3 ms·
I'm late to the party. But let me offer first the finite-differences linear equation A[n]=A[n-1]*k For A[0] = 1, the solution is the sequence 1, k, k
by thanatropism 7y ago
I'm late to the party. But let me offer first the finite-differences linear equation
A[n]=A[n-1]*k
For A[0] = 1, the solution is the sequence
1, k, k^2, k^3, k^4, k^5, k^6....
If we let k be the complex number i, this becomes
1, i, -1, -i, 1, i, -1, -i, 1...
As you can see, the introduction of imaginary parts extends the original idea of powers-of-k (which if real go exponentially to infinity if k>1 and to 0 if k<1) to an alternating oscillatory pattern both in the real parts
1, 0, -1, 0, 1, 0, -1, 0, 1
and the imaginary parts. What you're seeing there is almost the celebrated relationship between imaginary exponentials and sine functions.
Now: that was the finite-differences linear equation. If instead we take the differential linear equation
f'(x) = kf(x)
we get the exponential function. Again, if k=i, you get oscillatory behavior.