Let A = X3 and x be the issuer. Issuers have a standard formula: x (n+1) = xn+(a/x2-xn)1/3. (n, n+ 1 is the lower corner) For example, A=5, and 5 is between the third power of 1 and 2. The initial value X0 can be 1. 1, 1.5, 1.2, 1.3, 1.4,1.5,65438. For example, let's take X0 = 1.9, according to the formula: Step 1: x1=1.9+(5/1.9)1/3 =1.75. 1.38504 16- 1.9=-0.5 149584, -0.5 149584× 1/3=-0. 17 16528, 1.9+(-0. 17 16528)= 1.728。 That is, take a 2-digit value, namely 1.7. Step 2: x2 =1.7+(5/1.72; -1.7)1/3 =1.71.7 =1.730/kloc-0,/kloc-0. Take 3 digits, one digit more than the previous one. Step 3: x3 =1.71+(5/1.72; -1.71)1/3 =1.709. Step 4: x4 =1.709+(5/1.72; -1.709)1/3 =1.7099 This method can be adjusted automatically. The values in the first and third steps are too large, but the output value will automatically decrease after calculation; In the second and fourth steps, the input value is small and the output value automatically increases. That is, 5 =1.7099 3; Of course, the initial value X0 can also be 1. 1, 1.2, 1.3. . . Any one of 1.8 and 1.9 is x1=1.7 >; . Of course, we'd better use the intermediate value as the initial value in actual operation, that is, 1.5. 1.5+(5/ 1.5^2; - 1.5) 1/3= 1.7。 If you square with this formula, you can change 3 into 2, and 2 into 1. That is X(n+ 1) = Xn+(A/Xn? Xn) 1/2。 For example, a = 5: 5 is between 2 2; To 3 2; We can take the initial value 2. 1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, and it is better to take the middle value 2.5. Step 1: 2.5+(5/2.5-2.5)1/2 = 2.2; That is, 5/2.5=2, 2-2.5=-0.5, -0.5× 1/2=-0.25, 2.5+(-0.25)=2.25. Take two digits, 2.2. Step 2: 2.2+(5/2.2-2.2)1/2 = 2.23; That is, 5/2.2=2.272, 2.272-2.2=-0.072, -0.072× 1/2=-0.036, 2.2+0.036=2.23. Take three digits. Step 3: 2.23+(5/2.23-2.23)1/2 = 2.236. That is, 5/2.23 = 2.242, 2.242-2.23 = 0.0 12, 0.0 12× 1/2 = 0.006, 2.23+0.006 = 2.236.

Feedback prescription

Take one more digit with each step. This method is also called feedback root. It doesn't matter even if you enter an incorrect value. The output value will automatically adjust, close to the exact value. This method is based on Newton tangent method. It can also be derived from Newton's binomial theorem. A = (x y) k = expansion, and the derivation process is to add the descendants of A to the formula, that is, the (x y) k expansion. X is an imaginary number and y is an error value. x(n+ 1)=xn-(x^k-a)/kx^(k- 1)=xn-f(x)/f'(x)=xn+(a/x^(k- 1)-xn) 1/k(f(x)=x^k-a； f'(x)=kx^(k- 1)； Is to transform Newton's binomial theorem into Newton's tangent method in the process of square root. (Wang Xiaoming Wang Ruike) Description of this method; 1980, Wang Xiaoming deduced this formula with Newton binomial and founded Jiangxi Normal University. A professor felt familiar and deduced it again on the spot, just like Newton's tangent method. Fu from Anshan, Liaoning introduced in the book Mathematical Athena, Shaanxi Future Publishing House 1994, Tianjin Xin Lei Publishing House. Because it is Newton's formula, the author Wang Xiaoming dare not be greedy. Therefore, Mr. Fu also clearly stated in the introduction of the article that it was derived from Newton's tangent method.