What is the significance of 'junk' in the binomial theorem for derivatives?

[techieadmin]

I've started listening to the lectures for the MIT OpenCourseWare 18.01 Single Variable Calculus class. I understood all of it up until the teacher found the derivative of xⁿ. Here's what he wrote on the board:

$$\frac{d}{dx} x^{n} = \frac{\Delta f}{\Delta x} = \frac{(x+\Delta x)^{n} - x^{n}}{\Delta x}$$

That, I understand. Then we get to the binomial theorem to try to simplify $$(x+\Delta x)^{n}$$. The professor said that $$(x + \Delta x)$$ is multiplied by itself n times, which I understand. Then he wrote:

$$(x+\Delta x)^{n} = x^{n} + nx^{n-1}\Delta x + junk$$
$$junk = O((\Delta x)^{2})$$ ("big O of delta x squared")

What I'm confused about is the last line. Why is that what "junk" equals? I understand that the rest of the terms don't matter as $$\Delta x$$ approaches 0, but why are they equal to what he says they are equal to "big O of delta x squared," as he said?

 

[tiny-tim]

Big O

$$(x+\Delta x)^{n} = x^{n} + nx^{n-1}\Delta x + junk$$
$$junk = O((\Delta x)^{2})$$ ("big O of delta x squared")

What I'm confused about is the last line. Why is that what "junk" equals? I understand that the rest of the terms don't matter as $$\Delta x$$ approaches 0, but why are they equal to what he says they are equal to "big O of delta x squared," as he said?

Hi techieadmin!

"= O((∆x)²)" is shorthand for "is of the order of (∆x)²" …

for more details, see http://en.wikipedia.org/wiki/Big_O_notation

 

[HallsofIvy]

Well, it's exactly what your title implies- the binomial theorem.

$$(x+ y)^n= \sum_{i=0} \left(\begin{array}{c}n \\ i\end{array}\right)x^{n-i}y^i$$
When i= 0, $\left(\begin{array}{c}n \\ 0\end{array}\right)$ is 1 so the i=0 term is $x^n$. When i= 1, $\left(\begin{array}{c}n \\ 1\end{array}\right)$ is n so the i= 1 term is $nx^{n-1}y$. So
$$(x+ y)^n= x^n+ n x^{n-1}y+ \sum_{i= 2}^n\left(\begin{array}{c}n \\ i\end{array}\right)x^{n-i}y^i$$
It is that last sum that is the "junk" referred to.

f(x)= O(g(x)) means that f(x) and g(x) go to the same limit "at about the same rate" as x goes to some value- specifically, that f(x)/g(x) has a no-zero finite limit.

Here, the "junk" involves powers of $\Delta x$ of degree 2 and higher: you could write it as
$$\left(\begin{array}{c}n \\ 2\end{array}\right)x^{n-2}(\Delta x)^2+ "other junk"$$
where the "other junk" are the remaining terms: involving powers of $\Delta x$ of degree 3 and higher. Dividing that by $(\Delta x)^2$ will give $\left(\begin{array}{c}n \\ i\end{array}\right)x^{n-2}$ plus terms that still involve $\Delta x$ which will go to 0 as $\Delta x$ goes to 0 (which is, of course, the important value for a derivative).

 

[techieadmin]

So the "junk" eventually approaches some number which is multiplied by $$\Delta x^{2}$$ and since $$\Delta x$$ goes to 0, the "junk" eventually disappears?