View Full Version : Algebraic Sum
Alejandro Sanchez
Apr 21, 2010, 01:29 PM
f(x)=[(x+∆x )+2/((x+∆x))]-[((x)+2/((x) )) ]
ebaines
Apr 21, 2010, 02:29 PM
Two items:
1. Why do you have so many parentheses in this equation? It looks like what you mean is this:
f(x) = x + \Delta x + \frac x {(x + \Delta x)} - (x + \frac 2 x )
is that right?
2. Do you have a question?
Alejandro Sanchez
Apr 21, 2010, 02:41 PM
You are right, to many parenthesis. Here is the correct the equation.
f(x)=[(x+∆x+2)/(x+∆x)]-[(x+2)/(x)]
ebaines
Apr 21, 2010, 02:46 PM
OK great. That answers the first item I asked about. But how about the second - do you have a question?
Alejandro Sanchez
Apr 21, 2010, 02:51 PM
Ok. My question is, what it would be the final result if I make the algebraic sustraction of the first function minus the second function.
First function: [(x+∆x+2)/(x+∆x)] MINUS (-) Second function: [(x+2)/(x)]
ebaines
Apr 21, 2010, 03:06 PM
I get:
f(x) = \frac {-2 \Delta x} {x(x+ \Delta x)}
which for small values of \Delta x approaches:
f(x) = \frac {-2 \Delta x} {x^2}
Alejandro Sanchez
Apr 21, 2010, 03:16 PM
Ok, great. And for the following one:
First function: [(x+∆x)+(2/(x+∆x))] MINUS (-) Second function: [x+(2/x)]
Alejandro Sanchez
Apr 21, 2010, 03:42 PM
Thanks.