Suppose that the \(n\)th prime number is given by \(p_n\) and the number of primes that do not exceed \(n\) is \(\pi(n)\). Let two functions \(f\) and \(g\) be asymptotic to each other, denoted by \(f\sim g\), iff \(\lim_{x\rightarrow\infty}{\frac{f(x)}{g(x)}}=1\). The prime number theorem states
\[\pi(x)\sim \frac{x}{\log{x}}.\] This statement happens to be equivalent to \(p_n\sim x\log{x}\). I will show this equivalence, using the argument made in An Introduction to the Theory of Numbers, though with much of the gaps and leaps in reasoning filled in by myself. The first thing to note is that \(p_n\) and \(\pi(n)\) are inverse functions. That is, \(\pi(p_n)=n\) and \(p_{\pi(n)}=n\). Now, we consider the function \[y=\frac{x}{\log{x}}.\] Taking the logarithm of both sides, we obtain, \[\log{y}=\log{x}-\log{\log{x}}.\] Now, we show that \(\log{\log{x}}=o(\log{x})\). By L'Hôpital's rule, \[\lim_{x\rightarrow\infty}{\frac{\log{\log{x}}}{\log{x}}}=\lim_{x\rightarrow\infty}{\frac{1}{\log{x}}}=0,\] as desired. So now, we can divide the logarithm of \(y\) by the logarithm of \(x\) to obtain \[\frac{\log{y}}{\log{x}}=1-\frac{\log{\log{x}}}{\log{x}}.\] In the limit, this approaches 1, so that \(\log{y}\sim \log{x}\). Hence, when we have \[x=y\log{x},\] we have that the LHS is the inverse of \(y\), but \(\log{x}\) is asymptotic to \(\log{y}\), so the inverse of \(y\) is asymptotic to \(y\log{y}\). Now, it suffices to show that for two divergent functions \(f\sim g\), we must also have \(f^{-1}\sim g^{-1}\). Observe that \[f^{-1}(f(x))=x\Rightarrow\lim_{x\rightarrow\infty}{\frac{f^{-1}(f(x))}{x}}=1.\] Likewise, we have \[\lim_{x\rightarrow\infty}{\frac{g^{-1}(g(x))}{x}}=1.\] But since \(f\sim g\), we can replace \(g\) with \(f\) in the limit to obtain \[\lim_{x\rightarrow\infty}{\frac{g^{-1}(f(x))}{x}}=1.\] Dividing our two limits, we obtain \[\lim_{x\rightarrow\infty}{\frac{f^{-1}(f(x))}{g^{-1}(f(x))}}=1,\] and since \(f\) is divergent, we can write this as \[\lim_{f(x)\rightarrow\infty}{\frac{f^{-1}(f(x))}{g^{-1}(f(x))}}=1,\] so \(f^{-1}\sim g^{-1}\), as desired. Hence, since \(\pi(n)\sim y\), the inverse of \(y\) is asymptotic to \(n\log{n}\), and \(p_n\) is the inverse of \(\pi(n)\), we have \[\boxed{p_n\sim n\log{n}}.\]
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Continuing on an MIT pset, we have the following equation for fluid flow. Let \(\mathbf{F}(x,y,t)=\rho (x,y,t)\mathbf{v}(x,y,t)\), where \(\rho\) is density and \(\mathbf{v}\) is velocity. Then,
\[\frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf{F}=0.\] This is the so-called equation of continuity. As it turns out, it is simply equivalent to the conservation of mass. We can see this by integrating the equation over some region \(R\). We obtain \[\iint_{R}{\frac{\partial\rho}{\partial t}\textrm{ d}A}=-\iint_{R}{\nabla\cdot\mathbf{F}\textrm{ d}A}.\] But now, using Green's theorem in normal form, the RHS becomes a line integral over \(C\), which is the positively-oriented boundary of \(R\): \[\iint_{R}{\frac{\partial\rho}{\partial t}\textrm{ d}A}=-\oint_{C}{M\textrm{ d}y-N\textrm{ d}x},\] where \(M\) and \(N\) are the \(x\) and \(y\) components of \(\mathbf{F}\), respectively. Observe that the RHS is now the negative of the fluid flux through \(C\). That is, the change in pressure over a region is equal to the negative of the flux through the boundary of that region. Obviously. If mass wants to leave a region, it must pass through the boundary of that region. It cannot simply "vanish". Mass is conserved. In a previous blog post, we talked about the convective derivative, and how it is used in defining incompressible flow. In particular, a flow is incompressible when the convective derivative of the pressure is zero. We can combine this with the equation of continuity to find another necessary and sufficient condition for flow incompressibility. We have \[\begin{split} \frac{D\rho}{Dt}&=\frac{\partial\rho}{\partial t}+\mathbf{v}\cdot\nabla\rho\\ &=0. \end{split}\] But by rearranging the product rule, we have \(\mathbf{v}\cdot\nabla\rho=\nabla\cdot(\rho\mathbf{v})-\rho\nabla\cdot\mathbf{v}\). So our convective derivative becomes \[\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})-\rho(\nabla\cdot\mathbf{v})=0.\] The first two terms sum to zero by the equation of continuity, which leaves us with \(\boxed{\nabla\cdot\mathbf{v}=0}\). Flow is incompressible only when the divergence of the velocity is zero. My linear class is weird.
