# Useful Bounds via Taylor's Theorem

*Posted: 2017-10-06.*

In this post, we'll take a look at some common and useful bounds on the exponential and logarithm functions and see how they're derived using Taylor's Theorem.

Specifically, we'll show bounds like, for all $x \geq 0$, \begin{align*} 1-x &\leq e^{-x} \leq 1-x+\frac{x^2}{2} \\ x-\frac{x^2}{2} &\leq \ln(1+x) \leq x , \end{align*} and arbitrarily-tighter versions.

## The Workhorse

The key tool we're going to use is Taylor's Theorem, which deals with approximating a function by a polynomial made up of its derivatives of various orders.

**Taylor's Theorem (one useful form).**
Let $f: \mathbb{R} \to \mathbb{R}$ be $k+1$-times differentiable with all derivatives continuous.
Fix $x_0$ and let
\begin{align*}
P_k(x) &= f(x_0) + f'(x_0)(x-x_0) + \ldots + \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k \\
R_k(\xi;x) &= \frac{f^{(k+1)}(\xi)}{(k+1)!}(x-x_0)^{k+1}.
\end{align*}
Then
$$ f(x) = P_k(x) + R_k(\xi;x) $$
for some $\xi$ lying between $x_0$ and $x$.

*Note.* $P_k(x)$ is interpreted as an *approximation* to $f$, while $R_k(\xi;x)$ is some "remainder".

More traditionally, this is taught as useful when bounds can be given on $|R_k(\xi,x)|$, as this implies that $P_k(x)$ is a good approximation to $f$.
Here, we have a similar motivation but slightly different approach.
Our goal will be to find **upper or lower bounds** on $f$, though of course we would prefer that they be reasonably close.

**Corollary.**
If $R_k(\xi;x) \leq 0$ for all $\xi$ between $x_0$ and $x$, then $f(x) \leq P_k(x)$.
Symmetrically, if $R_k(\xi;x) \geq 0$ for all $\xi$ between $x_0$ and $x$, then $f(x) \geq P_k(x)$.

## The Exponential Function

Let's consider the function $e^x$, with Taylor series around $x_0=0$ of \[ 1 + x + \frac{x^2}{2} + \cdots + \frac{x^k}{k!} + \cdots \] as well as $e^{-x}$, which has \[ 1 - x + \frac{x^2}{2} - \cdots + (-1)^k\frac{x^k}{k!} + \cdots . \]

The first thing to figure out, when applying a bound or approximation, is the *regime* you're interested in.
If you're dealing with $x=1000$, $x=10000$, or similar, then $e^x$ is just big -- what else do you want to say? (In this post, the bounds we derive will actually hold for arbitrarily large $x$, but they won't be very meaningful for $x$ bigger than $1$.)

But for small $x$, say $x \approx 0$, the function $e^x$ isn't so scary and it makes lots of sense to try to upper-bound or lower-bound it by more prosaic functions that might be easier to deal with.

$1+x$ is not such a great approximation to $e^x$ on a large range, but is when $x \approx 0$:

Similarly for $1-x+\frac{x^2}{2}$ and $e^{-x}$:

Note that our previous choice of $x_0 = 0$ makes sense since we're approximating or bounding our functions near zero. For $e^x$, we have $R_k(\xi;x) = \frac{1}{(k+1)!}e^{\xi} x^{k+1}$. Notice that for $k$ odd, $R_k(\xi;x) \geq 0$ for all values of $x$ and $\xi$. Given the corollary, this immediately says:

**Bounds for $e^x$:**

**Fact.** For all $x$, $1+x \leq e^x$.

**Fact.** For all $x$ and all odd $k$, $\sum_{j=0}^k \frac{x^j}{j!} \leq e^x$.

(Note this is not immediately obvious when $x$ is negative, at least, not to me.)

The bound for $1+x$ is often used with probabilities as follows: Suppose we have $n$ independent trials each with a probability $p$ of failure. Then the chance that they all succeed is the product of the probabilities: \begin{align*} (1-p)^n &\leq \left(e^{-p}\right)^n \\ &= e^{-pn}. \end{align*}

Speaking of smooth segues, sometimes in the above scenario we want a bound going the other way, i.e. to say that $e^{-x}$ isn't much bigger than $1-x$, at least near zero. For $e^{-x}$, we have $R_k(\xi;x) = (-1)^{k+1}\frac{1}{(k+1)!}e^{-\xi}x^{k+1}$. Focusing on positive $x$, we get $R_k(\xi;x)$ is positive (negative) when $k$ is odd (even).

**Bounds for $e^{-x}$:**

**Fact.** For all $x \geq 0$,
\[ 1 - x \leq e^{-x} \leq 1 - x + \frac{x^2}{2} , \]
and more tightly
\[ 1 - x + \frac{x^2}{2} - \frac{x^3}{6} \leq e^{-x} \leq 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} , \]
and so on.

## Logs

Our other example for today is $\ln(1+x)$, which has Taylor series \[ x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots + (-1)^{k-1} \frac{x^k}{k} + \cdots . \] Again, we're interested in the regime where $x$ is close to zero, and this is the Taylor series around $x_0=0$; notice that the $k$th derivative of $\ln(1+\xi)$ is $(-1)^{k-1}\frac{(k-1)!}{(1+\xi)^k}$.

Here $R_k(\xi;x) = (-1)^{k}\frac{1}{(k+1)(1+\xi)^{k+1}} x^{k+1}$. Focusing on positive $x$ for simplicity (which implies $\xi$ is positive as well), we get $R_k(\xi;x)$ is negative for odd $k$ and positive for even $k$. So odd terms give upper bounds, and even terms give lower bounds.

**Bounds for $\ln(1+x)$:**

**Fact.** For all $x \geq 0$,
\[ x - \frac{x^2}{2} \leq \ln(1+x) \leq x , \]
and more tightly
\[ x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} \leq \ln(1+x) \leq x - \frac{x^2}{2} + \frac{x^3}{3} , \]
and so on.

*Exercise.* Show that the upper bounds on $\ln(1+x)$ above hold for all $x \gt -1$.

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