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Calculating Any Root


Date: 10/13/97 at 18:07:00
From: Lee Tat
Subject: Calculating any root

I need to find an algorithm to determine any root of a number, i.e.
the 3rd root, 4th root, 5th root, or any root.  I was told I could
determine the estimated value by using Newton's Method.  Can you
help? I don't remember what Newton's Method is. 

Any information you can provide would be appreciated.

Thanks,
Lee


Date: 10/14/97 at 13:54:24
From: Doctor Tom
Subject: Re: Calculating any root

Hi Lee,

There's a trivial way to take the n-th root of a number using
logarithms.

root = antilog(log(number)/n)

So to take the fifth root of 37, find:

antilog(log(37)/5) = antilog(3.61091791264/5)

                   = antilog(.722183582528) = 2.05892413648

(according to my calculator).  To check, when I multiply 2.05...
by itself 5 times, I get:  37.0000000002 - pretty good.

Newton's method works for any function, so it's more general than the 
method above. What you're trying to do is to solve the equation:

    x^n - number = 0

where you're trying to take the n-th root of "number".  For the 
example above, you're trying to solve:

    x^5 - 37 = 0.

Newton's method is used to solve equations of the form f(x) = 0,
so in this case, f(x) = x^5 - 37.

It works by taking an initial guess g-old and getting a new
guess g-new using this equation:

    g-new = g-old - f(g-old)/f'(g-old)

where f' is the derivative of f.

Repeat this operation over and over and the method will zero in on the 
root.

For the example x^5 - 37 = f(x),  f'(x) = 5x^4, so the iteration is:

    g-new = g-old - (g-old^5 - 37)/(5g-old^4).

Suppose (same example), that your first guess is 3.

The next guess is:

    3 - (3^5 - 37)/(5*3^4) = 2.49135802469.

Plug 2.49... in as the old guess and get the next approximation:

    2.18516863143

Next guess:

    2.07269258474

Next guess:

    2.05910584704

Next guess:

    2.05892416855

Next guess:

    2.05892413648

which is accurate to 10 places.

-Doctor Tom, The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    


Date: 05/02/2005 at 17:44:51
From: Jack
Subject: Dont understand antilog and relations to root calculations

When I did the log(37) in the link above, I got 1.568201724, not the
number you received. What am I doing that's different than your method?

Also, what is antilog? It's apparently important in the link and there's 
no antilog on my calculator. I would like it if you could tell me what 
it is or its numerical value so I can complete the problem.


Date: 05/02/2005 at 23:11:26
From: Doctor Peterson
Subject: Re: Dont understand antilog and relations to root calculations

Hi, Jack.

Dr. Tom apparently used the natural log ("ln" on the calculator) rather 
than the common log ("log" on the calculator) in his calculation. It 
doesn't matter which log you use, as long as you use the corresponding 
antilog; you'll get different intermediate numbers but the same end result.

The antilog of the common logarithm is 10^x, and the antilog of the
natural logarithm is e^x. (The Windows calculator accessory uses "inv
log" and "inv ln" for those functions, making it clearer that they are 
the inverses of the logs, or antilogs, but harder for someone who 
doesn't already know that to figure out how to do exponential functions.)

So Dr. Tom did the following (supposing he used the Windows
calculator, for the sake of choosing a readily available standard):

  37 ln / 5 = inv ln

but he could have done this to get the same final answer:

  37 log / 5 = inv log

In the latter case, the work as he wrote it would look like

  antilog(log(37)/5) = antilog(1.568201724/5)

                     = antilog(0.3136403448) = 2.05892413648

If you have any further questions, feel free to write back.


- Doctor Peterson, The Math Forum
  Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Logs
High School Square & Cube Roots

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