# Thread: Big-O Notation in Geometric Interpretation of Derivative

1. ## Big-O Notation in Geometric Interpretation of Derivative

I just watched the first lecture of , 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!

2. Originally Posted by fishcake
I just watched the first lecture of , 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!
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

3. Thanks for the reply! But it still doesn't answer my second question (or maybe I'm missing something ).
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?

4. Originally Posted by fishcake
Thanks for the reply! But it still doesn't answer my second question (or maybe I'm missing something ).
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

5. 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!

6. 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!
By the way all of your limits should be $\Delta x \to 0$ not $x \to \Delta x$

CB