It's very difficult to predict the impedance of an end-fed wire, other than to say it's high. Usually it's determined empirically.
You are looking for a theoretical formulation. Consider, the feedpoint is a voltage source which makes a difference in electric potential between to things. The end of the dipole, and...what?
simulate this circuit – Schematic created using CircuitLab
Maybe you could imagine the feedpoint connected to a theoretical shell of infinite radius, similar to how self-capacitance is calculated? I'd guess the result depends strongly on the geometry of the wire, but generally the thicker the wire, the lower the impedance. I don't know of what practical value this model would be since any real antenna has at least a feedline and a ground/aircraft/spacetraft nearby which would be more significant.
You can also calculate the impedance of an off-center fed dipole, very close to the end. You'll notice the limit of that function as the feedpoint approaches the end is infinity. This is of course an approximation assuming the dipole is relatively thin, and that the capacitance to the other half of the dipole is the most relevant factor.
So if there isn't another half of the dipole, what is the relevant factor? In practice it's going to be the ground, and the feedline. Neither is amenable to a simple expression. Simulation of your particular installation is your best bet.
We can make some generalizations though:
An end-fed dipole is resonant (or not) just like a more ordinary center-fed dipole. The feedpoint (in the middle, near the end, or somewhere between) is transparent to the resonance of the wire.
An exactly half-wave dipole isn't resonant in the sense that its feedpoint impedance has a reactive component. Theoretically, (73 + j42.5)Ω.
When the dipole is too short, its reactance will be capacitive. When it's too long, inductive. Since the exactly half-wave dipole has a slightly inductive reactance, shortening it can eliminate the reactance. The exact amount of shortening depends on the thickness of the wire. 0.41λ sounds like a reasonable estimate.
The feedpoint impedances repeat with every wavelength of length. That is, in terms of feedpoint impedance, 0.5λ, 1.5λ, 2.5λ, ... dipoles all look the same.
As the dipole becomes thicker, its bandwidth increases. That means for an equal change in length, the impedance of a thicker dipole will change less than a thinner dipole.
Since the impedance must repeat with every wavelength, this also means that the highest impedance (for example, at the ends of a half-wave dipole) is lower with a thicker wire.
Finally, a practical point: since the impedance at the end of a dipole is very high, the common-mode impedance looks relatively low. Successfully making an end-fed dipole then depends on very effective choking (a very high common-mode impedance), which is difficult to realize in practice. If the choking isn't very effective (meaning, the common mode impedance isn't much higher than the differential mode), then making the choking more effective will increase the impedance seen by the transmitter.