Solve Unit Step Function Algebraically - Help Needed

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Discussion Overview

The discussion revolves around algebraically performing operations on the Heaviside step function, specifically focusing on piecewise definitions and evaluations of expressions involving multiple Heaviside functions. Participants seek clarity on how to handle boundaries and special cases in the context of the unit step function.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants inquire about the algebraic manipulation of Heaviside functions, specifically in expressions like (x+1)*H(x+1) - x*H(x).
  • There is discussion on the boundaries where the Heaviside function switches values, particularly at x = -1 and x = 0.
  • Participants express confusion over how to define the piecewise function correctly and the implications of different definitions of H(0) from various sources.
  • Some suggest that the function can be defined as 0 for x <= -1, x for -1 < x < 0, and 1 for x >= 0, while others argue for different interpretations.
  • There is a challenge regarding the evaluation of the function at specific points, particularly at the boundaries, leading to questions about the correctness of certain piecewise definitions.
  • Participants note that the case -1 < x < 0 should read x + 1 instead of x, indicating a potential misunderstanding in earlier evaluations.
  • One participant mentions that the solutions manual they consulted appears to be incorrect, prompting further discussion on how to determine relevant boundaries in piecewise definitions.

Areas of Agreement / Disagreement

Participants express differing views on the correct piecewise definitions of the function involving Heaviside functions. There is no consensus on the final form of the piecewise function, and discussions remain unresolved regarding the treatment of boundaries and special values.

Contextual Notes

Participants highlight that different sources may define the Heaviside function differently, particularly at H(0), which adds complexity to the discussion. The relevance of boundary conditions and how to handle jumps in the function are also noted as important considerations.

Who May Find This Useful

This discussion may be useful for students and learners in mathematics, particularly those studying piecewise functions and the Heaviside step function, as well as individuals seeking to understand algebraic manipulations involving discontinuous functions.

paulmdrdo1
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How would I go about algebraically (not graphically) performing operations on two or more Heaviside function.H(x) = {0, if x<0
...{1, if x>=0

ex. define the function piecewise and graph.
a.) (x+1)*H(x+1)-x*H(x)
b.) (x+1)*H(x+1)

i'm having a hard time solving this please help me!
 
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Re: Unit step function. help!

paulmdrdo said:
How would I go about algebraically (not graphically) performing operations on two or more Heaviside function.H(x) = {0, if x<0
...{1, if x>=0

ex. define the function piecewise and graph.
a.) (x+1)*H(x+1)-x*H(x)
b.) (x+1)*H(x+1)

i'm having a hard time solving this please help me!

Welcome to MHB, paulmdrdo! :)

The Heaviside step function switches value at x=0.
Where does H(x+1) switch value?
So which are the relevant boundaries to consider for (a)?
Following up, what does (a) look like if x is below the lowest boundary?
 
Re: Unit step function. help!

I like Serena said:
Welcome to MHB, paulmdrdo! :)

The Heaviside step function switches value at x=0.
Where does H(x+1) switch value?
So which are the relevant boundaries to consider for (a)?
Following up, what does (a) look like if x is below the lowest boundary?

thanks for the reply! the point of interest in (a) is -1. the function values at x<-1 are all 0.
but I'm still confused with that kind of problem like in (a). could you provide me your solution in problem (a). i just want to compare my thought process with others when I'm solving that. thanks!
 
Re: Unit step function. help!

paulmdrdo said:
thanks for the reply! the point of interest in (a) is -1. the function values at x<-1 are all 0.
but I'm still confused with that kind of problem like in (a). could you provide me your solution in problem (a). i just want to compare my thought process with others when I'm solving that. thanks!