Consider two continuous functions \(f,g\colon [0,1]\to\mathbb{R}\). Prove that \[\left|\int_{0}^{1}{f(x)g(x)\textrm{ d}x}\right|\leq\sqrt{\int_{0}^{1}{f(x)^2\textrm{ d}x}}\sqrt{\int_{0}^{1}{g(x)^2\textrm{ d}x}}.\] This problem is pretty trivial. Simply apply Cauchy-Schwarz to the Riemann sum approximation of the LHS and take the limit as \(n\rightarrow\infty\). This was a starred (challenge) problem in my linear homework following the week where we discussed Cauchy-Schwarz. Strange. Anyway, I've encountered a lot of cute problems lately. Among these is the following. Problem (HMMT): 10 students take a test. Each problem is solved by precisely 7 people and nine of the students solve exactly 4 of the problems. How many problems does the 10th student solve? Solution: Let the number of problems on the test be \(P\). Consider a set of \(P\) bins, with each corresponding to a particular problem, and a set of 10 bins, with each corresponding to a particular student. For every student that solves a certain problem, we place a stone in that problem's bin and for every problem that a student solves, we place a stone in that student's bin. Note that if we place a stone in a problem's bin, we must also place a stone in a student's bin. That is, there exists a bijection between the stones in the problem bins and the student bins. The total number of stones in the problem bins is given to be \(7P\) and the total number of stones in the student bins is \(x+4\cdot9\), where \(x\) is the number of problems that the 10th person solves. Hence, \[x=7P-36.\] Since \(x\geq0\), we must have \(P\geq6\). Furthermore, since the 10th person cannot solve more problems than are on the test, we have \(x=7P-36<P\), which means \(P\leq6\). Therefore, \(P=6\) and \(x=\boxed{6}\). Back in my (early) wannabe days, in the summer between sophomore and junior year, I was studying single-variable calculus. From the AoPS book, I found this extra sidenote.
