all my creations

maths

VILOKANAM

The Sutra 'Vilokanam' means 'Observation'. Generally we come across problems which can be solved by mere observation. But we follow the same conventional procedure and obtain the solution. But the hint behind the Sutra enables us to observe the problem completely and find the pattern and finally solve the problem by just observation.

Let us take the equation x + ( 1/x ) = 5/2 Without noticing the logic in the problem, the conventional process tends us to solve the problem in the following way.

1 5
x + __ = __
x 2

x2 + 1 5
_____ = __
x 2

2x2 + 2 = 5x
2x2 – 5x + 2 = 0
2x2 – 4x – x + 2 = 0
2x (x – 2) – (x – 2) = 0
(x – 2) (2x – 1) = 0
x – 2 = 0 gives x = 2
2x – 1 = 0 gives x = Ѕ

But by Vilokanam i.e.,, observation

1 5
x + __ = __ can be viewed as
x 2

1 1
x + __ = 2 + __ giving x = 2 or Ѕ.
x 2

Consider some examples.

Example 1 :
x x + 2 34
____ + _____ = ___
x + 2 x 15

In the conventional process, we have to take L.C.M, cross-multiplication. simplification and factorization. But Vilokanam gives

34 9 + 25 3 5
__ = _____ = __ + __
15 5 x 3 5 3

x x + 2 3 5
____ + _____ = __ + __
x + 2 x 5 3

gives
x 3 5
_____ = __ or __
x + 2 5 3

5x = 3x + 6 or 3x = 5x + 10
2x = 6 or -2x = 10
x = 3 or x = -5


Example 2 :

x + 5 x + 6 113
____ + _____ = ___
x + 6 x + 5 56

Now,
113 49 + 64 7 8
___ = _______ = ___ + ___
56 7 x 8 8 7

x + 5 7 x+5 8
____ = __ or ____ = __
x + 6 8 x+6 7

8x + 40 = 7x + 42 7x + 35 = 8x + 48
or
x = 42 - 40 = 2 -x = 48 – 35 = 13
x = 2 or x = -13.

Example 3:

5x + 9 5x – 9 82
_____ + _____ = 2 ___
5x - 9 5x + 9 319

At first sight it seems to a difficult problem.

But careful observation gives

82 720 841 - 121 29 11
2 ___ = ___ = ________ = ___ - __
319 319 11 x 29 11 29

(Note: 292 = 841, 112 = 121)

5x + 9 29 -11
_____ = __ or ___
5x - 9 11 29

(Note: 29 = 20 + 9 = 5 x 4 + 9 ; 11 = 20 – 9 = 5 x 4 – 9 )

i.e.,
x = 4 or
5x + 9 -11
_____ = ___
5x - 9 29

145x + 261 = -55x + 99
145x + 55x = 99 – 261
200x = -162

-162 -81
x = ____ = ____
200 100

Simultaneous Quadratic Equations:

Example 1: x + y = 9 and xy = 14.

We follow in the conventional way that

(x – y)2 = (x + y)2 – 4xy = 92 – 4 (14) = 81 - 56 = 25
x – y = √ 25 = ± 5



x + y = 9 gives 7 + y = 9
y = 9 – 7 = 2.

Thus the solution is x = 7, y = 2 or x = 2, y = 7.

But by Vilokanam, xy = 14 gives x = 2, y = 7 or x = 7, y = 2 and these two sets satisfy x + y = 9 since 2 + 7 = 9 or 7 + 2 = 9. Hence the solution.

Example 2: 5x – y = 7 and xy = 6.

xy = 6 gives x = 6, y = 1; x = 1, y = 6;

x = 2, y = 3; x = 3, y = 2 and of course negatives of all these.

Observe that x = 6, y = 1; x = 1, y = 6: are not solutions because they do not satisfy the equation 5x – y = 7.

But for x = 2, y = 3; 5x – y = 5 (2) – 3 = 10 – 3 = 7 we have 5(3)–2≠7.

Hence x = 2, y = 3 is a solution.

For x = 3, y = 2 we get 5 (3) – 2 = 15 – 2 ≠ 7.

Hence it is not a solution.

