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Math in the Wild: Taming Differential Equations |
Math in the Wild: Taming Differential equations Computer Algebra Aided Chebyshev Method Michael Xue University of Indianapolis, Sept. 2024 DE in Practice Nonlinear PDE $latex \displaystyle \frac{\partial u(x,t)}{\partial t} = a(x)\frac{\partial^2 u(x,t)}{\partial x^2} +b(x)\frac{\partial u(x,t)}{\partial x} + g(u(x, t))$ Linear PDE $latex \displaystyle \frac{\partial u(x,t)}{\partial t} = a(x)\frac{\partial^2 u(x,t)}{\partial x^2} +b(x)\frac{\partial u(x,t)}{\partial x} +c(x)u(x,t) +…
Bridging the Gaps in Einstein’s Explanation of Mercury’s Perihelion Precession |
Mercury's perihelion precession refers to the gradual and predictable shift of the point at which Mercury is closest to the Sun in its elliptical orbit. This point, called the perihelion, slowly moves or precesses around the Sun over time. The perihelion precession is measured by the angular shift of the major axis of Mercury's elliptical…
Sandwich Theorem for Functions 2 |
For given functions $latex f(x), g(x),$ and $latex h(x),$ $latex \left(\forall x : f(x) \le g(x) \le h(x), \lim\limits_{x \rightarrow \infty} f(x) = \lim\limits_{x \rightarrow \infty} h(x)=L\right) \implies \lim\limits_{x \rightarrow \infty} g(x) = L.$ The given condition gives $latex \forall \epsilon > 0, \exists x_1^* \ni x>x_1^*, |f(x)-L|<\epsilon$ and $latex \forall \epsilon > 0, \exists…
Introducing Feynman’s Integral Method (MAA Meeting, Spring 2024) |
Introducing Feynman's Integral Method Michael Xue Indiana Section MAA Spring 2024 MEETINGMarian University - IndianapolisApril 5-6, 2024 "The first principle is that you must not fool yourself and you are the easiest person to fool." -- Richard Feynman Feynman's Integral Method Leibniz : "$latex \displaystyle \frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta) \;dx = \int\limits_{a}^{b}\frac{\partial}{\partial\beta}f(x, \beta)\;dx$'' Feynman: "To evaluate hard…
Harnessing Feynman’s Integral Technique |
Show that $latex \displaystyle\int_{-\infty}^{\infty}e^{-x^2}\;dx = \sqrt{\pi}$ This integral is renowned in mathematics as the Gaussian integral. Its evaluation poses a challenge due to the absence of an elementary antiderivative expressed in standard functions. Conventionally, one method involves "squaring the integral" and subsequently interpreting the resulting double integral in polar coordinates. However, an alternative approach, which…
An Analytic Proof of the Extraordinary Euler Sum |
In Deriving the Extraordinary Euler Sum , we derived one of Euler's most celebrated results: $latex \displaystyle 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^3} + ... = \frac{\pi^2}{6}\quad\quad\quad(1)$ Now, we aim to provide a rigorous proof of this statement. First, we expressed the partial sum of the left-hand side as follows: $latex \displaystyle \sum\limits_{i=1}\frac{1}{i^2} = \sum\limits_{i=1}\frac{1}{(2i)^2}…
Introducing Feynman’s Integral Method (MAA Meeting, Spring 2024) |
Feynman's trick, also known as parameter differentiation under the integral sign, is a powerful mathematical technique used for simplifying complex integrals. By introducing an auxiliary parameter into the integral and then differentiating with respect to that parameter, Feynman's method often transforms difficult integrals into more manageable forms. It serves as a versatile tool in physics…
