3 ms·
That's really freaky, though no more so than that for DE f'(x) = f(x), the solution is e^x The self-similarity in these two DEs is why Euler's formula works.
by hyperpallium2 5y ago
That's really freaky, though no more so than that for DE f'(x) = f(x), the solution is e^x
The self-similarity in these two DEs is why Euler's formula works. Are there any other such DEs?
- contravariant 5y agoIn general any such linear differential equation f'(t) = M f(t) can be solved as follows f(t) = e^Mt f(0) where M is a matrix and f(0) is a vector, and the matrix exponential is defined using the Taylor series of e^x. Something like f''(t) = -f(t) can be split up as follows: f(t) = h'(t) h'(t) = -f(t) which corresponds to the matrix M = [[ 0 1] [-1 0]] which not coincidentally is a matrix representation of 'i' (it satisfies M^2 = -I), so e^Mt basically generates the same values as e^it (except as 2d vectors instead of complex numbers). Since the above is also the differential equation for constant rotation in the 2D plane this is another way of deriving the relation between the complex exponential and the trigonometric functions.