Which is bigger: π^e or e^π?

The question of whether \pi^e or e^{\pi} is larger is a classic one.  Of course, with a calculator it is easy to see what the answer is.  But how would you answer the question without a calculator?

There are lots of interesting ways to tackle the question.  For this (short) post I would like to highlight one I saw on math.SE, given by Aryabhata.  We need the assumptions that \pi > e and e^x > x+1 for x > 0.  (One way to view the latter claim is that, for x > 0, e^x is larger than the first two terms of its Maclaurin series.  Since all the rest of the terms of the Maclaurin series are positive, and the Maclaurin series converges to e^x, we get that e^x > x+1 for x > 0.)

On to the proof.  Since \pi > e, \pi/e - 1 > 0.  Applying the second assumption, then, we get \displaystyle e^{\pi/e-1} > \pi/e.  This means that \displaystyle e^{\pi/e} > \pi, which in turn implies that \displaystyle e^{\pi} > \pi^e.

Advertisements
This entry was posted in real analysis. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s