Iterative Method for Square root & Cube root in Sadratnamālā

Here we will be knowing an another method (Iterative method) to find square root and cube root in Sadratnamālā.

Verse 19

इष्टाप्तेष्टैक्यार्धमिष्टविशिष्टं कृतेः पदम् ।

घनमूलं द्विराप्तेष्टयोगार्धमविशेषितम् ॥ १९ ॥

Translation

Divide the number (whose square root is to be found) by any arbitrary number and find half of the sum of the arbitrary number and the quotient. Divide the number again by this new divisor. The same process is continued until the quotient becomes equal to the divisor. In the case of cube root, the first quotient is divided again by the arbitrary number to get the second quotient. The half of the sum of the arbitrary number and the second quotient is found which is the second divisor. This is repeated until the divisor becomes equal to the second divisor.

Example: Square root of 625

Iteration 1	Divide the number (625) by an arbitrary number - 10	10)	6	2	5	(62
			6	2	0
	Divide the number (625) by ${\frac {(Previous\ divisor+Previous\ quotient)}{2}}=$ ${\frac {(10+62)}{2}}=36$	36)	6	2	5	(17
			3	6
			2	6	5
			2	5	2
Iteration 2	Divide the number (625) by ${\frac {(Previous\ divisor+Previous\ quotient)}{2}}=$ ${\frac {(36+17)}{2}}=26$	26)	6	2	5	(24
			5	2
			1	0	5
			1	0	4
Iteration 3	Divide the number (625) by ${\frac {(Previous\ divisor+Previous\ quotient)}{2}}=$ ${\frac {(26+24)}{2}}=25$	25)	6	2	5	(25
			5	0
			1	2	5
			1	2	5
					0

In the last iteration (Iteration 3) the divisor (25) and the quotient(25) are same & the remainder is zero . Hence

Square root of 625 = 25

Example: Cube root of 512

Iteration 1	Divide the number (512) by an arbitrary number - 10	10)	5	1	2	(51
			5	1	0
	Divide the previous quotient (51) by previous divisor (10)	10)	5	1	(5
			5	0
Iteration 2	Divide the number (512) by ${\frac {(Previous\ divisor+Previous\ quotient)}{2}}=$ ${\frac {(10+5)}{2}}=7$	7)	5	1	2	(73
			4	9
				2	2
				2	1
	Divide the previous quotient (73) by previous divisor (7)	7)	7	3	(10
			7	0
Iteration 3	Divide the number (512) by ${\frac {(Previous\ divisor+Previous\ quotient)}{2}}=$ ${\frac {(7+10)}{2}}=8$	8)	5	1	2	(64
			4	8
				3	2
				3	2
	Divide the previous quotient (64) by previous divisor (8)	8)	6	4	(8
			6	4
				0

In the last iteration (Iteration 3) the first divisor (8) and the second quotient(8) are same & the remainder is zero . Hence

Cube root of 512 = 8

See Also

सद्रत्नमाला में 'वर्गमूल और घनमूल के लिए पुनरावृत्तीय विधि'