First, define the Lorenz curve: it is the curve that consists of all points $(u,v)$ such that the poorest $u$ portion of population in the country owns $v$ portion of the total wealth.

The Gini coefficient $G/\mu$ is defined as the area between the Lorenz curve and the line $u=v$ divided by the area enclosed by the three lines $u=v$, $v=0$, and $u=1$.

Now, suppose the wealth distribution in the country is $p(X)$, where $p\!\left(x\right)\mathrm dx$ is the portion of population that has wealth in the range $[x,x+\mathrm dx]$.

Then, the Lorenz curve is the graph of the function $g$ defined as $$g(F(x))=\frac1\mu\int_{-\infty}^xtp\!\left(t\right)\mathrm dt,$$ where $$F\!\left(x\right)\coloneqq\int_{-\infty}^xp\!\left(t\right)\mathrm dt$$ is the cumulative distribution function of $p(X)$, and $$\mu\coloneqq\int_{-\infty}^{+\infty}tp\!\left(t\right)\mathrm dt$$ $$(1)$$ is the average wealth of the population, which is just $\mathrm E[\mathrm X]$ ($X$ is a random variable such that $X\sim p(X)$).

Then, the Lorenz curve is $$v=g(u)\coloneqq\frac1\mu\int_{-\infty}^{F^{-1}(u)}tp\!\left(t\right)\mathrm dt.$$

According to the definition of the Gini coefficient, $$\begin{align*} G&\coloneqq2\mu\int_0^1\left(u-g(u)\right)\mathrm du\\ &=\mu-2\mu\int_0^1g\!\left(u\right)\mathrm du\\ &=\mu-2\int_{u=0}^1\int_{t=-\infty}^{F^{-1}(u)}tp\!\left(t\right)\mathrm dt\,\mathrm du. \end{align*}$$ Interchange the order of integration, and we have $$\begin{align*} G&=\mu-2\int_{t=-\infty}^{+\infty}\int_{u=F(t)}^1tp\!\left(t\right)\mathrm dt\,\mathrm du\\ &=\mu-2\int_{-\infty}^{+\infty}\left(1-F(t)\right)tp\!\left(t\right)\mathrm dt. \end{align*}$$ Substitute Equation 1 into the above equation, and we have $$\begin{align*} G&=\int_{-\infty}^{+\infty}2tF\!\left(t\right)p\!\left(t\right)\mathrm dt-\mu\\ &=\int_{-\infty}^{+\infty}\left(2tF\!\left(t\right)-1\right)tp\!\left(t\right)\mathrm dt\\ &=\int_0^1\left(2u-1\right)F^{-1}\!\left(u\right)\mathrm du. \end{align*}$$ Now here is the neat part. Separate it into two parts, and write them in double integrals: $$\begin{align*} G&=\int_0^1uF^{-1}\!\left(u\right)\mathrm du-\int_0^1\left(1-u\right)F^{-1}\!\left(u\right)\mathrm du\\ &=\int_{u_2=0}^1\int_{u_1=0}^{u_2}F^{-1}\!\left(u_2\right)\mathrm du_1\,\mathrm du_2 -\int_{u_1=0}^1\int_{u_2=u_1}^1F^{-1}\!\left(u_1\right)\mathrm du_1\,\mathrm du_2. \end{align*}$$ Interchange the order of integration of the second term, and we have $$\begin{align*} G&=\int_{u_2=0}^1\int_{u_1=0}^{u_2}\left(F^{-1}\!\left(u_2\right)-F^{-1}\!\left(u_1\right)\right)\mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{u_2=0}^1\int_{u_1=0}^1\left|F^{-1}\!\left(u_2\right)-F^{-1}\!\left(u_1\right)\right|\mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\left|x_2-x_1\right|p\!\left(x_1\right)p\!\left(x_2\right)\mathrm dx_1\,\mathrm dx_2\\ &=\frac12\mathrm E\!\left[\left|X_2-X_1\right|\right], \end{align*}$$ where $X_1$ and $X_2$ are two independent random variables with $p$ being their respective distribution functions: $\left(X_1,X_2\right)\sim p\!\left(X_1\right)p\!\left(X_2\right)$.

By this result, we can easily see how the Gini coefficient represents the statistical dispersion.