Consider a positively oriented closed simple curve (a curve that is closed, non-self intersecting, and traced such that region enclosed by the curve is always on the left) parameterized by \((x(t),y(t))\) from \(t_0\) to \(t_1\). Then, the area of the enclosed region is given by: \[\frac{1}{2}\int_{t_0}^{t_1}{x(t)y'(t)-x'(t)y(t)\textrm{ d}t}.\] Back then, I could barely understand the statement, and once I did, I had no clue as to how it would be true. The sidenote said that it was a special case of Green's Theorem from vector calculus. Two years later, I can finally say that I know where this comes from. Green's theorem states that for a positively oriented closed simple curve \(C\) enclosing a region \(R\), and a vector field \(\mathbf{F}\), \[\oint_{C}{\mathbf{F}\cdot\textrm{d}\mathbf{r}}=\iint_{R}{\nabla\times\mathbf{F}\textrm{ d}A},\] where \(\nabla\times\mathbf{F}\) is the 2D curl (I know, it's a 3D thing so I'm abusing the notation a little). This is a cool result. For instance, it immediately establishes all the stuff about path-independence of line integrals over conservative fields which have to be curl-free and blah, blah, blah. If \(\nabla\times\mathbf{F}=1\), then the RHS evaluates to the area of \(R\). So we want a vector field, \(\mathbf{F}=\langle M,N\rangle\), such that \(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=1\). Perhaps the simplest \(\mathbf{F}\) satisfying this is \(\mathbf{F}=\left\langle -\frac{y}{2},\frac{x}{2}\right\rangle\). Then by Green's theorem, we have \[\begin{split} A&=\oint_{C}{-\frac{y}{2}\textrm{ d}x+\frac{x}{2}\textrm{ d}y}\\ &=\frac{1}{2}\oint_{C}{\left(x\frac{\textrm{d}y}{\textrm{d}t}-y\frac{\textrm{d}x}{\textrm{d}t}\right)\textrm{d}t}\\ &=\frac{1}{2}\int_{t_0}^{t_1}{x(t)y'(t)-x'(t)y(t)\textrm{ d}t}, \end{split}\] as desired. \(\square\) Cool beans. It's a shame that quantum and relativistic corrections are necessary. Though perhaps it's self-entitled of me to wish that the universe would be simple enough for me to understand it.
It turns out that current flows in such a way that power loss in resistors is minimized in parallel. This claim comes from problem 2 from a pset here. Consider two resistors in parallel. The current into a terminal is \(I\). Let the current the resistors be \(I_1\) and \(I_2\), and let the resistances be \(R_1\) and \(R_2\), respectively. Then by conservation of charge: $$I=I_1+I_2,$$ and since power dissipated through a resistor is \(I^2R\), we have a total power loss of: $$P(I_1,I_2)=I_1^2R_1+I_2^2R_2.$$ Now we have a simple optimization problem. We must minimize the function above subject to the constraint \(S(I_1,I_2)=I_1+I_2=I\). We proceed with Lagrange multipliers: $$\nabla P=\lambda\nabla S\Rightarrow\left<2I_1R_1,2I_2R_2\right>=\left<\lambda,\lambda\right>.$$ From this, we immediately obtain: $$I_1R_1=I_2R_2,$$ which if Ohm's law holds, is simply \(V\), the potential difference across the resistors! This physically makes sense. Potential difference is constant over resistors in parallel since all of the resistors of coinciding terminals and potential difference is path-independent. The calculation above easily generalizes to an arbitrary number of resistors. I am awaiting a response on StackExchange for a greater physical insight into why currents arrange themselves such that power dissipation is minimized. This seems to be a manifestation of a common theme throughout physics. For instance, Fermat's principle from optics states that light will always choose a path that minimizes travel time. This is an interesting principle, but it is actually due to the Huygens principle, in which waves are framed as propagating by new wavelets emanating from each wavefront. I am simply wondering if there is a similar underlying principle behind why currents travel so that power dissipation is minimized in parallel resistors. Here are two problems in a pset from MIT 18.02 (multivariable calculus).
