead of $y$). Then the complete solution is straightforward.


TITLE: Lottery game w/ replacement
QUESTION [0 upvotes]: My professor has given us a question that is confusing me. Here it is verbatim:

A state lottery game is played by selecting three numbers between 0 and 9, inclusive, with replacement. The winning numbers are selected in the same way. You win if your numbers match the winning numbers in any order.Find the total number of ways the winning numbers can be selected.

He is telling us the answer is 220, but with how this is worded, it differentiates "winning numbers" or numbers drawn and "your numbers". It's asking how many ways can the winning number be selected. I see how if the winning numbers were selected, then there would be 6 ways for YOUR number to be selected, but I dont understand how 220 is selected. 10^3 = 1,000, but I'm not sure how to adjust for duplicates.

REPLY [1 votes]: There are $\binom{10}3$ ways of choosing 3 distinct numbers, each of which can be drawn in 6 equally probable ways
There are $2\binom{10}2$ ways of having two of one number and one of another each of which can be drawn in 3 equally probable ways
There are $\binom{10}1$ ways to have three of a kind each of which can be drawn in 1 equally probable way
$$ \binom{10}3+2\binom{10}2 + \binom{10}1=220$$
$$6 \times \binom{10}3+3\times2\binom{10}2 + 1 \times\binom{10}1=1000$$


TITLE: The derivative of $\int_0^xxf(t)dt$ and why we cannot treat $xf$ as continuous.
QUESTION [1 upvotes]: This comes from Calculus by Michael Spivak, 3rd ed., chapter 14, Problem 11.

Find $F'(x)$ if $F(x) = \int_0^x xf(t)dt$. The answer is not $xf(x)$; you should perform an obvious manipulation on the integral before trying to find $F'$.)

Here, Spivak fails to include the assumption that $f$ is continuous.
With this assumption, the answer is $\int_0^xf(t)dt + xf(x)$, and is solved here, here, and here.
I understand the solution. I was able to do it myself. It makes sense to me. I also solved it using the $\lim_{h\to 0}$ approach, and that worked too.
Yet, naively, it seems like the answer should be $xf(x)$. It seems like since $x$ "behaves like a constant" in that integration, and because $f$ is continuous, then $xf$ must also be continuous, and so $F'(c) = cf(c)$.
I'm hoping to build better intuition about why this is wrong.
I was thinking perhaps one way of looking at it is that $xf$ cannot be assumed to be continuous because we are free to change the value of $x$ somewhat arbitrarily. In fact, at any fixed point $c$, $xf(c)$ varies with $x$, so certainly we cannot guarantee anything about the value of $xf(t)$ simply by restricting $t$ to certain values.
Does this make any sense? I'm sorry to use such sloppy reasoning/language/arguments.
There's a good chance I just don't yet have the tools to fully "get it" yet, but I appreciate any insight you might be able to share.
Edit
Ian's answer helps explain something that confused me in Spivak's solution to a later problem.
The next two problems are:

Prove that if $f$ is continuous, then
$$\int_0^xf(u)(x-u)du = \int_0^x\left(\int_0^uf(t)dt\right)du.$$
Hint: Differentiate both sides, making use of Problem 11.

Use Problem 12 to prove that
$$\int_0^xf(u)(x-u)^2du = 2\int_0^x\left(\int_0^{u_2}\left(\int_0^{u_1}f(t)dt\right)du_1\right)du_2.$$

These each require fairly straightforward calculations, the details of which we can skip here. However, in the published Answer Book solution for the latter, problem 13, Spivak does something that initially I found baffling. He begins:

Applying Problem 12 to $g(u) = f(u)(x-u)$, we obtain
$$\int_0^xf(u)(x-u)^2du =\int_0^x[f(u)(x-u)](x-u)du$$
$$=\int_0^x\left(\int_0^uf(t)(x-t)dt\right)du$$

This substitution is misleading. The result he's using from problem 12. applies only to functions that do not depend on $x$, while Spivak's $g$ here clearly does depend on $x$.
I think he gets away with it here only because $\frac{\partial}{\partial x}[f(u)(x-u)] = f(u).$

REPLY [1 votes]: In this particular case, you can solve the problem as
$$\frac{d}{dx} \int_0^x x f(t) dt = \frac{d}{dx} \left ( x \int_0^x f(t) dt \right ) = \int_0^x f(t) dt + x f(x)$$
by the product rule together with FTC.
More generally you have the Leibniz rule:
$$\frac{d}{dx} \left ( \int_{a(x)}^{b(x)} f(t,x) dt \right ) = f(b(x),x) b'(x) - f(a(x),x) a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f}{\partial x}(t,x) dt.$$
The Leibniz rule captures the dependence of the integral on $x$ through both the limits of integration and the integrand. In your problem, giving the answer $x f(x)$ would only encapsulate the dependence on the limits of integration.


TITLE: Where do electrons get or lose the energy that allows them to change energy levels?
QUESTION [0 upvotes]: How do electrons gain energy and hop levels?  Does it come from shooting it with a laser or introducing heat, or some such method?  By the same token, why do electrons lose energy?  I assume they don't say to themselves, "I'm feeling lazy" and move down a few levels.
Also, what is the energy called that they gain and lose?  Is it "quantum" energy, radiation, EM energy, or some other type of energy?
An answer that is simpler would be best. Thanks!

REPLY [1 votes]: You can think of an electron's energy level like a person on earth. You go up in elevation and you have more potential energy, go down in elevation and you release it.
In this analogy, the higher energy level is the higher elevation.
With electrons, it is from getting hit by a photon. A photon is the particle that carries energy in the electromagnetic spectrum. This includes radio waves, heat, light, UV, and more. In order for the electron to change energy levels the photon it absorbs must carry at least enough energy for it to jump upwards