I have heard that said, and it sounds vaguely plausible, but my maths is not up to proving it. Here is why: The Ideal n-th order Butterworth response, in terms of S-parameters, is $$ |s21|^2 = \frac{1}{1 + \omega^{2n}} $$ On the other hand, the Chebychev ideal resonse is $$ |s21|^2 = \frac{1}{1 + \epsilon^2 T_n^{2}(\omega)} $$ As I see it, as \$\epsilon\$ approaches zero, \$|s21|^2\$ approaches one identically. Is my initial assumption wrong, or is there some mathematical magic that can come up with a different form of the limit for Chebychev with \$\epsilon\$ approaching zero?
For example, I can imagine the (semi-)circular pole-zero plot for Butterworth "morphing" from a circle into an ellipse, but I can't relate that to the transfer function.
[EDIT] On thinking about this more, perhaps the comparison can only be made when the cutoff specifications for the two types are identical (3db frequency, 1db etc). In this case, the "cutoff frequency" for zero ripple Chebychev would be \$\infty\$, making a direct comparison impossible.
I think this just needs a restatement of the Butterworth power transfer function as follows $$ |s21|^2 = \frac{1}{1 + k\omega^{2n}} $$ to allow a cutoff value other than 3db and enable the direct comparison where appropriate.