The first step in the piecewise solution of (a) is:
$$
(x+1) H(x+1)-x H(x) = \left\{ \begin{aligned}
(x+1) H(x+1)-x H(x) & \qquad \text{if } x< -1 \\
(x+1) H(x+1)-x H(x) & \qquad \text{if } -1 < x< 0 \\
(x+1) H(x+1)-x H(x) & \qquad \text{if } x > 0 \\
\text{special value} & \qquad \text{if } x = -1 \vee x = 0 \\
\end{aligned}\right.
$$

Note that with the Heaviside step function there are a couple more boundary conditions that I'm going to ignore for now, since H(0) is defined to be 1/2.

Can you simplify the piecewise solution?
 
Re: Unit step function. help!

I like Serena said:
The first step in the piecewise solution of (a) is:
$$
(x+1) H(x+1)-x H(x) = \left\{ \begin{aligned}
(x+1) H(x+1)-x H(x) & \qquad \text{if } x< -1 \\
(x+1) H(x+1)-x H(x) & \qquad \text{if } -1 < x< 0 \\
(x+1) H(x+1)-x H(x) & \qquad \text{if } x > 0 \\
\text{special value} & \qquad \text{if } x = -1 \vee x = 0 \\
\end{aligned}\right.
$$

Note that with the Heaviside step function there are a couple more boundary conditions that I'm going to ignore for now, since H(0) is defined to be 1/2.

Can you simplify the piecewise solution?

(x+1) H(x+1)-x H(x) = {0, if x<-1
........{x, if -1<x<0
........{1, x>0

but why did you consider x=-1 and x=0 special value?
 
Re: Unit step function. help!

paulmdrdo said:
(x+1) H(x+1)-x H(x) = {0, if x<-1
........{x, if -1<x<0
........{1, x>0

Can you reevaluate the case -1<x<0?

but why did you consider x=-1 and x=0 special value?

Hmm, I just found that different sources define H(x) differently.
According to Wolfram $H(0)=\frac 1 2$, but according to wiki $H(0)=1$.
 
Re: Unit step function. help!

I like Serena said:
Can you reevaluate the case -1<x<0?
Hmm, I just found that different sources define H(x) differently.
According to Wolfram $H(0)=\frac 1 2$, but according to wiki $H(0)=1$.

oh! is there another answer to that? how to do that?
 
Last edited:
Re: Unit step function. help!

paulmdrdo said:
oh! is there another answer to that? how to do that?

hey i just noticed something here
if x=-1 the function value is 0 and when x=0 the fuction value is 1

so we can define the function as
(x+1) H(x+1)-x H(x) = {0, if x<=-1
......{x, if -1<x<0
......{1, x=>0
am I correct?
 
Re: Unit step function. help!

paulmdrdo said:
oh! is there another answer to that? how to do that?

After closer inspection, in this particular case both answers turn out to be the same:
$$
(x+1) H(x+1)-x H(x) = \left\{ \begin{array}{ll}
0 & \qquad \text{if } x< -1 \\
x+1 & \qquad \text{if } -1 \le x < 0 \\
1 & \qquad \text{if } x \ge 0 \\
\end{array}\right.
$$

Note that the case $-1 \le x < 0$ is different from what you have.
 
  • #10
Re: Unit step function. help!

I like Serena said:
After closer inspection, in this particular case both answers turn out to be the same:
$$
(x+1) H(x+1)-x H(x) = \left\{ \begin{array}{ll}
0 & \qquad \text{if } x< -1 \\
x+1 & \qquad \text{if } -1 \le x < 0 \\
1 & \qquad \text{if } x \ge 0 \\
\end{array}\right.
$$

Note that the case $-1 \le x < 0$ is different from what you have.

uhm, please bear for my being ignorant here.

when i evaluated the function at x=-1 it gives 0. why is that different to -1<=x<0?
can you give me example of numerical computation of it?
 
Last edited:
  • #11
Re: Unit step function. help!

paulmdrdo said:
when i evaluated the function at x=-1 it gives 0.

That is correct.
why is that different to -1<=x<0?

It isn't.
But you specified for $-1 \le x<0$ the value $x$.
That implies that at $x=-1$ it gives $-1$, which is incorrect.
 
