root question


by Holocene
Tags: root
Holocene
Holocene is offline
#1
Nov21-07, 07:32 PM
Holocene's Avatar
P: 231
Is there any "pencil and paper" method to find the nth root of a number?

Since multiplying a number by itself any number of times quickly yeilds extremely large numbers, trial and error might seem to pinpoint the root of a number, so long as it is a perfect square or cube or whatever.

But, is there any real way to pinpoint the root of a number without using a calculator or trial and error?
Phys.Org News Partner Mathematics news on Phys.org
Researchers help Boston Marathon organizers plan for 2014 race
'Math detective' analyzes odds for suspicious lottery wins
Pseudo-mathematics and financial charlatanism
Count Iblis
Count Iblis is offline
#2
Nov21-07, 08:05 PM
P: 2,159
To compute the nth root of a number Y, just make some guess X and then improve that guess using the formula:

[tex]\left(1-\frac{1}{n}\right)X + \frac{Y}{n X^{n-1}}[/tex]

You can iterate this to make further improvements. E.g. suppose you want to estimate the cube root of 10. Then you can take X = 2. the formula gives: 4/3 + 10/(3*4) = 2 + 1/6

If you then take X = 2+1/6 and insert that in the formula to get 2.1545. Iterating again gives 2.15443469224. Now, believe it or not but:

2.15443469224^3 = 10.0000000307
slider142
slider142 is offline
#3
Nov21-07, 08:12 PM
P: 876
If you want to know the general theory behind the above method, see Newton's method.

Count Iblis
Count Iblis is offline
#4
Nov21-07, 08:13 PM
P: 2,159

root question


So, this is still trial and error, but it converges very fast. At each step you double to correct number of digits. You go from a wild guess to a number that is correct to ten significant digits in about four iterations.
Holocene
Holocene is offline
#5
Nov21-07, 08:40 PM
Holocene's Avatar
P: 231
Quote Quote by Count Iblis View Post
To compute the nth root of a number Y, just make some guess X and then improve that guess using the formula:

[tex]\left(1-\frac{1}{n}\right)X + \frac{Y}{n X^{n-1}}[/tex]

You can iterate this to make further improvements. E.g. suppose you want to estimate the cube root of 10. Then you can take X = 2. the formula gives: 4/3 + 10/(3*4) = 2 + 1/6

If you then take X = 2+1/6 and insert that in the formula to get 2.1545. Iterating again gives 2.15443469224. Now, believe it or not but:

2.15443469224^3 = 10.0000000307
wow, that's pretty neat. Thanks!
Count Iblis
Count Iblis is offline
#6
Nov21-07, 08:44 PM
P: 2,159
The case n = -1 is also very useful. In that case X = 1/Y but Newton's method gives:

[tex]2X - X^{2} Y[/tex]

Since there are no divisions in here, you can use it to do divisions. It's much faster than long division.


Register to reply

Related Discussions
Root Locus question General Engineering 3
Root Question Calculus & Beyond Homework 7
Polynomial Root Question Precalculus Mathematics Homework 2
TI-83 root question Introductory Physics Homework 1
Root sequence question Linear & Abstract Algebra 11