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#### paulmdrdo

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- May 13, 2013

- 386

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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- Thread starter
- #1

- May 13, 2013

- 386

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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- #2

- Mar 5, 2012

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Welcome to MHB, paulmdrdo!

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!

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?

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- May 13, 2013

- 386

thanks for the reply! the point of interest in (a) is -1. the function values at x<-1 are all 0.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?

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!

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- Mar 5, 2012

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The first step in the piecewise solution of (a) is: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!

$$

(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?

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- May 13, 2013

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

..............................{1, x>0

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

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- Mar 5, 2012

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Can you reevaluate the case -1<x<0?(x+1) H(x+1)-x H(x) = {0, if x<-1

..............................{x, if -1<x<0

..............................{1, x>0

Hmm, I just found that different sources define H(x) differently.but why did you consider x=-1 and x=0 special value?

According to Wolfram $H(0)=\frac 1 2$, but according to wiki $H(0)=1$.

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- May 13, 2013

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- May 13, 2013

- 386

hey i just noticed something hereoh! is there another answer to that? how to do that?

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?

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- #9

- Mar 5, 2012

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After closer inspection, in this particular case both answers turn out to be the same:oh! is there another answer to that? how to do that?

$$

(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.

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- #10

- May 13, 2013

- 386

uhm, please bear for my being ignorant here.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.

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?

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That is correct.when i evaluated the function at x=-1 it gives 0.

It isn't.why is that different to -1<=x<0?

But you specified for $-1 \le x<0$ the value $x$.

That implies that at $x=-1$ it gives $-1$, which is incorrect.

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- #12

- May 13, 2013

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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!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.

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- May 13, 2013

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i have tried other answersThat 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.

(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:

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- #14

- Mar 5, 2012

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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.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?

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- #15

- May 13, 2013

- 386

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

(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.

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- #16

- Mar 5, 2012

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Then the solutions manual is wrong.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.

Sorry.

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- #17

- May 13, 2013

- 386

yeah. i also suspected that. now it is confirmed. you told about relevant boundaries I should consider, how would I determine those relevant boundaries?Then the solutions manual is wrong.

Sorry.

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- #18

- Mar 5, 2012

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Just check what your function does at x=-1 and x=0.yeah. i also suspected that. now it is confirmed. you told about relevant boundaries I should consider, how would I determine those relevant boundaries?

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.

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- #19

- May 13, 2013

- 386

can you spot what i'm doing wrong.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.

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.