  • #12
Re: Unit step function. help!

I like Serena said:
That is correct.

It isn't.
But you specified for $-1 \le x<0$ the value $x$.
That implies that at $x=-1$ it gives $-1$, which is incorrect.

thank you! I'm so amazed! you're so generous in answering me queries! I've never been in a forum like this one. hope you'll continue helping a math learners like me. I can't learn math on my own thanks to MHB! more power. by the way I'm a 6th grader. learning this is so challenging!
 
  • #13
Re: Unit step function. help!

I like Serena said:
That is correct.

It isn't.
But you specified for $-1 \le x<0$ the value $x$.
That implies that at $x=-1$ it gives $-1$, which is incorrect.

i have tried other answers

(x+1) H(x+1)-x H(x) = {0, if x<-1
........{x, if -1<=x<=0
........{1, if x>0

and

(x+1) H(x+1)-x H(x) = {0, if x<=-1
......{x+1, if -1<=x<=0
......{1, if x>0

also,
(x+1) H(x+1)-x H(x) = {0, if x<=-1
......{x, if -1<x<0
......{1, x=>0
are these correct?
 
Last edited:
  • #14
Re: Unit step function. help!

paulmdrdo said:
i have tried other answers

(x+1) H(x+1)-x H(x) = {0, if x<-1
........{x, if -1<=x<=0
........{1, if x>0

and

(x+1) H(x+1)-x H(x) = {0, if x<=-1
......{x+1, if -1<=x<=0
......{1, if x>0

also,
(x+1) H(x+1)-x H(x) = {0, if x<=-1
......{x, if -1<x<0
......{1, x=>0
are these correct?

The place where you put the equalities are not particularly relevant, but the case -1<x<0 really has to read x+1 instead of x.
 
  • #15
Re: Unit step function. help!

I like Serena said:
The place where you put the equalities are not particularly relevant, but the case -1<x<0 really has to read x+1 instead of x.

actually this is the answer i saw on my solution's manual

(x+1) H(x+1)-x H(x) = {0, if x<-1
........{x, if -1<=x<=0
........{1, if x>0

but I still don't fully understand what are the steps used in solving this.
 
  • #16
Re: Unit step function. help!

paulmdrdo said:
actually this is the answer i saw on my solution's manual

(x+1) H(x+1)-x H(x) = {0, if x<-1
........{x, if -1<=x<=0
........{1, if x>0

but I still don't fully understand what are the steps used in solving this.

Then the solutions manual is wrong.
Sorry.
 
  • #17
Re: Unit step function. help!

I like Serena said:
Then the solutions manual is wrong.
Sorry.

yeah. i also suspected that. now it is confirmed. you told about relevant boundaries I should consider, how would I determine those relevant boundaries?
 
  • #18
Re: Unit step function. help!

paulmdrdo said:
yeah. i also suspected that. now it is confirmed. you told about relevant boundaries I should consider, how would I determine those relevant boundaries?

Just check what your function does at x=-1 and x=0.
When there are "jumps", you should consider what happens exactly at those jumps.
In your case (a) there aren't any jumps, meaning it doesn't matter.

In practice those jumps are usually not particularly relevant for the Heaviside step function.
 
  • #19
Re: Unit step function. help!

I like Serena said:
Just check what your function does at x=-1 and x=0.
When there are "jumps", you should consider what happens exactly at those jumps.
In your case (a) there aren't any jumps, meaning it doesn't matter.

In practice those jumps are usually not particularly relevant for the Heaviside step function.

can you spot what I'm doing wrong.

what i did is I plugged x-values to the function
it turns out that on the interval (-∞ ,-1) the function value is 0. and then on the interval (-1, 0) the function values are the same as the x-values (on the interval(-1,0)) that i started with. and then i graphed it. and on the interval (0,+∞ ) the function values is 1.
the graph of the function looks like continuous. when i try to define it piecewise I don't know how to put my cases correctly.
 

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