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Linear Algebra 14 Online
OpenStudy (anonymous):

how ? The square root of 2 is 1.141

OpenStudy (anonymous):

It's not. It's about 1.414.

OpenStudy (anonymous):

what is the appropriate method to get that answer?

OpenStudy (anonymous):

Start with a number \[a_{0} = 1\] Use the recursive formula \[a_{n+1} = a_n / 2 + 1 / a_n\] For each iteration, you will get a better approximation: \[a_1 = 1 / 2 + 1 / 1 = 3 / 2\] \[a_2 = (3 / 2) / 2 + 1 / (3 / 2) = 3 / 4 + 2 / 3 = 17 / 2\] \[a_3 = (17 / 12) / 2 + 1 / (17 / 12) = 17 / 24 + 12 / 17 = 577 / 408\]

OpenStudy (anonymous):

\[a_3 = 577 / 408 \approx 1.414\]

OpenStudy (anonymous):

Too see that the formula is reasonable, assume that a_n converges and show that its square converges to 2. To do this, substitute a for a_n and a_(n+1) in the formula, multiply by 2a, substract a^2 and you'll arrive at a^2 = 2.

ganeshie8 (ganeshie8):

nice :)

OpenStudy (anonymous):

thanks man

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