Question

is |(-1,-2)| = (|-1|,|-2|)

Answer 1

is that a point on a graph

Answer 2

on properties of absolute value are or not?

Answer 3

oops are they equal or not?

Answer 4

I ment is (-1,-2) a point on a graph or an equation because if its a point on a graph then you take the absolute value of both numbers so yeah basically they would be equal (1 , 2)

Answer 5

(1 , 2) would be the actual point

Answer 6

let h(x)=(-sin(x),-cos(x)) then is |h(x)|=|(-sin(x),-cos(x))|=(sin(x),cos(x))?

Answer 7

I would imagine that |(-1, -2)| means "what is the magnitude of the vector (-1, -2)

Answer 8

are you studying vectors?

Answer 9

yeah this question is very vague

Answer 10

no ,is real analysis

Answer 11

the notation used still looks like it is asking for the magnitude of a vector.
so, if h(x) = (-sin(x), -cos(x))

then:\[|h(x)|=|(-\sin(x), -\cos(x))|=\sqrt{\sin^2(x)+\cos^2(x)}=1\]

Answer 12

actually i am trying to solve this

Answer 13

wait a minute

Answer 14

\[s={(x,y) \epsilon R ^{2}:x ^{2}+y ^{2}}\] S is the unit circle in the plane define
\[h:[0,2\pi)\rightarrow S\] by
\[h(\theta)=(\cos \theta , \sin )\]

Answer 15

@jacobian you meant \(S=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) anyways yes that is a valid coordinate map

Answer 16

i've try it like this
Let ϵ>0 and a∈R be given. Choose δ=ϵ. Then we have that \[|\sin \theta - \sin a|\le 2|\frac{ \theta - a }{ 2 }|\\ = |\theta - a| < \]

Answer 17

this is also similar for cosine they are in the form of \[|\theta - a |<epsilon \]

Answer 18

so if we choose \[\delta = \min \left\{ \delta _{1},\delta _{2}\ \right\} \] we get \[\left| h(\theta) - h(a)\right| = \left| (\cos \theta,\sin \theta) - (\cos a , \sin a)\right|\] then i don't know how to continue

Answer 19

@asnaseer @amistre64 @KingGeorge

Answer 20

are you trying to prove this function \(h:\mathbb{R}\to\mathbb{R}^2\) is continuous...?

Answer 21

nop ,h:[0,2pi)->S