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

The Bell Curve:

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

Improper Integral vs. Infinite Series

November 26, 2018Chinoiseries2014 Elementary Math 1 Comment

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

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

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

October 11, 2018Chinoiseries2014 Elementary Math 1 Comment

(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:

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

WiFi Password = Integral Answer

September 10, 2018tomcircle Elementary Math, Uncategorized 2 Comments

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