What is greater: e^pi or pi^e?

You can answer almost immediately using calculator, but what if you don’t have it?

I don’t remember the exact origin of this meme, but this situation was almost every time when we played with numbers.

– Propose a number.

– e^π.

This even might appear in usual conversations:

– What is her size?

– e^π

– So small. I prefer π^e.

And a following arguing about what is greater? Until we simply compare results using calculator.

Let’s see how this problem can be solved without using a calc.

Warning: Goal of this article not only to show correct solution to the problem, but provide other solutions (even incorrect!) that can convince somebody.

Origins

π is the mathematical constant. It’s just a number that represents ratio between circle circumference and diameter `C/d`. Approximately, equal to `3.1415`. Get more decimal places for FREE!

e is the harder one. If you sum infinite sequence below, you’ll get `e`. Approximately, equal to `2.7183`. FREE decimal places without registration!

I’m pretty sure that’s more than enough you need to know to solve the problem.

Rough estimate

We can use exponent function on integers, so let’s just round the numbers. `3.1415` becomes `3`, `2.7183` becomes… `3` also.

``````3^3 vs 3^3
27 = 27
``````

Let’s use floor and ceiling functions.

Floor: `floor(e) = 2`, `floor(π) = 3`

``````2^3 vs 3^2
8 < 9
``````

Ceiling: `ceil(e) = 3`, `ceil(π) = 4`

``````3^4 vs 4^3
81 > 64
``````

Two different results depend on round function that we use. Can’t be sure. Need another method for verification.

Calculus

This needs basic understanding of logarithm and derivative

``````e^π vs π^e
``````

Use logarithm property `a < b => log(a) < log(b)`, and take `log` with base `e` (`ln`). Honestly, it’s not logarithm property, but monotonic function property. Who cares?

``````ln(e^π) vs ln(π^e)
``````

Use logarithm property `log_a(a^b) = b`

``````π vs ln(π^e)
``````

Use logarithm property `log_a(b^c) = c * log_a(b)`

``````π vs e * ln(π)
``````

Now, people who know value of `ln(π)` can answer the question. We don’t know. Move `ln(π)` to the left side

``````π/ln(π) vs e
``````

Take a look at expression `π/ln(π)`. Change `π` to `x`. Got `x/ln(x)`. What is it? It’s a function!

Let’s find extrema of this function `f(x)`. Take first derivative.

``````f'(x) = (ln(x) - 1)/(ln^2(x))
``````

We get extrema at the points where derivative equals to zero

``````(ln(x) - 1)/(ln^2(x)) = 0
``````

Ratio number equals to zero if its numerator equals to zero.

``````ln(x) - 1 = 0
``````

Move `1` to the right

``````ln(x) = 1
``````

Logarithm equals to `1` if base equals to argument. Therefore

``````x = e
``````

And if you know `1/ln(x)` behavior, you get that extremum is a minimum of function `f(x)`. Substitute `x` with `e`

``````e/ln(e) = e
``````

Function minimum is `e` where `x = e`. If `x = π`, due to minimum, function value will be greater than `e`. Remember that we do this to get estimate of `π/ln(π)` which was on the left side of expression `π/ln(π) vs e` and if its greater than e, `e^π` greater than `π^e`. Huh. Done.

Note: We skip mathematical proof that extrema `x = e` is minimum. Not hard to show.

Manual computations

Why don’t we perform calculation manually?

Ok, we have `π`, which is `3.1415` or `22/7` its famous simplest ratio approximation. On the other side we have `e`, which is `2.7183`. But I’m not aware of any `e` approximation. Let’s create one from the definition of `e`. Take and sum first 5 elements.

``````1 + 1 + 1/2 + 1/6 + 1/24 = 65/24
``````

Excellent. Now we need just compare `2.7183^(22/7)` and `3.1415^(65/24)`. What can be simpler?