A Gift That Keeps On Giving |
We see from "Seek-Lock-Strike!" Again that given the missile's position $latex (x, y),$ $latex \frac{dx}{dy}=\frac{a+v_a t-x}{b-y}$ where $latex x$ and $latex y$ themselves are functions of time $latex t.$ It means $latex \frac{dx}{dy} = \frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{a+v_a t-x}{b-y}\implies \frac{dx}{dt} = \frac{a+v_a t-x}{b-y}\cdot\frac{dy}{dt}.$ That is, let $latex \kappa(x,y,t) = \frac{a+v_a t-x}{b-y},$ $latex \frac{dx}{dt}=\kappa\cdot \frac{dy}{dt}.\quad\quad\quad(1)$ We also have (see…
“Seek-Lock-Strike!” Animated |
In "Seek-Lock-Strike!" Again, we obtained the missile's trajectory. Namely, $latex x = \frac{1}{2} \left(\frac{(b-y)^{r+1}}{b^r \cdot f \cdot (r+1)}-\frac{b^r \cdot f \cdot (b-y)^{1-r}}{1-r}\right) -k\quad\quad\quad(*)$ where $latex f = \frac{a}{b}+\sqrt{1+(\frac{a}{b})^2},$ $latex k = \frac{b}{2}\left(\frac{1}{f \cdot (r+1)}-\frac{f}{1-r}\right).$ Since the fighter jet maintains its altitude ($latex y= b$), the missile must strike it at $latex (x_s, b)$. Setting $latex…
“Seek-Lock-Strike!” Animated |
In "Seek-Lock-Strike!" Again, we obtained the missile's trajectory. Namely, $latex x = \frac{1}{2} \left(\frac{(b-y)^{r+1}}{b^r \cdot f \cdot (r+1)}-\frac{b^r \cdot f \cdot (b-y)^{1-r}}{1-r}\right) -k\quad\quad\quad(*)$ where $latex f = \frac{a}{b}+\sqrt{1+(\frac{a}{b})^2},$ $latex k = \frac{b}{2}\left(\frac{1}{f \cdot (r+1)}-\frac{f}{1-r}\right).$ Since the fighter jet maintains its altitude ($latex y= b$), the missile must strike it at $latex (x_s, b)$. Setting $latex…
“Seek-Lock-Strike!” Again |
We can derive a different governing equation for the missile in "Seek-Lock-Strike!". Fig. 1 Looking from a different viewpoint (Fig. 1), we see $latex \frac{dx}{dy} = \frac{a+v_at-x}{b-y}.\quad\quad\quad(1)$ Solving (1) for $latex t$, $latex t = -\frac{\frac{dx}{dy}y-b\frac{dx}{dy}-x+a}{v_a}.\quad\quad\quad(2)$ We also have $latex v_m t = \int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2} \implies t = \frac{\int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2}\;dy}{v_m}.\quad\quad\quad(3)$ Equate (1) and (2) gives $latex -\frac{\frac{dx}{dy}y-b\frac{dx}{dy}-x+a}{v_a}-\frac{\int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2}\;dy}{v_m}…
“Seek-Lock-Strike!” Again |
We can derive a different governing equation for the missile in "Seek-Lock-Strike!". Fig. 1 Looking from a different viewpoint (Fig. 1), we see $latex \frac{dx}{dy} = \frac{a+v_at-x}{b-y}.\quad\quad\quad(1)$ Solving (1) for $latex t$, $latex t = -\frac{\frac{dx}{dy}y-b\frac{dx}{dy}-x+a}{v_a}.\quad\quad\quad(2)$ We also have $latex v_m t = \int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2} \implies t = \frac{\int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2}\;dy}{v_m}.\quad\quad\quad(3)$ Equate (1) and (2) gives $latex -\frac{\frac{dx}{dy}y-b\frac{dx}{dy}-x+a}{v_a}-\frac{\int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2}\;dy}{v_m}…
“Seek-Lock-Strike!” Simplified |
There is an easier way to derive the governing equation ((5), "Seek-Lock-Strike!") for the missile. Solving $latex \frac{dy}{dx}(a+v_at-x) = b-y$ for $latex t, $ we have $latex t = \frac{(x-a)\frac{dy}{dx}-y+b}{v_a\frac{dy}{dx}}.\quad\quad\quad(1)$ From $latex v_m t = \int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2},$ we also have $latex t = \frac{\int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx}{v_m}.\quad\quad\quad(2)$ Equate (1) and (2) gives $latex \frac{\int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx}{v_m}-\frac{(x-a)\frac{dy}{dx}-y+b}{v_a\frac{dy}{dx}}=0.\quad\quad\quad(3)$ Differentiate (3) with repect to…
