Bi-quadratic Equation by Vedic Math

Solve
$latex (x+7)^4+(x+5)^4=706$
Let y = x + 6 = average of (x+5, x+7)
$latex (y+1)^4+(y-1)^4=706$
Cancel terms y, y³:
$latex 2y^4+12y^2+2=706$
$latex y^4+6y^2-352=0$
$latex y^2=16$ or
$latex y^2=-22$
y= ±4 or ±$latex sqrt{-22}$
y=x+6
x=-2, -10, ± $latex sqrt{-22}-6$

“Transire suum pectus mundoque potiri”
[To transcend human limitations & to master the universe]

Amateur versus Professional

1. Amateur is at liberty to study only those things he likes.
2. Professional must also study what he doesn’t like.
3. Conclusion: Most famous theorems are found by Amateurs.

Examples:
Fermat = Judge (Number Theory, Probabilty),
Venn = Anglican Pastor (Venn Diagram),
Ramanujan = Railway clerk (Number Theory)
Cayley = Lawyer (Group),
Leibniz = Diplomat (Calculus, Binary 0 & 1)

This is an old arabic problem:

An old man had 11 horses. When he died, his will stated the following distribution to his 3 sons:
1/2 gives to the eldest son,
1/4 for 2nd son,
1/6 for 3rd son.

Find: how many horses each son gets ?

There are 2 methods to solve: first using simple arithmetic trick without knowing the theory behind; the second method will explain the first method “from an advanced standpoint” – Number Theory (Felix Klein’s Vision )

1) Arithmetic trick:

11 is odd, not divisible by 2, 4 and 6.

Loan 1 horse to the old man:
11+1 = 12

1st son gets: 12/2 = 6 horses
2nd son gets:12/4 = 3 horses
3rd son gets: 12/6 = 2 horses

Total = 6+3+2=11 horses

Up to you if you want the old man to return the 1 loan horse 🙂

Strange! WHY ?

2)

Prove: (Euler Gamma Γ Function)
$latex displaystyle n! = int_{0}^{infty}{x^{n}.e^{-x}dx}$

Proof:
∀ a>0
Integrate by parts:

$latex displaystyleint_{0}^{infty}{e^{-ax}dx}=-frac{1}{a}e^{-ax}Bigr|_{0}^{infty}=frac{1}{a}$

∀ a>0
$latex displaystyleint_{0}^{infty}{e^{-ax}dx}=frac{1}{a}$ …[1]

Feynman trick: differentiating under integral => d/da left side of [1]

$latex displaystylefrac{d}{da}displaystyleint_{0}^{infty}e^{-ax}dx= int_{0}^{infty}frac{d}{da}(e^{-ax})dx=int_{0}^{infty} -xe^{-ax}dx$

Differentiate the right side of [1]:
$latex displaystylefrac{d}{da}(frac{1}{a}) = -frac{1}{a^2}$
=>
$latex a^{-2}=int_{0}^{infty}xe^{-ax}dx$

Continue to differentiate with respect to ‘a’:
$latex -2a^{-3} =int_{0}^{infty}-x^{2}e^{-ax}dx$
$latex 2a^{-3} =int_{0}^{infty}x^{2}e^{-ax}dx$
$latex frac{d}{da} text{ both sides}$
$latex 2.3a^{-4} =int_{0}^{infty}x^{3}e^{-ax}dx$
…
…
$latex 2.3.4dots n.a^{-(n+1)} =int_{0}^{infty}x^{n}e^{-ax}dx$
Set a = 1
$latex boxed{n!=int_{0}^{infty}x^{n}e^{-x}dx}$ [QED]

Another Example using “Feynman Integration”:

$latex displaystyle text{Evaluate }int_{0}^{1}frac{x^{2}-1}{ln x} dx$

$latex displaystyle text{Let I(b)} = int_{0}^{1}frac{x^{b}-1}{ln x} dx$ ; for b > -1

$latex displaystyle text{I'(b)} = frac{d}{db}int_{0}^{1}frac{x^{b}-1}{ln x} dx = int_{0}^{1}frac{d}{db}(frac{x^{b}-1}{ln x}) dx$

$latex x^{b} = e^{ln x^{b}} = e^{b.ln x} $

$latex frac{d}{db}(x^{b}) = frac{d}{db}e^{b.ln x}=e^{b.ln x}.{ln x}= e^{ln x^{b}}.{ln x}=x^{b}.{ln x}$

$latex text{I'(b)}=int_{0}^{1} x^{b} dx=frac{x^{b+1}}{b+1}Bigr|_{0}^{1} = frac{1}{b+1}$
=>
$latex…

The derivative of a function can be thought of as:

(1) Infinitesimal: the ratio of the infinitesimal change in the value of a function to the infinitesimal change in a function.