Rational exponent can be expanded as nth-root of some power:

To take n-th root manually we use a special case of Newton’s method. Just one iteration:

Estimate for `e^π`:

``````2.7183^(22/7)
sqrt_7(2.7183^22)
``````

…Multiplying 22 times…

In fact, not all so bad. To calculate `a^22` we need only 6 multiplications:

``````a^1 * a^1 = a^2
a^2 * a^2 = a^4
a^4 * a^4 = a^8
a^8 * a^8 = a^16
a^16 * a^4 = a^20
a^20 * a^2 = a^22
``````

Nth-root we obtain (decimal dropped):

``````sqrt_7(3585440111)
``````

Applying Newton’s method. Select closest root:

``````1^7 = 7 < 3585440111
2^7 = 128 < 3585440111
3^7 = 2187 < 3585440111
...
23^7 = 3404825447 < 3585440111
24^7 = 4586471424 > 3585440111
``````

Good. Most closest value is `23`. Then

``````sqrt_7(3585440111.25)
sqrt_7(23^7 + 180614664)
23 + 180614664/(7 * 23^6)
23 + 180614664/1036251223
23 + 0.17
23.17
``````

And the same for `π^e`:

``````3.1415^(65/24)
sqrt_24(3.1415^65)
``````

This power `65` much larger than `22`. But it needs just 7 multiplications.

We can cheat here. We get rough result `2 * 10^32`. Now we test `23^24` (most closest root from previous) and its value `4.8 * 10^32`. So, definitely nth-root will be lower than `23`. Solved.

Don’t ask how many basic arithmetic operations we performed. A lot.

Liar approach

Actually, it’s offensive to use the word “liar”. There is a whole technique called Mathematical Fallacy. It based on some illegal math transformation (division by zero, roots from negative numbers, etc.) to get any desired result.

This is powerful instrument for mathematicians to show magic for not mathematicians.

Suppose, we don’t know exact value of π. Not a problem, but we know complex numbers. And also the fact that `sqrt(-1) = i`. Where `i` is imaginary unit

We have bad memory and can’t remember π value, but we know Euler’s formula

``````e^(i * x) = cos(x) + i * sin(x)
``````

x is a variable. Take `x = 2π`:

``````e^(2π * i) = cos(2π) + i * sin(2π)
``````

From the school math course we know: `cos(2π) = 1`, `sin(2π) = 0`. Just substitute:

``````e^(2π * i) = 1 + i * 0
``````

Everything multiplied by `0` is `0`. Next step:

``````e^(2π * i) = 1
``````

`a^b` gives `1` if `b = 0`. That we also know from logarithms.

``````2π * i = 0
``````

It means either `2` or `π` or `i` is zero. `2` is not zero. `i` is not zero. `π` we don’t know. Ok, that means `π` equals to `0`.

Returning to comparing our constants:

``````e^0 vs 0^e
``````

Left term is `1`, right is `0`. That means `e^π` greater than `π^e`.

The trick that we used hard to spot. Look carefully again through the proof.

The fallacy was where we implicitly took a logarithm and said that `2π * i = 0`. The number `e^(2π * i)` is not real and we not allowed to logarithm it. But if person who listens you don’t know this fact… This is an excellent proof.

Programming approach

You not allowed to use calculator. Nothing said about programming language.

One-liner in Java:

``````System.out.println(Math.pow(Math.E, Math.PI) > Math.pow(Math.PI, Math.E));
``````

Fun approach

Derived from joke:

Scientific lab wasted a lot of money for calculating decimal places of π. They come up with idea, let’s π will be equal to 4, to save money.

– But why 4 instead of 3?

– Because 3 reserved for e.

That way `e^π` equal to `81`, and `π^e` equal to `64`. `81 > 64`. Simple, concise way.

Conclusion

There are much more fun about these irrational bastards. Math is fun.

If you were asked question of such type on interview, 99% that interviewer is joking and want to test your reaction. Last solution would be good. But if he or she really expect solution with using calculus, nothing to lose.

06 November 2012