Characterization of Euler's Number by Inequality
Theorem
Let $a$ be a (strictly) positive real number.
Then:
- $a = e \iff \forall x \in \R: a^x \ge x + 1$
where $e$ denotes Euler's number.
Proof
Forward Implication
Proved in Exponential Function Inequality.
$\Box$
Reverse Implication
Consider $\map {f_a} x = a^x - x - 1$.
Then we need to prove:
- $a = e \impliedby \forall x \in \R: \map {f_a} x \ge 0$
By Linear Combination of Derivatives, Derivative of Power of Constant, Derivative of Identity Function, and Derivative of Constant:
- $\map {f'_a} x = a^x \ln a - 1$
By Linear Combination of Derivatives, Derivative of Power of Constant, Derivative of Constant Multiple, and Derivative of Constant:
- $\map {f_a} x = a^x \paren {\ln a}^2$
Now, we divide the theorem into two cases:
Case 1: $0 < a \le 1$
Consider $x = 1$.
Then:
| \(\ds \map {f_a} 1\) | \(=\) | \(\ds a^1 - 1 - 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds a - 2\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds 1 - 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -1\) | ||||||||||||
| \(\ds \) | \(<\) | \(\ds 0\) |
Thus the right hand side is not true.
$\Box$
Case 2: $a > 1$
We have:
| \(\ds \ln a\) | \(>\) | \(\ds 0\) | Logarithm of 1 is 0 and Logarithm is Strictly Increasing | |||||||||||
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \paren {\ln a}^2\) | \(>\) | \(\ds 0\) | Square of Non-Zero Real Number is Strictly Positive | |||||||||
| \(\text {(2)}: \quad\) | \(\ds a^x\) | \(>\) | \(\ds 0\) | Power of Positive Real Number is Positive over Reals | ||||||||||
| \(\ds \vdash \ \ \) | \(\ds a^x \paren {\ln a}^2\) | \(>\) | \(\ds 0\) | Real Number Axiom $\R \text O2$: Usual Ordering is Compatible with Multiplication, using $(1)$ and $(2)$ | ||||||||||
| \(\ds \map {f_a} x\) | \(>\) | \(\ds 0\) |
Hence, by Real Function with Strictly Positive Second Derivative is Strictly Convex, we have that $f_a$ is strictly convex.
Trying to find its minimum, we solve:
- $\map {f'_a} x = 0$
Therefore:
| \(\ds a^x \ln a - 1\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds a^x \ln a\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds a^x\) | \(=\) | \(\ds \frac 1 {\ln a}\) | ||||||||||||
| \(\ds x\) | \(=\) | \(\ds \map {\log_a} {\frac 1 {\ln a} }\) | ||||||||||||
| \(\ds x\) | \(=\) | \(\ds - \map {\log_a} {\ln a}\) |
At this point:
| \(\ds \map {f_a} x\) | \(=\) | \(\ds \map {f_a} {- \map {\log_a} {\ln a} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {\ln a} + \map {\log_a} {\ln a} - 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {\ln a} + \frac {\map \ln {\ln a} } {\ln a} - 1\) | Change of Base of Logarithm | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {1 + \map \ln {\ln a} } {\ln a} - 1\) |
It is to be noted that when $a = e$, the minimum is:
- $\dfrac {1 + \map \ln {\ln e} } {\ln e} - 1 = 0$
meaning that the only solution to $\map {f_e} x = 0$ is $x = 0$.
Also, from Exponential Function Inequality, we have:
- $\forall x \in \R: e^x \ge x + 1$
Substituting $x = \map \ln {\ln a}$:
- $\ln a \ge \map \ln {\ln a} + 1$
From Logarithm is Strictly Increasing and Logarithm of 1 is 0:
- $\ln a > 0$
Thus:
- $\dfrac {\ln a} {\ln a} \ge \dfrac {\map \ln {\ln a} + 1} {\ln a}$
Therefore:
- $\dfrac {1 + \map \ln {\ln a} } {\ln a} - 1 \le 0$
Then, we solve the case of equality:
| \(\ds \frac {1 + \map \ln {\ln a} } {\ln a} - 1\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \frac {1 + \map \ln {\ln a} } {\ln a}\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds 1 + \map \ln {\ln a}\) | \(=\) | \(\ds \ln a\) |
From earlier, we have that the only solution is $\map \ln {\ln a} = 0$, when $a = e$.
Therefore, for other values of $a$, the minimum is negative.
$\blacksquare$