# Why do Maclaurin series approximate a function for negative domain values?

A common analogy used as an intuitive explanation for a Maclaurin series is that of a car. If you know the position, velocity, acceleration, jerk etc. of a car at time zero, you are able to predict its position at any time after that moment. Simarly, a Maclaurin has the same derivative, second derivative etc. as the function, which is why it approximates the function and can be used to find the range value of a function at any domain value.

Knowing the position, velocity, acceleration, jerk etc. of a car at time zero allows you to figure out its position at any time after time zero. It tells you nothing about its position before time zero. Similarity, shouldn't a Maclaurin series approximate a function for all $x > 0$? Why does it approximate the function for negative domain values?

I would appreciate it if answers made reference to the analogy of the car. Thank you.

• I don't feel this analogy is apt. Some Maclaurin series do not converge to the actual function. – George V. Williams Apr 14 '13 at 0:20
• @GeorgeV.Williams Hey George, I wasn't aware of that. Could you please provide an example? Thanks. – user70518 Apr 14 '13 at 0:22
• Consider a function $g(x)$. Let $g(x) = e^{-1/x^2}$ if $x\neq 0$, and $g(x) = 0$ if $x = 0$. Then the Maclaurin series of $g(x) = 0$, but the function is not actually $0$. (This is a classical example.) – George V. Williams Apr 14 '13 at 0:26

## 1 Answer

Yes it does. Let's for the time being, be crude and just use velocity.

$t$ amount of time after time 0, the position is roughly position at time $t = \text{position at 0} + t \cdot \text{velocity}$.

This is because we assume, velocity does not change much for small value of $t$. Then, we can turn the argument around to $\text{position at time } {-t} = \text{position at 0} + t \cdot (-\text{velocity})$

This is because if you are walking with a constant speed down a road, I would know where you were 2 minutes ago. Anyhow, while this does serve as a good intuition (it is the first time I have heard of it, and I think it is good), do not use this a rule for 'which value of $x$ should this work'.