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Square roots

Table of contents: Definitions - Illustrations - Elementary functions - Specific values - Quadratic equations - Functional equations - Complex parts - Bounds and inequalities - Derivatives and integrals - Series expansions

Definitions

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Symbol: Sqrt — z\sqrt{z} — Principal square root

Illustrations

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Image: X-ray of z\sqrt{z} on z∈[−3,3]+[−3,3]iz \in \left[-3, 3\right] + \left[-3, 3\right] i

Elementary functions

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z=exp⁡ ⁣(12log⁡ ⁣(z))\sqrt{z} = \exp\!\left(\frac{1}{2} \log\!\left(z\right)\right)
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z=z1/2\sqrt{z} = {z}^{1 / 2}

Specific values

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Table of n\sqrt{n} to 50 digits for 0≤n≤500 \le n \le 50
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x∈[0.707106781186547524400844362105±1.51⋅10−31]   where x=12=12=22x \in \left[0.707106781186547524400844362105 \pm 1.51 \cdot 10^{-31}\right]\; \text{ where } x = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}
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−1=i\sqrt{-1} = i
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i=12(1+i)\sqrt{i} = \frac{1}{\sqrt{2}} \left(1 + i\right)
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∞~=∞~\sqrt{{\tilde \infty}} = {\tilde \infty}
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∞=∞\sqrt{\infty} = \infty
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eiθ∞=eiθ/2∞\sqrt{{e}^{i \theta} \infty} = {e}^{i \theta / 2} \infty

Quadratic equations

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zeros⁡ z∈C(z2−c)={c,−c}\mathop{\operatorname{zeros}\,}\limits_{z \in \mathbb{C}} \left({z}^{2} - c\right) = \left\{\sqrt{c}, -\sqrt{c}\right\}
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zeros⁡ z∈C(z2−c)={i−c,−i−c}\mathop{\operatorname{zeros}\,}\limits_{z \in \mathbb{C}} \left({z}^{2} - c\right) = \left\{i \sqrt{-c}, -i \sqrt{-c}\right\}
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zeros⁡ z∈C(az2+bz+c)={−b+b2−4ac2a,−b−b2−4ac2a}\mathop{\operatorname{zeros}\,}\limits_{z \in \mathbb{C}} \left(a {z}^{2} + b z + c\right) = \left\{\frac{-b + \sqrt{{b}^{2} - 4 a c}}{2 a}, \frac{-b - \sqrt{{b}^{2} - 4 a c}}{2 a}\right\}

Functional equations

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(z)2=z{\left(\sqrt{z}\right)}^{2} = z
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z2=z\sqrt{{z}^{2}} = z
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x2=∣x∣\sqrt{{x}^{2}} = \left|x\right|
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−z=iz\sqrt{-z} = i \sqrt{z}
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−z=−iz\sqrt{-z} = -i \sqrt{z}
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z2=z2\sqrt{\frac{z}{2}} = \frac{\sqrt{z}}{\sqrt{2}}
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ab=ab\sqrt{a b} = \sqrt{a} \sqrt{b}
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1z=1z\sqrt{\frac{1}{z}} = \frac{1}{\sqrt{z}}
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ab=ab\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}
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reiθ=reiθ/2\sqrt{r {e}^{i \theta}} = \sqrt{r} {e}^{i \theta / 2}
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z−cz2=z1−cz\sqrt{z - c {z}^{2}} = \sqrt{z} \sqrt{1 - c z}
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zz+c=zz+c\sqrt{\frac{z}{z + c}} = \frac{\sqrt{z}}{\sqrt{z + c}}
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zz−c=−zc−z\sqrt{\frac{z}{z - c}} = \frac{\sqrt{-z}}{\sqrt{c - z}}
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zc−z=z1c−z\sqrt{\frac{z}{c - z}} = \sqrt{z} \sqrt{\frac{1}{c - z}}

