# Is e^pi is bigger than pi^e ?

“Is e^pi is bigger than pi^e ?” is a question every maths nerd, that the comparison between e^pi and pi^e. We can calculate using the calculator, but most of the peoples don’t know the exact reason behind it. In this article, we will see both the rough proof and the conceptual reason behind it. Many of the quora users proved it in the right way but it would be difficult for most of the students who are looking for much easier proof. So let’s have a match between e^{\pi} vs \pi^{e} .

## Is e^pi is bigger than pi^e?

e^{\pi} vs \pi^{e} ?

As we know:
e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{2}}{3 !}+\cdots

Here, e^{x} \geq 1+x for all x \geqslant 0 Equality only holds for x=0.
Otherwire, e^{x}>1+x for all \neq 0.
Let, take x=\frac{\pi}{e}-1 Here x \neq 0, because if x=0 \Rightarrow \pi=e (Not Possible)
Put x=\frac{\pi}{e}-1 in the equation
e^{x}>1+x, \quad \forall x \neq 0 . We get e^{\frac{\pi}{2}-1}>\frac{\pi}{e} e^{\pi / e} \cdot e^{-1}>\frac{\pi}{e} \frac{e^{\frac{\pi}{2}}}{e}>\frac{\pi}{e} \Rightarrow e^{\pi / e}>\pi \Rightarrow\left(e^{\pi / e}\right)^{e}>\pi^{e} \Rightarrow e^{\pi}>\pi^{e}

Here is one more conceptual proof of Is e^pi is bigger than pi^e?

Let, f(x)=x^{1 / x} ; x \in(0, \infty) Taking derivatives:
\begin{aligned} &\frac{1}{f(x)} f^{\prime}(x)=\frac{1}{x} \cdot \frac{1}{x}-\frac{1}{x^{2}} \log x \\ &f^{\prime}(x)=f(x) \frac{1}{x^{2}}(1-\log x) \\ &f^{\prime}(x)=\frac{x^{1 / x}}{x^{2}}(1-\log x)\end{aligned} \text { Critical Points: } \quad f^{\prime}(x)=0 \begin{aligned} &\frac{x^{\frac{1}{x}}}{x^{2}}(1-\log x)=0 \\ &1-\log x=0 \end{aligned} x=e is the critical points.

At x=e f^{\prime \prime}(e)<0 (which is negative)
Here, f^{\prime}(x)>0, \forall x \in(0, e) (since \log x<0 \quad \forall x \in(0, e)).
\Rightarrow f(x) is increasing in (0, e].
We know that:
If x_{1}<x_{2} then f\left(x_{1}\right)<f\left(x_{2}\right) \forall x_{1}, x_{2} \in(0, e].
Taking; e, \pi \in[e, \infty).

As e<\pi \Rightarrow f(e)>f(\pi) . \Rightarrow e^{\frac{1}{e}}>\pi^{\frac{1}{\pi}} Taking e\pipower on both sides.
\Rightarrow\left(e^{\frac{1}{e}}\right)^{e \pi}>\left(\pi^{\frac{1}{\pi}}\right)^{e \pi} \Rightarrow e^{\pi}>\pi^{e}

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