Negative values of the above are also not the solutions. Thus one set of the solutions i.e., x = 2, y = 3 can be found. Of course the other will be obtained from solving 5x – y = 7 and 5x + y = -13.

i.e., x = -3 / 5, y = -10.

Partial Fractions:

Example 1: Resolve

2x + 7
___________ into partial fractions.
(x + 3) (x + 4)

2x + 7 A B
We write ____________ = ______ + ______
(x + 3)(x + 4) (x + 3) (x + 4)

A (x + 4) + B (x + 3)
= __________________
(x + 3) (x + 4)

2x + 7 ≡ A (x + 4) + B (x + 3).

We proceed by comparing coefficients on either side

coefficient of x : A + B = 2 ..........(i) X 3

Independent of x : 4A + 3B = 7 .............(ii)

Solving (ii) – (i) x 3 4A + 3B = 7
3A + 3B = 6
___________
A = 1

A = 1 in (i) gives, 1 + B = 2 i.e., B = 1

Or we proceed as

2x + 7 ≡ A (x + 4) + B (x + 3).
Put x = -3, 2 (-3) + 7 ≡ A (-3 + 4) + B (-3 + 3)
1 = A (1) ... A = 1.

x = -4, 2 (- 4) + 7 = A (-4 + 4) + B (-4 + 3)
-1 = B(-1) ... B = 1.

2x + 7 1 1
Thus ____________ = _____ + _____
(x + 3) (x + 4) (x + 3) (x + 4)

2x + 7
But by Vilokanam ____________ can be resolved as
(x + 3) (x + 4)

(x + 3) + (x + 4) =2x + 7, directly we write the answer.

Example 2:

3x + 13
____________
(x + 1) (x + 2)

from (x + 1),(x + 2) we can observe that

10 (x + 2) – 7(x + 1) = 10x + 20 – 7x – 7 = 3x + 13

3x + 13 10 7
Thus ____________ = _____ - _____
(x + 1) (x + 2) x + 1 x + 2

Example 3:

9
________
x2 + x - 2

As x2 + x – 2 = (x – 1) (x + 2) and
9 = 3 (x + 2) – 3 (x – 1)
(3x + 6 – 3x + 3 = 9)


9 3 3
We get by Vilokanam, ____________ = ____ - ____
x2 + x – 2 x - 1 x + 2


--------------------------------------------------------------------------------

I. Solve the following by mere observation i.e. vilokanam

1. 2.
1 25 1 5
x + __ = __ x - __ = __
x 12 x 6

3.
x x + 1 1
_____ + _____ = 9 __
x + 1 x 9

4.
x + 7 x + 9 32
____ - ____ = ___
x + 9 x + 7 63

II. Solve the following simultaneous equations by vilokanam.

1. x – y = 1, xy = 6 2. x + y = 7, xy = 10

3. 2x + 3y = 19, xy = 15

4. x + y = 4, x2 + xy + 4x = 24.

III. Resolve the following into partial fractions.

1.
2x - 5
____________
(x – 2) (x – 3)

2.
9
____________
(x + 1) (x – 2)

3.
x – 13
__________
x2 - 2x - 15

4.
3x + 4
__________
3x2 + 3x + 2

what all gods have common

what all gods have common,very interesting
on numerology
1 2 3 4 5 6 7 8 9
A B C D E F G H I
J K L M N O P Q R
S T U V W X Y Z


1. HINDU ( SHREE KRISHNA)
1+8+9+5+5+2+9+9+1+8+5+1=63=6+3=9
2. MUSLIM (MOHAMMED)
4+6+8+1+4+4+5+4=36=3+6=9
3. SHIK (GURU NANAK)
7+3+9+3+5+1+5+1+2=36=3+6=9
4. PARSI (ZARA THUSTRA )
8+1+9+1+2+8+3+1+2+9+1=45=4+5=9
5.BUDH (GAUTAM)
7+1+3+2+1+4=18=1+8=9
6.JAIN (MAHAVIR)
4+1+8+1+4+9+9=36=3+6=9
7. ESAI (ESA MESSIAH)
5+1+1+4+5+1+1+9+1+8=36=3+6=9
8. SAI NATH
1+1+9+5+1+2+8=27=2+7=9