“Seek-Lock-Strike!” Animated |
In "Seek-Lock-Strike!", we obtained the missile's trajectory. Namely, $latex x = \frac{1}{2} \left(\frac{f\cdot (b-y)^{r+1}}{r+1}-\frac{f^{-1}\cdot (b-y)^{1-r}}{1-r}\right) -k\quad\quad\quad(*)$ where $latex f = \frac{1}{b^r\left(\frac{a}{b}+\sqrt{1+(\frac{a}{b})^2}\right)},$ $latex k = \frac{1}{2}\left(\frac{f\cdot b^{r+1}}{r+1}-\frac{f^{-1}\cdot b^{1-r}}{1-r}\right).$ Since the fighter jet maintains its altitude ($latex y= b$), the missile must strike it at $latex (x_s, b)$. Setting $latex y=b$ in $latex (*)$ gives $latex x_s…
Newton’s Pi Simplified |
We know from "arcsin" : $latex \frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}}.$ Integrate from $latex 0$ to $latex x\; (0<x<1):$ $latex \int\limits_{0}^{x}\frac{d}{dx}\arcsin(x)\;dx = \int\limits_{0}^{x}\frac{1}{\sqrt{1-x^2}}\;dx$ gives $latex \arcsin(x)\bigg|_{0}^{x} = \int\limits_{0}^{x}\frac{1}{\sqrt{1-x^2}}\;dx.$ i.e., $latex \arcsin(x) = \int\limits_{0}^{x}\frac{1}{\sqrt{1-x^2}}\;dx.\quad\quad\quad(\star)$ Rewrite the integrand $latex \frac{1}{\sqrt{1-x^2}}$ as $latex (1-x^2)^{-\frac{1}{2}}= (1+(-x^2))^{-\frac{1}{2}}$ so that by the extended binomial theorem (see "A Gem from Issac Newton"), $latex…
Fast Pi |
There is a recipe for rapidly calculating the digits of $latex \pi$. Consider the two-term iteration with initial values $latex a_0 = \sqrt{2}, \quad b_0 = 1\quad\quad\quad(1)&s=1$ given by $latex a_n = \frac{a_{n-1}+b_{n-1}}{2}, \quad b_n = \sqrt{a_{n-1} b_{n-1}}, \quad n \ge 1.\quad\quad\quad(2)&s=1$ Then $latex \pi_{n} = \frac{a_n^2}{1-\sum\limits_{i=0}^{n}2^{i-1}\left(a_n^2-b_n^2\right)}\quad\quad\quad(3)&s=2$ converges to $latex \pi.$ We have implemented (1),…
Faster Pi |
There is a more astonishing algorithm than what is described in "Fast Pi" for rapidly calculating $latex \pi.$ Let us consider the three-term iteration with initial values $latex a_0 = \sqrt{2}, \quad b_0 =0, \quad \pi_0 = 2+\sqrt{2}$ given by $latex a_n = \frac{1}{2}\left(\sqrt{a_{n-1}} + \frac{1}{\sqrt{a_{n-1}}}\right), \quad b_{n} = \sqrt{a_{n-1}}\left(\frac{b_{n-1}+1}{b_{n-1} + a_{n-1}}\right)&s=1, \quad \pi_{n} =…
Sandwich Theorems and Their Proofs |
Prove: $latex \begin{cases} \exists N\in \mathbb{N} \ni n > N, a_n \le b_n \le c_n\quad\quad(1)\\ \lim\limits_{n \rightarrow \infty}a_n=\lim\limits_{n\rightarrow \infty}c_n=L \quad\quad\quad\quad\quad\quad(2) \end{cases}\implies \lim\limits_{n\rightarrow \infty}b_n = L.$ We express (1) as $latex n > N \implies a_n \le b_n \le c_n$ and (2): $latex \forall \epsilon >0, \exists n_1 \ni n >n_1 \implies |a_n-L|< \epsilon.$ $latex \forall…
A Mind Unleashed |
In 1665, following an outbreak of the bubonic plague in England, Cambridge University closed its doors, forcing Issac Newton, then a college student in his 20s, to go home. Away from university life, and unbounded by curriculum constraints and tests, Newton thrived. The year-plus he spent in isolation was later referred to as his annus…
arcsin |
Consider set $latex S = \{(x,y) | \sin(y)=x, -1<x<1\}.$ Since $latex \sin(\underbrace{\frac{\pi}{2}}_{y}) = \underbrace{1}_{x}, \quad \sin(\underbrace{\frac{5\pi}{2}}_{y}) = \underbrace{1}_{x},$ it does not define a function. However, redefine $latex S$ as $latex S = \{(x,y) | \sin(y)=x, -1 < x < 1, \boxed{-\frac{\pi}{2} \le y \le \frac{\pi}{2}}\},\quad\quad\quad(1)$ we have $latex \forall (x, y_1), (x, y_2) \in S,…