We can apply similar tricks to the variance $\sigma_X^2$. $$\begin{align*} \sigma_X^2&=\mathrm E\!\left[X^2\right]-\mathrm E\!\left[X\right]^2\\ &=\int_{-\infty}^{+\infty}t^2p\!\left(t\right)\mathrm dt -\left(\int_{-\infty}^{+\infty}tp\!\left(t\right)\mathrm dt\right)^2\\ &=\int_0^1F^{-1}\!\left(u\right)^2\,\mathrm du -\left(\int_0^1F^{-1}\!\left(u\right)\mathrm du\right)^2. \end{align*}$$ Separate the first into two halves, and write the altogether three terms in double integrals: $$\begin{align*} \sigma_X^2&=\frac12\int_0^1F^{-1}\!\left(u_2\right)^2\,\mathrm du_2\int_0^1\mathrm du_1\\ &\phantom{=~}{}-\int_0^1F^{-1}\!\left(u_1\right)\mathrm du_1\int_0^1F^{-1}\!\left(u_2\right)\mathrm du_2\\ &\phantom{=~}{}+\frac12\int_0^1F^{-1}\!\left(u_1\right)^2\,\mathrm du_1\int_0^1\mathrm du_2\\ &=\frac12\int_0^1\int_0^1 \left(F^{-1}\!\left(u_2\right)^2-2F^{-1}\!\left(u_1\right)F^{-1}\!\left(u_2\right)+F^{-1}\!\left(u_1\right)^2\right) \mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \left(x_2-x_1\right)^2p\!\left(x_1\right)p\!\left(x_2\right)\mathrm dx_1\,\mathrm dx_2\\ &=\frac12\mathrm E\!\left[\left(X_2-X_1\right)^2\right]. \end{align*}$$ Then we can derive the relationship between the Gini coefficient and the variance: $$2\sigma_X^2-4G^2=\sigma_{\left|X_2-X_2\right|}^2.$$

The aliens intiated their attack to the earth! They shoot bullets with mass $m$ and speed $v$ from a far-awar planet. To defend, humans built a field $U=\alpha/r$ that can repel the bullets. What regions are safe?

Every possible trajectory of the bullet is parameterized by $b$, the impact parameter. The bullet has energy $E=\frac12mv^2$ and angular momentum $M=mvb$, which are conserved. According to the well-known results of Kepler problem, the trajectory is a hyperbola $$-\frac pr=1+e\cos\varphi,$$ where $$p\coloneqq\frac{M^2}{m\alpha},\quad e\coloneqq\sqrt{1+\frac{2EM^2}{m\alpha^2}}.$$ For convenience, denote the radius of the hyperbola as $$a\coloneqq\frac\alpha{2E},$$ then we can write the equation of the trajectory as $$-\frac{b^2}{ar}=1+\sqrt{1+\frac{b^2}{a^2}}\cos\varphi.$$ Rotate the trajectory so that the incident direction is always towards the positive $x$ direction: $$0=F\!\left(r,\varphi,b\right)\coloneqq\frac{b^2}{ar}+1+\cos\varphi+\frac ba\sin\varphi.$$ $$(1)$$ To find the envelope of the family of trajectories, solve $$0=\frac{\partial F}{\partial b}=\frac{2b}{ar}+a\sin\varphi,$$ and we have $$b=-\frac12r\sin\varphi.$$ Substitute back into Equation 1, and we have finally the equation of the envelope: $$\frac{4a}r=1-\cos\varphi,$$ which is a parabola with the the semi-latus rectum being $4a$. Therefore, the safe regions are the interior of a circular paraboloid.

Each hair grows at constant speed $v$. However, they cannot grow indefinitely because there is a probability of $\lambda$ per unit time for a hair to be lost naturally (this is the same assumption as the exponential decay). After a hair is lost, it restarts growing from zero length.

Suppose that at $t=0$ you have got a haircut so that the hair length distribution becomes $$f(l,0)=f_0(l).$$ $$(1)$$ Then, how does the distribution evolve with time?