AT LAST BY CAHANCE WHEN I TRY TO CALCULATE MY NAME "PIYUSHDADARIWALA" AS PER THIS METHOD ,GOT NINE......BUT I AM NOT GOD.......BUTI BELIEVE,WITHOUT GOD ,I AM DOG
9. PIYUSHDADRIWALA
7+9+7+3+1+8+4+1+4+9+9+5+1+3+1=72=7+2=9

I THINK YOU ALL ENJOY,I HAVE MORE,NEXT TIME
LOVE TO ALL
PIYUSHDADRIWALA
www.piyush-g.741.com
www.piyushdadriwalamaths.co.in

sum of each digit remain same(piyush constant)

sum of each digit remain same(piyush constant)
piyush constant
SUM OF EACH DIGIT REMAINS "9"(PIYUSH CONSTANT)
Fri, 2006-07-21 22:34 — piyushdadriwala
SUM OF EACH DIGIT REMAINS SAME(9),NINEI AM VERY MUCH FOND OF MATHS ,WHATEVER I AM WRITING HERE IS AMAZING,INTERESTING,LEARN IT,VERY SIMPLE.(FOR ANY NO OF DIGITS)NOW,I HAVE 25 AND 32, MULTIPLE THEM ,NOW YOU CAN MULTIPLE THEM IN FOUR WAYS LIKE THAT(just changing the position)25*32=80052*32=166425*23=57552*23=1196now substract any bigger to any lower you will always get sum of each digit nine.1664-1196=468(4+6+8=18=1+8=9)1664-800=864(8+6+4+18=1+8+=9)1664-575=1089(1+0+8+9=18=1+8=9)1196-800=396(3+9+6=18=1+8=9)1196-575=621(6+2+1=9)800-575=225(2+2+5=9).this i called "piyush contant"with lot of regardspiyushdadriwalawww.piyush-g.741.compkgdwala@rediffmail.comwww.piyushdadriwalamaths.co.in

piyush euquation

piyush equations
it is unique search in maths,and copyrighted to me

piyush discovered equations (cube and square),designing points on pyramid,no body has done so far ,it is my challenge.
Fri, 2006-07-21 08:34 — piyushdadriwala
piyushdadriwala discovered equations (cube and square),designing points on pyramid ,piyush also got that people used counting from 1 to 9 that is wrong,we should start from 0 to 9 instead from 1 to 9, that piyush got from these equations ,EQUATIONS ARE A CUBE =A(3A-2)+A(A-1)(A-2)AND A SQUARE =A(3A-2)-2A(A-1),AND ALSO FROM PYTHAGORES AND PASCAL TRIANGLE TOO ON THE BASE OF SYMETRY.PIYUSH ALSO GOT TAHT ABOVE VERTEX THER IS ALWAYS ZERO (space),which is coming from infinity.Zero means a lot of energy,whch goes to PYRAMID through vertex.Pyramid are still miracle.besides these piyush also obtain a new formula nd table for two digit square and a new triangle A.P. RIGHT ANGLE TRIANGLE IN WHICH A THEORUM ,BY THIS THEORY WE CAN SOLVE A PARTICULAR TYPE QUESTION WITHIN SECONDS ,OTHERWISE IT TAKES TO SOLVE 15 -20 MINUTES.HOW DID I GET THESE EQUATIONS?VERY AMAZING,WHENEVER I FEEL FREE USED TO WORK WITH MATHS,ONE DAY I JUST PUT THREE POINTS IN A ROW ,GET TWO MIDDLE POINTS AND THEN ONE,I DID IT AGAIN FOR MORE POINTS MY IMAGINATIONS AND CREATIVITY BRING ME THERE WHERE THERE WAS VERY INTERESTING AND AMAZING RESULT LYING,I WAS SO HAPPY BUT STILL NOT RECOGNISED, NOW I AM NEAR ABOUT 40 ,I WANT SHARE ALL THESE WITH MY LOVELY FRIENDS AND BLOGGERS,MY MOTTO "WHATEVER I HAVE (knowledge),is not mine ,it is for all ,and of GOD.AND "GOD CREATED,WE DO".NEXT MORE IN .......piyush dadriwalawww.piyush-g.741.compkgdwala@rediffmail.COMpiyushdadriwala@gmail.com
www.piyushdadriwalamaths.co.in
.