2 ms·
d/dt e^t = e^t. So while vapid and pop-sci, technically true. Of course, the exponent is constant or we'd be talking about the Ackermanity instead.
by sterlind 8y ago
d/dt e^t = e^t. So while vapid and pop-sci, technically true. Of course, the exponent is constant or we'd be talking about the Ackermanity instead.
- grumdan 8y agoTechnically true, but the quote makes it sound as if this were not necessarily the case for some kinds of exponential growth, which is clearly false.