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Approximating square roots using binomial expansion. And one can quickly check that $(x_3)^2=2.000006007\dots$, which is pretty much the square root of $2$. Share.
Integral $\\int \\sqrt{1+x^2}dx$ - Mathematics Stack Exchange 21 Feb 2018 · I was trying to do this integral $$\\int \\sqrt{1+x^2}dx$$ I saw this question and its' use of hyperbolic functions. I did it with binomial differential method since the given integral is in a form o...
Is $i$ equal to $\\sqrt{-1}$? - Mathematics Stack Exchange 22 Jul 2017 · Indeed we can extend the definition of the square root to any complex number, setting it as the principal value of $\sqrt z:=e^{1/2\ln z}$ with the same result. However is true that $(-i)^2=-1$. Share
complex numbers - What is $\sqrt {i}$? - Mathematics Stack … -1 is 1 rotated over $\pi$ radians. The square root of a number on the unit circle is the number rotated ...
Why the square root of any decimal number between 0 and 1 … 24 Jan 2018 · Think about a decimal number between 0 and 1 as a fraction with its numerator GREATER than its denominator. Say you are taking the square root of the number $1/25$. So, you acquire $\sqrt{1/25}$ as the expression which you have to evaluate. This becomes $\sqrt{1}/\sqrt{25}$, or $1/5$. $1/5 > 1/25$.
algebra precalculus - Square root inside a square root 24 Jun 2014 · The square root of the square root of x is therefore $$\sqrt{\sqrt x} = (\sqrt x)^{1/2} = (x^{1/2})^{1/2} = x^{1/4} = \sqrt[\large 4] x$$ Since the domain of $\sqrt x$ is $[0, + \infty)$ , this is also the domain of $\sqrt{\sqrt x} = x^{1/4}$ .
Is the square root of negative 1 equal to i or is it equal to plus or ... 25 Nov 2017 · The answer is that there are two square roots of $-1$. This is no different than with real numbers; for example, there are two square roots of $4$: $2$ and $-2$. The main difference is that the complex numbers don't have a good way to single out one of the two square roots as the "special" one.
What's bad about calling $i$ "the square root of -1"? 28 Mar 2015 · There is a bit more complicated, but more thorough explanation, however, involving complex analysis. The problem lies in trying to take fractional exponents of negative numbers, e.g. $(-1)^{1/2}$.
complex numbers - why is $\sqrt {-1} = i$ and not $\pm i ... 9 Jan 2015 · To get a continuous "branch of square root" it's necessary to remove enough of the plane that "the domain of the square root doesn't encircle the origin". The customary choice is to remove the non-positive reals. (Ironically, this explicitly excludes $-1$ from the domain of the square root.) A common alternative choice is to remove the non ...
How to compute $\\sqrt{i + 1}$ - Mathematics Stack Exchange Also a wanted property is that it is continuous except for the non-positive reals. And I guess it's also a wanted property that for all numbers except of the negative reals, the square root of the conjugate is the conjugate of the square root (for the negative reals, it's not possible to achieve that). $\endgroup$ –