I How is Euler's Identity proven using differential equations?

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Euler's Identity, expressed as e^πi = -1, can be proven using differential equations by defining the function f(x) = e^(ix) and g(x) = cos(x) + i sin(x). Both functions satisfy the same differential equation, f'(x) = i f(x) and g'(x) = i g(x), along with the initial condition f(0) = g(0) = 1. By the uniqueness of solutions to differential equations (Picard's theorem), f(x) and g(x) must be identical. Evaluating both functions at x = π confirms that f(π) = g(π), leading to the conclusion that e^πi = -1. This proof illustrates the relationship between complex exponentials and trigonometric functions.
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How do we know that e^πi= -1 if all numbers here are basically undefined/irrational?
 
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Hey, I'm not irrational! :smile: Anyway, start from$$e^{iz} = \sum_{k=0}^{\infty} \frac{i^k z^k}{k!} $$and split this sum into odd and even terms. See if you can get it in terms of sines and coses.
 
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ILoveParticlePhysics said:
How do we know that e^πi= -1 if all numbers here are basically undefined/irrational?
Generalisations of the exponential function can usually be defined via the power series method above. That's how we can also define the exponential of a matrix.
 
In case you are more familiar with differential equations, there's a proof of the fact that ##e^{ix}=\cos{x}+i\sin{x}## that it's really nice. If we define
$$f(x)=e^{ix}, \qquad g(x)=\cos{x}+i\sin{x}$$
It is really easy to see that both satisfy the differential equation
$$f'(x) = i f(x), \qquad g'(x) = i g(x)$$
Furthermore, both satisfy the initial condition
$$f(0)=g(0)=1$$
Because ##f(0)=e^{0}=1##.
Then, since these two functions satisfy the same differential equation with the same initial condition, by the uniqueness of the solutions (Picard's theorem) they must be the same function. QED

Then just evaluate ##f(\pi)=g(\pi)## to obtain Euler's identity.
 
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