Complex parts

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∣z∣=∣z∣\left|\sqrt{z}\right| = \sqrt{\left|z\right|}
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arg⁡ ⁣(z)=arg⁡ ⁣(z)2\arg\!\left(\sqrt{z}\right) = \frac{\arg\!\left(z\right)}{2}
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sgn⁡ ⁣(z)=sgn⁡ ⁣(z)\operatorname{sgn}\!\left(\sqrt{z}\right) = \sqrt{\operatorname{sgn}\!\left(z\right)}
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Re⁡ ⁣(z)=∣z∣+Re⁡ ⁣(z)2\operatorname{Re}\!\left(\sqrt{z}\right) = \sqrt{\frac{\left|z\right| + \operatorname{Re}\!\left(z\right)}{2}}
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Im⁡ ⁣(z)=sgn⁡ ⁣(Im⁡ ⁣(z))∣z∣−Re⁡ ⁣(z)2\operatorname{Im}\!\left(\sqrt{z}\right) = \operatorname{sgn}\!\left(\operatorname{Im}\!\left(z\right)\right) \sqrt{\frac{\left|z\right| - \operatorname{Re}\!\left(z\right)}{2}}
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z‾=z‾\sqrt{\overline{z}} = \overline{\sqrt{z}}

Bounds and inequalities

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∣z∣=∣z∣\left|\sqrt{z}\right| = \sqrt{\left|z\right|}
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∣x+a−x∣≤x(1−1−∣a∣x)\left|\sqrt{x + a} - \sqrt{x}\right| \le \sqrt{x} \left(1 - \sqrt{1 - \frac{\left|a\right|}{x}}\right)
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∣x+a−x∣≤x2(∣a∣x+∣a∣2x2)\left|\sqrt{x + a} - \sqrt{x}\right| \le \frac{\sqrt{x}}{2} \left(\frac{\left|a\right|}{x} + \frac{{\left|a\right|}^{2}}{{x}^{2}}\right)
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∣1x+a−1x∣≤∣a∣2(x−∣a∣)3/2\left|\frac{1}{\sqrt{x + a}} - \frac{1}{\sqrt{x}}\right| \le \frac{\left|a\right|}{2 {\left(x - \left|a\right|\right)}^{3 / 2}}

Derivatives and integrals

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ddz z=12z\frac{d}{d z}\, \sqrt{z} = \frac{1}{2 \sqrt{z}}
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d2dz2z=−14z3/2\frac{d^{2}}{{d z}^{2}} \sqrt{z} = -\frac{1}{4 {z}^{3 / 2}}
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drdzrz=(−1)r(−12)rzr−1/2\frac{d^{r}}{{d z}^{r}} \sqrt{z} = {\left(-1\right)}^{r} \left(-\frac{1}{2}\right)_{r} {z}^{r - 1 / 2}
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∫abz dz=23(b3/2−a3/2)\int_{a}^{b} \sqrt{z} \, dz = \frac{2}{3} \left({b}^{3 / 2} - {a}^{3 / 2}\right)

Series expansions

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z+x=z∑k=0∞(−1)k(−12)kzkk!xk\sqrt{z + x} = \sqrt{z} \sum_{k=0}^{\infty} \frac{{\left(-1\right)}^{k} \left(-\frac{1}{2}\right)_{k}}{{z}^{k} k !} {x}^{k}
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1z+x=1z∑k=0∞(−1)k(12)kzkk!xk\frac{1}{\sqrt{z + x}} = \frac{1}{\sqrt{z}} \sum_{k=0}^{\infty} \frac{{\left(-1\right)}^{k} \left(\frac{1}{2}\right)_{k}}{{z}^{k} k !} {x}^{k}
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cn=1na0∑k=1n(3k2−n)akcn−k   where cn=[xn]A, an=[xn]A{c}_{n} = \frac{1}{n {a}_{0}} \sum_{k=1}^{n} \left(\frac{3 k}{2} - n\right) {a}_{k} {c}_{n - k}\; \text{ where } {c}_{n} = [{x}^{n}] \sqrt{A},\,{a}_{n} = [{x}^{n}] A
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cn=1na0∑k=1n(k2−n)akcn−k   where cn=[xn]1A, an=[xn]A{c}_{n} = \frac{1}{n {a}_{0}} \sum_{k=1}^{n} \left(\frac{k}{2} - n\right) {a}_{k} {c}_{n - k}\; \text{ where } {c}_{n} = [{x}^{n}] \frac{1}{\sqrt{A}},\,{a}_{n} = [{x}^{n}] A

Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2019-06-18 07:49:59.356594 UTC