Difference between revisions of "Square root (real function)"
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Latest revision as of 05:46, 10 April 2017
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- Note: For other uses of square root see square root (disambiguation) - this page only covers the square root as a function on the non-negative reals.
Contents
Definition
Let [ilmath]x\in\mathbb{R}_{\ge 0} [/ilmath] be given, then to say [ilmath]y[/ilmath] is a square root of [ilmath]x[/ilmath] means that [ilmath]y^2\eq x[/ilmath], note that if [ilmath]y^2\eq x[/ilmath] then [ilmath]-y[/ilmath] is also a square root of [ilmath]x[/ilmath]:
- Notice: [ilmath](-y)^2\eq(-1)^2 y^2\eq (-1)^2 x[/ilmath] and that [ilmath]-1\times -1\eq 1[/ilmath], so [ilmath](-y)^2\eq x[/ilmath] also.
Note also however:
- If [ilmath]a^2\eq 0[/ilmath] then we must have [ilmath]a\eq 0[/ilmath], for if [ilmath]a\neq 0[/ilmath] then [ilmath]aa\neq 0[/ilmath] - contradicting that [ilmath]a[/ilmath] is a square-root.
- As such [ilmath]0[/ilmath] has only one square root, [ilmath]0[/ilmath] itself.
Given [ilmath]x\in\mathbb{R}_{\ge 0} [/ilmath] we write:
- [ilmath]\sqrt{x}\in\mathbb{R}_{\ge 0} [/ilmath] - called the principle square root - a number such that [ilmath]\sqrt{x}\times\sqrt{x}\eq x[/ilmath] and such that [ilmath]\sqrt{x}\in\mathbb{R} [/ilmath] and [ilmath]\sqrt{x}\ge 0[/ilmath]
- [ilmath]-\sqrt{x}\in\mathbb{R}_{\le 0} [/ilmath] - called the negative square root - this is just [ilmath](-1)\sqrt{x} [/ilmath]
- [ilmath]\pm\sqrt{x} [/ilmath] to emphasise there are possibly two of them, if there is one this becomes [ilmath]\pm 0[/ilmath], and [ilmath]x\pm 0\eq x[/ilmath] so it doesn't matter.
Evaluating the square root
TODO: Flesh this out
- [ilmath]\sqrt{x}\eq e^{\frac{1}{2}\text{ln}(x)} [/ilmath] where [ilmath]e[/ilmath] is Euler's number and [ilmath]\text{ln}(x)[/ilmath] is the natural logarithm of [ilmath]x[/ilmath]
Properties of the square-root function
Let [ilmath]f:\mathbb{R}_{\ge 0}\rightarrow\mathbb{R} [/ilmath] be a function given by [ilmath]f:x\mapsto \sqrt{x} [/ilmath], then we claim:
- [ilmath]f\big\vert_{\mathbb{R}_{>0} }:\mathbb{R}_{>0}\rightarrow\mathbb{R} [/ilmath] is smooth
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Presley's Elementary Differential Geometry claims it can be done easily by induction on page 15. It claims:
- [math]\frac{\mathrm{d}^n f}{\mathrm{d}x^n}\eq(-1)^{n-1}\cdot\frac{1\times 3\times 5\times\cdots\times(2n-1)}{2^n}\cdot x^{-\frac{2n+1}{2} } [/math]
References
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Tough one, it's just known!
