I'm reading CLRS and there is the following:

When x→0, the approximation of $e^x$ by $1+x$ is quite good: $$e^x=1+𝑥+Θ(𝑥^2)$$

I suppose I understand what means this equation from math perspective and, also, there is an answer in another cs question. But I don't understand some things, so have a few questions.

  1. Why do they use $Θ$ here and why do they use $=$ sign?
  2. Is it possible to explain how the notation here is related to the author's conclusion that $e^x$ is very close to $1 + x$ when $x \rightarrow 0 $?
  3. Also, how is it important here that $x$ tends to $0$ rather than to $\infty$ as we usually use asymptotic notations?

I'm sorry if there are a lot of questions and if they are stupid, I'm just trying to master this topic.

  • 1
    $\begingroup$ Note that the quoted paragraph continues: "In this equation, the asymptotic notation is used to describe the limiting behavior as $x\to 0$ rather than as $x\to \infty$". The way $\Theta$ is used here is completely different from how it is used anywhere else in the book. $\endgroup$ – Tom van der Zanden Jul 4 '19 at 8:26

What the equation means is that there exist constant $A>0$ and $B,C$ such that $$ |x| \leq A \Longrightarrow Bx^2 \leq e^x-1-x \leq Cx^2. $$ In particular, for small $x$, $e^x$ is very close to $1+x$, since the error is only $O(x^2)$ (when $x$ is small, $x^2$ is much smaller than $x$).

The same estimate doesn't hold for large $x$ — indeed, $e^x$ grows faster than $x^2$, indeed faster than any fixed power of $x$. So the quoted estimate only holds for small $x$.

Here is a graphic example. Below is the plot of $\frac{e^x-1-x}{x^2}$ for $|x| \leq 0.1$. You can see that it is very close to $0.5$. This is consistent with the Taylor expansion of $e^x$, which is $1 + x + x^2/2 + O(x^3)$.

enter image description here

In contrast, here is the same plot for $10 \leq x \leq 20$. You can see that the ratio shoots up to infinity.

enter image description here

Credit: plots executed using Desmos.

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  • $\begingroup$ I think I got it. Thank you very much for your detailed answer :) $\endgroup$ – E. Shcherbo Jul 5 '19 at 20:05
  • $\begingroup$ @YuvalFilmus I found you answer great.. Now I felt like clarifying a few points, (1) the authors have used the asymptotic notation in a completely different meaning instead of $n\geq n_0$ they have changed the definition to $n\leq\epsilon$ where $\epsilon$ is a small positive quantity. (2) assuming (1) and as you showed $e^x-x-1$ is tight bounded by $x^2$ for very small $x$. Is this so? $\endgroup$ – Abhishek Ghosh Jun 19 at 16:02
  • $\begingroup$ Right, usually big Theta is with respect to $n \to \infty$, here it is used for $x \to 0$. $\endgroup$ – Yuval Filmus Jun 19 at 16:48
  • $\begingroup$ Taylor's theorem shows that for any $k$, we have $e^x = 1 + x + \cdots + x^k/k! + e^{\theta x} x^{k+1}/(k+1)!$, where $\theta$ is between $0$ and $x$. $\endgroup$ – Yuval Filmus Jun 19 at 16:50

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