07/17/2010, 01:50 AM
Hmm. We have
\( f(z) = f(z - 1) + f(z + i) \)
If we assume a solution of the form \( f(z) = r^z \) exists, like in the Fibonacci numbers, we can get the equation
\( r^z = r^{z-1} + r^{z+i} \)
Dividing both sides by \( r^{z-1} \) gives
\( r = 1 + r^{1+i} \)
or
\( r^{1+i} - r = -1 \).
There is a caveat, however: \( r^{1+i} \) is multivalued. It is more useful, then, to recast this equation in terms of \( u = \log( r ) \),
\( e^{(1+i)u} - e^u = -1 \)
and then the solutions for the functional equation are given by \( f(z) = e^{uz} \) for any \( u \)-value satisfying the above exponential equation. The functional equation is linear, so any linear combination of such solutions will be another solution, and since there are infinitely many such \( u \)-values, we can even consider infinite sums
\( f(z) = \sum_{n=0}^{\infty} C_n e^{u_n z} \)
with arbitrary \( C_n \), provided this sum converges. Since there are infinitely many constants \( C_n \), one could say the equation is like it has "infinitely many initial conditions".
This graph shows the function \( e^{(1+i)u} - e^u + 1 \) on the complex plane. You can see the roots, the values of \( u \) used to construct solutions of the functional equation. The scale runs between \( \pm30 \) on each axis (x = real, y = imag). One set of roots seems to lie along the line \( t + it \), while the other seems to lie along a slight curve (curving not visible here) that is asymptotic to the imaginary axis \( it \).
For example, we could take two terms with the roots given by \( u_0 \approx 0.5397851608092811048455891598 + 0.5397851608092811048455891598i \) and \( u_1 \approx -1.453673666461041618684343568 - 1.453673666461041618684343568i \) and coefficients \( C_0 = C_1 = 1 \). This function is plotted below at the same scale. Numerical calculation can be done to verify it really does solve \( f(z) = f(z-1) + f(z+i) \). I do not believe there is a closed form solution for these \( u \)-values in terms of any conventional special functions, but I could be wrong (and if I am, I'd like to know what the closed solution is.).
Note that these may not be the only possible solutions -- remember that the very simple case \( f(z) = f(z-1) \) has all 1-periodic functions as solutions. Not sure what the appropriate analogy is here. But the above could be thought of as a sort of "canonical" solution like how Binet's formula solves the Fibonacci numbers.
\( f(z) = f(z - 1) + f(z + i) \)
If we assume a solution of the form \( f(z) = r^z \) exists, like in the Fibonacci numbers, we can get the equation
\( r^z = r^{z-1} + r^{z+i} \)
Dividing both sides by \( r^{z-1} \) gives
\( r = 1 + r^{1+i} \)
or
\( r^{1+i} - r = -1 \).
There is a caveat, however: \( r^{1+i} \) is multivalued. It is more useful, then, to recast this equation in terms of \( u = \log( r ) \),
\( e^{(1+i)u} - e^u = -1 \)
and then the solutions for the functional equation are given by \( f(z) = e^{uz} \) for any \( u \)-value satisfying the above exponential equation. The functional equation is linear, so any linear combination of such solutions will be another solution, and since there are infinitely many such \( u \)-values, we can even consider infinite sums
\( f(z) = \sum_{n=0}^{\infty} C_n e^{u_n z} \)
with arbitrary \( C_n \), provided this sum converges. Since there are infinitely many constants \( C_n \), one could say the equation is like it has "infinitely many initial conditions".
This graph shows the function \( e^{(1+i)u} - e^u + 1 \) on the complex plane. You can see the roots, the values of \( u \) used to construct solutions of the functional equation. The scale runs between \( \pm30 \) on each axis (x = real, y = imag). One set of roots seems to lie along the line \( t + it \), while the other seems to lie along a slight curve (curving not visible here) that is asymptotic to the imaginary axis \( it \).
For example, we could take two terms with the roots given by \( u_0 \approx 0.5397851608092811048455891598 + 0.5397851608092811048455891598i \) and \( u_1 \approx -1.453673666461041618684343568 - 1.453673666461041618684343568i \) and coefficients \( C_0 = C_1 = 1 \). This function is plotted below at the same scale. Numerical calculation can be done to verify it really does solve \( f(z) = f(z-1) + f(z+i) \). I do not believe there is a closed form solution for these \( u \)-values in terms of any conventional special functions, but I could be wrong (and if I am, I'd like to know what the closed solution is.).
Note that these may not be the only possible solutions -- remember that the very simple case \( f(z) = f(z-1) \) has all 1-periodic functions as solutions. Not sure what the appropriate analogy is here. But the above could be thought of as a sort of "canonical" solution like how Binet's formula solves the Fibonacci numbers.