Seek-Lock-Strike! |
A guided missile is launched to destroy a fighter jet (Fig. 1). Fig. 1 We introduce a coordinate axes such that at $latex t=0,$ the missile is at origin $latex (0, 0)$ and the jet at $latex (a,b)$. The jet flies parallel to the x-axis with constant speed $latex v_a$. The missile has locked onto…
Have we a new proof ? |
For a right triangle: On one hand, its area is $latex \frac{1}{2}ab.$ On the other hand, according to Heron's formula (see "An Algebraic Proof of Heron’s Formula"), $latex \sqrt{s(s-a)(s-b)(s-c)}\quad\quad\quad(*)$ where $latex s=\frac{a+b+c}{2}.$ Hence, $latex \frac{1}{2} ab=\sqrt{s(s-a)(s-b)(s-c)}.$ Squaring it gives $latex \frac{1}{4}a^2b^2=s(s-a)(s-b)(s-c).$ Using a CAS, we obtain $latex \frac{(c^2-b^2-a^2)^2}{16} = 0 \implies c^2-b^2-a^2=0$ from which the Pythagorean…
All Roads Lead to Rome |
"Keep hitting the square root button of a calculator after entering any positive number, "1" will be the eventual result displayed." This is observed by many when playing around with the calculator. To explain this "phenomena", we shall show that $latex \forall a >0, \lim\limits_ {n \rightarrow \infty} a^{\frac{1}{n}} = 1.$ There are three cases…
A Mind Unleashed |
In 1665, following an outbreak of the bubonic plague in England, Cambridge University closed its doors, forcing Issac Newton, then a college student in his 20s, to go home. Away from university life, and unbounded by curriculum constraints and tests, Newton thrived. The year-plus he spent in isolation was later referred to as his annus…
A Mind Unleashed |
In 1665, following an outbreak of the bubonic plague in England, Cambridge University closed its doors, forcing Issac Newton, then a college student in his 20s, to go home. Away from university life, and unbounded by curriculum constraints and tests, Newton thrived. The year-plus he spent in isolation was later referred to as his annus…
Newton’s Pi |
Fig. 1 Shown in Fig. 1 is a semicircle centered at C $latex (\frac{1}{2}, 0)$ with radius = $latex \frac{1}{2}$. Its equation is $latex (x-\frac{1}{2})^2+y^2=(\frac{1}{2})^2, 0 \le x \le 1, y \ge 0.$ Simplifying and solving for $latex y$ gives $latex y = x^{\frac{1}{2}}\cdot(1-x)^{\frac{1}{2}}.\quad\quad\quad(1)$ We see that Area (sector OAC) = Area (sector OAB) +…
Generating Integrals |
It is well known that $latex \displaystyle\int \frac{1}{a^2+x^2}\;dx =\frac{1}{a}\arctan\left(\frac{x}{a}\right).\quad\quad\quad(1)$ Let $latex \displaystyle\int \frac{1}{(a^2+x^2)^k}\; dx=I_k, \;\;k=1, 2, 3, ...\quad\quad\quad(2)$ we have $latex I_1 = \frac{1}{a}\arctan\left(\frac{x}{a}\right).\quad\quad\quad(3)$ Differentiate (2) with respect to $latex a:$ $latex \frac{\partial}{\partial a}\displaystyle\int\frac{1}{(a^2+x^2)^k}\;dx = \frac{\partial}{\partial a}I_k.$ By “Differentiation under the integral sign”(see Playing “Feynman’s Trick” on Indefinite Integrals – Tongue in Cheek), $latex \displaystyle \int…
A Gem from Isaac Newton |
The Binomial Theorem (see "Double Feature on Christmas Day", "Prelude to Taylor's theorem") states: $latex (x+y)^n = \sum\limits_{i=0}^{n} \binom{n}{i}x^{n-i}y^i, \quad n \in \mathbb{N}.$ Provide $latex x$ and $latex y$ are suitably restricted, there is an Extended Binomial Theorem. Namely, $latex (x+y)^r = \sum\limits_{i=0}^{\infty} \binom{r}{i}x^{r-i}y^i, \underline{|\frac{x}{y}|<1}, r \in \mathbb{R}$ where $latex \binom{r}{i} \overset{\Delta}{=} \frac{r(r-1)(r-2)...(r-i+1)}{i!}.$ Although Issac Newton…
