Re: querry

(1): Powers are multiplied and in the meantime exponents are added. Where 1 is the neutral number in the multiplication of positive numbers, 0 is the neutral number in the addition of any real numbers.

And remember: logarithms are exponents.

In professional math wordings:
The additive group (R, +) of reals and the multiplicative group (R+, *) of positive reals are isomorphic.
The continuous isomorphims from R onto R+ are the exponential functions x -> a^x, where a > 0 and a <> 1.
The continuous isomorphims from R+ onto R are all multiples of each other. They are the logarithmic functions y -> log_a (y), where a is the base of this particular logarithm, defined by log_a (a) = 1.

(2) No way!

(3)
The fastest way is Newton-Raphson iteration. Let Xn be an approximation to sqrt(A) with a relative error of epsilon, then X(n+1) = (Xn + A/Xn)/2 is a closer approximation, with a relative error of epsilon squared. I.e. at each iteration the number of correct decimal places is doubled.
For instance: 1.4 is already close to sqrt(2), as is seen in 1.4 * 1.4 = 1.96; relative error is 2/100. One Newton-Raphson step yields 1.41428571, which squares to 2.0002020408, relative error 1/10000.

A most insightful, but cumbersome way is an exhaustion process that much resembles long division. I will give an example: (typeset yourself the calculation below in fixed-space font in case my typesetting does not survive the e-mail transmission.)

Take for instance the number 88229. You do not yet know that 88229 = 297x297 + 20, but you can figure out that sqrt (88229) is somewhere between 200 and 300, far closer to 300 than to 200. (because 88229 is rather close to 300x300 = 90000).

You need to put 88229 = 8-88-29 (make pairs of digits from the end, or from the decimal point in both directions in case you have a non-integer at hand.)
(One could also say: rewrite the number in the centesimal number system, with digits 00, 01, ..., 99 instead of 0, 1, ..., 9)


So you put 88229 = (200 + B)^2 + remainder = (A + B)^2 + r = AA + 2AB + BB + r.

You know already that A=200 will do the job. (A=300 is too large.) The first digit of the result is 2.

Then you subtract 200x200 = 40000 and you are left with
48229 = 2AB + BB + r = B x (2A + B) + r.
This is best illustrated by preparing a square piece of paper and cutting out a smaller square at a corner, so that the remainder looks like a carpenter's square.

Now we go for the second digit. This is the same as trying in turn B = 00, 10, 20, ..., 90. It is found that B=90 will do the job.

The calculation is 48229 = 90 x 490 + 4129. But you only need to write down 482 = 9 x 49 + 41, and the digit 9 as the second digit of the result.

Then you append the last pair of digits 29, so you are left with
4129 = (290 + C)^2 - 290^2 + r = C x (580 + C) + r.

Finally you will find C = 7 as the third digit of the result: 4129 = 7x587 +20.

The overall result is 88229 = 297x297 + 20.

Of course you may calculate decimals of sqrt(88229) by processing the remainder 20 in the same way as you processed previously the leftovers 48229 and 4129.

The key idea of this exhaustion process is to take out successively larger squares until the remainder is 0 (for perfect squares) or is small enough for your situation (for imperfect squares). In the example we successively took out 200x200, 290x290, 297x297 and decided to be content witha remainder of 20.


)

sqrt | 08 82 29 | = 297
subtract 4
remainder 4 82
subtract . x 4. = 4 41 (. = 9 is the largest integer that does not yet exhaust the remainder)
remainder 41 29
subtract 7 x 587 = 41 09
remainder 20

an encore: two decimals        297.03

                                    20.00 00
3 x 59403 =                  17.82 09
remainder                       2.17 91

Now you see that taking square roots is essentially different from power manipulation!

Happy studies: Johan E. Mebius
cupidisdangerous@xxxxxxxxxxx wrote:

hi
someone please tell me in a very layman's language about following
problem
1.why anything raised to the power of zero i equal to one (x^0=1)
2.How can i resolve x^1/2 without going through root. i mean resolve it
through only the power manipulation.
3.Easiest way of getting to the square root of any figure (no use of
calculator or any other computerized help).
thanx in advance



.

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