Vedic MatheMatics
Guess and find the Cube Root!!
Let the no., whose cube root is to be calculated, be "a"
For example, let a=6
Begin with a guess, say x, to be its cube root.
In this case x=3/2
*** This guess is based on simple reasoning that cube of 2 is 8 and therefore 2 is too high and cube of 1 is 1 which means 1 is very low.The better the guess, the accurate the result ***
Now, we have a formula-
Y = 1/3[2x + a/(x*x)]
Y = 1/3[2*3/2 + 6/(9/4)] = 1/3[3 + 24/9] = 51/27 = 1.888888
Repeat the process by taking x = 51/27
The process can be repeated successively to get more accurate result, by taking Y=x.
Example2 : to find cube root of a=89
let x=9/2 then
Y = 1/3[2*9/2 + 89/(81/4)]
= 1/3[9 + 4.3950]
= 3 + 1.4650
= 4.4650
Repeat the process by taking x = 4.4650
N!$H@NT
Let the no., whose cube root is to be calculated, be "a"
For example, let a=6
Begin with a guess, say x, to be its cube root.
In this case x=3/2
*** This guess is based on simple reasoning that cube of 2 is 8 and therefore 2 is too high and cube of 1 is 1 which means 1 is very low.The better the guess, the accurate the result ***
Now, we have a formula-
Y = 1/3[2x + a/(x*x)]
Y = 1/3[2*3/2 + 6/(9/4)] = 1/3[3 + 24/9] = 51/27 = 1.888888
Repeat the process by taking x = 51/27
The process can be repeated successively to get more accurate result, by taking Y=x.
Example2 : to find cube root of a=89
let x=9/2 then
Y = 1/3[2*9/2 + 89/(81/4)]
= 1/3[9 + 4.3950]
= 3 + 1.4650
= 4.4650
Repeat the process by taking x = 4.4650
N!$H@NT
posted by N!$H@NT at
11:42 pm
6 Comments:
sahi...but do u know the logic behind this approximation method??
This is called NEWTON-RAPHSON METHOD
hey its a good method, but dont you think its really difficult to go on doing the same thing to get accurate results??
its a good approximating method...no doubt about it...but i want to know why it's like this..thelogic not the name of the method....
Can i know how it is derived? What is logic behind recursive calling of the function? Is this based on the expansion method. Like that of cos and sin expansion, the more no of terms you solve the more you are closer to the answer
Secondly, why can't we simply calculate it directly, instead of using this formula and trying to approch the answer.
there is no method as such like one of calculating square root to calculate the cube root..
however there's another method of vedic maths that calculates the square root...i'll put it here as soon as i have some time...
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