# Question about Taylor series and big Oh notation

1. Jun 19, 2009

### AxiomOfChoice

Question about Taylor series and "big Oh" notation

Can someone please explain WHY it's true that

$$e^x = 1 + x + \frac{x^2}{2} + \mathcal{O}(x^3)$$

I'm somewhat familiar with "big Oh" notation and what it stands for, but I'm not quite sure why the above statement is true (or statements like it). Thanks!

2. Jun 19, 2009

### Pengwuino

Re: Question about Taylor series and "big Oh" notation

I know big O notation goes beyond what it's used for here but in this case, it means the remaining terms in the taylor expansion of that exponential are of the order x^3 and higher.

3. Jun 19, 2009

### AxiomOfChoice

Re: Question about Taylor series and "big Oh" notation

Given what "big Oh" notation usually means, I think that, more precisely, it means that the absolute value of the difference between $e^x$ and the first few terms of the Taylor series is bounded by some constant times $|x^3|$ for $x$ sufficiently close to zero.

I, too, used to think that $\mathcal{O}(x^3)$ was just a stand-in for the sentence "terms of order $x^3$ and higher." But that's not all it says, apparently.

4. Jun 19, 2009

### Pengwuino

Re: Question about Taylor series and "big Oh" notation

Yah it says something more but I first learned it in our computer science courses which were simply god awful (god awful department as well) so we never knew what it meant beyond "of order and higher".

5. Jun 20, 2009

### Dragonfall

Re: Question about Taylor series and "big Oh" notation

This is only true if x<1. It means that e^x is 1+x+x/2 plus a function that is bounded above by a Kx^3 for some K, for sufficiently small x.

If x>1, you'd have to write $$e^x = 1 + x + \frac{x^2}{2} + \Omega(x^3)$$

Which means the same except a function bounded BELOW by Kx^3, for sufficiently large x.

Check out http://en.wikipedia.org/wiki/Big-oh#The_family_of_Bachmann.E2.80.93Landau_notations

Last edited: Jun 20, 2009