- #1
ILoveParticlePhysics
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How do we know that e^πi= -1 if all numbers here are basically undefined/irrational?
How is Euler's Identity proven using differential equations?
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- Thread starter ILoveParticlePhysics
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In summary, the exponential function can be defined through power series and differential equations. By setting two functions, f(x) and g(x), that satisfy the same differential equation and initial condition, we can prove that e^ix = cosx + isinx. This is known as Euler's identity.
- #2
etotheipi
Hey, I'm not irrational! 
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.
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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- #3
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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.ILoveParticlePhysics said:
- #4
Gaussian97
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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.
$$f(x)=e^{ix}, \qquad g(x)=\cos{x}+i\sin{x}$$
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.
FAQ: How is Euler's Identity proven using differential equations?
How did mathematicians discover that e^πi = -1?
Mathematicians discovered this relationship through the use of complex numbers and the study of exponential functions. In particular, the French mathematician Leonhard Euler is credited with proving this identity in the 18th century.
Can you explain the concept of e^πi and its relationship to -1?
The concept of e^πi involves using the imaginary number i, which represents the square root of -1, in an exponential function. When πi is used as the exponent, it results in a value of -1. This can be proven using Euler's formula: e^(ix) = cos(x) + i*sin(x). When x = π, cos(π) = -1 and sin(π) = 0, resulting in e^(πi) = -1.
How is e^πi related to the unit circle and trigonometric functions?
The unit circle is a circle with a radius of 1, centered at the origin on a Cartesian plane. When e^πi is graphed on the unit circle, it traces a path that corresponds to the cosine and sine functions. This is because of Euler's formula, which relates the exponential function to trigonometric functions.
What are the practical applications of e^πi = -1?
This identity has many practical applications in mathematics, physics, and engineering. It is used in signal processing, quantum mechanics, and electrical engineering, among other fields. It is also used in the study of waves and oscillations, as well as in solving differential equations.
Is e^πi = -1 always true, or are there exceptions?
Yes, e^πi = -1 is always true. This is because it is a mathematical identity, meaning it is true for all values of e and π. It is not an equation that can be solved for a specific value, but rather a relationship that holds true regardless of the values used for e and π.
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