(2) Symbolic: The derivative of
$Latex x^{n} = nx^{n-1} $
the derivative of sin(x) is cos(x),
the derivative of f°g is f’°g*g’,
etc.

(3) Logical:
$Latex boxed{text{f'(x) = d}} $
$Latex Updownarrow $
$latex forall varepsilon, exists delta, text{ such that }$
$latex boxed{
0 < |Delta x| < delta,
implies
Bigr|frac{f(x+Delta x)-f(x)}{Delta x} – d Bigr| < varepsilon
}$

(4) Geometric: the derivative is the slope of a line tangent to the graph of the function, if the graph has a tangent.

(5) Rate: the instantaneous speed of f(t), when t is time.

(6) Approximation: The derivative of a function is the best linear approximation to the function near a point.

(7)

The French method of drawing curves is very systematic:

“Pratique de l’etude d’une fonction”

Let f be the function represented by the curve C

Steps:

1. Simplify f(x). Determine the Domain of definition (D) of f;
2. Determine the sub-domain E of D, taking into account of the periodicity (eg. cos, sin, etc) and symmetry of f;
3. Study the Continuity of f;
4. Study the derivative of fand determine f'(x);
5. Find the limits of fwithin the boundary of the intervals in E;
6. Construct the Table of Variation;
7. Study the infinite branches;
8. Study the remarkable points: point of inflection, intersection points with the X and Y axes;
9. Draw the representative curve C.

Example:

$latex displaystyletext{f: } x mapsto frac{2x^{3}+27}{2x^2}$
Step 1: Determine the Domain of Definition D
D = R* = R –…

Riemann intuitively found the Zeta Function ζ(s), but couldn’t prove it. Computer ‘tested’ it correct up to billion numbers.

$latex zeta(s)=1+frac{1}{2^{s}}+frac{1}{3^{s}}+frac{1}{4^{s}}+dots$

Or equivalently (see note *)

$latex frac {1}{zeta(s)} =(1-frac{1}{2^{s}})(1-frac{1}{3^{s}})(1-frac{1}{5^{s}})(1-frac{1}{p^{s}})dots$

ζ(1) = Harmonic series (Pythagorean music notes) -> diverge to infinity
(See note #)

ζ(2) = Π²/6 [Euler]

ζ(3) = not Rational number.

1. The Riemann Hypothesis:
All non-trivial zeros of the zeta function have real part one-half.

ie ζ(s)= 0 where s= ½ + bi

Trivial zeroes are s= {- even Z}:
s(-2) = 0 =s(-4) =s(-6) =s(-8)…

You might ask why Re(s)=1/2 has to do with Prime number ?

There is another Prime Number Theorem (PNT) conjectured by Gauss and proved by Hadamard and Poussin:

π(Ν) ~ N / log N
ε = π(Ν) – N / log N
The error ε hides in the Riemann Zeta Function’s non-trivial zeroes, which all lie on the Critical…

A—————C———-B

$Latex frac {AB}{AC} = frac{AC}{CB}$
= 1.61803… = Φ
= $Latex frac {1+ sqrt{5}} {2}$

$Latex frac {6}{5} Phi^2$
= ∏ = 3.14159…

Donald Knuth (Great Computer Mathematician, Stanford University, LaTex inventor) noted the Bible uses a phrase like:
“as my Father is to me, I am to you”
=> F= Father = line AB
I (or me) = AC
U = You = CB
=> F/I = I/U = Φ
Note: Φ = 1.61803 = – 2 sin 666°

$Latex i^{i } = 0.207879576…$
$latex i = sqrt{-1}$

If a is algebraic and b is algebraic but irrational then $latex a^b $ is transcendental. (Gelfond-Schneider Theorem)

Since i is algebraic but irrational, the theorem applies.

