# Is this correct method to aproximate values of roots of positive integers?

• Nov 23rd 2013, 02:04 AM
kpkkpk
Is this correct method to aproximate values of roots of positive integers?
EXAMPLE:

Square root 2:

First, I pick DOUBLES of positive integer raisings of base number 2:

1,1,2,2,4,4,8,8,16,16,...

Then I take some fixed number of them from left...in this example I choose to take 7 of them:

1,1,2,2,4,4,8

Then, from Pascalīs triangle, I choose horizontal row where there are the same number of binomial coefficients, like that:

1,6,15,20,15,6,1

After that I multiply these two number groups pair by pair and sum them up:

1*1 + 6*1 + 15*2 + 20*2 + 15*4 + 6*4 + 1*8 = 169

Then I do the same trick, but this time I forget the first "1" from raising group of base number two:

1*1 + 6*2 + 15*2 + 20*4 + 15*4 + 6*8 + 1*8 = 239

Aproximation of square root 2 is then 239/169 = 1,414201183 (says my pocket calculator)

If I want a better aproximation, then I simply use longer list of base number 2 raisings and appropriate horizontal row from Pascalīs triangle, for example:

(1*1 + 8*2 + 28*2 + 56*4 + 70*4 + 56*8 + 28*8 + 8*16 + 1*16) / (1*1 + 8*1 + 28*2 + 56*2 + 70*4 + 56*4 + 28*8 + 8*8 + 1*16) = 1393/985 = 1,414213198...

...which is quite close to the value my pocket calculator gives to the screen: 1,414213562...

If that is a correct method, then I can anticipate that by using corresponding raisings of base number 3 in pairs, I can find square root 3...for example:

1,1,3,3,9,9,27,27,81...

(1*1 + 5*3 + 10*3 + 10*9 + 5*9 + 1*27) / (1*1 + 5*1 + 10*3 + 10*3 + 5*9 + 1*9) = 208/120 = 1,7333... which is already quite close to pocket calculator answer for square root of 3: 1,732050808...

If these two examples are correct, perhaps I can then approximate all square roots of positive integers with this method?

How about bigger roots then? Is my assumption correct that third root of 2 is possible to approximate by using triples of raisings of base number 2 like that:

1,1,1,2,2,2,4,4,4,8,8,8,16,16,16,32,...

(1*1 + 7*1 + 21*2 + 35*2 + 35*2 + 21*4 + 7*4 + 1*4) / (1*1 + 7*1 + 21*1 + 35*2 + 35*2 + 21*2 + 7*4 + 1*4) = 306/264 = 1,159090909...

...where multipliers come from Pascalīs triangle horizontal row 7, or...

(1*1 + 8*1 + 28*2 + 56*2 + 70*2 + 56*4 + 28*4 + 8*4 + 1*8) / (1*1 + 8*1 + 28*1 + 56*2 + 70*2 + 56*2 + 28*4 + 8*4 + 1*4) = 693/549 = 1,262295082...

Third root of base number 2 is 1,25992105...in calculator screen.

If my guessings are correct, then, is this a method computers use while calculating these values?