# Basic Derivation with negative exponents

1. Jul 9, 2009

### fyamnky

So I'm trying to teach mysel calculus with Thompson's Calculus made easy.
I'm still in the very early stages and I got stuck with negative exponents.
I'd basically like an explanation fof what's going on here.

http://www.flickr.com/photos/26906498@N03/3705112800/ [Broken]

I've tried using online sources like:
http://www.purplemath.com/modules/exponent.htm
http://www.algebrahelp.com/lessons/simplifying/negativeexponents/

I also looked up Binomial Theorem on wikipedia which just confused me more.
http://en.wikipedia.org/wiki/Binomial_theorem

Last edited by a moderator: May 4, 2017
2. Jul 9, 2009

### rock.freak667

Well basically, the binomial theorem is valid for only n being an integer such that

$$(a+b)^n = b^n +^nC_1 b^{n-1}a + ^nC_2 b^{n-2}a^2 +...+ ^nC_r b^{n-r}a^r + a^n$$

But if you use the definition of nCr being n!/(n-r)!r!. Then terms like nC1 works out as

$$^nC_1 = \frac{n!}{(n-1)!1!} = \frac{n(n-1)!}{(n-1)!}=n$$

$$^nC_2 = \frac{n!}{(n-2)!1!} = \frac{n(n-1)(n-2)!}{(n-1)!2!}= \frac{n(n-1)}{2}$$

So (a+b)n simplifies to this

$$(a+b)^b = b^n + nb^{n-1}a + \frac{n(n-1)}{2!}b^{n-2}a^2 + \frac{n(n-1)(n-2)}{3!}b^{n-3}a^3 + ...$$

so if ever you put n as a negative number or a fractional number then you'd have an infinite number of terms which is valid for |b/a| < 1.

so in the chapter, the expansion is valid for

$$\left | \frac{dx}{x} \right | < 1 \Rightarrow -1< \frac{dx}{x}<1$$

or simply dx/x is small, that is why they neglected the terms after x-3

EDIT: Normally you won't need to do the y+dy = x +dx thing always, you can just usually apply y=xn => dy/dx=nxn-1

3. Jul 9, 2009

### HallsofIvy

Staff Emeritus
rockfreak667, if you go down far enough on the wikipedia page cited, you find "Newton's Generalized Binomial Formula" which does NOT require positive integers.

For any non-negative integers, n and m, $n\ge m$
$$\left(\begin{array}{c}n \\ m\end{array}\right)= \frac{n!}{m!(n-m)!}= \frac{n(n-1)(n-2)\cdot\cdot\codt\(n-m+1)}{m(m-1)(m-2)\cdot\cdot\cdot(3)(2)(1)}$$
In that final form, we can, in fact, write it out for n and m any numbers, not just non-negative integers. Of course,
[tex]\left(\begin{array}{c} n \\ n\end{array}\right)= 1[/itex]
and
[tex]\left(\begin{array}{c}n \\ n-1\end{array}\right)= n[/itex]
for any n, non-negative integer or not. Thus, we can write
(x+ h)n= (1)xn+ (n)x{n-1}h+ terms involving h2 or higher, for any number n, positive integer or not. Then, (x+h)n- xn= n xn-1h+ terms involving h2 or higher and, dividing by h, [(x+h)n- xn]/h= n xn-1+ terms involving h or powers of h. Taking the limit as h goes to 0, all those "terms involving h or powers of h" go to 0 leaving only nxn-1.

4. Jul 9, 2009

### rock.freak667

That's how I learned the binomial theorem, first with n being a positive integer and then the generalized form for fractions and negative numbers. Most likely due to working out 50C34 and such we just used a calculator instead of writing it out...things like -3C1 on a calculator would make me go