3 ms·
I understand Taylor expansion and that we're not talking rigorous maths here. However, what I don't get in your example is the implied connection between the t
by aix1 4y ago
I understand Taylor expansion and that we're not talking rigorous maths here.
However, what I don't get in your example is the implied connection between the timescale and the order of the Taylor polynomial. Could you elaborate?
- j7ake 4y agoTruncating the Taylor expansion after the first term says exponential y = exp(t) can be approximated as y_approx = 1 which is a constant. This is true if you zoom in around t=0 far enough. Then as you zoom out, you have to include higher order terms to account for the exponential increase.
- froj 4y agoThe whole point of exp(t) is that the derivative at t is equal to the value at t. Thus if y_approx = 1 then the derivative is also 1 and not 0 (i.e. constant). If you're exponentially growing, but currently it's approximately constant, then you're at t = -inf and you'll probably be dead by the time you achieve something significant.
- deleted 4y ago[deleted]
- j7ake 4y agoYeah the constant is a poor approximation of the exponential. Maybe the analogy works better if I start at the second term . Zeroth order approximations are rarely useful because the dont capture the local gradient. But in the life analogy, there are people who assume where they are in life will be same in future (eg the “new” normal during pandemic). So maybe it’s not so terrible after all.
- froj 4y agoWell, 0th order approximations are fine if the gradient is indeed about zero for all intents and purposes. It's just very much not the case for exponential functions. Maybe your analogy could work if you want to say that exponential growth might feel like linear growth locally?