Have we a new proof ? |
For a right triangle: On one hand, its area is $latex \frac{1}{2}ab.$ On the other hand, according to Heron's formula (see "An Algebraic Proof of Heron’s Formula"), $latex \sqrt{s(s-a)(s-b)(s-c)}\quad\quad\quad(*)$ where $latex s=\frac{a+b+c}{2}.$ Hence, $latex \frac{1}{2} ab=\sqrt{s(s-a)(s-b)(s-c)}.$ Squaring it gives $latex \frac{1}{4}a^2b^2=s(s-a)(s-b)(s-c).$ Using a CAS, we obtain $latex \frac{(c^2-b^2-a^2)^2}{16} = 0 \implies c^2-b^2-a^2=0$ from which the Pythagorean…
Gotta Catch ‘Em All ! |
Solving $latex \frac{d^2y}{dx^2} = \frac{y}{x}\cdot\frac{dy}{dx}.&s=2$ Even though 'contrib_ode', Maxima's ODE solver choked on this equation (see "An Alternate Solver of ODEs"), it still can be solved as demonstrated below: multiplied by $latex x,$ $latex x\frac{d^2y}{dx^2} = y\frac{dy}{dx},$ i.e., $latex x\frac{d}{dx}\left(\frac{dy}{dx}\right) = y\frac{dy}{dx}.$ Integrate it, we have $latex \int x\frac{d}{dx}\left(\frac{dy}{dx}\right)\;dx = \int y \frac{dy}{dx} \;dx \implies \int…
How i Was Born |
In Memory of Johann Weilharter (1953-2021) Girolamo Cardano (1501-76) was an Italian intellect whose interests and proficiencies ranged through those of mathematician, physician, biologist, physicist, chemist, astrologer, astronomer, philosopher, writer, and gambler. While conducting research on solving algebraic equations, Cardano discovered that by means of a suitable substitution, the general cubic equation $latex y^3+by^2+cy+d=0$ can…
A Computer Algebra Aided Proof in Plane Geometry |
Given $latex \Delta ABC$ and two squares $latex ABEF, ACGH$ in Fig. 1. The squares are sitting on the two sides of $latex \Delta ABC, AB$ and $latex AC$, respectively. Both squares are oriented away from the interior of $latex \Delta ABC$. $latex \Delta BCP$ is an isosceles right triangle. $latex P$ is on the…
Integral: I vs. CAS |
Evaluate $latex \displaystyle\int\frac{\sqrt{1+p^2}}{p}\;dp$ Let $latex u = \sqrt{1+p^2},\quad\quad\quad(1)$ we have $latex u^2=1+p^2 \implies p^2=u^2-1 \implies p =\pm\sqrt{u^2-1} \quad\quad\quad(2)$ and $latex \frac{dp}{du}= \pm\frac{1}{2}\cdot\frac{2u}{\sqrt{u^2-1}} =\pm \frac{u}{\sqrt{u^2-1}}.\quad\quad\quad(3)$ Consequently, $latex \int\frac{\sqrt{1+p^2}}{p}\;dp$ $latex \overset{(1), (2)}{=} \int \frac{u}{\pm\sqrt{u^2-1}}\cdot\frac{dp}{du}\;du$ $latex \overset{(3)}{=}\int \frac{u}{\pm\sqrt{u^2-1}}\cdot(\pm\frac{u}{\sqrt{u^2-1}})\;du$ $latex = \int\frac{u^2}{u^2-1}\;du$ $latex = \int \frac{u^2-1+1}{u^2-1}\;du$ $latex = \int du + \int \frac{1}{u^2-1}\;du$ $latex = u +\int \frac{1}{2}\left(\frac{1}{u-1}-\frac{1}{u+1}\right)\;du$ $latex…
My Pi |
Besides "Wallis' Pi", there is another remarkable expression for the number $latex \pi$ as an infinite product. We derive it as follows: From the trigonometric identity $latex \sin(2x) = 2\sin(x)\cos(x)$, we have $latex \sin(x) = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$ $latex = 2\cdot 2\sin(\frac{x}{4})\cos(\frac{x}{4}) \cdot\cos(\frac{x}{2})$ $latex = 2\cdot 2\cdot 2\sin(\frac{x}{8})\cos(\frac{x}{8})\cdot \cos(\frac{x}{4})\cdot \cos(\frac{x}{2})$ $latex \ddots$ $latex = 2^n\sin(\frac{x}{2^n})\cdot \prod\limits_{i=1}^{n}\cos(\frac{x}{2^i})$. That is, $latex…