Posted August 5, 2013 By Presh Talwalkar. Read about me , or email me .

You could solve today’s puzzle with a calculator, but that is missing the point. Can you figure it out without doing a numerical computation?

Which quantity is greater: eπ or πe?

(I thank Mahesh for telling me about this delightful problem.).

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"All will be well if you use your mind for your decisions, and mind only your decisions." Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon. .

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Answer to e^pi versus pi^e

The answer is eπ is larger. There are several ways you can solve this problem. I’ve listed a few proofs below. I feel the first proofs are the most intuitive methods to approach the problem, but they are relatively involved proofs.

Method 5 is by far the most elegant but had me wondering “how the heck did they think of that?” Be sure to read method 5.

Also, I suggest reading method 6 as it’s a proof I developed after seeing the answer in method 5. The slight difference is my proof could be presented in pre-calculus.

Full disclosure: I will not pretend I solved this problem on my own. I tried and I could not figure out the answer. All of these are proofs I read and spent time understanding so I could explain them here.

Method 1: comparing ex to xe

In one expression, the value of e is in the base and in the other it is in the exponent. One way to approach the problem is to compare the two quantities.

Define a function

f(x) = ex – xe

We can evaluate this function at the values of interest.

f(e) = ee – ee = 0 f(π) = eπ – πe

We are basically interested in finding out if f(π) is positive or negative. We know that f(e) is equal to zero, so we might get an idea of whether the function is increasing or decreasing by considering its derivative. So we calculate.

f(x) = ex – xe f'(x) = ex – exe-1

Now comes a bit of mathematical reasoning. From this website: “We observe that f‘(1) = 0 and f‘(e) = 0, so 1 and e are the two points of intersection. The exponential function [e^x] grows to infinity faster than the power function [x^e], so f'(x) > 0 for x > e.”

Since f'(x) is positive for x > e, this implies that f(x) is increasing for values x > e.

Since π > e, we have f(π) > f(e) = 0. This will lead to the result.

f(π) = eπ – πe > 0 hence eπ > πe

Method 2: π to e(ln(π))

The natural logarithm is a monotonically increasing function. This means we can take the natural log of two expressions and compare the size of the resulting values. Thus we can compare:

ln(eπ) = π ln(πe) = e ln(π)

Our attack will be similar to method 1. Notice that in the above expressions, the value of π appears once as its own value and once as an argument to the natural log function. We can create the following function to compare the quantities.

f(x) = x – e ln(x) Note that f(e) = e – e ln(e) = 0 f(π) = π – e ln(π) = ??

If we can figure out whether f(π) is positive or negative, we can solve our problem. We proceed by looking at the derivative.

f'(x) = 1 – e/x When y > e, we will have e/y < 1, so f'(y) > 0

We have just reasoned that f'(x) is positive for values larger than e, therefore f(x) will be increasing for values larger than e, such as π.

We can finish the proof

f(π) = π – e ln(π) > 0 This implies π > e ln(π) And using each side as an exponent to the base e, we get eπ > πe

Method 3: consider π/ln(π)

This is very similar to method 2 that I found on Ask Dr. Math. When we take the natural logarithms of both sides we get the following.

ln(eπ) = π ln(πe) = e ln(π)

Let’s say that we divide the first equation by the second. Then we end up with

π/(e ln π) = π/(ln π) * (1/e)

We now have an equation that involves the ratio of π to the natural log of π. As in the first two methods, we will define a function that will yield light on the size of the quantities involved.

Let f(x) = x/(ln x), then we find the derivative (skipping a few steps) f'(x) = (-1+ ln x)/(ln x)2 f'(x) = 0 when x = e and f'(x) > 0 when x > e and f'(x) < 0 when x > e

Therefore we have found that f(x) has a local minimum at x = e. So we can conclude:

f(e) < f(π) e/ln(e) < π/(ln(π)) e < π/(ln(π)) e ln(π) < π And using each side as an exponent to the base e, we get πe < eπ

Method 4: exponentiating to the reciprocal

This is the most popular answer on Math Stackexchange.

Let f(x) = x1/x Differentiating (by taking the natural log of both sides and using implicit differentiation and then doing a lot of algebra), this cleans up to be the expression f'(x) = x1/x(1/x2)(1 – ln x)

Clearly f'(e) = 0. We can also do a bit of reasoning to conclude this is a global maximum.

(When x < e, all terms in the derivative are positive so the function is increasing. But when x > e the first two quantities are positive but (1 – ln x) will be negative so the function in increasing. In other words, the function increases up to x = e and then decreases, so x = e is a global maximum).

We use this to derive the result.

f(e) = e1/e > f(π) = π1/π Now we raise both sides to the power of e to get e1 > πe/π Now we raise both sides to the power of π eπ > πe

Method 5: Taylor series

This is a solution I found on Math Stackexchange. We start with the Taylor series expansion of ex

ex = 1 + x + x2/2! + x3/3! + … For positive values of x, we know the partial sum has to be less than the infinite sum, so we can conclude the partial sum of just the first two terms will be less than the entire sum: ex > 1 + x

Believe it or not, this is all you really need to know in this elegant proof! We will pick a clever choice of value for x, namely x = π/e – 1. When we substitute we find that

eπ/e – 1 > (π/e – 1) + 1 = π/e We can re-write this as the following since e-1 = (1/e) (1/e) eπ/e > π/e Now we can cancel out the (1/e) on both sides eπ/e > π Now we raise both sides to the power of e eπ > πe

And magically we are done!

Method 6: my own proof

I came up with this after seeing the steps used in method 5.

Let’s say we have a bank account that offers an interest rate π/e – 1. If the bank compounds the interest once per year, then in one year a $1 investment in the bank account will result in the balance:

End of year = 1 + (π/e – 1) = π/e

What if the bank compounds the interest n times per year? We know two things: first, that compounding more will end up with more money than just compounding once. Second, we can write out the balance by using the compound interest formula.

Compounding n times > Compounding 1 time Compounding n times > π/e (1 + (π/e – 1)/n)n > π/e

Now we use a trick where we compound an infinite number of times per year, and we let the limit of n go to infinity. It is a known result that when compounding an interest rate x, the continuous compounding result is ex. In our case we have x = π/e – 1, so we get the following.

Compounding infinitely many times > Compounding 1 time e(π/e – 1) > π/e

The rest of the proof follows the steps in method 5. The steps are to: (a) cancel the 1/e term on both sides, and (b) raise both sides to the power of e. The resulting expression is the answer.

eπ > πe

As you can see this method is pretty much like method 5, but the advantage is the result is derived using methods taught through “pre-calculus” (as Taylor series is definitely a calculus topic).

Just for the record

eπ = 23.14069…

πe = 22.45915…

So you could do this on a calculator, but it’s oh so much more fun to solve it using other methods 😉