Comment Comment by qnleigh Uuuuuum no?e^(ix) = cos(x) + isin(x). In particular e^(ipi) = -1(1 + 1/n)^n = e. This is part of what makes e such a uniquely useful exponent base.Not applied enough? What about:d/dx e^x = e^x. This makes e show up in the solutions of all kinds of differential equations, which are used in physics, engineering, chemistry...The Fourier transform is defined as integral e^(iomega*t) f(t) dt.And you can't just get rid of e by changing base, because you would have to use log base e to do so.Edit: how do you escape equations here? Lots of the text in my comment is getting formatted as italics. parent qnleigh 10 months ago
e^(ix) = cos(x) + isin(x). In particular e^(ipi) = -1
(1 + 1/n)^n = e. This is part of what makes e such a uniquely useful exponent base.
Not applied enough? What about:
d/dx e^x = e^x. This makes e show up in the solutions of all kinds of differential equations, which are used in physics, engineering, chemistry...
The Fourier transform is defined as integral e^(iomega*t) f(t) dt.
And you can't just get rid of e by changing base, because you would have to use log base e to do so.
Edit: how do you escape equations here? Lots of the text in my comment is getting formatted as italics.