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Root question

by Holocene
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Holocene
#1
Nov21-07, 07:32 PM
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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?
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Count Iblis
#2
Nov21-07, 08:05 PM
P: 2,157
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
#3
Nov21-07, 08:12 PM
P: 898
If you want to know the general theory behind the above method, see Newton's method.

Count Iblis
#4
Nov21-07, 08:13 PM
P: 2,157
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
#5
Nov21-07, 08:40 PM
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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
#6
Nov21-07, 08:44 PM
P: 2,157
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.


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