Derivatives that cycle?

We are looking for solutions to y^(n) = y.

==> y^(n) - y = 0, a homogeneous linear differential equation of order n.

This has characteristic equation r^n - 1 = 0.

==> r = ω^k, where ω = e^(2πi/n) and k = 0, 1, 2, ..., n-1.

[These are the n- nth roots of unity.]

So, the general solution is y = C₁ e^x + C₂ e^(ωx) + ... + C_n e^(ω^(n-1) x).

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Let's look at a few special cases:

(i) n = 1 ==> y' = y.

This has general solution y = Ce^x (as we already know).

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(ii) n = 2 ==> y'' = y. (Note that the square roots of unity are -1 and 1.)

So, this has general solution y = C₁ e^x + C₂ e^(-x).

(That is, e^x and e^(-x) have cycle length 2.)

However, since sinh x = (1/2)(e^x - e^(-x)) and cosh x = (1/2)(e^x + e^(-x)),

we are just rewriting the general solution above in terms of sinh x and cosh x:

y = A cosh x + B sinh x.

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(iii) n = 3 ==> y''' = y. (Note that the cube roots of unity are 1, (-1 ± i√3)/2)

So, this has general solution y = C₁ e^x + C₂ e^((-1 + i√3)x/2) + C₃ e^((-1 - i√3)x/2).

(In particular, e^((-1 ± i√3)x/2) have cycle length 3.)

However, note that by Euler's Identity

e^((-1 ± i√3)x/2)

= e^(-x/2) * e^(±ix√3/2)

= e^(-x/2) * [cos(x√3/2) ± i sin(x√3/2)]

= e^(-x/2) cos(x√3/2) ± i e^(-x/2) sin(x√3/2)

So, we can rewrite the general solution as

y = Ae^x + Be^(-x/2) cos(x√3/2) + Ce^(-x/2) sin(x√3/2).

So, e^(-x/2) cos(x√3/2) and e^(-x/2) sin(x√3/2) are real function which cycles in 3.

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(iv) n = 4 ==> y'''' = y. (Note that the 4th roots of unity are 1, -1, i, and -i.)

So, this has general solution y = C₁ e^x + C₂ e^(-x) + C₃ e^(ix) + C₄ e^(-ix).

(In particular, e^(ix) and e^(-ix) have cycle length 4.)

However, since sin x = (1/(2i))(e^(ix) - e^(-ix)) and cos x = (1/2)(e^(ix) + e^(-ix)), along with the hyperbolic functions, we can rewriting the general solution as

y = A cosh x + B sinh x + C cos x + D sin x.

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We can keep looking at this for n = 5, 6, ... for more functions of this special type.

(n = 6, 8, 12 have nice answers, too.)

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Another approach is to use power series:

If y = Σ(k = 0 to ∞) x^(nk+r)/(nk+r)! for any fixed r = 0, 1, 2, ..., n-1,

then y^(n) = y as well.

In fact, one can check that

Σ(k = 0 to ∞) x^(nk)/(nk)! = (1/n)(e^x + e^(ωx) + e^(ω^2 x) + ... + e^(ω^(n-1) x)).

Its derivatives will give the other n-1 series above.

(These series give another way of expressing the general solution to y^(n) = y

but with real coefficients as the imaginary portions of the exponentials cancel out.)

I hope this helps!

As you know, the e^x cycles in a pattern of 1.

e^-x is is compound function (more specifically a horizontal reflection) involving e^x.

sin(x), cos(x) as well as sinh(x) and cosh(x), can all be written as a function involving e^x.

Example: sin(x) = (e^ix - e^-ix) / 2i

Notice that the involvement of e^ix, and of e^-ix, and of i, is what makes it a four-cycle pattern.

sinh(x) = (e^x - e^-x) / 2 is a two-pattern.

All because of the properties of e^x, and hence there is nothing unique about sin(x) that is not already accounted for by e^x.

One could conceivable construct any number of functions involving the sums and differences of e^ax, for various values of 'a', to create unique cycling patterns.

Notice that f(x) = 0 is the most trivial case of a 1-pattern-cycle.

I would also like to introduce you to a new topic called Fractional Calculus, if you dont already know about it. Imagine taking a half-derivative. Or a third-derivative. Such that it takes two half-derivatives, or three third-derivatives, to create a full derivative. Interesting idea? By taking half-derivatives, you double the cycling pattern... or third-derivatives, you triple the pattern... e^x can be third-differentiated three times in order to arrive back at the original function. For sin(x), it would take twelve third-derivatives.