What happens in this step function?

In summary, the conversation discusses the definition of the integral and its limits, specifically when the lower limit is 0. It is agreed that the limit for the integral should be a closed interval, but there is a difference in interpretation when the lower limit is 0. The conversation also touches on the value of the function at a single point not affecting the value of the integral.
  • #1
garylau
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3
when the x>0,then the theta is equal to 1
So the theta =0 when x=0

in the second term of the integral,
it starts to integrate the function from "0 "to infinity(see the yellow loop inside)Since the" 0 "should not be included ,otherwise theta(0)=0 and (0*df/dt)=0

but why the 0 is still included in the integral to integrate the integrand!??

Sorry
i got confused
can anyone help me

thank you!??
 

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  • #2
An open interval does not include its endpoints, and is indicated with parentheses. For example, (0,1) means greater than 0 and less than 1. A closed intervalincludes its endpoints, and is denoted with square brackets. For example, [0,1] means greater than or equal to 0 and less than or equal to 1.

however
in the definition of the intregral ,the lower and upper limit should be closed interval rather than open interval

So did they do something wrong?

however,the step function does not include x=0(which is the end point) (or x<0) So it is open interval rather than closed interval
So it didn't fit the definition of an integral ?
 

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  • #3
The value of theta(0) is not important it is often taken to be 0, 1/2, or 1; but it does not matter.
The value of a function at one point does not effect an integral.
 
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  • #4
garylau said:
Since the" 0 "should not be included ,otherwise theta(0)=0 and (0*df/dt)=0

but why the 0 is still included in the integral to integrate the integrand!??
  • [itex]\int_{a}^{\infty}f(x)dx [/itex] is defined as [itex]\lim_{u\rightarrow\infty}\int_{a}^{u}f(x)dx [/itex]
  • In the same manner, if [itex]\lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx [/itex] exists, [itex]\int_{0}^{u}f(x)dx = \lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx [/itex].
  • Informally: Changing the value of the integrand at a single point does not affect the value of the integral.
 
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  • #5
lurflurf said:
The value of theta(0) is not important it is often taken to be 0, 1/2, or 1; but it does not matter.
The value of a function at one point does not effect an integral.
it may be affect so much if f(0) is very large

therefore the difference between f(0)*1 and f(0)*0 will be large?
 
  • #6
Svein said:
  • [itex]\int_{a}^{\infty}f(x)dx [/itex] is defined as [itex]\lim_{u\rightarrow\infty}\int_{a}^{u}f(x)dx [/itex]
  • In the same manner, if [itex]\lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx [/itex] exists, [itex]\int_{0}^{u}f(x)dx = \lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx [/itex].
  • Informally: Changing the value of the integrand at a single point does not affect the value of the integral.
Svein said:
  • [itex]\int_{a}^{\infty}f(x)dx [/itex] is defined as [itex]\lim_{u\rightarrow\infty}\int_{a}^{u}f(x)dx [/itex]
  • In the same manner, if [itex]\lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx [/itex] exists, [itex]\int_{0}^{u}f(x)dx = \lim_{p\rightarrow 0}\int_{p}^{u}f(x)dx [/itex].
  • Informally: Changing the value of the integrand at a single point does not affect the value of the integral.
But why the definition from the website that i quote is quiet different

The limit is belong to closed interval rather than open interval except the (+ / )infinite large's one

Where is your definition coming from(the one which imclude limit u-->0?
the improper integral is for infinite limit only but not for u=0 ?
 
  • #7
First: See https://en.wikipedia.org/wiki/Riemann_integral.
Second: Take the integral [itex]\int_{\epsilon}^{1}\theta(x)dx [/itex]. Since θ(x) = 1 for x>0, the value is 1 - ε. Therefore (since ε is arbitrary) the integral is 1.
garylau said:
Where is your definition coming from(the one which imclude limit u-->0?
the improper integral is for infinite limit only but not for u=0 ?
An integral is defined as some limit (see the link above). The only problem with Riemann integrals are that there may not be a limit. In this case there is a limit.

That aside, I definitely do not like the paragraph b) in your quote. It assumes that ∞ - ∞ = 0, which is not necessarily true. In fact, ∞ - ∞ has no definite value.
 
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Related to What happens in this step function?

1. What is a step function?

A step function is a mathematical function that produces an output based on different inputs. It is a special type of function that only changes its output at specific points or "steps" along the input axis.

2. What happens in each step of a step function?

In each step of a step function, the output value remains constant until it reaches the next step or transition point. At this point, the output value jumps to a new constant value, and the process repeats for each subsequent step.

3. How is a step function different from other types of functions?

A step function is different from other types of functions because it has a discontinuous graph. This means that the graph of a step function is not a smooth, continuous curve, but rather a series of connected horizontal lines.

4. What are some real-world applications of step functions?

Step functions are commonly used in economics, finance, and computer science. In economics, they can be used to model the behavior of consumers and businesses. In finance, they are used to calculate the value of options and other financial derivatives. In computer science, they are used in algorithms and data structures.

5. How are step functions helpful in scientific research?

Step functions are helpful in scientific research because they can be used to represent and analyze complex systems with discrete inputs and outputs. They can also be used to model and study the behavior of natural phenomena, such as population growth and chemical reactions.

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