Algebric Proof of Ekadhikena Purvena
a). Consider (Ax+y)^2 = A^2 * X^2 + 2Axy + y^2
for x=10 and y=5 becomes
(10A + 5)^2 = A^2 * 10^2 + 2*10A*5 + 5^2
= A^2 * 10^2 + 10^2 * A + 5^2
= 10^2 (A^2 + A) + 5^2
= A(A+1) * 10^2 + 25
It represent two digit number Like 25,35,45,55......
L.H.S. is A(A+1)
R.H.S. is 25
So A(A+1)/25
It can be used for Two digit numbers only.
Algebric Proof of Ekadhikena Purvena - 2nd
b). Consider (Ax^2+Bx+C)^2 = A^2 *x^4 + B^2 * x^2 + C^2 + 2ABx^3 + 2BCx + 2CAx^2
= A^2 * x^4 + 2ABx^3 + (B^2 + 2CA)x^2 + 2BCx + C^2
for x = 10 and C = 5 becomes
(A*10^2 + B*10 + 5)^2 = A^2*10^4+2AB10^3+(B^2 + 2*5A)10^2+B*10^2 + 5^2
= A^2 * 10^4 + 2AB10^3 + (B^2 + 10A)10^2 + B*10^2 + 5^2
= A^2 * 10^4 + 2AB10^3 + B^2 * 10^2 + A10^3 + B*10^2 + 5^2
= A^2 * 10^4 + (2AB + A) 10^3 + (B^2 + B) 10^2 + 5^2
= (A^2 * 10^2 + 2AB*10 + A*10 + B^2 + B) 10^2 + 5^2
= (10A+B) (10A+B+1) 10^2 + 25
= Z(Z+1) 10^2 + 25 where Z = 10A+B
It represent three digit number Like 125,135,145,155......
L.H.S. is Z(Z+1)
R.H.S. is 25
So Z(Z+1)/25
= A^2 * x^4 + 2ABx^3 + (B^2 + 2CA)x^2 + 2BCx + C^2
(A*10^2 + B*10 + 5)^2 = A^2*10^4+2AB10^3+(B^2 + 2*5A)10^2+B*10^2 + 5^2
= (A^2 * 10^2 + 2AB*10 + A*10 + B^2 + B) 10^2 + 5^2
= (10A+B) (10A+B+1) 10^2 + 25
= Z(Z+1) 10^2 + 25 where Z = 10A+B
Algebraic Proof of Multiplication Method
Any Fraction of 1/A9
1/A9 = 1/ ((A + 1) x - 1) where x = 10
= 1/((A + 1)*(1 - 1/(A + 1)x ))
= (1 - 1/(A + 1)x)^(-1) * (A + 1)x
= (1 + 1/(A + 1)x + 1/(A + 1)x^2 + ........)/((A + 1)x)
= 1/(A + 1)x + 1/(A + 1)^2*x^2 + 1/(A + 1)^3*x^3 + ..........
x = 10
= 10^(-1) (1/(A + 1) + 1/(A + 1)^2*10^(-2) + 1/(A + 1)^3*10^(-3) + .....)
This Series explain the process of ekadhiken.
Hi. I’m Designer of Blog Magic. I’m CEO/Founder of ThemeXpose. I’m Creative Art Director, Web Designer, UI/UX Designer, Interaction Designer, Industrial Designer, Web Developer, Business Enthusiast, StartUp Enthusiast, Speaker, Writer and Photographer. Inspired to make things looks better.
0 comments:
Post a Comment