# Math with conditions

1. Dec 13, 2013

### Jhenrique

Given
$$x = \left\{\begin{matrix} y \;\;\; case\;A\\ z \;\;\; case\;B\\ \end{matrix}\right.$$
So, operate x means to operate the 2 cases of right side? For example:
$$\int x\;dx = \left\{\begin{matrix} \int y\;dx \;\;\; case\;A\\ \int z\;dx \;\;\; case\;B\\ \end{matrix}\right.$$
Correct?

2. Dec 13, 2013

### 1MileCrash

Sure.

3. Dec 13, 2013

### Jhenrique

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

4. Dec 13, 2013

### R136a1

Could you give a concrete example?

5. Dec 13, 2013

### Mandelbroth

Are you intending for the $x$ in your $\mathrm{d}x$ to be different?

6. Dec 13, 2013

### Jhenrique

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

7. Dec 13, 2013

### Number Nine

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.

8. Dec 13, 2013

### Student100

$$x = \left\{\begin{matrix} y \;\;\; case\;A\\ z \;\;\; case\;B\\ \end{matrix}\right.$$

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

Is this what you mean?

9. Dec 14, 2013

### Jhenrique

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

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.