# Walter Bradley Center Fellow Discovers Longstanding Flaw in Elementary Calculus

… The paper on the proof of the second derivative (Quotient rule) :

https://www.dropbox.com/s/mv6npb3tpe76xjl/4.pdf?dl=0

Reference:

Faà di Bruno Fomula

$\boxed{ (f\circ H)^{(n)} =\displaystyle\sum_{\sum_{j=1}^n j\,m_j=n} \frac{n!}{m_!\ldots m_n!}\, \bigl(f^{(m_1+\ldots+m_n)})\circ H\bigr)\, \prod_{j=1}^n \left(\frac{H^{(j)}}{j!}\right)^{m_j} }$

# The Error Function & The Integral of e^(-x^2)

$\boxed{ \displaystyle erf(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} {e^{-{t}^2}dt}}$

The Bell Curve:

$\boxed{\displaystyle y = e^{-x^2}}$

# Improper Integral vs. Infinite Series

$\displaystyle \int_{2}^{\infty}{\frac{1}{x^{2} - 1}dx} =\boxed{ \ln {3}} = 1.09$

$\displaystyle \sum_{n=2}^{\infty}{\frac{1}{n^{2} - 1}} = \boxed{\frac{3 } {2 }} = 1.5$

To test if the infinite series $\displaystyle \sum_{n=2}^{\infty}{\frac{1}{n^{2} - 1}}$ converges,
provided the function $\displaystyle f(n) = {\frac{1}{n^{2} - 1}}$
in the interval [2, ∞[ is:

• positive &
• decreasing function

you can convert it to $\displaystyle \int_{2}^{\infty}{\frac{1}{x^{2} - 1}dx}$
(although they converge to different values.)

# Integration by parts, DI method

(Traditional Method learnt at O-Level) Integration by Parts:

NEW Technique taught by this UC-Berkley Chinese Mathematician: “DI” Method

First Stop: “D column at 0

2nd Stop:

3rd Stop:

$\displaystyle \int {e^{x}}\sin {x} dx$

China 南京航空航天大学 Nanjing University of Aeronautics and Astronautics set the WiFi password as the answer of this integral (first 6 digits).

Can you solve it?

(If can’t, please revise GCE “A-level” / Baccalaureate / 高考 Calculus 微积分)

Answer : Break the integral (I) into 2 parts:

I = A(x) + B(x)

$\displaystyle A(x) = \int_{-2 }^ {2} x^{3}. \cos \frac{x}{2}.\sqrt{4-x^2}dx$

$\displaystyle B(x) = \int_{-2 }^ {2} \frac{1}{2}\sqrt{4-x^2}dx$

A(x) = – A(-x) => Odd function
=> A(x) = 0 since its area canceled out over [-2, 2]

B(x) = B(-x) => Even function
$\displaystyle\implies B(x) = 2\int_{0 }^ {2} \frac{1}{2}\sqrt{4 - x^2}dx$
$\displaystyle\implies B(x) = \int_{0 }^ {2} \sqrt{4 - x^2}dx$

Let x = 2 sin t => dx = 2 cos t. dt

x = 2 = 2 sin t => sin t = 1 => t = π / 2

x = 0 = 2 sin t => sin t = 0 => t = 0

$\displaystyle B(x) = \int_{0 }^ {\pi/2} \sqrt{4 - 4.\sin^{2} {t} }. (2 \cos t. dt)$

$\displaystyle \implies B(x) = \int_{0 }^ {\pi/2} 2.\cos t. (2 \cos t. dt)$

$\displaystyle\implies B(x) = \int_{0 }^ {\pi/2} 4 \cos^{2} t. dt$

$\displaystyle \cos ^{2} t = \frac {1 + \cos 2t} {2 }$

$\displaystyle\implies B(x) = \int_{0 }^ {\pi/2} (2 + 2\cos 2t) . dt$

$\displaystyle\implies B(x) = (2 t) \Bigr|_{0 }^ {\pi/2} + (2. \frac{1}{2} \sin 2t) \Bigr|_{0 }^ {\pi/2}$

$\displaystyle\implies B(x) = (\pi ) + \sin \pi = \pi$

$\boxed{ I = \pi = 3.14159}$

A smarter method using Analytic Geometry: A circle of radius 2 is

$x^2 + y^2 = 4 \implies y = \sqrt {4 -x^2}$