# Big-O Notation in Geometric Interpretation of Derivative

• Jan 25th 2010, 03:12 AM
fishcake
Big-O Notation in Geometric Interpretation of Derivative
I just watched the first lecture of MIT 18.01 Single Variable Calculus, Fall 2006, and in that video, the lecturer explained about the geometric interpretation of derivative. A derivative of a function $f(x)$ can be defined as:

$f'(x) = \lim_{x \to \Delta x} \frac{f(x+\Delta x) - f(x)}{\Delta x}$

The lecturer then used this to find the derivative of the function $f(x) = x^n$, and this is where it starts to get confusing for me. Using the definition above, the derivate of the function $f(x) = x^n$ is equivalent to:

$f'(x) = \lim_{x \to \Delta x} \frac{(x+\Delta x)^n - x^n}{\Delta x}$

The part which confuses me is when the term $(x+\Delta x)^n$ is evaluated using the Binomial Theorem:

$(x+\Delta x)^n = x^n + n(\Delta x)x^{n-1}+\mathit{O}((\Delta x)^2)$

Why are the rest of the terms replaced by the Big-O Notation $\mathit{O}((\Delta x)^2)$? And why the term $(\Delta x)^2$ specifically?

Thanks! (Happy)
• Jan 25th 2010, 05:23 AM
CaptainBlack
Quote:

Originally Posted by fishcake
I just watched the first lecture of MIT 18.01 Single Variable Calculus, Fall 2006, and in that video, the lecturer explained about the geometric interpretation of derivative. A derivative of a function $f(x)$ can be defined as:

$f'(x) = \lim_{x \to \Delta x} \frac{f(x+\Delta x) - f(x)}{\Delta x}$

The lecturer then used this to find the derivative of the function $f(x) = x^n$, and this is where it starts to get confusing for me. Using the definition above, the derivate of the function $f(x) = x^n$ is equivalent to:

$f'(x) = \lim_{x \to \Delta x} \frac{(x+\Delta x)^n - x^n}{\Delta x}$

The part which confuses me is when the term $(x+\Delta x)^n$ is evaluated using the Binomial Theorem:

$(x+\Delta x)^n = x^n + n(\Delta x)x^{n-1}+\mathit{O}((\Delta x)^2)$

Why are the rest of the terms replaced by the Big-O Notation $\mathit{O}((\Delta x)^2)$? And why the term $(\Delta x)^2$ specifically?

Thanks! (Happy)

What big-O means in this case is that there exists a positive constant $K$ (which may depend on $x$)such that for $|\Delta x|$ small enough:

$
\left|(x+\Delta x)^n - [x^n + n(\Delta x)x^{n-1}]\right| < K |\Delta x^2|
$

The remaining terms are replaced by $O(\Delta x^2)$ because of the existance of the bound that this implies, and that is all thet is needed for the rest of the argument. That is it saves us writing out a lot of detailed stuff that is not needed.

CB
• Jan 25th 2010, 06:07 AM
fishcake
Thanks for the reply! But it still doesn't answer my second question (or maybe I'm missing something (Thinking)).
Why is the function $\Delta x^2$ chosen in the first place? Why not $\Delta x^3$? I mean, how is the statement (in your reply) deduced?
• Jan 25th 2010, 08:52 AM
CaptainBlack
Quote:

Originally Posted by fishcake
Thanks for the reply! But it still doesn't answer my second question (or maybe I'm missing something (Thinking)).
Why is the function $\Delta x^2$ chosen in the first place? Why not $\Delta x^3$? I mean, how is the statement (in your reply) deduced?

By looking at the form of the omitted terms, one of which is a multiple of $\Delta x^2$ which for small $\Delta x$ dominates the other omitted terms.

CB
• Jan 25th 2010, 03:04 PM
fishcake
Quote:

Originally Posted by CaptainBlack
By looking at the form of the omitted terms, one of which is a multiple of $\Delta x^2$ which for small $\Delta x$ dominates the other omitted terms.

CB

Ah, now I understand. I got it the other way round, since I was considering for large $\Delta x$ and was wondering why it's not $\Delta x^n$. Thanks for the help! (Happy)
• Jan 25th 2010, 09:39 PM
CaptainBlack
Quote:

Originally Posted by fishcake
Ah, now I understand. I got it the other way round, since I was considering for large $\Delta x$ and was wondering why it's not $\Delta x^n$. Thanks for the help! (Happy)

By the way all of your limits should be $\Delta x \to 0$ not $x \to \Delta x$

CB