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Thank you for explaining this! To elaborate for anyone who’s curious: the square root function is 𝑓(𝑥) = 𝑥^½, so its derivative is 𝑓′(𝑥) = ½𝑥^-½ (power r

Thank you for explaining this! To elaborate for anyone who’s curious: the square root function is 𝑓(𝑥) = 𝑥^½, so its derivative is 𝑓′(𝑥) = ½𝑥^-½ (power rule) = 1 / 2𝑥^½ = 1 / 2𝑓(𝑥). If we know the value of 𝑓 for some nearby input 𝑎, we can approximate its value at 𝑥 by using 𝑓’s derivative (i.e. rate of change) at 𝑎 to guess how much the value of 𝑓(𝑎) has increased or decreased by the time it becomes 𝑓(𝑥). We can calculate that guess by multiplying 𝑓′(𝑎) by the difference between 𝑎 and 𝑥. It’s only a guess because it assumes that 𝑓 has a constant rate of change from 𝑎 to 𝑥, which the square root function doesn’t (it’s a curve, not a straight line), but if 𝑎 and 𝑥 are close together then the error isn’t too large. e.g. 𝑓(67) ≈ 𝑓(64) + 𝑓′(64) × (67 - 64) = 8 + (1/16) × 3 = 8.1875

credit_guy 5y ago
There's a different way to get to this, without using calculus. If we call x = sqrt(67), then the first approximation is n = sqrt(64) = 8. We then use the formula for the difference of squares x^2 - n^2 = (x-n)(x+n) to get x-n = (x^2-n^2)/(x+n), or x = n + (x^2-n^2)/(n+x). In our case x = 8 + 3/(8+x). This is circular, but we can just plug in the right hand side the first approximation for x, which is 8, and we end up with 8 + 3/16. The beauty of this formula is that you can continue to iterate. Once you have x ~ 8.1875, you can plug again in the right hand side and get x ~ 8 + 3/16.1875, which is exact to the first 4 decimals. Of course, you can continue, and you end up with the (aptly named) continued fraction method for the square root [1]. [1] https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Continued_fraction_expansion https://en.wikipedia.org/wiki/Methods_of_computing_square_ro...

credit_guy 5y ago

There's a different way to get to this, without using calculus. If we call x = sqrt(67), then the first approximation is n = sqrt(64) = 8. We then use the formula for the difference of squares x^2 - n^2 = (x-n)(x+n) to get x-n = (x^2-n^2)/(x+n), or x = n + (x^2-n^2)/(n+x). In our case x = 8 + 3/(8+x). This is circular, but we can just plug in the right hand side the first approximation for x, which is 8, and we end up with 8 + 3/16. The beauty of this formula is that you can continue to iterate. Once you have x ~ 8.1875, you can plug again in the right hand side and get x ~ 8 + 3/16.1875, which is exact to the first 4 decimals. Of course, you can continue, and you end up with the (aptly named) continued fraction method for the square root [1]. [1] https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Continued_fraction_expansion https://en.wikipedia.org/wiki/Methods_of_computing_square_ro...