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!
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?
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!
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?
paulmdrdo said:(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?
paulmdrdo said:oh! is there another answer to that? how to do that?
paulmdrdo said:oh! is there another answer to that? how to do that?
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.
paulmdrdo said:when i evaluated the function at x=-1 it gives 0.
why is that different to -1<=x<0?
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 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.
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?
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.
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.
I like Serena said:Then the solutions manual is wrong.
Sorry.
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?
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.
A unit step function is a mathematical function that takes a value of 0 for all negative inputs and a value of 1 for all positive inputs. It is often represented using the notation u(t) or θ(t).
To solve a unit step function algebraically, you need to first identify the interval in which the function is defined. Then, you can use the definition of a unit step function to write out the function for that interval. Finally, you can use basic algebraic techniques to simplify the function and solve for the desired value.
Yes, the unit step function can be used to model and solve various real-world problems, such as modeling the response of an electrical circuit or the growth of a population over time. It is a useful tool in engineering, economics, and other fields.
While the unit step function is a useful tool, it does have some limitations. It is a discontinuous function, meaning it has abrupt changes at specific points, which may not accurately reflect real-life situations. Additionally, it can only take on two values (0 and 1), so it may not be suitable for modeling more complex systems.
Yes, there are alternative methods for solving unit step functions, such as using graphical techniques or using software programs like MATLAB or Wolfram Alpha. These methods may be more efficient and accurate for more complex functions and equations.