Problem 2: Let \(f(x,y,z,t)\) be a smooth function, and let \(\nabla f=\left<f_x,f_y,f_z\right>\) be the gradient in space variables only. Let \(\mathbf{r}=\mathbf{r}(t)=\left<x(t),y(t),z(t)\right>\) be a smooth curve, and \(\mathbf{v}=\mathbf{r}'(t)\); and suppose we use the notation \(\frac{\textrm{D}f}{\textrm{D}t}=\frac{\textrm{d}}{\textrm{d}t}f(\mathbf{r}(t),t)\). Use the Chain Rule to show that \(\frac{\textrm{D}f}{\textrm{D}t}=\frac{\partial f}{\partial t}+\mathbf{v}\cdot\nabla f\). Solution: We have: \[f(\mathbf{r}(t),t)=f(x(t),y(t),z(t),t)\] By the Chain Rule: \[\frac{\textrm{d}}{\textrm{d}t}f(x(t),y(t),z(t),t)=\frac{\partial f}{\partial x}\frac{\textrm{d}x}{\textrm{d}t}+\frac{\partial f}{\partial y}\frac{\textrm{d}y}{\textrm{d}t}+\frac{\partial f}{\partial z}\frac{\textrm{d}z}{\textrm{d}t}+\frac{\partial f}{\partial t}\] Which becomes: \[\frac{\textrm{d}}{\textrm{d}t}f(x(t),y(t),z(t),t)=\frac{\partial f}{\partial t}+\left<\frac{\textrm{d}x}{\textrm{d}t},\frac{\textrm{d}y}{\textrm{d}t},\frac{\textrm{d}z}{\textrm{d}t}\right>\cdot\left<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right>\] Which is: \[\frac{\partial f}{\partial t}+\mathbf{v}\cdot\nabla f\] As desired. \(\square\) This function, \(\frac{\textrm{D}f}{\textrm{D}t}\), is called the convective derivative or the material derivative. In fact, there are quite a few names for this. It is important to realize that \(\mathbf{r}\) defines a path or trajectory through space. The function \(f\) then describes something that is changing along a trajectory with time. The next problem makes this clear, letting \(f=\rho\), the density of a fluid. When \(\rho\) is constant in \(t\), the flow is termed steady. Unsteady flow, as one can imagine by this definition, must be enormously complicated, and it includes phenomena such as turbulence. In the case of steady flow, each trajectory is called a streamline. A fluid flow is called incompressible if the convective derivative of \(\rho\) is zero. In steady flow, this means that there is no pressure change along a streamline (which makes sense!). Problem 3a: Suppose that the density function depends only on time \(t\) but is constant in the space variables \((x,y,z)\), that is, \(\rho=\rho(t)\). Then show that the flow is incompressible if and only if the density \(\rho(t)\) is constant in all the variables \((x,y,z,t)\) (in other words, the flow must be steady). Solution: We want: \[\frac{\partial \rho}{\partial t}+\mathbf{v}\cdot\nabla\rho=0\] But since \(\rho\) does not depend on spatial variables, \(\nabla\rho=0\) and \(\frac{\partial \rho}{\partial t}=\frac{\textrm{d}\rho}{\textrm{d}t}\). Hence: \[\frac{\textrm{d}\rho}{\textrm{d}t}=0\] Integrating both sides WRT \(t\): \[\int{\frac{\textrm{d}\rho}{\textrm{d}t}\textrm{ d}t}=\rho(t)=C\] Hence, the flow is steady. \(\square\) Problem 3b: Next suppose instead that the density depends only on the space variables \((x,y,z)\) but not (explicitly) on \(t\), so that \(\rho=\rho(x,y,z)\). An incompressible flow in this case is called stratified. Use the result of problem 2 to give the condition on \(\rho\) and \(\mathbf{v}\) for stratified flow. Solution: In this case: \[\mathbf{v}\cdot\nabla\rho=0\] So the velocity is always orthogonal to direction in which \(\rho\) changes the most. But recall that \(\nabla\rho\) is always orthogonal to the contour surfaces of \(\rho\). It follows that the velocity must always be parallel, and hence tangential to, surfaces of equal density. \(\square\) I was up till 3:30. There was no way that I would sleep before the problem would.
