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As we have the unwritten index $2$ for the sqare root, we multiply it by the index of the root inside the first root. $$\sqrt{\sqrt{x}} = \sqrt[2\times2]{x}= \sqrt[4]{x}$$ Another example is: $$\sqrt[3]{\sqrt[4]{x}} = \sqrt[3\times4]{x}= \sqrt[12]{x}$$
As traditional known the square root of any number would have two result, its cube root would have three ...
2. There are only two square roots of ii (as there are two square roots of any non-zero complex number), namely ± (1 + i) / √2. In the context of your answer, what happens is that the different values are e (πi / 2 + 2πik) / 2 = eπi / 4 + πik; but the value of this depends only on the parity of k, and so gives just two values, namely ± ...
The suggested way, is to choose a value for x so that (1 − 2x) has the form 2 ∗ 'a perfect square'. This can be done by taking x = 0.01. Thus, (1 − 2x) = (1 − 2 ∗ 0.01) = 0.98 = 2 ∗ 0.72. And (1 − 2x)1 2 = 0.981 2 = 0.7√2 Which is equal to the previously established expansion, so we can now go ahead and find √2.
To find a square root of a given complex number z, you first want to find a complex number w which has half the argument of z (since squaring doubles the argument). Compute r = |z| and let w = z + r; thus w lies r steps to the right of z in the complex plane. Draw a picture of this, and it should be clear that the points 0, z and w form an ...
Calculating the square root of a number is one of the first problems tackled with numerical methods, known I think to the ancient Babylonians. The observation is that if x, y> 0 and y ≠ √x then y, x / y will be on opposite sides of √x, and we could try averaging them. So try y0 = 1, yn + 1 = 1 2(yn + x yn).
This method is second in the class of Householder's methods. Halley's method is: xn+1 =xn −2f(. If we let f(x) = x2 − a which meets the criteria, (continuous second derivative) Then Halley's method is: xn + 1 = xn − (2x3n − 2axn) 3x2n + a Which has the simplification: xn + 1 = x3n + 3axn 3x2n + a I also will add this document which ...
Calculators probably use some form of Newton-Raphson, while arbitrary precision libraries will probably use a much faster algorithm to calculate $\ln$ and Newton-Raphson to compute $\exp$ from inverting $\ln$, and then use $\sqrt[n]{a} = e^\frac{\ln(a)}{n}$. $\endgroup$
Newton's Method is based upon finding roots of a function f(x). To see how this applies to square or cube roots, suppose that y = √n for some fixed n. Well, then this y would be a root of the equation f(x) = x2 − n. Similarly, f(x) = x3 − n would provide us with a way to calculate the cube root of n. Newton's Method works as follows ...
But now, I forget one. Here is a simple method that I know. Find the prime divisors of the number. Omit the half of numbers that have been appeared even times. multiply the rest. For example you want to find square root of 36. You find the divisors. They are 2x2x3x3. In step 2 they appeared as 2x3.