Here is that funny long division-like method for finding square and cube roots generalized to nth roots. These are all really extensions of the Binomial Theorem.
Steps
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Partition your number. Separate the number you want to find the nth root of into n-digit intervals before and after the decimal. If there are fewer than n digits before the decimal, then that is the first interval. And if there are no digits or fewer than n digits after the decimal, fill in the spaces with zeroes.
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Find an initial estimate. Find a number (a) raised to the nth power closest to the first n digits (or the fewer than n digits before the decimal) as a base-ten number without going over. This is the first and only digit of your estimate so far.Advertisement
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Modify the difference. Subtract your estimate to the nth power (a n ) from those first n digits and bring down the next n digits next to that difference to form a new number, a modified difference. (Or multiply the difference by 10 n and add the next n digits as a base-ten number.)
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Find the second digit of your estimate. Find a number b such that ( n C 1 a n - 1 (10 n-1 ) + n C 2 a n - 2 b (10 n - 2 ) ) + . . . + n C n - 1 a b n - 2 (10 ) + n C n b n - 1 (10 0 ) )b is less than or equal to the modified difference above (10 n (d ) + d 1 d 2 . . . d n ). This becomes the second digit of your estimate so far.
- The combinations notation n C r represents n! divided by the product of (n - r)! and r!, where n! = n(n - 1)(n - 2)(n - 3) . . . (3)(2)(1). The notation n C r is sometimes expressed as n over r within tall parentheses without a division bar, and it can be calculated simply as the first r factors of n! divided by r!, which is often written as n P r divided by r!
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Find your new modified difference. Subtract the two quantities in the last step above (10 n (d ) + d 1 d 2 . . . d n minus n C 1 a n - 1 (10 n-1 ) + n C 2 a n - 2 b (10 n - 2 ) ) + . . . + n C n - 1 a b n - 2 (10 ) + n C n b n - 1 (10 0 ) )b) to form your new modified difference by bringing down the next set of n digits next to that result. (Or multiply the difference by 10 n and add the next n digits as a base-ten number.)
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Find the third digit of your estimate. Find a new number c and use your estimate so far, a (which is now 2 digits), such that ( n C 1 a n - 1 (10 n - 1 ) + n C 2 a n - 2 c (10 n - 2 ) + . . . + n C n - 1 a c n - 2 (10 ) + n C n c n - 1 (10 0 ) ) c is less than or equal to the new modified difference in above (10 n (d ) + d 1 d 2 . . . d n ). This becomes the third digit of your estimate so far.
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Repeat. Keep repeating the last two steps above to find more digits of your estimate.
- This is basically a rolling binomial expansion minus the lead term, where the two terms involved are the prior estimate multiplied by 10 and the next digit to improve the estimate.
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Community Q&A
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QuestionHow do I find the 7th root of 4 without using a calculator?Community AnswerYou must follow the steps for the seventh root, grouping 4 as 4. 0000000 0000000, etc. And you'll likely need a 4-function calculator. "By Hand" here really means, without using the nth-root function on a scientific calculator. Though, with a lot of work, it could be done by hand.
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QuestionThe equation to find the second digit of your estimate as described in step 4 is too vaguely defined for me to even devise a clearly defined equation from it. Could you define it as a summation?Keith RaskinCommunity AnswerYes, we can. As a sum: Sum 1 through k of n_C_k a^(n - k)b^(k - 1)10^(n - k).
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