Neat! I wish they'd teach the old pen and paper square root technique again. It required a fair amount of brute force mathin', but it got me results that I could feel confident were accurate to whatever arbitrary degree of precision.
For the numerator, it is okay, so that we got 65-63=2 For the denominator, your approach is rather complicated, confusing and inaccurate. I have a version that is simpler. Sqrt(65)+sqrt(63) =sqrt(64+1)+sqrt(64-1) =sqrt(8+a)^2+sqrt(8-a)^2 So (8+a)^2=8^2+2x8xa+a^2 where 16a+a^2=1 …….i Similarly (8-a)^2=8^2-2x8xa+a^2 where -16a+a^2=-1 …..ii Minus i with ii, we got 32a=2 so a = 1/16 The denominator will be Sqrt(8+1/16)^2+sqrt(8-1/16)^2=8+1/16+8-1/16=16 Final result = 2/16 = 1/8 ______________________ Same approach can be applied to similar question. For example, Sqrt(66) - sqrt(62) = 4/16=1/4 Sqrt(82) - sqrt(80)=2/18=1/9
*@ Math Window* Instead of your method, you should use the binomial theorem. Write it as 8*sqrt(1 + 1/64) - 8*sqrt(1 - 1/64), and just use two terms in each. You will up with an approximation of *0.125* which is relatively very close.
By Taylor formula for sqrt(x+1)-sqrt(x-1) at x=64 up to first derivative, you get (8+1/16)-(8-1/16)=1/8=0,125 that is a good approximation even though the intervals appears to be large
S^2 is closer to 256 than to 255 as 15x15 is only 225, you could interpolate a more accurate estimation. Also 128+2x63=254. But this does not invalidate your method. Usualy one uses the conjugate product to get rid of the squre roots at the denominator, this is a nice example just doing the opposite.
Since S is very very close to 16, you can estimate a lot closer to 0.25 than to 0.3333. So in stead of 0.3 we could say 0.26. I really like this example!
@ Math Window -- Or, as the radicands are 8^2 + 1 and 8^2 - 1, use the approximation a + b/(2a). This will work well, because the absolute values of 1 and -1 are relatively small compared to 64. For example, for the first number, a = 8 and b = 1. When you use this approximation on both numbers, and then take their difference, the result is *0.125.*
since you aren't eliminate ing the square roots its b any better than the orginal. If you figure 65^1/2 + 63^1/2 it becomes half of the orginal answer and you can approximate bu doing 1/64^1/2 which 1/8 fir aorioamately 0.125 and exactly 0.12500381510477788086176596938599 Wich is ~ the right answer
Or approximate that sqrt(65) + sqrt(63) = 2 * sqtr(64). The difference is about 0.0005. This gives us that sqrt(65) - sqrt(63) = 1/8, which is a much better approximation than 1.3
8 + (1/16) = 8.0625. √65 = 8.0622 . . . Pretty close. 8 - (1/16) = 7.9375. (Notice the minus sign because you're looking upstairs.) √63 = 7.9372 . . . Pretty close. Ya got really close approximations. Take the difference. There's your answer. The Aussie on the Tech Math channel has an excellent video on estimating square roots. But he doesn't cover the minus technique for perfect squares above. It's a simple three step process. 😃
Expand and simplify. The result is the difference between 65 and 63. This works with all square roots. I had to try is for myself to find out the general rule for this.
Question By my thinking That result is 6.7% higher than the true value And the lower range in that estimation (0.125) is less than 0.0031% lower than the true value. Is there somewhere in your estimation process that could be tweaked to get it closer?
When doing approximation, it is good to have a sandwich as you can tell the range of the approximation. Knowing the answer is 0.125 is good but honestly not enough. However, there must be a better way for the upper bound as it varies too much.
Write 65 = 64+1 and 63 = 64-1. Factor by 8 out of the radicals. A=8*[ (1+x)^1/2 - (1-x)^1/2 ] where x = 1/64 Then use the Taylor series of (1+x)^1/2: Stopping at order 1, we get the approximate value = 1/8. Easy. But how good the approx is? Go to the 3rd order given order 0 and 2 cancel because of the minus sign of the 2 radicals. The 2 3rd order terms = 2*(1/2*-1/2*-3/2*1/6!)*x^3 = 1/8*x^3 Multiplied by 8 in front = x^3 = (1/64)^3 = 4/2^20 2^20 being approx 10^6: 1/8 approximates the correct value with an approx error of 4*10^-6. Good enough. Won't use the Lagrange reminder formula of the Taylor series to get a rigorous upper bound of the approx. A quick calculation in Excel gives A = 0.125004
When you multiply by the conjugate, the "middle terms" disappear. Here we have (√65-√63)(√65+√63) which expands to: (√65)² - √65√63 + √65√63 - (√63)². The middle terms cancel, leaving: (√65)² - (√63)², or simply 65 - 63, which is just 2.
I am with you until you round all the way up to 0.13. It is OBVIOUS that the answer is going to be much closer to .125 than to .133 so a reasonable "guestimate" would be .126. In fact, if you WERE to use a calculator it comes out to 0.1250038. The only justification for rounding to .13 is if you are limiting the answer to 2 decimal places.
Huuuu ,,,, this is really much too complicated (at least for the resolution reached). Take the nearest squarenumbers ... 64 and 81 in the case of 65 and 49 and 64 in the case of 63. Now approximate the sqrt(65) by sqrt(64) + (65-64)/(81-64) which is 8 + 1/17 Approximate sqrt(63) by sqrt(64) + (63-64)/(64-49) which is 8 - 1/15 Finally sqrt(65)-sqrt(63) is approximately 8+1/17 - (8-1/15) = 1/17 + 1/15 = 32/255 = 0.12549.. The real value would be 0.1250038...
The task should be estimate, not calculate. Before trying to teach ask mathematicians how to do it in a much simpler and more accurate way. Also you could use calculator and check your answer, it’s .125!
BETTER RESULT : "s" = 1/8 + (1/64)^3 = .125003815 ( correct to 9 decimal places ). For a deeper ( and incredibly accurate ) result use the binomial expansions here: "s" = sqrt(64+1) - sqrt(64-1) = 8 x [ sqrt(1 + 1/64) + sqrt(1 - 1/64) ]. Then using the binomial expansions for ( 1 + x )^1/2 , with x = 1/64 and x = -1/64, we find that all of the odd terms cancel out. The two second terms add to give : 8 x (1/64) = 1/8 = .125 which is the rough approximation answer in this video and most comments. The two 3rd and 5th terms cancel out, but the 4th terms add to : (1/64)^3 . Using the calculator this is : .000003815 ( to 9 dec places ). The 6th terms add to a number much too small to affect the first 9 digits. So, here's a good easy approximation ( correct to 9 decimal places ) "s" = 1/8 + (1/64)^3 = 2^(-3) + 2^(-18) !! Check this out on your calculator on both sides : "s" = sqrt(65) - Sqrt(63) = .125003815 = 1/8 + (1/64)^3 .
