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Math with conditions 
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#1
Dec1313, 12:21 PM

P: 686

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? 


#3
Dec1313, 01:01 PM

P: 686

I don't known that it's operable. It's cool, will be very important to me in engineering!



#4
Dec1313, 01:13 PM

Newcomer
P: 341

Math with conditions
Could you give a concrete example?



#5
Dec1313, 02:13 PM

P: 615




#6
Dec1313, 02:46 PM

P: 686

x, y and z are only a representative symbol...



#7
Dec1313, 07:41 PM

P: 772




#8
Dec1313, 09:39 PM

P: 560

[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? 


#9
Dec1413, 04:02 AM

P: 686

[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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