Because $f$ is a distribution function, There is a normalization restriction on $f$: $$\int_0^\infty f\!\left(l,t\right)\mathrm dl=1,\quad t\ge0.$$ $$(2)$$ This normalization condition also applies to the initial condition ($t=0$). This means that the function $f_0$ also satisfies the normalization restriction (Equation 2) $$\int_0^\infty f_0\!\left(l\right)\mathrm dl=1.$$ $$(3)$$ Because of the natural loss of hair, only a $1-\lambda\,\mathrm dt$ portion of hair will survive $\mathrm dt$. According to this, we can construct the following equation: $$f\!\left(l+v\,\mathrm dt,t+\mathrm dt\right)= \left(1-\lambda\,\mathrm dt\right)f\!\left(l,t\right).$$ This equation can be reduced to a first-order linear PDE: $$v\frac{\partial f}{\partial l}+\frac{\partial f}{\partial t}+\lambda f=0.$$ $$(4)$$

Define $\theta\coloneqq l-vt$, and define a new function $$g\!\left(\theta,t\right)\coloneqq f\!\left(\theta+vt,t\right).$$ Then, $$\frac{\partial g}{\partial t}=v\frac{\partial f}{\partial l}+\frac{\partial f}{\partial t}.$$ Then, Equation 4 can be reduced to $$\frac{\partial g}{\partial t}+\lambda g=0.$$ The solution is $$g\!\left(\theta,t\right)=\Phi\!\left(\theta\right)\mathrm e^{-\lambda t},$$ where $\Phi$ is an arbitrary function defined on $\mathbb R$. Therefore, the general solution to Equation 4 is $$f\!\left(l,t\right)=\Phi\!\left(l-vt\right)\mathrm e^{-\lambda t}.$$ $$(5)$$

By utilizing Equation 1, we can find $\Phi(\theta)$ for $\theta>0$. Substitute $t=0$ into Equation 5 and compare with Equation 1, and we have $$\Phi(\theta>0)=f_0(\theta).$$ $$(6)$$ This only gives $\Phi(\theta)$ for $\theta>0$ because $f_0$ is not defined on negative numbers. The rest of $\Phi$, however, may be deduced from Equation 2.

Substitute Equation 5 into Equation 2, and we have $$\begin{align*} 1&=\int_0^\infty\Phi\!\left(l-vt\right)\mathrm e^{-\lambda t}\,\mathrm dl\\ &=\mathrm e^{-\lambda t}\left( \int_0^{vt}\Phi\!\left(l-vt\right)\mathrm dl +\int_{vt}^\infty\Phi\!\left(l-vt\right)\mathrm dl \right)\\ &=\mathrm e^{-\lambda t}\left( \int_{-vt}^0\Phi\!\left(\theta\right)\mathrm d\theta +\int_0^\infty f_0\!\left(\theta\right)\mathrm d\theta \right)\\ &=\mathrm e^{-\lambda t}\left(1-\int_0^{-vt}\Phi\!\left(\theta\right)\mathrm d\theta\right). \end{align*}$$ Therefore, we have the integral of $\Phi$ on negative intervals: $$\int_0^{-vt}\Phi\!\left(\theta\right)\mathrm d\theta =1-\mathrm e^{\lambda t}.$$ Find the derivative of both sides of the equation w.r.t. $t$, and we have $$-v\Phi\!\left(-vt\right)=-\lambda\mathrm e^{\lambda t}.$$ In other words, $$\Phi\!\left(\theta<0\right)=\frac\lambda v\mathrm e^{-\frac\lambda v\theta}.$$ $$(7)$$

Combining Equation 6 and 7 and substituting back to Equation 5, we can find the special solution to Equation 4 subject to restrictions Equation 2 and 1: $$f(l,t)=\begin{cases} \frac\lambda v\mathrm e^{-\frac\lambda vl},&0<l<vt,\\ \mathrm e^{-\lambda t}f_0(l-vt),&l>vt. \end{cases}$$ $$(8)$$

This is our final answer.

This result is interesting in that any distribution will finally evolve into an exponential distribution with the rate parameter being $\frac\lambda v$ as $t\to\infty$ no matter what the initial distribution is: $$f(l,\infty)=\frac\lambda v\mathrm e^{-\frac\lambda vl}.$$ This distribution is the stationary solution to Equation 4. This is actually a normal behavior for first-order PDEs. For example, the thermal equilibrium state is the stationary solution to the heat equation, and any other solution approaches to the stationary solution over time.

