Calculating 4th Roots, 5th Roots...Date: 05/05/99 at 03:44:44 From: Mark Smith Subject: Calculation of Roots Is there an EASY way to calculate roots of any given depth? For example, if I wanted to solve for this: z = y _______ Where: y=2 would indicate I need the Square root of X \ / Where: y=3 would indicate I need the Cube root of X \/ X Where: y=4 etc..... Is there an easy way to solve for z, and is there an easy way to solve for X ? Please help. Date: 05/05/99 at 08:29:08 From: Doctor Jerry Subject: Re: Calculation of Roots Hi Mark, Do you want to calculate roots by hand? If so, then there are several methods. 1. Use logs. z = x^{1/y} log(z) = (1/y)log(x) Look up the log of x and divide by y. Then look up the antilog of the result. This was used in the past, before scientific calculators, which do the calculation using logarithms (if y = 2, they use another method). 2. Use Newton's method. Suppose we want a^{1/n}, where n is a positive integer. Look at f(x) = x^n-a. The x-intercept of this function, that is, where x^n-a = 0, is the value of x you are looking for. Let x1 be a guess of this value. Improvements x2,x3,... on this guess are calculated from Newton's formulas: x2 = x1-(x1^n-a)/(n*x1^{n-1}) x3 = x2-(x2^n-a)/(n*x2^{n-1}) etc. For example, suppose we want 5^{1/4}, the fourth root of 5. Let x1 = 1.5; then x2 = 1.495370 x3 = 1.495349 If you take the fourth power of x3 you'll obtain something close to 5. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ Date: 05/05/99 at 08:56:22 From: Doctor Rob Subject: Re: Calculation of Roots Thanks for writing to Ask Dr. Math. When you say "solve for z" and "solve for X," I assume from your first paragraph that you mean find numerical values. I also assume from the message that you want y to be an integer. The following algorithm will find z very quickly: 0. Pick a level of approximation you are willing to live with, that is, a number e > 0 such that your answer and the actual answer differ by less than e. 1. Let z(0) = a first guess at the value of the root. 2. Let n = 0. 3. Compute z(n+1) = (1-1/y)*z(n) + (1/y)*x/z(n)^(y-1). 4. If |z(n+1)-z(n)| < e, stop and declare that z = z(n+1). 5. Replace n with n+1 and go to Step 3. This is called Newton's Method, after Sir Isaac Newton. It converges on the right answer very quickly, and more so if your guess z(0) is a good one. Example: Find the fifth root of 11 to 8 decimal places. 0. e = 1/10^8 1. z(0) = 2 2. n = 0 3. z(1) = (4/5)*2 + (1/5)*11/2^4 = 1.6 + 0.1375 = 1.7375 4. |z(1)-z(0)| = 0.2625 > e 5. n = 1 3. z(2) = (4/5)*1.7375 + (1/5)*11/1.7375^4 = 1.631392308 4. |z(2)-z(1)| = 0.106107692 > e 5. n = 2 3. z(3) = (4/5)*1.631392308 + (1/5)*11/1.631392308^4 = 1.615704970 4. |z(3)-z(2)| = 0.015687338 > e 5. n = 3 3. z(4) = 1.615394386 4. |z(4)-z(3)| = 0.000310584 > e 5. n = 4 3. z(5) = 1.61539426620220. 4. |z(5)-z(4)| = 0.000000119475 > e 5. n = 5 3. z(6) = 1.61539426620218 4. |z(6)-z(5)| = 0.00000000000002 < e Stop. The fifth root of 11 is 1.61539427 to 8 decimal places. Now to solve for X, you have X = z^y. Since y is an integer, this is just a matter of repeated multiplication. You can shortcut this a little if y is of a fair size in the following way. Compute z^2, z^4, z^8, z^16, ..., z^(2^n), where 2^n <= y < 2^(n+1). You do this by squaring each term to get the next one. Now figure out which of these to multiply together to get z^y = X. That is done by figuring out how to write y as a sum of powers of 2, that is, how to write y in base 2 (or binary). Example: What number has 20-th root 3? Here z = 3, y = 20. Writing y in binary, y = 10100, y = 2^4 + 2^2, z^20 = z^16 * z^4. Then z^2 = 9, z^4 = 81, z^8 = 6561, and z^16 = 43046721, so z^20 = 81*43046721 = 3486784401. This reduces the calculation from y - 1 = 19 multiplications to just 5. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/