According-to-Wikipedia-and-WolframAlpha-the-sign-function-sgn-x-is-defined-as-sgn-x-x-x-x-x-for-x-0-and-satisfies-sgn-x-x-1-x-but-sgn-0-0-In-short-sgn-x-1-f

Question Number 2400 by Filup last updated on 19/Nov/15

According to Wikipedia and WolframAlpha,

the sign function, sgn (x), is defined as :

sgn (x) \equiv \frac{x}{∣ x ∣} = \frac{∣ x ∣}{x} for x \neq 0

and satisfies :

sgn (x) = \sqrt{x} \sqrt{\frac{1}{x}}

b u t

sgn (0) = 0

In short :

sgn (x) = {\begin{cases} 1 for x > 0 \\ 0 for x = 0 \\ - 1 for x < 0 \end{cases}

Why does sgn (0) = 0 ?

Commented by Rasheed Soomro last updated on 19/Nov/15

H o w d o e s s g n (x) s a t i s f y

sgn (x) = \sqrt{x} \sqrt{\frac{1}{x}} ?

F o r e x a m p l e s g n (- 6) = - 1

S o - 1 = \sqrt{- 6} \sqrt{\frac{1}{- 6}}

= i \sqrt{6} \times \frac{1}{i \sqrt{6}}

= 1 ?

Commented by Filup last updated on 20/Nov/15

I^{'} m not sure how it satisfies . That was

Just what I found online

Commented by Rasheed Soomro last updated on 20/Nov/15

Why does sgn (0) = 0 ?

A n I n t e r p r e t a t i o n f o r s g n (0) = 0

H e r e 1, 0, - 1 a r e \underline{r e p r e s e n t a t i v e} e l e m e n t s o n l y .

∙^{'} 1^{'} r e p r e s e n t s t h e s i t u a t i o n x > 0

∙ O p p o s i t e s i t u a t i o n x < 0 i s r e p r e s e n t e d b y^{'} - 1^{'} (w h i c h i s

o p p o s i t e o f^{'} 1^{'})

∙ x = 0 i s n e u t r a l s i t u a t i o n w h i c h s h o u l d b e r e p r e s e n t e d

b y n e u t r a l e l e m e n t t h a t i s^{'} 0^{'}

Commented by prakash jain last updated on 20/Nov/15

sgn (x) = \sqrt{x} \sqrt{\frac{1}{x}} is correct

sgn (- 6) = \sqrt{- 6} \times \sqrt{- \frac{1}{6}} = - 1 (and not 1)

\sqrt{- 6} = i \sqrt{6}

\sqrt{- \frac{1}{6}} = i \frac{1}{\sqrt{6}}

Answered by 123456 last updated on 19/Nov/15

for a certain things and simetry, read

the definition for complex sign : D

- - - - curiosity

it as generalized to C by /

sgn z = \frac{z}{∣ z ∣}

this act like a vercsor for complex number

also

sgn z = e^{- ı \arg z}

\arg 0 is undefined because unlike

z = 1 = e^{2 π k ı}, k \in Z (k = 0 is principal argument, 0 ⩽ θ < 2 π)

z = 0 = 0 e^{ı θ}, \forall θ \in R

you can pick all angle you want .

Commented by Rasheed Soomro last updated on 19/Nov/15

I {d i d n}^{'} t u n d e r s t a n d s o m e t h i n g s :

∙ this act like a vercsor for complex number

W h a t i s vercsor ?

∙ also if you take \arg 0 = 0 (wich is undefined)

its flows natural W h a t d o e s t h i s m e a n ?

O v e r a l l t h e c o m e n t i s k n o w l e d g e - i n c r e a s i n g .

T h e i d e a o f g e n e r a l i z a t i o n o f s i g n t o c o m l e x

n u m b e r s i s i n s p i r i n g!

Commented by Filup last updated on 19/Nov/15

What is \arg (z) ?

I^{'} ve seen it before

Commented by 123456 last updated on 19/Nov/15

the argument of the complex number

if you consider the complex a vector

it is the angle of it and the real axis

Commented by Filup last updated on 19/Nov/15

I see . I {haven}^{'} t leant much on the complex

plane, so this interests me . I^{'} ll have to

take a look at complex mathematics!

Commented by Rasheed Soomro last updated on 19/Nov/15

TH {a n k}^{S}

Commented by 123456 last updated on 19/Nov/15

vecsor is a vector with unitary module

its ussefull if you want only the direction

of a thing

f_{g} = G \frac{m_{1} m_{2}}{∣ {\vec{r}}_{12} ∣} {\hat{r}}_{12}

the vecsor {\hat{r}}_{12} only give the direction,

this is the gravitational force into body

1

for the segund thinked wrong XD

sorry for that

Commented by Rasheed Soomro last updated on 20/Nov/15

I w a n t t o u n d e r s t a n d c l e a r l y :

I f z = a + i b t h e n

s g n (a + i b) = \frac{a + i b}{∣ a + i b ∣} = \frac{a}{\sqrt{a^{2} + b^{2}}} + \frac{i b}{\sqrt{a^{2} + b^{2}}}

C l e a r l y a l l z h a v i n g s a m e ∣ z ∣ h a v e s a m e s g n (z)!

D o a l l c o m p l e x n u m b e r s h a v i n g s a m e a b s o l u t e v a l u e

h a v e s a m e s i g n ?

Commented by Filup last updated on 20/Nov/15

A b s o l u t e V a l u e

z_{1} = a_{1} + b_{1} i ∣ z_{2} = a_{2} + b_{2} i

∣ z_{1} ∣= \sqrt{a_{1}^{2} + b_{1}^{2}} ∣ ∣ z_{2} ∣= \sqrt{a_{2}^{2} + b_{2}^{2}}

∣ z_{1} ∣⩾ 0 ∣ ∣ z_{2} ∣⩾ 0

a s s u m e ∣ z_{1} ∣=∣ z_{2} ∣

S i g n

sgn (z_{1}) = e^{i \arg (z_{1})}

sgn (z_{2}) = e^{i \arg (z_{2})}

if when complex functions with equal abs . values

are equal, the sgn are equal, t h e n :

f o r ∣ z_{1} ∣=∣ z_{2} ∣

e^{i \arg (z_{1})} = e^{i \arg (z_{2})}

∴ \arg (z_{1}) = \arg (z_{2})

\arg (a_{1} + b_{1} i) = \arg (a_{2} + b_{2} i)

(?)

Commented by 123456 last updated on 20/Nov/15

z_{1} and z_{2} have same sign only if

\exists λ \in R, 0 < λ < + \infty such that

z_{2} = λ z_{1}

Commented by prakash jain last updated on 20/Nov/15

z_{1} = a_{1} + {i b}_{1}

\arg (z) = \arctan (a_{1}, b_{1})

Individual sign of a_{1} and b_{1} are important .

On {Rasheed}^{'} s comment

all z having the same ∣ z ∣ does not mean

same sgn (z)

z_{1} = 3 + 4 i, z_{2} = 4 + 3 i

sgn (z_{1}) = . 6 + . 8 i, sgn (z_{2}) = . 8 + . 6 i

Commented by Rasheed Soomro last updated on 20/Nov/15

THANKS T o A l l o f Y o u!

I W a s I n M i s U n d e r s t a n d i n g .