Here's a way to get very good approximations:
Step I: Find a number close to the square root. In our case, the square root of 2 is between 1 and 2, so we use x=1 as our first approximation.
Step II: compute (x^2 +2)/(2x). This is a better approximation.
Repeat step II as needed.
So, with x=1, the second approximation will be (1^2 +2)/(2*1)=3/2.
The next approximation will be [(3/2)^2 +2]/(2* 3/2)=17/12.
The next will be [(17/12)^2 +2]/(2 *17/12) =577/408.
The good thing about this method is that the number of decimal places of accuracy doubles every time step II is used.
This can also be modified as follows (it's the same algorithm, just simplified a bit):
Let p/q be your first approximation. Then [p^2 +2q^2]/(2pq) is the next approximation.
So, with x=1/1, p=1 and q=1, so the next is [1^1 +2* 1^2]/(2*1*1)=3/2. The next is [3^2+2*2^2]/(2*3*2)=17/12. Etc.
2007-01-01 07:35:47
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answer #1
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answered by mathematician 7
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(1+1)^.5
Use the binomial theorem for fractional powers.
1^(1/2)+(1/2)*1^(-1/2)*1-(1/2)*(-1/2)/2*1^(-3/2)*1^2+........
These terms can be described by the recursive equations:
X(n+1) = X(n)*Y(n)/(n+1)
Y(n+1) = Y(n) - 1
Where X(0) = 1, Y(0) =1/2,
X represents the coefficient and Y represents the leading power.
This gives the sum to be:
1 + 1/2 - 1/8 + 1/16 - 5/128 + 7/256 - .......
That is one way. Another way is probably much better by hand is just take a guess. Divide the guess into two. Then average the first guess and the result. This is your new guess. Now keep doing that until you get as close as you want to the real answer.
Neither one would be much fun to do without the aid of a calculator.
2007-01-01 15:51:46
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answer #2
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answered by adrian b 3
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Yes, I learned how to do this in school when I was 12 years old. We did it over and over again in class, and in examinations, and I can still do it more than 50 years later. I can knock out sqrt(2) to 12 decimal places, with nothing but pencil and paper, in two or three minutes. It's just a modified form of long division.
I can't show you the working in Yahoo! Answers because it would come out all misaligned, in this Arial font. But there's a small example and a complete explanation at this web site.
2007-01-01 15:31:11
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answer #3
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answered by bh8153 7
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The classical algorithm is explained fairly well at:
http://jwilson.coe.uga.edu/EMT668/EMAT4680.folders/Nowlen/squareroot.html
(If this URL doesn't all appear, just do a Yahoo search on "square root algorithm." You'll find lots more explanatory sites from this search.)
I don't really think you can expect someone to go through it all and explain each step, just for you. Simply check this web-site and then do it for yourself --- that way, you're more likely to learn how to do it.
I recall my father teaching me this method when I was about 9 years old.
Live long and prosper.
POSTSCRIPT The method by "mathematician," below, is EXTREMELY GOOD AND FAST. For example, 577/408 is ~ 1.000001502xsqrt(2), and therefore quite remarkably close to sqrt(2). Hats off to "mathematician"!
2007-01-01 15:30:40
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answer #4
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answered by Dr Spock 6
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huh. Took me a long time.
The answer is 1.414
2007-01-01 15:37:08
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answer #5
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answered by Anonymous
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log(base 0.5)2=x meaning 2^0.5=x=1.41421356
2007-01-01 16:28:36
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answer #6
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answered by christianlecorps 1
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oh, i , well, just keep it in mind- the square root of 2 is 1.414...haha
2007-01-01 15:21:28
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answer #7
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answered by Miss P 1
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One way is called linear approximation.
L(x) = f(a) + f^1(a)*(x-a)
2007-01-01 15:24:05
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answer #8
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answered by Anonymous
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