This behavior can explain why human body hair tends to grow to only a certain length instead of being indefinitely long. You may try shaving your leg hair and wait for some weeks. You can observe that they grow to approximately the original length but not any longer. It is similar for your hair (on top of your head), but $\lambda$ of hair is so small that it can hardly reach its terminal length if you get haircuts regularly.

Another thing to note is that this may explain a phenomenon that we may observe: the longer your hair is, the more slowly it grows, and your hair no longer seems to grow when it reaches a certain length. If the length of hair that we observe is actually the mean length of the hair, then it is $$\mu(t)=\int_0^\infty lf\!\left(l,t\right)\mathrm dl =\left(\mu_0-\frac v\lambda\right)\mathrm e^{-\lambda t}+\frac v\lambda,$$ where $\mu_0$ is the mean of the distribution $f_0$. It can be seen that the growth rate of the mean length of hair varies exponentially.

Suppose we have an ODE (with initial conditions) $$x'\!\left(t\right)=f\!\left(x\!\left(t\right),t\right), \quad x\!\left(t_0\right)=C,$$ $$(1)$$ where $x$ is the unknown function, and $f$ is Lipschitz continuous in its first argument and continuous in its second argument. By Picard–Lindelöf theorem, we can seek the unique solution within $t\in\left[t_0-\varepsilon,t_0+\varepsilon\right]$.

Here, I propose the following method: we can write out a sequence of functions defined by $$x_0\!\left(t\right)\coloneqq C,$$ $$(2)$$ $$x_{n+1}\!\left(t\right)\coloneqq\int_{t_0}^tf\!\left(x_n\!\left(s\right),s\right)\,\mathrm ds+C$$ $$(3)$$ (the properties of $f$ guarantee that the integral is well-defined). Then, if the sequence $\left(x_n'\right)$ converges uniformly on $\left[t-\varepsilon,t+\varepsilon\right]$ (question: can this condition actually be proved?), then the sequence $\left(x_n\right)$ converges uniformly to a function $x$ on $\left[t-\varepsilon,t+\varepsilon\right]$, which is the unique solution to Equation 1.

Proof

The proof is easy. Note from Equation 3 that $$x_{n+1}'\!\left(t\right)=f\!\left(x_n\!\left(t\right),t\right).$$ Then, take the limit $n\to\infty$. By the uniform convergence, we recovers Equation 1.

An example

Suppose $f\!\left(x,t\right)\coloneqq x$ and $t_0\coloneqq0$, then $$x_n\!\left(t\right)=C\sum_{j=0}^{n-1}\frac{t^j}{j!}.$$ This is because $$\int_0^tx_n\!\left(s\right)\,\mathrm ds+C=C\sum_{j=0}^{n-1}\frac{t^{j+1}}{j!\left(j+1\right)}+C=x_{n+1}\!\left(t\right).$$ By taking the limit $n\to\infty$, we get $x\!\left(t\right)=C\mathrm e^t$, which means that the unique solution to $x'\!\left(t\right)=x\!\left(t\right)$ with initial condition $x\!\left(0\right)=C$ is $x\!\left(t\right)=C\mathrm e^t$, expectedly.

Approximation

This method is good because integration is sometimes much easier than solving ODE. We can use integration to get functions that are close to the exact solution. A natural question to ask is how close $x_n$ is to the exact solution $x$.

If Equation 1 is defined on $\mathbb R^n$ (where we may define differences and infinitesimals), it is clear that $x_n-x$ is an infinitesimal of higher order than $\left(t-t_0\right)^n$.

Some stipulations:

• Without special statements, all vectors appearing in this article are $n$-dimensional vectors, $n\in\mathbb N$;
• Iteration variable $k$ always iterates over $\left[0,n\right)\cup\mathbb Z$;
• $\operatorname{sum}\vec\xi\coloneqq\sum_k\xi_k$;
• $\operatorname{prod}\vec\xi\coloneqq\prod_k\xi_k$;
• If the independent and dependent variables of function $f$ are both scalars, then define $f\!\left(\vec\xi\right)\coloneqq\left(f\!\left(\xi_0\right),f\!\left(\xi_1\right),\ldots,f\!\left(\xi_n\right)\right)$;
• $\vec\xi^{\vec\eta}\coloneqq\prod_k\xi_k^{\eta_k}$;
• $\min\vec\xi\coloneqq\min_k\xi_k$;
• $\max\vec\xi\coloneqq\max_k\xi_k$;
• $\delta_{\xi,\eta}\coloneqq\begin{cases}1,&&\xi=\eta,\\0,&&\xi\ne\eta;\end{cases}$
• By saying $\vec\xi$ is congruent, all components of $\vec\xi$ are equal to each other.

