Square root (real function)

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Made this page to prove smoothness really Alec (talk) 05:46, 10 April 2017 (UTC)

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.

Definition

Let $x\in\mathbb{R}_{\ge 0}$ be given, then to say $y$ is a square root of $x$ means that $y^2\eq x$, note that if $y^2\eq x$ then $-y$ is also a square root of $x$:

• Notice: $(-y)^2\eq(-1)^2 y^2\eq (-1)^2 x$ and that $-1\times -1\eq 1$, so $(-y)^2\eq x$ also.

Note also however:

• If $a^2\eq 0$ then we must have $a\eq 0$, for if $a\neq 0$ then $aa\neq 0$ - contradicting that $a$ is a square-root.
• As such $0$ has only one square root, $0$ itself.

Given $x\in\mathbb{R}_{\ge 0}$ we write:

• $\sqrt{x}\in\mathbb{R}_{\ge 0}$ - called the principle square root - a number such that $\sqrt{x}\times\sqrt{x}\eq x$ and such that $\sqrt{x}\in\mathbb{R}$ and $\sqrt{x}\ge 0$
• $-\sqrt{x}\in\mathbb{R}_{\le 0}$ - called the negative square root - this is just $(-1)\sqrt{x}$
• $\pm\sqrt{x}$ to emphasise there are possibly two of them, if there is one this becomes $\pm 0$, and $x\pm 0\eq x$ so it doesn't matter.

Evaluating the square root

TODO: Flesh this out

• $\sqrt{x}\eq e^{\frac{1}{2}\text{ln}(x)}$ where $e$ is Euler's number and $\text{ln}(x)$ is the natural logarithm of $x$

Properties of the square-root function

Let $f:\mathbb{R}_{\ge 0}\rightarrow\mathbb{R}$ be a function given by $f:x\mapsto \sqrt{x}$, then we claim:

• $f\big\vert_{\mathbb{R}_{>0} }:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ is smooth

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Presley's Elementary Differential Geometry claims it can be done easily by induction on page 15. It claims:

• $$\frac{\mathrm{d}^n f}{\mathrm{d}x^n}\eq(-1)^{n-1}\cdot\frac{1\times 3\times 5\times\cdots\times(2n-1)}{2^n}\cdot x^{-\frac{2n+1}{2} }$$

References

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Tough one, it's just known!