# Big-O Notation in Geometric Interpretation of Derivative

• Jan 25th 2010, 02:12 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 $\displaystyle f(x)$ can be defined as:

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

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

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

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

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

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

Thanks! (Happy)
• Jan 25th 2010, 04:23 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 $\displaystyle f(x)$ can be defined as:

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

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

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

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

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

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

Thanks! (Happy)

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

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

The remaining terms are replaced by $\displaystyle O(\Delta x^2)$ 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, 05:07 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 $\displaystyle \Delta x^2$ chosen in the first place? Why not $\displaystyle \Delta x^3$? I mean, how is the statement (in your reply) deduced?
• Jan 25th 2010, 07:52 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 $\displaystyle \Delta x^2$ chosen in the first place? Why not $\displaystyle \Delta x^3$? 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 $\displaystyle \Delta x^2$ which for small $\displaystyle \Delta x$ dominates the other omitted terms.

CB
• Jan 25th 2010, 02:04 PM
fishcake
Quote:

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

CB

Ah, now I understand. I got it the other way round, since I was considering for large $\displaystyle \Delta x$ and was wondering why it's not $\displaystyle \Delta x^n$. Thanks for the help! (Happy)
• Jan 25th 2010, 08:39 PM
CaptainBlack
Quote:

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

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

CB