Recall that: $ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots $

Therefore, $ \forall x > 0 $, $ e^x > 1 + x $.

Substituting $x = \frac{\pi}{e} - 1$ in the above inequality, we get:

$ e^{\frac{\pi}{e} - 1} > \frac{\pi}{e}

\implies e^{\frac{\pi}{e}} > \pi

\implies e^{\pi} > \pi^{e} $

Therefore, $ \forall x > 0 $, $ e^x > 1 + x $.

Substituting $x = \frac{\pi}{e} - 1$ in the above inequality, we get:

$ e^{\frac{\pi}{e} - 1} > \frac{\pi}{e}

\implies e^{\frac{\pi}{e}} > \pi

\implies e^{\pi} > \pi^{e} $

thanks

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