square ROOT

[made_of_asp]

Hello,

Is there some sort of a formula to calculate square root ( with your brain . I dont know how to do log or E

[sql_lall]

Yes, there is (though i don't have it on me right now). If you search for one you should be able to find it. If not, then all you have to do to get one to 2 dec. pl. is multiply it by 100^2, find the integer square root of this, then divide it by 100.
(This can be done for more than 2 dec. pl. as well.)

I'm 15 so i won't

[Guv]

If you are a good guesser and can do simple arithmetic, you can do square root.

Make a guess, divide the guess into the number, use the average of the guess and the quotient as the next guess. I will illustrate by giving an example.

Suppose you did not know that the square root of 100 is 10, and wanted to calculate it. You are a bad guesser and decide that 20 is a reasonable guess.

100/20 = 5
Average(5, 20) = 12.5

100/12.5 = 8
Average(8, 12.5) = 10.25 (We are getting closer)

100/10.25 = 9.756 097 561
Average(9.756 097 561, 10.25) = 10.003 048 751

100/10.003 048 751 = 9.996 952 178
Average(9.996 952 178, 10.003 048 751) = 10.000 000 465

Keep going until you think you are close enough.

Note that as you get closer, the guesses do not change much. This is how you know you are close.

There is a method which is a bit like long division, but I find the above easier. I make good first guesses.

[twanvl]

I think dividing is really hard without a calculator, I use this method:

try to find sqr(2):

guess 1: 1.5
1.5 ^ 2 = 1*1.5 + .5*1.5 = 2.25
That is to high

guess 2: 1.4
1.4^2 = 1.5^2 - 2*1.5*(1.5-1.4) + (1.5-1.4)^2 = 2.25 - 0.3 + 0.01 = 1.96 (I use this method, because it's much easier then recalculating the entire thing)

1.96 is to low, so the answer is between 1.4 and 1.5

guess 3: 1.45
1.45^2 = 1.4^2 + 2*1.4*0.05 + 0.05^2 = 1.96 + 0.14 + .0025 = 2.1025

etc.

The formulas I use are:
(A+B)^2 = A^2 + 2*A*B + B^2
and
(A-B)^2 = A^2 - 2*A*B + B^2

[Guv]

Twanvl: Note how fast the guess divide and average method converges. See previous post.

That method got 8 digits of precision in 4 iterations, starting with a terrible first guess (20 as guess for Square Root 100).

The method you posted will take a lot more work.

[DavidHooper]

Lighten up Guv, it was his seventh post!

[nishantp]

Originally posted by Guv
If you are a good guesser and can do simple arithmetic, you can do square root.

Make a guess, divide the guess into the number, use the average of the guess and the quotient as the next guess. I will illustrate by giving an example.

Suppose you did not know that the square root of 100 is 10, and wanted to calculate it. You are a bad guesser and decide that 20 is a reasonable guess.

100/20 = 5
Average(5, 20) = 12.5

100/12.5 = 8
Average(8, 12.5) = 10.25 (We are getting closer)

100/10.25 = 9.756 097 561
Average(9.756 097 561, 10.25) = 10.003 048 751

100/10.003 048 751 = 9.996 952 178
Average(9.996 952 178, 10.003 048 751) = 10.000 000 465

Keep going until you think you are close enough.

Note that as you get closer, the guesses do not change much. This is how you know you are close.

There is a method which is a bit like long division, but I find the above easier. I make good first guesses.

Guv, Would i be right in saying that with that method, you would get infinitesimaly close to the answer, but never reach it exactly?

[riis]

Guv, Would i be right in saying that with that method, you would get infinitesimaly close to the answer, but never reach it exactly?

Nishantp, isn't that the case with most square roots? Even if you know sqrt(2) in 300 decimals, you still don't have the exact value. So, I see no problem with Guv's method.

[Guv]

Nishantp: Right you are. You will never get an exact square root.

The remark by Riis is pertinent.

If you try the method on exact sqaures like 1.44, 4, 9, 256, et cetera, you might guess that you know what the exact root is and stop the process.

[lloydie]

There are loads of ways to do this... and quite a few formula too!

There are not really any that stick out as being the easiest, but I suppose that doesn't matter anyway... we all have different ways of doing something!

Well, this is the point where I tell you how to do it - but I'm not going to!

It's quite easy to understand and do, but a complete pain in the ass to describe. It took me a while to find somewhere on the Interent that already describes this (save me the trouble of writing my own explaination!), but I found this:

http://mathforum.org/dr.math/tocs/sq...ot.middle.html

There are quite a few examples.
For some more examples (or ones of a higher/lower calibre of maths) try these links:

http://mathforum.org/dr.math/tocs/squareroot.high.html
http://mathforum.org/dr.math/tocs/squareroot.elem.html

Hope this helps.

Regards,
Lloyd

[made_of_asp]

Thanks everyone for your answers

There is also this forumula if anyone solves it???


Exp(Log(x) / 2)

meaning...

2.71828182845905 ^ (Log(x) / 2)

[thinktank2]

e^(Log_eA) = A

and [Log A]/2 = 1/2* Log A = Log A^1/2


Therefore e^((Log_ex)/2) = e^(Log_ex1/2) = x^1/2

[sql_lall]

That's right , but only one slight technical hitch.
You have written log(x), which generally suggests log 10(x).
If you want to make sure you are using the natural logarithm -e- then put ln(x). This always means log e(x).

N.B. Sorry, i don't know how to subscript

The equation corrected:

x ^1/2 =
e/2*ln(x)

I'm not sure, but does
x ^ 1/y =
e/y * ln(x)

Maybe you can even say:
x ^ y/z =
e*y/z*ln(x) =
10*y/z*log(x)

Though I haven't checked these last two.

[thinktank2]

Originally posted by sql_lall
The equation corrected:

x ^1/2 = e/2*ln(x)

I'm not sure, but does
x ^ 1/y = e/y * ln(x)

Maybe you can even say:
x ^ y/z = e*y/z*ln(x) = 10*y/z*log(x)

Though I haven't checked these last two.

Your equations are wrong...

Start from here ..

ln[x^(1/2)] = (1/2) * ln[x]

Raising both sides to the power of e

e^( ln[x^(1/2)] ) = e^( (1/2) * ln[x] )

Which means

x^(1/2) = e^( (1/2) * ln[x] ) --------------Eq (1)


Similarly,

log₁₀[x^(1/2)] = (1/2) * log₁₀[x]

Raising both sides to the power of 10

10^( log₁₀[x^(1/2)] ) = 10^( (1/2) * log₁₀[x] )

Which means,

x^(1/2) = 10^( (1/2) * log₁₀[x] ) --------Eq (2)

From Eq1 and Eq 2 you should see that...

x^(1/y) = e^( (1/y) * ln[x] ) = 10^( (1/y) * log₁₀[x])

and

x^(y/z) = e^( (y/z) * ln[x] ) = 10^( (y/z) * log₁₀[x])

[sql_lall]

Yeah, that's what i meant. Don;t know why i used times instead of ^ though. Maybe just tired

[maelkyzzydec]

There are quite a few ways to figure out a root. The most popular that I can think of is newton's method that is performed on :
Say you want the sqrt(7). You know that y = 7-x^2 has roots at plus or minus sqrt(7). Use newton's method to find those roots and BAM you've got your answer

Continued fractions from number theory is another - more intersting way

later