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Math with conditions

by Jhenrique
Tags: conditions, math
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Jhenrique
#1
Dec13-13, 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?
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1MileCrash
#2
Dec13-13, 12:37 PM
1MileCrash's Avatar
P: 1,289
Sure.
Jhenrique
#3
Dec13-13, 01:01 PM
P: 686
I don't known that it's operable. It's cool, will be very important to me in engineering!

R136a1
#4
Dec13-13, 01:13 PM
Newcomer
P: 341
Math with conditions

Could you give a concrete example?
Mandelbroth
#5
Dec13-13, 02:13 PM
Mandelbroth's Avatar
P: 615
Quote Quote by Jhenrique View Post
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?
Jhenrique
#6
Dec13-13, 02:46 PM
P: 686
x, y and z are only a representative symbol...
Number Nine
#7
Dec13-13, 07:41 PM
P: 772
Quote Quote by Jhenrique View Post
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.
Student100
#8
Dec13-13, 09:39 PM
Student100's Avatar
P: 547
Quote Quote by Jhenrique View Post
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?
Jhenrique
#9
Dec14-13, 04:02 AM
P: 686
Quote Quote by Student100 View Post
[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?
Quote Quote by Jhenrique View Post
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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