As it turns out, the integral equation that I discussed earlier is well-known. It is an example of a linear Volterra equation of the first kind. \[f(c)=\int_{a}^{c}{K(c,x)\rho(x)\textrm{ d}x}\] The function \(K\) is known as the kernel. In our case, the kernel is: \[K(c,x)=c+b-x\] The idea here is to use the Leibniz integral rule (which follows from Fundamental Theorem). By differentiating both sides with respect to \(c\), we obtain: \[\frac{\textrm{d}}{\textrm{d}c}\left(\int_{c}^{\infty}{(c+b-x)\rho(x)\textrm{ d}x}\right)=\frac{\textrm{d}R}{\textrm{d}t}\lim_{R\rightarrow\infty}{\left[(c+b-R)\rho(R)\right]}-b\rho(x)+\int_{c}^{\infty}{\rho(x)\textrm{ d}x}\] Observe that if \(G(c, R)=\int_{c}^{R}{(c+b-x)\rho(x)\textrm{ d}x}\) is well-behaved, then we may interchange the limit and derivative above to obtain: \[\frac{\textrm{d}R}{\textrm{d}t}\lim_{R\rightarrow\infty}{\left[(c+b-R)\rho(R)\right]}=\lim_{R\rightarrow\infty}{\left[(c+b-R)\rho(R)\frac{\textrm{d}R}{\textrm{d}t}\right]}=0\] As cited, the conditions we want for \(G\) for this manipulation to be valid are:
Anyway, we have: \[b\rho(c)=\int_{c}^{\infty}{\rho(x)\textrm{ d}x}\] Let \(P'=\rho\). Then, by the Fundamental Theorem: \[bP'(c)=\lim_{R\rightarrow\infty}{P(R)}-P(c)\] Let \(L=\lim_{R\rightarrow\infty}{P(R)}\). Then, the equation above rearranges to: \[P'+\frac{1}{b}P=\frac{L}{b}\] This is simply a first-order linear ODE. We solve this by using an integrating factor of \(\exp{\int{\frac{1}{b}\textrm{ d}{x}}}\). This yields: \[P(x)=L+Ce^{-x/b}\] Differentiating this, we obtain: \[\boxed{\rho(x)=Ce^{-x/b}}\] Obviously with \(C>0\). Uniform convergence is forced by the fact that this solution satisfies our Volterra equation. It can probably also be proven with \(\epsilon\)-\(\delta\) calculations. That's for another time. And I'm done with mechanics for this weekend I think. Here's what I do know.
For brevity, let \(K=c+b\). Let \(u=\rho(x)\) and \(\textrm{d}v=K-x\textrm{ d}x\). Then, integrating by parts: \[\int{(K-x)\rho(x)\textrm{ d}x}=\rho(x)\left(Kx-\frac{1}{2}x^2\right)-\int{\left(Kx-\frac{1}{2}x^2\right)\rho'(x)\textrm{ d}x}\] We continue integrating by parts, letting the polynomial in each new integrand be \(\frac{\textrm{d}v}{\textrm{d}x}\), and the derivatives of \(\rho(x)\) be \(u\). This then yields: \[\int{(K-x)\rho(x)\textrm{ d}x}=\sum_{n=0}^{\infty}{(-1)^n\rho^{(n)}(x)\left(\frac{Kx^{n+1}}{(n+1)!}-\frac{x^{n+2}}{(n+2)!}\right)}\] Ok Andrew, cool. Now what? Ah, simply observe. Our desired integral is the improper integral evaluated from \(c\) to \(\infty\). This means that the limit of the antiderivative (above) as \(x\) approaches \(\infty\) must converge! In other words: \[\lim_{x\rightarrow\infty}{\sum_{n=0}^{\infty}{(-1)^n\rho^{(n)}(x)\left(\frac{Kx^{n+1}}{(n+1)!}-\frac{x^{n+2}}{(n+2)!}\right)}}\] Must converge! For this to occur, every single term must converge! But observe that the polynomials in each term of the summation tend to \(\infty\), hence for convergence to occur, every derivative of \(\rho(x)\), along with \(\rho(x)\) itself, must tend to \(0\) (this follows plainly from L'Hôpital's rule). In other words, we have derived: \[\lim_{x\rightarrow\infty}{\rho^{(n)}(x)}=0\textrm{ }\forall n\in\mathbb{N}_0\] This in and of itself hints at possible functions. For instance, an exponential decay function, or something of the form \(Ax^n\) for some \(n<0\). Ok sure, we can substitute these functions into our original integral equation and find that one of them actually works, but I want a little more insight. And rigor. And what sort of lame solution is that? Deriving one property and guessing and checking? Ew. The next thing I will use is the Leibniz integral rule. That is: \[\frac{\textrm{d}}{\textrm{d}x}\left(\int_{a(x)}^{b(x)}{f(x,t)\textrm{ d}t}\right)=f(x,b(x))\cdot\frac{\textrm{d}}{\textrm{d}x}b(x)-f(x,a(x))\cdot\frac{\textrm{d}}{\textrm{d}x}a(x)+\int_{a(x)}^{b(x)}{\frac{\partial}{\partial x}f(x,t)\textrm{ d}t}\] While this looks complicated, it just follows from the Fundamental Theorem of Calculus. And while I'd love to continue this post with the application of this rule, I should probably do my history for once. The next post will include this, and discussion of uniform continuity, and uniform and pointwise convergence. Stay tuned. I'm stuck on a Morin problem.
