How do you integrate and differentiate a function with steps and impulses?

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Discussion Overview

The discussion revolves around the integration and differentiation of piecewise functions, particularly those involving step functions and impulses. Participants explore the implications of operating on such functions and the mathematical correctness of their approaches.

Discussion Character

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

Main Points Raised

  • One participant proposes that operating on a piecewise function involves integrating each case separately, questioning if this is correct.
  • Another participant expresses enthusiasm about the operability of such functions in engineering contexts.
  • A request for a concrete example is made to clarify the concept further.
  • There is a repeated emphasis on the potential confusion arising from using the same symbol (x) in different contexts within the discussion.
  • One participant suggests that the differentiation of a step function, represented as H(x), leads to a contradiction with the identity involving the Dirac delta function, indicating a potential flaw in the proposed technique.
  • Participants are trying to understand how to integrate and differentiate functions that exhibit step changes and impulses, both graphically and algebraically.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the proposed methods for integrating and differentiating piecewise functions. There are competing views regarding the implications of using the same variable in different contexts and the validity of the mathematical techniques discussed.

Contextual Notes

The discussion highlights the limitations in clarity regarding the use of symbols and the mathematical steps involved, particularly in relation to the differentiation of the step function and its implications.

Jhenrique
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Given
[tex]x = \left\{\begin{matrix} y \;\;\; case\;A\\ z \;\;\; case\;B\\ \end{matrix}\right.[/tex]
So, operate x means to operate the 2 cases of right side? For example:
[tex]\int x\;dx = \left\{\begin{matrix} \int y\;dx \;\;\; case\;A\\ \int z\;dx \;\;\; case\;B\\ \end{matrix}\right.[/tex]
Correct?
 
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Sure.
 
I don't known that it's operable. It's cool, will be very important to me in engineering!
 
Could you give a concrete example?
 
Jhenrique said:
Given
[tex]x = \left\{\begin{matrix} y \;\;\; case\;A\\ z \;\;\; case\;B\\ \end{matrix}\right.[/tex]
So, operate x means to operate the 2 cases of right side? For example:
[tex]\int x\;dx = \left\{\begin{matrix} \int y\;dx \;\;\; case\;A\\ \int z\;dx \;\;\; case\;B\\ \end{matrix}\right.[/tex]
Correct?
Are you intending for the ##x## in your ##\mathrm{d}x## to be different?
 
x, y and z are only a representative symbol...
 
Jhenrique said:
x, y and z are only a representative symbol...

The problem is that you're using x in two different places, and it's not clear whether or not you intend them to mean the same thing.
 
Jhenrique said:
Given
[tex]x = \left\{\begin{matrix} y \;\;\; case\;A\\ z \;\;\; case\;B\\ \end{matrix}\right.[/tex]
So, operate x means to operate the 2 cases of right side? For example:
[tex]\int x\;dx = \left\{\begin{matrix} \int y\;dx \;\;\; case\;A\\ \int z\;dx \;\;\; case\;B\\ \end{matrix}\right.[/tex]
Correct?


[tex]x = \left\{\begin{matrix} y \;\;\; case\;A\\ z \;\;\; case\;B\\ \end{matrix}\right.[/tex]

[tex]\int x\;dx = \left\{\begin{matrix} \int y\;dy \;\;\; case\;A\\ \int z\;dz\;\;\; case\;B\\ \end{matrix}\right.[/tex]

Is this what you mean?
 
Student100 said:
[tex]x = \left\{\begin{matrix} y \;\;\; case\;A\\ z \;\;\; case\;B\\ \end{matrix}\right.[/tex]

[tex]\int x\;dx = \left\{\begin{matrix} \int y\;dy \;\;\; case\;A\\ \int z\;dz\;\;\; case\;B\\ \end{matrix}\right.[/tex]

Is this what you mean?
Jhenrique said:
operate x means to operate the 2 cases of right side?

I'm trying find... I think that it's math isn't correct, because:
[tex]H(x) = \left\{\begin{matrix} 0\;\;\; x<0\\ 1\;\;\; x=0\\ 1\;\;\; x>0\\ \end{matrix}\right.[/tex]
[tex]\frac{\mathrm{d} }{\mathrm{d} x}H(x) = \left\{\begin{matrix} \frac{\mathrm{d} }{\mathrm{d} x}0\;\;\; x<0\\ \frac{\mathrm{d} }{\mathrm{d} x}1\;\;\; x=0\\ \frac{\mathrm{d} }{\mathrm{d} x}1\;\;\; x>0\\ \end{matrix}\right. = \left\{\begin{matrix} 0\;\;\; x<0\\ 0\;\;\; x=0\\ 0\;\;\; x>0\\ \end{matrix}\right.[/tex]

is wrong, 'cause contradicts the identity d/dx H(x) = δ(x). So, this "technic" above is not useful, although it seems make sense, theoretically... In other words, I'm trying know how do to integrate and derivative, graphically and algebraically, a function with steps and impulses.
 

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