hello!dear friends i want to know that How to find square root of non perfect numbers manually or by hand?i mean how to find square root of 10,6,5 or 15i want to know that what are the rules? what are the principles which should be followed?
hello!dear friends i want to know that How to find square root of non perfect numbers manually or by hand?i mean how to find square root of 10,6,5 or 15i want to know that what are the rules? what are the principles which should be followed?
There's a technique/algorithm for doing it by hand that I've long since forgotten, thanks to calculators.
Sometimes, I've done it to a few decimal places by hand - but in those cases, I haven't used whatever that technique it was that they taught us in school. Instead, I just make good guesses and zero in on the correct answer decimal by decimal. ("Try that - it's too low. So try the next one - still too low. So try the next one - too high. I've now found that decimal, so on to the next one.") I know it's not the optimal method, but it gets the job done.
There are two math thingies that aid my guesses.
1) $\displaystyle f(x) = \sqrt{x} \rightarrow f'(x) = \frac{1}{2\sqrt{x}}$, so $\displaystyle f(x) \approx f(x_0) + f'(x_0)(x-x_0)$ gives:
$\displaystyle \sqrt{x} \approx \sqrt{x_0} + \frac{x-x_0}{2\sqrt{x_0}}$.
2)Taylor Polynomials: $\displaystyle \sqrt{x} \approx \sqrt{x_0} + \frac{x-x_0}{2\sqrt{x_0}} - \frac{(x-x_0)^2}{8\sqrt{x_0^3}} + \frac{(x-x_0)^3}{16\sqrt{x_0^5}}$.
3) From $\displaystyle (x+a)^2 = x^2 + 2ax + a^2 \approx x^2 + 2ax$ for a small. Thus if x is "close" to the square root, and a is small, then adding a to x will increase the result by about 2ax.
Example: $\displaystyle \sqrt{7}$
$\displaystyle \sqrt{7} \approx \sqrt{9} + \frac{7-9}{2\sqrt{9}} = 3 + \frac{-2}{6}} = 2 2/3 = 8/3$, so my 1st guess is 8/3
$\displaystyle (8/3)^2 = 64/9 = 7+1/9$ so it's over by a little bit. Using observation 3, being high by 1/9 means I want [/tex]2ax = -1/9[/tex], so $\displaystyle 2a(8/3) = -1/9$, so $\displaystyle a = (-1/9)(3/16) = -1/(3*16)$.
So my next try is 8/3 + a = 8/3-1/(3*16) = (8*16 - 1)/48 = (80 + 48 - 1)/48 = 1 + (80 - 1)/48 = 1 + 79/48 = 2 + 31/48.
Hand long division: 48 into 31.000: 0.6 -> 28.8 R2.20. then 0.64-> 28.8+1.92 R0.280. I'm, happy with 0.64 as my guess.
Current guess is 2+31/48, so about 2.64. Hand multiply to 2.64-squared = 6.9696.
Now I'd just go through decimal by decimal. 2.64-squared= too low. 2.65-squared = 7.0225. Too high. So The answer is 2.64****.
Now a "binary search", so test 2.645-squared = 6.996025 = too low. 2.646-squared = 7.001316 = too high. So The answer is 2.645***.
Now a "binary search" again (pesky voice in head: "why not just use a calculator?"). Try 2.6455. Anyway, you get the idea.
Here's the Taylor polynomials approach, to power 3:
$\displaystyle \sqrt{x} \approx \sqrt{x_0} + \frac{x-x_0}{2\sqrt{x_0}} - \frac{(x-x_0)^2}{8\sqrt{x_0^3}} + \frac{(x-x_0)^3}{16\sqrt{x_0^5}}$, so
$\displaystyle \sqrt{7} \approx \sqrt{9} + \frac{-2}{2\sqrt{9}} - \frac{(-2)^2}{8\sqrt{9^3}} + \frac{(-2)^3}{16\sqrt{9^5}}$.
$\displaystyle \sqrt{7} \approx 3 - \frac{1}{3} - \frac{1}{2*3^3} - \frac{1}{2*3^5}$.
$\displaystyle \sqrt{7} \approx 3 - \frac{2*3^4}{2*3^5} - \frac{3^2}{2*3^5} - \frac{1}{2*3^5} = 3 - \frac{2*3^4 + 3^2 + 1}{2*3^5}= 3 - \frac{2(81) + 10}{2*3^5}$
$\displaystyle = 3 - \frac{172}{2*3^5}= 3 - \frac{86}{3^5} = 3 - \frac{86}{243}$. Now do long division, only out to a couple decimals.
Note that by analysis you can say how far off your Taylor method approach is (i.e. you can bound the remainder term).
OK - that's it. I'm sure someone else has a better approach to offer, but that's at least a way to proceed. But I'd rather us a calculator.