Consider a semi-infinite stick (one that extends infinitely in one direction) with a linear mass density \(\rho(x)\). The stick enjoys the property that it can be cut anywhere, have a fulcrum placed at a distance \(b\) away from the cut, and balance on the fulcrum. Find \(\rho(x)\). Here is what I have so far. Suppose that the end of the stick is at \(x=0\). Let a cut be made at \(x=c\). Then, the fulcrum is placed at \(x=c+b\). Balancing torques about the fulcrum: \[\int_{c}^{c+b}{(c+b-x)\rho(x)\textrm{ d}x}=\int_{c+b}^{\infty}{(x-c-b)\rho(x)\textrm{ d}x}\] This can be rearranged to: \[\int_{c}^{\infty}{(c+b-x)\rho(x)\textrm{ d}x}=0\] This is an integral equation. I have no ideas as to how to solve it. I refuse to look at the solution just yet. This is from David Morin's Introductory Classical Mechanics.
The problem is to consider a rope of linear mass density \(\rho\) and length \(\ell\) lying on an inclined plane with an initial angle of \(0\) which is then raised to some angle \(\theta\). Suppose a coefficient of friction, \(\mu\) exists between the plane and the rope, and that the topmost end of the rope is nailed to the plane. What is the tension at the top of the rope? Morin approaches this by first parameterizing the position along the rope as \(z=x\sec{\theta}=y\csc{\theta}\), where the plane is oriented with the bottommost point of the plane (also the bottom end of the rope) at the origin, and the plane extends into the first quadrant. Next, he considers an infinitesimal segment of the rope, from \(z\) to \(z+\textrm{d}z\). The upward forces on this segment are \(T(z+\textrm{d}z)\), the tension on the upper end of the rope, and \(F_f(z)\textrm{d}z\), the force of friction on the segment, where the function \(F_f(z)\) gives the force of friction per unit length for a given \(z\) coordinate. The downward forces on the segment are \(T(z)\), the tension on the lower end of the rope, and \(\rho g\sin{\theta}\textrm{ d}z\), the weight down the plane of the segment. Since the rope is at rest everywhere: \[T(z+\textrm{d}z)+F_f(z)\textrm{d}z=T(z)+\rho g\sin{\theta}\textrm{ d}z\] Morin then states: "Expanding this to first order in \(\textrm{d}z\) gives" \[\frac{\textrm{d}T}{\textrm{d}z}=\rho g\sin{\theta}-F_f(z)\] Now I don't know what it means to expand something to first order in some differential unit, but I can arrive where Morin did by rearranging the first equation as: \[\frac{T(z+\textrm{d}z)-T(z)}{\textrm{d}z}=\rho g\sin{\theta}-F_f(z)\] And then observing that the LHS is equivalent to the limit definition of the derivative: \[\frac{T(z+\textrm{d}z)-T(z)}{\textrm{d}z}=\lim_{h\rightarrow 0}{\frac{T(z+h)-T(z)}{h}}=\frac{\textrm{d}T}{\textrm{d}z}\] Is that what it means to "expand to the first order"? Anyway, Morin continues by asserting that if there is to be any tension at all in the top portion of the rope (or any portion of the rope for that matter), we must have \(\tan{\theta}>\mu\) (which is a well-known, easily obtainable result). In that case, frictional force is maximized: \[F_f(z)=\rho g\mu\cos{\theta}\] The result follows upon integration: \[\int{\textrm{d}T}=\int{\rho g\sin{\theta}-\rho g\mu\cos{\theta}\textrm{ d}z}\] Using the initial condition \(T(0)=0\), we conclude: \[T(\ell)=\rho g(y_0-\mu x_0)\] Where \((x_0,y_0)\) is the location of the top end of the rope. I went about this problem in an almost correct way. I forgot to account for the fact that the tension in the bit of rope above each infinitesimal segment pulled upward on that segment. :(. Hey, now I know better. Yes, solutions to problems from Introductory Classical Mechanics will be up soon. I will present solutions to problems that I have solved myself correctly (so not this one). Otherwise, I'm effectively regurgitating the solutions manual. |
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