"Exponential Madness" (Gauss's challenge)

Question

From Euler's identity, we see that

$e^{i\pi}=-1$

$\Rightarrow e^{2ik\pi}=1$ [squaring both sides].

This equation surely holds for all integers $k$.

EDIT: From the second equation we get $e^{1+i2k\pi}=e$

So, $e^{1+i2k\pi}=\left (e^{1+i2k\pi}\right )^{(1+i2k\pi)}$

and $e=e^{(1+i2k\pi)^2}=e^{1+i4k\pi}e^{-4k^2\pi^2}$ $ ...(*)$

But $e^{i4k\pi}=1$ from Euler's identity and $e^{1+i4k\pi}=e$

So $e=ee^{-4k^2\pi^2}$ and hence $e^{-4k^2\pi^2}=1$ $[$from $(*)$$]$

But this is true only for $k=0$ whereas we started out with a statement true for all $k\in \mathbb{Z}$. What is the catch here?

Answer 1

I think that MrYourMath has isolated one way to see the main issue, which is about multivaluedness of complex functions, but I would like to see if I can bring this point into more explicit contact with your reasoning.

I think the problem is ultimately one about the scope of the definitions and properties you are using. In particular, you are assuming that $(e^a)^b = e^{ab}$ for all $a,b\in\mathbb{C}$, but I think this is only valid if $b\in\mathbb{Z}$.

You begin with deriving $e^{1+2\pi k i} = e$. This is true, but let me take a moment to be careful about what we mean. The left side is, by definition, the value at $z=1+2\pi k i$ of $e^z$. In turn, $e^z$ is, by definition, the limit of the series

$$1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots $$

which is convergent for all $z$ in the complex plane. Thus your equation $e^{1+2\pi k i} = e$ is the statement that when $z$ has the form $1+2\pi k i$, for $k$ an integer, then this series converges to the limit $e$, i.e. the same limit it does when $z=1$.

The critical move in your derivation is raising both sides of the equation $e^{1+2\pi k i} = e$ to the power of $1+2\pi k i$, and then replacing $(e^{1+2\pi i k})^{1+2\pi i k}$ with $e^{(1+2\pi i k)^2}$. I think that the problem lies here. The question is, what does it mean to do this? And once we settle on a meaning, does the familiar fact that $(x^a)^b = x^{ab}$ (which we are used to from the case when $x$ is real and $a,b$ are integers) remain true in this new setting?

So consider the equation $e^{1+2\pi i k} = (e^{1+2\pi i k})^{1+2\pi i k}$. What should we take this equation to be saying? Well, the left side is the limit of the above series at the value $z=1+2\pi i k$ for $k$ some integer. The right side is this same limit, raised to some power. Once we recognize that the limit in the parentheses is $e$, we can use the same series definition to understand what the right side is. I want to point out that we would have a bit of a problem if the limit inside the parentheses had come out to anything except $e$. Because what does $a^b$ mean when $b$ is complex? The usual definition is to take $a^b = e^{(\log a)\cdot b}$, but the problem is that $\log a$ is not single-valued when we work with complex numbers. It can differ by integer multiples of $2\pi i$, which won't affect $e^{\log a}$ but can affect $e^{(\log a)\cdot b}$. So, the right side of your equation would already be ambiguous if it didn't so happen that the parenthetical expression already evaluates to $e$. I say this just to point out that the familiar behavior of exponential expressions is pretty much out the window once we want to involve complex exponents.

But to really get to the point - the above paragraph gives a (grantedly convoluted) definition that makes the right side of the equation $e^{1+2\pi i k} = (e^{1+2\pi i k})^{1+2\pi i k}$ meaningful and makes the equation true. But on this definition, we have no reason to expect the familiar identity $(e^a)^b = e^{ab}$ to hold. In fact, as discussed in the last paragraph, if $a$ is not such that $e^a = e$, and $b$ is complex, it seems to me the left side of this identity cannot be defined unambiguously, while the right side can! So it can't hold in that generality.

If $b$ is an integer, then we can assign $(e^a)^b$ a different meaning. Use the series to define $e^a$, but then just use the repeated multiplication definition to raise to the $b$ power. In this situation, we can prove that $(e^a)^b = e^{ab}$ always, using a calculation with power series (to show that $e^x\cdot e^y = e^{x+y}$) and then induction on the size of $|b|$. But this proof will only work if $b$ is an integer!

Answer 2

NOTE:

$$(x^k)^2 \not = x^{k^2}, \text{ instead} \quad (x^k)^2 = x^{2k}$$

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For the second update note that the summation/ product of exponents is valid only for real exponent. In other words $(e^{1+2ki\pi})^{1+2ki\pi} \not = e^{(1+2ki\pi)^2}$

Answer 3

Great question, it was fun to think of an explanation :D! The problem with your calculations is that you are using multivaluedness in the wrong way.

Your mistake is in the last lines

$e^{1+i2k\pi}=\left (e^{1+i2k\pi}\right )^{(1+i2k\pi)}$

$e=e^{(1+i2k\pi)^2}=e^{1+i4k\pi}e^{-4k^2\pi^2}$

But $e^{i4k\pi}=1$ from Euler's identity and $e^{1+i4k\pi}=e$

So $e=ee^{-4k^2\pi^2}$ and hence $e^{-4k^2\pi^2}=1$.

First you are introducing multivaluedness and then removing it. If you are very careful you have to write $e^{2\pi k i}$ and not $1$ when working with complex numbers.

Imagine a situation where you take the logarithm of one

$\ln 1=0$ and $\ln (e^{2\pi i})=2\pi i$ but clearly $0\neq 2\pi i$. If you use $\ln(e^{2\pi ik})=2\pi i k$ you can see that for $k=0$ you get $0$ and for $k=1$ you get $2\pi i$. So when working with complex numbers you are not allowed to first use multivaluedness and then pretend as it wasn't multivalued.

Answer 4

Thank you for an entertaining and instructive paradox. Most of your humourless commenters have missed the point, because the point involves the use of reason rather than faith in scriptural authority.

The catch is that the only values of $z$ for which $e^z$ has one value are those for which the real part of $z$ is an integer. Thus $$e^{(1-4k^2\pi^2)+4k\pi{i}}$$ is single-valued only when $1-4k^2\pi^2$ is an integer: that is, when $k=0$.

I am not going to quote formulae to prove this, because formula don't provide truth, only reflect it. But here is some motivation.

• $1^{1/1}$ has just one value: $1$.
• $1^{1/2}$ has two values: $1, -1$.
• $1^{1/4}$ has four values: $1, -1, i, i$.
• $1^{1/n}$ has $n$ values.

For irrational exponents $x$, $1^x$ has infinitely many values, one of which is $1$.

As long as a family of values all behave in the same way, this multi-valuedness doesn't much matter. What your paradox has done is point out a sequence of operations where the multiplicity of the values causes trouble if you ignore it. Well done for pointing it out! To be brutal about it, not many people would think that (for example) $1^\pi$ has infinitely many values. (In general they all have the form $e^{2k\pi^2{i}}$ for some integer $k$.