Definition 1. Suppose we have samples $\vec x\in\left(\mathbb R^+\right)^n$, weights $\vec w\in\left\{\vec\xi\in\left(\mathbb R^+\right)^n\,\middle|\,\operatorname{sum}\vec\xi=1\right\}$, and parameter $p\in\left[-\infty,+\infty\right]$. Define the Hölder mean by $$M_{p,\vec w}\!\left(\vec x\right)\coloneqq\left(\vec w\cdot\vec x^p\right)^{\frac 1p}.$$

Note. The function is indefinite when $p\in\left\{-\infty,0,+\infty\right\}$, but actually there exist limits $$\lim_{p\to0}M_{p,\vec w}\!\left(\vec x\right)=\vec x^{\vec w},$$ $$\lim_{p\to-\infty}M_{p,\vec w}\!\left(\vec x\right)=\min\vec x,$$ $$\lim_{p\to+\infty}M_{p,\vec w}\!\left(\vec x\right)=\max\vec x.$$ The limits are to be proved as theorems later. We can use them to define the Hölder mean for $p\in\left\{-\infty,0,+\infty\right\}$.

Theorem 1. $$\lim_{p\to0}M_{p,\vec w}\!\left(\vec x\right)=\vec x^{\vec w}.$$

Proof. $$\begin{aligned} \lim_{p\to0}M_{p,\vec w}\!\left(\vec x\right) &=\lim_{p\to0}\left(\vec w\cdot\vec x^p\right)^{\frac 1p} &\text{(Definition 1)}\\ &=\lim_{p\to0}\exp\frac{\ln\!\left(\vec w\cdot\vec x^p\right)}p\\ &=\exp\lim_{p\to0}\frac{\ln\!\left(\vec w\cdot\vec x^p\right)}p\\ &=\exp\lim_{p\to0}\frac{\vec w\cdot\left(\vec x^p\ln\vec x\right)}{\vec w\cdot\vec x^p} &\text{(L'H\^opital's rule)}\\ &=\exp\!\left(\vec w\cdot\ln\vec x\right)\\ &=\vec x^{\vec w}. \end{aligned}$$ $\square$

Theorem 2. $$M_{p,\vec w}\!\left(\vec x\right)=M_{-p,\vec w}\!\left(\vec x^{-1}\right)^{-1}.$$

Proof. $$\begin{aligned} M_{p,\vec w}\!\left(\vec x\right) &=\left(\vec w\cdot\vec x^p\right)^{\frac 1p} &\text{(Definition 1)}\\ &=\left(\left(\vec w \cdot\left(\vec x^{-1}\right)^{-p}\right)^{-\frac1p}\right)^{-1}\\ &=M_{-p,\vec w}\!\left(\vec x^{-1}\right)^{-1} &\text{(Definition 1)} \end{aligned}$$ $\square$

Theorem 3. $$\lim_{p\to+\infty}M_{p,\vec w}\!\left(\vec x\right)=\max\vec x.$$

Proof. Because $\forall k:\frac{x_k}{\max\vec x}\le1$, then $\lim_{p\to+\infty}\left(\frac{\vec x}{\max\vec x}\right)^p=\delta_{\max\vec x},\vec x$. $$\begin{aligned} \lim_{p\to+\infty}M_{p,\vec w}\!\left(\vec x\right) &=\lim_{p\to+\infty}\left(\vec w\cdot\vec x^p\right)^{\frac 1p} &\text{(Definition 1)}\\ &=\left(\max\vec x\right)\lim_{p\to+\infty}\left(\vec w\cdot\left(\frac{\vec x}{\max\vec x}\right)^p\right)^{\frac 1p}\\ &=\max\vec x\left(\vec w\cdot\lim_{p\to+\infty}\left(\frac x{\max\vec x}\right)^p\right)^{\lim_{p\to+\infty}\frac 1p}\\ &=\left(\max\vec x\right)\left(\vec w\cdot\delta_{\left(\max\vec x\right),\vec x}\right)^0\\ &=\max\vec x. \end{aligned}$$ $\square$

