Is $e^{\pi}$ greater than $\pi^{e}$?

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    2015
In summary, $e^{\pi}$ is greater than $\pi^{e}$, as shown by the proof that $e^{\pi} > 1 + \pi > 1 + e > \pi^{e}$. This number is significant because it is a transcendental number and has important applications in mathematics and science. The inequality $e^{\pi} > \pi^{e}$ cannot be reversed, and the value of $e^{\pi}$ is larger than other famous mathematical constants.
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Ackbach
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Here is this week's POTW:

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Prove that $e^{\pi}>\pi^{e}$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to Rido12 for his correct solution! It is posted below.

To prove $e^{\pi}>\pi^e$, let us first generalize it: $e^x>x^e \iff x>e \ln x \iff \frac{x}{\ln x}-e>0$.

Consider the function $f(x)=\frac{x}{\ln x}-e$ with derivative $f'(x)=\frac{\ln x -1}{\ln^2 x}$. The critical values of $f$ are at $x=1$ and $x=e$, and $f'(x)<0$ for $x \in (0,e)$ and $f'(x)>0$ for $x>e$. The function obtains a minimum at $x=e$ and hence is monotonically increasing for $x>e$.
Since $f(x)>f(e)$ for $x>e$ and $\pi>e$, then $f(\pi)>f(e)=0$.
\(\displaystyle f(\pi)>0 \iff \frac{\pi}{\ln \pi}>e\iff e^{\pi}>\pi^e$$ \), as required.
 

FAQ: Is $e^{\pi}$ greater than $\pi^{e}$?

Is $e^{\pi}$ greater than $\pi^{e}$?

Yes, $e^{\pi}$ is greater than $\pi^{e}$.

How can you prove that $e^{\pi}$ is greater than $\pi^{e}$?

The proof involves using the fact that $e^x$ is always greater than $1 + x$ for any real number $x$. By substituting $x = \pi$, we get $e^{\pi} > 1 + \pi$. Similarly, $\pi^e < 1 + e$. Therefore, $e^{\pi} > 1 + \pi > 1 + e > \pi^{e}$, proving that $e^{\pi}$ is greater than $\pi^{e}$.

What is the significance of the number $e^{\pi}$?

The number $e^{\pi}$ is significant because it is a transcendental number, meaning it cannot be expressed as a finite algebraic combination of integers and the constant $e$. It also has important applications in mathematics and science, such as in complex analysis, number theory, and physics.

Can the inequality $e^{\pi} > \pi^{e}$ be reversed?

No, the inequality $e^{\pi} > \pi^{e}$ cannot be reversed. This can be proven by using the same logic as in the proof for the original inequality. By substituting $x = e$, we get $e^{e} > 1 + e$, and $\pi^{e} < 1 + \pi$. Therefore, $e^{e} > 1 + e > 1 + \pi > \pi^{e}$, showing that the inequality cannot be reversed.

How does the value of $e^{\pi}$ compare to other famous mathematical constants?

The value of $e^{\pi}$ is approximately 23.14069, which is greater than the value of other famous mathematical constants such as $\sqrt{2} \approx 1.41421$, $\pi \approx 3.14159$, and $e \approx 2.71828$. This shows that $e^{\pi}$ is a relatively large number in comparison to other well-known constants.

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