Square roots -- manually

[Ben]

Does anyone know how to find square roots manually? It's for extra credit in math, and I need this to go from a 69 (F) to the 70s (70=D, 71-79=C) if I don't want to go to summer school. I'm searching google right now, soooo .... help :(

*Looks at conkermaniac*

[add]
OK, I think I found it in one of the search results:


Ancient Square Roots

The ancient Babylonians had a nice method of computing square roots
that can be applied using only simple arithmetic operations. To find
a rational approximation for the square root of an integer N, let k
be any number such that k^2 is less than N. Then k is slightly less
than the square root of N, and so N/k is slightly greater than the
square root of N. It follows that the average of these two numbers
gives an even closer estimate

k + N/k
k_new = -------
2

Iterating this formula leads to k values that converge very rapidly
on the square root of N. This formula is sometimes attributed to
Heron of Alexandria, because he described it in his "Metrica", but
it was evidently known to the Babylonians much earlier.
Can anyone verify this? Conker?

[Gayowulf]

how the hell is 69% an F.

where I come from that's a C+. I strive for those.

I can't verify what you've got there, but I do know that it can be done with infinite series (binomial theorum), approximated with simple arithmetic and one other way I think.

I'll get my math notes, and see if I was paying attention that day.

edit:

I can verify your method. We called it, guess, divide, and average

to put it simply, guess the square root value of your number. integral is easiest, and a crappy guess will do.

ie guess that 2 is the square root of 12.

divide 12 by our guess of 2 and get 6.

average 6 and 2: (6+2)/2 = 4

repeat with the new guess of 4. So 12/4 = 3

average 4 and 3: (4+3)/2 = 3.5

Repeat: 12/3.5 = 3.43

Average: (3.5 + 3.43)/2 = 3.465

keep going till you're tired of it. That's the plain english version of what you quoted.

as for the binomial theorum one, the one your teacher is probably looking for, It's a bit more complex. To tell you the truth I don't completely understand it anymore. (should have done the exercises)

If I have a moment of significant realization, or the sudden, uncontrollable urge to do some math I'll try to explain it. ;)

[Owen]

I just wrote a quick C++ program to see if that forumla works, and it is pretty close to the actual answer the calculator gives. So that method is pretty good.

However, when you get into big numbers, unless you want to work problem 100s of times through, it is not a great method, but adequate for what you are probably doing(although I dont know for sure).

There is a method called Newton's Method, which involves the tangent line of a equation at certain points (derviatives), but I am sure that is not what you are looking for. However, what may take me many times to do with that old formula, may take me only 6 or 7 times to get a precise answer. IF you want Newton's method, I will give it to you, but it is pretty complex if you have never taken calculus.

[Ben]

Thanks Gayowulf!

Owen: You can try, I hope I can understand it :eek:

[anhedonia]

hehe, trusting your grade to an online forum. Interesting.

[Ben]

Originally posted by anhedonia
hehe, trusting your grade to an online forum. Interesting.
I already found the info before anyone posted, and asked two people I know in real life if it was correct. I was just interested in the second method Owen mentioned.

[Owen]

Its a real "fun" method of determining roots. It is fairly simple if you know what derivatives are(don't worry though, I will do that work for you in this problem). However, it does have its drawbacks, but it isn't prevalent in finding square roots, cube roots, etc.

First off, here is the basic forumla.

new_n = n - F(n) / F'(n)

Now lets take it piece by peice.

n - a number you think it is close to correct answer the first time through. Through every time you do it afterwards, it is was you got in the previous iteration as new_n.

F(n) - Function of n. For finding the sqaure root, this would be:
F(n) = n^2 - x

x if the root you are looking for. So if you are looking for the square root of 2 x is 2, 3, r is 3, 10, r is 10, and so on.

F'(n) - The derivative of F(n). I am not going to bore you with what it is, but for what you need, it will always be 2n.

Simply do the algebra, and repeat the process to get more accurate roots. Generally, the level of precision for the decimal places doubles through each use (for Instance, 1 decimal place the 1st time, then 2 then 2nd, then 4 the 3rd, then 8, 16, and so on). Not always true, but generally.

Also, if you are interesting in finding the cube root, 4th root, or the nth root(or, in this case, the r-th root), the forumla is fairly easy to change.

new_n = n - (n^r - x) / (r * n^(r - 1))

Where r is the root you are looking for (3 for 3rd root, 6 for 6th root, and so on) and x is the number you want to find the root of.

Now you may want to test all that out to make sure what I gave you is correct. I am pretty sure it is, but I may have made an error in what I wrote.

Now here is somewhat of a suprise. Both the ancient method and Newton's method are the same equation in the end. However, with the ancient equation, it is hard to see how to do adjust it to get a 3rd root, a 4th root, a 10th root, etc. With Newton's method, it is a fairly easy to see how to adjust it.

That leads me to think that at least part of Newton's calculus came from the Egyptians. Hmm....

Anyways, the compact formula for finding the nth root of a number is

new_n = ( (r-1) * n^r + x ) / (r * n^(r - 1) )

[Gayowulf]

Here's the binomial theorem method. I found my math notes, but I also found this site: http://www.krysstal.com/binomial.html

it would make me feel really smart and proud of myself if I were to explain this to you, but why waste time when there's a site that does a better job?

[conkermaniac]

We never learn this kind of stuff in school because our teachers believe that manual math is totally useless in this day and age. ;)

Interesting read...thanks Gayowulf.

[Gayowulf]

Originally posted by conkermaniac
We never learn this kind of stuff in school because our teachers believe that manual math is totally useless in this day and age.

Lucky you.

We aren't allowed to use calculators

[conkermaniac]

Actually, Ben, I thought up another way that you might do this, but you would have to be familiar with logarithms (I think you mentioned that you were in Alg I?)...

Let's say we wanted to find the square root of 5.35. We would look this up on the log table (you have to have one handy). In this case:

log 5.35 = 0.7284

Now, you might now that sqrt(x) is the same thing as saying x^(1/2). The logarithm of the square root of 5.35 can thus be written as such:

log 5.35^(1/2)

Using the properties of logs, this is the same thing as saying:

(1/2) log 5.35

We already know that log 5.35 = 0.7284, so:

(1/2) log 5.35 = (1/2)(0.7284)
(1/2) log 5.35 = 0.3642

But this isn't the answer -- we haven't gotten rid of the "log" function. So in order to do that, we now have to search for something close to 3642 on the log table, and use the rows and columns to determine the square root of 5.35. The closest one I found was 3636, and that corresponds to 2.31. Therefore, 2.31 is a close approximation of sqrt(5.35).

My calculator gives me 2.3130 as an approximation, so our estimate was very close! :)

[Ben]

Originally posted by conkermaniac
Actually, Ben, I thought up another way that you might do this, but you would have to be familiar with logarithms (I think you mentioned that you were in Alg I?)...
Yes, I'm in Algebra I, but I have no idea what a logarithm is. :confused2

[CareBear]

Originally posted by Ben
Yes, I'm in Algebra I, but I have no idea what a logarithm is. :confused2 it's the inverse of a^x

a^x = b <=> x = loga b

[conkermaniac]

It's nothing more than a different way to write exponents. So if I felt like it, I could rewrite 5^2 = 25 as log_5 25 = 2. It comes in very handy in certain calculations. :)

[Ben]

Ooh ok, thanks for the explanation.:)

[tandoc]

Originally posted by anhedonia
hehe, trusting your grade to an online forum. Interesting.

hasn't failed me yet