Theorem 4. $$\lim_{p\to-\infty}M_{p,\vec w}\!\left(\vec x\right)=\min\vec x.$$

Proof. $$\begin{aligned} \lim_{p\to-\infty}M_{p,\vec w}\!\left(\vec x\right) &=\lim_{p\to-\infty}M_{-p,\vec w}\!\left(\vec x^{-1}\right)^{-1} &\text{(Theorem 2)}\\ &=\lim_{p\to+\infty}M_{p,\vec w}\!\left(\vec x^{-1}\right)^{-1}\\ &=\max\left(\vec x^{-1}\right)^{-1} &\text{(Theorem 3)}\\ &=\min\vec x. \end{aligned}$$ $\square$

Theorem 5. If $p>q$, then $$M_{p,\vec w}\!\left(\vec x\right)\ge M_{q,\vec w}\!\left(\vec x\right),$$ where the equality holds iff $\vec x$ is congruent.

Proof. Case 1: $p>q>0$.

Let $f:\mathbb R^+\to\mathbb R^+:\xi\mapsto\xi^{\frac pq}$, then it has second derivative $$\frac{\mathrm d^2f\!\left(\xi\right)}{\mathrm d\xi^2}=\frac pq\left(\frac pq-1\right)\xi^{\frac pq-2}.$$ Because $p>q>0$, then $\frac pq\left(\frac pq-1\right)>0$, and then $\frac{\mathrm d^2f}{\mathrm d\xi^2}>0$, i.e. $f$ is convex. Therefore, according to Jensen’s inequality, $$\vec w\cdot f\!\left(\vec x^q\right)\ge f\!\left(\vec w\cdot\vec x^q\right),$$ i.e. $$\vec w\cdot\vec x^p\ge\left(\vec w\cdot\vec x^q\right)^{\frac pq}.$$ Take $\frac1p$th power to both sides of the equation. Without changing the direction of the inequality sign, we have $$\vec w\cdot\vec x^p\ge\vec w\cdot\vec x^q,$$ i.e. (according to Definition 1) $$M_{p,\vec w}\!\left(\vec x\right)\ge M_{q,\vec w}\!\left(\vec x\right).$$ According to the condition for the equality to hold in Jensen’s inequality, the equality holds iff $\vec x$ is congruent.

Case 2: $p>q=0$.

Because the logarithm function is concave, according to Jensen’s inequality, $$\ln\!\left(\vec w\cdot\vec x^p\right)\ge\vec w\cdot\ln\vec x^p.$$ Take exponential on both sides of the equation, and we have $$\vec w\cdot\vec x^p\ge\vec x^{p\vec w}.$$ Take $\frac1p$th power to both sides of the equation. Without changing the direction of the inequality sign, we have $$\left(\vec w\cdot\vec x^p\right)^{\frac1p}\ge\vec x^{\vec w},$$ i.e. (according to Definition 1) $$M_{p,\vec w}\!\left(\vec x\right)\ge M_{q,\vec w}\!\left(\vec x\right).$$ According to the condition for the equality to hold in Jensen’s inequality, the equality holds iff $\vec x$ is congruent.

Case 3: $p=0>q$.

$$\begin{align*} M_{q,\vec w}\!\left(\vec x\right) &=M_{-q,\vec w}\!\left(\vec x^{-1}\right)^{-1} &\text{(Theorem 2)}\\ &\le M_{0,\vec w}\!\left(\vec x^{-1}\right)^{-1} &\text{(Case 2)}\\ &=M_{0,\vec w}\!\left(\vec x\right). &\text{(Theorem 2)} \end{align*}$$ The equality holds iff $\vec x$ is congruent (Case 2).

Case 4: $0>p>q$.

Because $-q>-p>0$, we have $$\begin{align*} M_{q,\vec w}\!\left(\vec x\right) &=M_{-q,\vec w}\!\left(\vec x^{-1}\right)^{-1} &\text{(Theorem 2)}\\ &\le M_{-p,\vec w}\!\left(\vec x^{-1}\right)^{-1} &\text{(Case 1)}\\ &=M_{p,\vec w}\!\left(\vec x\right). &\text{(Theorem 2)} \end{align*}$$ The equality holds iff $\vec x$ is congruent (Case 1).