1. We know
$latex e^{ix}= cos x + i sin x$

Let $latex x = pi/2 $

2. $latex e^{i pi/2} = cos pi/2 + i sin pi/2 $

$latex cos pi/2 = cos 90^circ = 0 $

$latex sin 90^circ = 1 $
$latex i sin 90^circ = (i)*(1) = i $

3. Therefore
$latex e^{ipi/2} = i$
4. Take the ith power of both sides, the right side being $latex i^i $ and the left side =
$latex (e^{ipi/2})^{i}= e^{-pi/2} $
5. Therefore
$latex i^{i} = e^{-pi/2} = .207879576…$

Just can’t imagine how strange a plane MH370 could just disappear in the air, no explosion, no terrorists (?) although 2 Iranian passengers with stolen passports from an Italian and an Austrain.

Malaysian Flight: MH 370

Departure : Passengers, among them the majority are 153 Chinese, boarded on 3.7 (March 7) around 11 PM at Kuala Lumpur International Airport, disappeared 1 hour later in the air.

http://www.nst.com.my/latest/font-color-red-missing-mh370-font-timeline-of-flight-mh370-1.507516

Just notice 370 is a strange number:

$latex boxed { (3)^{3} + (7)^{3}+ (0)^{3} = 370}$

A lot of mystery numbers have such behaviors when decompose the digit, then each powered by 3, sum them up, you get back the mystery number itself.

Bible Math: 153 St. Peter Fish
[John 21:3-11]
3  So they went out and got into the boat, but that night they caught nothing.
6 He said, ”Throw your net on the right side of the boat and you will find some.” When…

This will be my final post associated with the Analysis I course, for which the last lecture was yesterday. It’s possible that I’ll write further relevant posts in the nearish future, but it’s also possible that I won’t. This one is a short one to draw attention to other material that can be found on the web that may help you to learn the course material. It will be an incomplete list: further suggestions would be welcome in the comments below.

A good way to test your basic knowledge of (some of) the course would be to do a short multiple-choice quiz devised by Vicky Neale. If you don’t get the right answer first time for every question, then it will give you an idea of the areas of the course that need attention.

Terence Tao has also created a number of multiple-choice quizzes, some of which are relevant…

One of the hardest questions for many math teachers to answer in a way that is relevant to students is: “why do I need to know this?”  “For the next course you take”, the easiest answer in many cases, does not answer the question that was usually being asked.

My answers to this question obviously depend on the topic being studied at moment, and I don’t have “good” answers for all topics…  but here is my list of key quantitative life skills I learned directly or indirectly from math class, with

I didn’t encounter the Quadrilateral Midpoint Theorem (QMT) until I had been teaching a few years.  Following is a minor variation on my approach to the QMT this year plus a fun way I leveraged the result to introduce similarity.

In case you haven’t heard of it, the surprisingly lovely QMT says that if you connect, in order, the midpoints of the four sides of a quadrilateral–any quadrilateral–even if the quadrilateral is concave or if its sides cross–the resulting figure will always be a parallelogram.

This is a cool and easy property to explore on any dynamic geometry software package (GeoGebra, TI-Nspire, Cabri, …).

SKETCH OF THE TRADITIONAL PROOF:  The proof is often established through triangle similarity:  Whenever you connect the midpoints of two sides of a triangle, the resulting segment will be parallel to and half the length of the triangle’s third side…

Note:
China: 九章算术， 秦九韶
Persia : Algebra, Algorithm
Italy: Fibonacci

The millennium bug or the Y2K bug was going to cause planes to fall from the sky, bank accounts to be wiped out, electricity grids to cease functioning, trains to crash, cars to collide as stop lights stopped functioning, life support units to malfunction and computers to crash around the globe. For years leading up to midnight on New Years eve 1999 consults were paid extraordinary amounts of money to solve the problem.  When the clock ticked over to 1 ^st Jan 2000 nothing much happened. It was, indeed, a non-event, an error in logic.  (Pic from, appropriately, digyourowngrave.com)  

The Maths Error: Guessing the Answer.

Computer programmers represented the year in the date of many programs using two digits but claimed logical errors would arise upon “rollover” from x99 to x00.

While consultants claimed their advice saved the world from catastrophe countries that spent very little on the…

http://en.m.wikipedia.org/wiki/Ng%C3%B4_B%E1%BA%A3o_Ch%C3%A2u

Interview (in Chinese):

http://wk.baidu.com/view/2919f063caaedd3383c4d3a4?pcf=2#page/1/1388828375964