# Thread: Negative Index & Binomial Theorem

1. ## Negative Index & Binomial Theorem

hey everyone, I'm teaching myself calculus, because I felt it would be useful, even if I'm not going to use it in accounting. I am learning about the binomial theorem and differentiating negative indexes.

Here is the current problem I'm working on
1.) y=x^-2
2.) y+dy=(x+dx)^-2
3.) =x^-2(1+ dx/x)^-2
4.) =x^-2 [1-2dx/x + 2(2-1)/1*2 * (dx/x)^2 - etc. ]
5.) =x^-2 -2x^-3*dx + 3x^-4(dx)^2 -4x^-5(dx)^3 +etc...
Ignoring small units
6.) y+dy=x^-2 -2x^-3*dx
Subtracting original y=x^-2
7.) dy= -2x^-3*dx
8.) dy/dx= -2x^-3

the binomial theorem as it is explained in the book follows,

(a+b)^n=a^n + n(a^n-1b/1) + n(n-1)(a^n-2b^2/2) + n(n-1)(n-2)(a^n-3b/3) + etc....

I understand how one moves from Step 1 to 2, that being said how does (x+dx)^-2 become x^-2(1+dx/x)^-2? Secondly, could someone how one goes about applying the binomial theorem to a negative index?

any help would be greatly appreciated

2. Originally Posted by Mausoldj
....

I understand how one moves from Step 1 to 2, that being said how does (x+dx)^-2 become x^-2(1+dx/x)^-2? Secondly, could someone how one goes about applying the binomial theorem to a negative index?

any help would be greatly appreciated
$(x+dx)^{-2} = \left(x \cdot \left(1+\frac{dx}x \right) \right)^{-2} = x^{-2} \cdot \left(1+\frac{dx}x \right)^{-2}$

For example:

$(a+b)^{-2} = \frac1{(a+b)^2} = \frac1{a^2+2ab+b^2}$

3. Following the rule wouldn't it be,

1.) (x+dx)'2
2.) 1/(x'2) + 1/(2x*dx) + (dx/1)'2
minus insignificant value
3.) 1/x'2 + 2x*dx

In your equation (x*(1+dx/x))'-2 shouldn't it be 1 over x? The rule you used doesn't seem to work for the problem.

4. Originally Posted by Mausoldj
Following the rule wouldn't it be,

1.) (x+dx)'2
2.) 1/(x'2) + 1/(2x*dx) + (dx/1)'2
minus insignificant value
3.) 1/x'2 + 2x*dx

In your equation (x*(1+dx/x))'-2 shouldn't it be 1 over x? The rule you used doesn't seem to work for the problem.
You can continue the first example using the second example:

$(x+dx)^{-2} = \left(x \cdot \left(1+\frac{dx}x \right) \right)^{-2} = x^{-2} \cdot \left(1+\frac{dx}x \right)^{-2}$ $~ = ~$

$\frac1{x^2} \cdot \dfrac1{\left(1+\dfrac{dx}x \right)^{2}} = \frac1{x^2} \cdot \dfrac1{1+\dfrac{2dx}x + \dfrac{(dx)^2}{x^2}}$

But you can't split the last fraction into three different fractions with a single summand as denominator. Remember: The line of a fraction sets automatically parantheses - even though you can't see them.