By all 4 cases, the original proposition is proved. $\square$

Corollary (HM-GM-AM-QM inequalities). $$\min\vec x\le n\left(\sum\vec x^{-1}\right)^{-1} \le\left(\prod\vec x\right)^\frac1n \le\frac{\sum\vec x}n \le\sqrt{\frac{\sum\vec x^2}{n}} \le\max\vec x,$$ where the equality holds iff $\vec x$ is congruent.

Proof. Let $\vec w=\left(\frac1n,\dots,\frac1n\right)$. Then according to Theorem 5, $$M_{-\infty,\vec w}\!\left(\vec x\right) \le M_{-1,\vec w}\!\left(\vec x\right) \le M_{0,\vec w}\!\left(\vec x\right) \le M_{1,\vec w}\!\left(\vec x\right) \le M_{2,\vec w}\!\left(\vec x\right) \le M_{+\infty,\vec w}\!\left(\vec x\right),$$ i.e. (according to Definition 1) $$\min\vec x\le n\left(\sum\vec x^{-1}\right)^{-1} \le\left(\prod\vec x\right)^\frac1n \le\frac{\sum\vec x}n \le\sqrt{\frac{\sum\vec x^2}{n}} \le\max\vec x,$$ where the equality holds iff $\vec x$ is congruent (Theorem 5). $\square$

Notice: Because this article was written very early, there are many mistakes and inappropriate notations.

(I’m going to challenge writing articles without sum symbols!)

In this article, functions all refer to unary functions; both independent and dependent variables are scalars; and differentials refer to ordinary differentials.

Definition 1 (linear differential operator). Let $L\!\left(\mathrm D,x\right)$ be a differential operator. If for any two functions $F_1\!\left(x\right),F_2\!\left(x\right)$ and constants $C_1,C_2$, the operator satisfies $$L\left(C_1F_2+C_2F_2\right)=C_1LF_1+C_2LF_2,$$ then $L$ is a linear differential operator.

Lemma 1. The sufficient and necessary condition for $L$ to be a linear differential operator is that for any tuple of functions $\vec F\!\left(x\right)$ and any tuple of constants $\vec C$ (the dimensions of the two vectors are the same), the operator satisfies $$L\left(\vec C\cdot\vec F\right)=\vec C\cdot L\vec F.$$

Brief proof. Directly letting $\vec C$ and $\vec F$ be 2-dimensional vectors and using Definition 1, the sufficiency can be proved; by using mathematical induction on the dimension of $\vec C$ and $\vec F$, the necessity can be proved. $\square$

Definition 2. $\overrightarrow{a_n}_{n=0}^m\coloneqq\left(a_0,a_1,\ldots,a_m\right)$.

Lemma 2. Suppose the $\left(m+1\right)$-dimensional vector $\vec a$ is independent to $x$, and $\left\{f_n\!\left(x\right)\right\}_{n=0}^s$ is a sequence of functions, then $$\mathrm D^k\left(\vec a\cdot\overrightarrow{f_n\!\left(x\right)}_{n=0}^s\right) =\vec a\cdot\overrightarrow{\mathrm D^kf_n\!\left(x\right)}_{n=0}^s.$$

Brief proof. By mathematical induction. $\square$

Lemma 3. Suppose $\vec P$ is a tuple of functions w.r.t. $x$, and the dimension of $\vec P$ is $m+1$, then the differential operator $$L\!\left(\mathrm D,x\right)\coloneqq\vec P\cdot\overrightarrow{\mathrm D^k}^m_{k=0}$$ is a linear differential operator.

Brief proof. By Definition 1. $\square$

Corollary of Lemma 3. Suppose $\vec p$ is a constant $\left(m+1\right)$-dimensional vector, then the differential operator $$L\!\left(\mathrm D\right)\coloneqq\vec p\cdot\overrightarrow{\mathrm D^k}^m_{k=0}$$ $$(1)$$ is a linear differential operator.

