This idea actually came to me once, quite by accident, as I was doodling in class. I came across this series of fractions that equals one third: one half minus one eighth minus etc. In short, you take one half and subtract one divided by 2 taken to odd numbered powers: one divided by 2 to the third power, then to the fifth power, etc. When I got home and tested this, and it seems to converge pretty quickly. That will be important, as we will see in a minute. Anyways, this isn’t the only thing that occurred to me. I was thinking of how someone could get a cube root using an ordinary, non-scientific calculator. The ordinary square root function you find on an ordinary calculator only works for even numbered roots. Then I thought, if you treat the square root function as taking something to the one half power, and if you treat division as subtraction of exponents, you find you can easily use these two principles plus the series I discovered for getting the cube root on ordinary calculators.
Here is how it works:
EXAMPLE. Find the cube root of 2 using this method.
Simply take 2 and square root it. Then divide it by 2 square rooted three times. Then divide this result by 2 square rooted five times and so forth, always square rooting an odd number of times. After you do this about 5 to ten times, you get the undeniable result of about 1.26, which a quick check on the calculator tells you is indeed the correct answer for the cubed root of 2.
(Note. For more complex numbers, simply store the number you wanted the cube root of in memory, and use the above technique.)
Anyways, my reason for putting it here in MPSIMS is basically twofold. I wish to share it. I also welcome any criticisms anyone may have. I have done it alot of times, so I know the method works as planned. But is there perhaps a series of fractions that converge even more quickly than the above method? Also, am I really the only one to come up with this idea. Does anyone know anyone else who ever came up with an idea like mine.
I don’t know how you would go about proving your proposed theorem (actually, given enough time I could probably do it, but I’m too tired to do it right now)…but did you realize the cube root of n is the same as n[sup]1/3[/sup]? (Not all calculators have the little “x^y” button, of course.)
Regarding your first part, you’ve just found the sum of an infinite geometric series that starts with 1/8 with common ratio 1/4. In other words,
1/2[sup]3[/sup] + 1/2[sup]5[/sup] + 1/2[sup]7[/sup] + … = 1/8 + 1/32 + 1/128 + …
This sum is just 1/6, which you can get quite readily by using a known formula or by just working out the formula yourself. Look here for more info.
And your discovery follows immediately since 1/3 = 1/2 - 1/6.
You don’t have to restrict yourself to using only the odd powers of two. You can also use the following series to approximate 1/3 = 1/2 - 1/4 + 1/8 - 1/16 + …
This series (like the one you used) can be used to approximate the cube root of a number by only using the square root key on a calculator. From a practical standpoint, most calculators have a power key (x[sup]y[/sup]) that can be used to get the cube root, and if not, using the bisection method will find the cube root of a number fairly rapidly.
