Written calculation Open method

Written calculation prescribe

Written calculation Kaiping method, this method can be used to find out any positive arithmetic square root.

(Not an approximation, but an exact value)

Principle:

the method can be précis-writers as-- 20 times times the beginning of business plus trial business

Prepare:

Here is a detailed calculation step:

1. Starting with a decimal point, the integer and fractional parts of the radicals are separated by 2 bits (integers from right to left, decimals from left to right, and 0 for less), divided into n segments, indicating that the square root is n-digit
2. Determines the first bit of the square root: The maximum number of squares. The maximum number of X (1..9) is not more than the first group, and X is the quotient of the divisor and the first group, and the root remainder is obtained (the first group of Remainders + the second group)
3. The second position of the square root: Assuming that the second bit is a, the calculation of the maximum value of a (this is a trial), so that the results are not greater than the root of the remainder. The new divisor is multiplied with the new quotient, and the new remainder is calculated.
4. After completing the next set of numbers to Xinyu number, repeat step 3
5. Decimal alignment, arithmetic to the desired precision

The narrative looks pale (with a general impression) and we look directly at the example:

For example, to prescribe 105625:

First, 105625 is segmented, from right to left, each two digits are divided into 1 segments, that is, 10,56,25 three segment numbers. First calculate the square root of the first digit, in the square not more than 10 of the number to take the largest, such as 1 squared for the square of 4,3 squared for the square of 9,4 is 16, 16 has more than 10, the square is 10 smaller, the square root of the first digit 3, because of the 3 largest
10-3 squared = 1, radicals the second paragraph of the number, and get 156. Now the square root is the second digit. Assuming that the second digit is a, take the equation A * (20*3+a), the equation 20 is a fixed number (regardless of radicals is how much) 3 is just the first number of square root calculated. The value of a is estimated so that a * (20*3+a) does not exceed 156. Take A=1,a (20*3+a) =61,a=2 A (20*3+a) =124,a=3 when A * (20*3+a) =189,189 has more than 156, so a in 1, 2 to take the largest number, that is, 2, the second digit of the square root is 2.
A (20*3+a) =124,62 multiplied by the square root second digit, namely 62*2=124,156-124=32, will radicals the third paragraph number to fill up, obtains 3225, is similar to the front, takes the equation B (20*32+b), the formula 20 or is fixed constant number, 32 is just the square root of the first two digits, the B value, so B (20*32+b) not more than 3225, by the calculation of the b=5, the square root of the third digit is 5
If the square root has a fourth digit, or more, assuming that there is a fourth digit after 325, calculate the fourth digit when the equation A (20*325+a), the 325 of the equation is the square root already calculated several numbers, the subsequent algorithm is similar to the previous. For radicals is a decimal, the segment should be noted, such as the square root of the calculation of 1.323, its decimal place has 3 bits, the number of bits is odd number, to fill a 0 up, that is 1.3230, and then from left to right every two digits into a paragraph. For example, the root number is 3.7478, the decimal place has 4 digits, is an even number of digits, do not need to fill 0, can be directly segmented, fractional and integer root calculation method is the same.

Practice:

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The above-mentioned written calculation method is the method that most of us give to the textbook appendix when we go to school, which is too troublesome in actual operation. We can take the following approach, the actual calculation is not afraid of a certain step wrong!!! And the above method is not.??

For example, 136161 this number, first we find a square root and 136161 is closer to the number, any one, for example, 300 to 400 of any number, here choose 350, as the representative.?? We calculate 0.5* (350+136161/350) get 369.5??

Then we calculate 0.5* (369.5+136161/369.5) to get 369.0003, we find that 369.5 and 369.0003 are almost the same, and that the 369^2 end number is 1. We have reason to conclude that 369^2=136161?? Generally can open the best, using the above method to calculate one or two times the basic results came out. Let's take another example: calculate the square root of 469225. First we find 600^2<469225<700^2, we can pick 650 as the number for the first calculation. That's the count??

0.5* (650+469225/650) gets 685.9. And 685 near the 685^2 end number is 5, so 685^2=469225??

For those who open countless number, using this method to calculate two or three times the accuracy is very considerable, generally reached several points after the decimal.

In practice, this algorithm is also used by computer to prescribe

Written calculation Open method