By substituting #x=1/7#, how can I show that #sqrt7=261/98#?

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Answer 1 · 2018-02-19T14:47:52

A guess at what this might refer too, but it gives #sqrt(7) ~~ 337/127#

I am not sure to what the question refers.

Since it is asked under precalculus, here's a roughly precalculus level way of doing things that involves putting #x=1/7# ...

The Maclaurin expansion for #(9+x)^(1/2)# tells us:

#(9+x)^(1/2) = 3 + x/6 - x^2/216 + x^3/3888 - (5 x^4)/279936 + O(x^5)#

Putting #x=1/7# we find:

#8/sqrt(7) = sqrt((63+1)/7) = (9+1/7)^(1/2)#

#= 3 + 1/42 - 1/10584 + 1/1333584 - O(x^4)#

So truncating, we have:

#8/sqrt(7) ~~ 3+1/42 = 127/42#

So:

#sqrt(7) ~~ (8 * 42)/127 = 337/127#

Hmmm - Not the same as the question, but similar accuracy.