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There might be an incorrect formula mentioned in the article Kronecker delta on Wikipedia.

The formula I'm talking about is this:

$\delta_{nm}=\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{k=1}^Ne^{2\pi i\frac{k}{N}\left(n-m\right)}$

After I simplified this formula (the right side), because I saw that there was $e^{\pi i}$ which could be turned into $-1$, it came out so that $\delta_{nm}=1$ which is obviously incorrect. So I'd like you guys to look into it just in case I'm wrong.

ssvv
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    The exponent also contains $2 \cdot \pi \cdot i$, so $e^{2 \cdot \pi \cdot i} = 1$ and $1^{z} = 1$. The problem here is that for complex $z_{1}$ and $z_{2}$ it is not always true that $e^{z_{1} \cdot z_{2}} = \left( e^{z_{2}} \right)^{z_{2}}$ applies. – Kevin Dietrich Mar 10 '24 at 11:33
  • Please see MSE Question 3219025. Identities of powers that hold over real numbers do not generally hold over the complex numbers. In particular, $(x^a)^b=x^{(a\cdot b)}$ does not generally hold in the complex numbers. – Jam Mar 17 '24 at 09:58

2 Answers2

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The formula is fine, the limit is the limit of a Riemann sum, and it equals $$ \int_0^1e^{2\pi i t(n-m)}\,dt, $$ which indeed gives $\delta_{mm}$.

It is not true in general, for $z,w\in\mathbb C$, that $e^{zw}=(e^z)^w$.

Martin Argerami
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It is like the classic example $$ i=i^1=i^{\frac{4}{4}}=(i^4)^{\frac{1}{4}}=(1)^{\frac{1}{4}}=1 $$ Whilst actually $$ i^{\frac{4}{4}}\neq(i^4)^{\frac{1}{4}} $$ It matters on the $\gcd$ of the two terms. For example: $$ i^{\frac{3}{4}}=(i^3)^{\frac{1}{4}} $$ Since $\gcd(3,4) = 1 $, but $\gcd(4,4)=4$.

Masd
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