Plotting heaviside(x^2-1)

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23
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Hello,

I am not where this question goes, its not part of a homework either!
I am trying to figure out how to plot the heaviside (unit step) with such an expression
H(x^2-1)
so I do this:H(x^2-1) = 1 for x^2-1>0 -> x>+- 1
and H(x^2-1) = 0 for x^2-1<-0 -> x<-+1

But this tells me only that it equals 1 when it is x>1 because the rest would be zero! any clarification would be appreciated
 

pasmith

Homework Helper
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Hello,

I am not where this question goes, its not part of a homework either!
I am trying to figure out how to plot the heaviside (unit step) with such an expression
H(x^2-1)
so I do this:H(x^2-1) = 1 for x^2-1>0 -> x>+- 1
and H(x^2-1) = 0 for x^2-1<-0 -> x<-+1

But this tells me only that it equals 1 when it is x>1 because the rest would be zero! any clarification would be appreciated
If [itex]x^2 - 1 > 0[/itex] then either [itex]x < -1[/itex] or [itex]x > 1[/itex].
 
23
0
so for x^2-1>0 I have
H(x^2-1) = 1 for x>1 and x>-1 ( so its equal to 1 from -1 to infinity)
and
H(x^2-1) = 0 for x<1 and x<-1 ( so its equal to 0 from 1 to negative infinity)

so why would I choose only : equals 1 for x>1 and x<-1, what about the other conditions?
 
32,852
4,573
so for x^2-1>0 I have
H(x^2-1) = 1 for x>1 and x>-1 ( so its equal to 1 from -1 to infinity)
and
H(x^2-1) = 0 for x<1 and x<-1 ( so its equal to 0 from 1 to negative infinity)

so why would I choose only : equals 1 for x>1 and x<-1, what about the other conditions?
Let's back up a bit, since you're confused about the solution to x2 - 1 > 0.
x2 - 1 ≥ 0 ==> x ≥ 1 or x ≤ -1.
So H(x2 - 1) = 1 for x ≥ 1 or x ≤ -1.
 
23
0
Thanks Mark,
I think I am missing a basic principle here about the ≥ relations

if I have a [tex] x^2 - 1 \geq 0 [/tex] then solving for x is
[tex] x^2 - 1 = 0 \rightarrow x = \pm 1 [/tex]
so in the case of positive one
[tex] x\geq 1 [/tex] and for negative one it changes to
[tex] x \leq -1 [/tex] and the positive it remains the same, is that a property of the relation?
also why do we choose ≥ instead of just >
 
32,852
4,573
Thanks Mark,
I think I am missing a basic principle here about the ≥ relations
Yes, I think so as well.
if I have a [tex] x^2 - 1 \geq 0 [/tex] then solving for x is
[tex] x^2 - 1 = 0 \rightarrow x = \pm 1 [/tex]
There's a big difference between solving an equation (as above) and solving an inequality.
What you've written below doesn't make any sense to me.
so in the case of positive one
[tex] x\geq 1 [/tex] and for negative one it changes to
[tex] x \leq -1 [/tex] and the positive it remains the same, is that a property of the relation?
also why do we choose ≥ instead of just >
Because H(x) = 1 if x ≥ 0 and H(x) = 0 for x < 0. That's why.

For your function, H(x2 - 1) = 1 for x2 - 1 ≥ 0, so we need to find the intervals for which x2 - 1 ≥ 0.

The part on the left side of the inequality equals 0 for x = -1 or x = 1. These two numbers divide the number line into three intervals: (-∞, -1), (-1, 1), and (1, ∞). By taking any number in each interval you can verify that the solution to x2 - 1 ≥ 0 is (-∞, -1] U [1, ∞). This is the same as saying x ≤ -1 or x ≥ 1.

It would be helpful for you to review the technique of solving inequalities, especially quadratic inequalities.
 
23
0
Thanks so much Mark, this clarified it
 

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