Lemma 4 (associativity). $$\overrightarrow{a_kb_k}_{k=0}^m\cdot\overrightarrow{c_k}_{k=0}^m =\overrightarrow{a_k}_{k=0}^m\cdot\overrightarrow{b_kc_k}_{k=0}^m.$$

Proof is omitted.

Definition 3 (linear differential operator with constant coefficients). Linear differential operators with form as Equation 1 are called linear differential operators with constant coefficients.

Definition 4 (linear ODE). Suppose $L$ is a linear differential operator. Then the ODE w.r.t. the function $y\!\left(x\right)$ $$Ly=f$$ is called a linear ODE, where $f\!\left(x\right)$ is a function.

Specially, if $f=0$, the ODE is called a homogeneous linear ODE. If $L$ is a linear differential operator with constant coefficients, then the ODE is called a linear ODE with constant coefficients.

Definition 5 (generating function). For a sequence $\left\{a_n\right\}_{n=0}^\infty$, the function $$G\!\left(x\right)\coloneqq\overrightarrow{a_n}_{n=0}^\infty\cdot\overrightarrow{x^n}_{n=0}^\infty$$ is called the (ordinary) generating function (OGF) of the sequence.

Note. here we do not introduce vectors with infinite dimensions. Actually,

$$G\!\left(x\right)\coloneqq\lim_{s\to\infty}\overrightarrow{a_n}_{n=0}^s\cdot\overrightarrow{x^n}_{n=0}^s.$$

Definition 6 (exponential generating function). For a sequence $\left\{a_n\right\}_{n=0}^\infty$, the OGF of the sequence $\left\{\frac{a_n}{n!}\right\}_{n=0}^\infty$ is called the exponential generating function (EGF) of the sequence $\left\{a_n\right\}_{n=0}^\infty$. In other words, $$G\!\left(x\right)\coloneqq\overrightarrow{\frac{a_n}{n!}}_{n=0}^\infty\cdot\overrightarrow{x^n}_{n=0}^\infty.$$

Lemma 5 (differential of power functions). Suppose $n,k\in\mathbb N$, then $$\mathrm D^k\left(x^n\right)=\frac{n!}{\left(n-k\right)!}x^{n-k}.$$

Note. it is stipulated that factorials of negative integers are infinity, so $\mathrm D^k\left(x^n\right)=0$ when $n<k$.

Brief proof. By mathematical induction. $\square$

Lemma 6 (differential of EGF). If $G\!\left(x\right)$ is the EGF of a sequence $\left\{a_n\right\}_{n=0}^\infty$, then $\mathrm D^kG$ is the EGF of $\left\{a_{n+k}\right\}_{n=0}^\infty$.

Proof. $$\begin{align*} \mathrm D^kG&=\mathrm D^k\left(\overrightarrow{\frac{a_n}{n!}}_{n=0}^\infty\cdot\overrightarrow{x^n}_{n=0}^\infty\right) &\text{(Definition 6)}\\ &=\overrightarrow{\frac{a_n}{n!}}_{n=0}^\infty\cdot\overrightarrow{\mathrm D^k\left(x^n\right)}_{n=0}^\infty& \text{(Lemma 2)}\\ &=\overrightarrow{\frac{a_n}{n!}}_{n=0}^\infty\cdot\overrightarrow{\frac{n!}{\left(n-k\right)!}x^{n-k}}_{n=0}^\infty& \text{(Lemma 5)}\\ &=\overrightarrow{\frac{a_n}{\left(n-k\right)!}}_{n=0}^\infty\cdot\overrightarrow{x^{n-k}}_{n=0}^\infty& \text{(Lemma 4)}\\ &=\overrightarrow{\frac{a_{n+k}}{n!}}_{n=0}^\infty\cdot\overrightarrow{x^n}_{n=0}^\infty.& \text{(note in Lemma 5)} \end{align*}$$ Then the result can be proved by Definition 6. $\square$

Corollary to Lemma 6. $$\overrightarrow{\mathrm D^k}_{k=0}^m\left(\overrightarrow{\frac{a_n}{n!}}_{n=0}^\infty\cdot\overrightarrow{x^n}_{n=0}^\infty\right) =\overrightarrow{\overrightarrow{\frac{a_{n+k}}{n!}}_{n=0}^\infty\cdot\overrightarrow{x^n}_{n=0}^\infty}_{k=0}^\infty.$$