Unjustified upper bound of Taylor expansion of exponential

Question

It's simple to see that, for $x>0$, the exponential function is lower bounded by its truncated Taylor expansion:

$$e^x = \sum_{j=0}^\infty \frac{x^j}{j!} $$

If we truncate this at $k< \infty$, we are getting a lower bound, as we are just summing over positive quantities. This gives us:

$$\sum_{j=0}^k \frac{x^j}{j!} < e^x $$

However, in this paper at page 3, after equation 2.4, I have found the following bound. Let $\epsilon \in (0,1]$, and $a \in [-1,1]$

$$e^{\epsilon a/2} \leq 1+ \frac{\epsilon a}{2} + \frac{\epsilon^2}{6}$$

This works numerically, but I don't have any intuition for the reason it works. Let's analyze the last term, which is the most troublesome one. They obviously removed the $a$ term from the "Taylor" expansion. However it's not clear how they played with the $/2$ at the denominator. The right term I expect from Taylor would be $$\frac{(\epsilon a /2)^2}{2!} = \frac{(\epsilon a)^2}{4*2!} = \frac{(\epsilon a)^2}{8} < \frac{(\epsilon )^2}{8} < \frac{(\epsilon)^2}{6}$$

What this bound is saying is that it bounds the remaining terms of the Taylor expansion centered in 0 of $e^{\epsilon a /2}$ $$\frac{\epsilon}{6} \geq \sum_{k=0}^\infty \frac{1}{k!}(\epsilon a/2)^k $$

From here I don t know how to proceed. I wish I could apply the convergence criteria of the gometric series, but there is $1/k!$ at the denominator that prevents me from doing it, and I don t have other ideas.

Numerically, I also checked if this works also removing the $/2$ at the denominator of the exponent. To my surprise, also the bound they propose don't work anymore.

epsilon = 1
ra=np.linspace(-1, 1, 3000)

one, two, three, four = [], [], [], []
for a in ra:
    one.append(    np.exp(epsilon*a) <= 1 + epsilon*a + (epsilon)**2/3 )  # original 
    two.append(    np.exp(epsilon*a) <= 1 + epsilon*a + (epsilon*a)**2/3 )  # taylor but with 6
    three.append(  np.exp(epsilon*a) <= 1 + epsilon*a + (epsilon*a)**2/2 )  # taylor right
    four.append(  np.exp(epsilon*a) <= 1 + epsilon*a + (epsilon*a)**2/2 )  # taylor right


print("orig \t\t\t", np.all(one))
print("taylor but with 6 \t", np.all(two))
print("taylor right \t\t", np.all(three))
print("taylor but with 4 \t\t", np.all(four))

Can you help me please understand the idea behind this step? Thank you.

Answer 1

It seems that looking at the remainder of the first-degree Taylor polynomial is not good enough to get your inequality. Looking at the second-degree polynomial seems to be barely enough to get your bound.

The Lagrange form of the remainder term $R(x)$ in $e^x = 1 + x + \frac{x^2}{2} + R(x)$ is $R(x) = \frac{1}{6} e^\xi x^3$ where $\xi$ is between $0$ and $x$.

Plugging in $x=\epsilon a / 2$ yields $e^{\epsilon a / 2} = 1 + \frac{\epsilon a}{2} + \frac{\epsilon^2 a^2}{8} + \frac{1}{48} e^{\xi} \epsilon^3 a^3$ for some $\xi$ between $0$ and $\epsilon a/2$.

It now suffices to show $$\frac{\epsilon^2 a^2}{8} + \frac{1}{48} e^{\xi} \epsilon^3 a^3 \le \frac{\epsilon^2}{6}.$$ Or equivalently, it suffices to show $$\frac{a^2}{8} + \frac{1}{48} e^{\xi} \epsilon a^3 \le \frac{1}{6}.$$ This holds because $\xi \le \epsilon |a|/2 \le 1/2$ and so the left-hand side can be bounded as $$\frac{a^2}{8} + \frac{1}{48} e^{\xi} \epsilon a^3 \le \frac{1}{8} + \frac{1}{48} e^{1/2}$$ which is just barely smaller than $1/6$.

Answer 2

The goal is to find $c$ such that if $|x| \leq 1$ then $\left | e^{x/2} - 1 - (x/2) \right | \leq cx^2$. (This whole mess with $\epsilon$ and $a$ is really just obfuscating that.)

From the Lagrange remainder, you have

$$\left | e^{x/2} - 1 - {x/2} \right | = \left | \sum_{k=2}^n \frac{1}{k!} \frac{x^k}{2^k} + \frac{1}{(n+1)!} e^{\xi/2} \frac{x^{n+1}}{2^{n+1}} \right |$$

where $\xi$ is between $0$ and $x$ and $n \geq 1$. (With $n=1$ the first term on the RHS is just $0$ since it is an empty sum.) So you can write

$$\left | e^{x/2} - 1 - {x/2} \right | \leq \sum_{k=2}^n \frac{1}{k!} \frac{|x|^k}{2^k} + \frac{1}{(n+1)!} e^{|x|/2} \frac{|x|^{n+1}}{2^{n+1}}.$$

You can try to increase $n$ until this works out as you want it to. With $n=1$ you have $\frac{1}{8} e^{|x|/2} x^2$ but $\frac{e^{1/2}}{8}>1/6$ so this is not good enough for your goal.

With $n=2$ you end up looking at $\frac{1}{8} x^2 + \frac{e^{1/2}}{48} |x|^3$, so it suffices to show $\frac{e^{1/2}}{48} \leq \frac{1}{6}-\frac{1}{8}=\frac{1}{24}$ i.e. $e^{1/2} \leq 2$. This follows from the same kind of argument, just rearrange $e^{1/2} \leq 1 + \frac{1}{2} e^{1/2}$.

As an aside, if $x \geq 0$ is small enough that $x^{n+1} \leq (n+1)!$, then:

$$\sum_{k=0}^n \frac{x^k}{k!} \leq e^x \leq \frac{\sum_{k=0}^n \frac{x^k}{k!}}{1-\frac{x^{n+1}}{(n+1)!}}.$$

One convenient sufficient condition for this is $x \leq \frac{n+1}{e}$.