Is this a research-level question...? Or did this site devolve?

@TerryTao: May I ask what the rationale for the implication is? I ask because I just learned about a differential equation that appears to be similarly "universal" (in the sense that any continuous function can be approximated by some solution of it arbitrarily well), but nevertheless autonomous (the equation being 3x'⁴x''x''''² − 4x'⁴x'''²x'''' + 6x'³x''²x'''x'''' + 24x'²x''⁴x'''' − 12x'³x''x'''³ − 29x'²x''³x'''² + 12x''⁷ = 0).

What I find mindblowing about the linearity of expectation is that it still holds when the variables are neither independent nor identically distributed. For the other cases it's way more plausible.

I"m a little confused (I'm not familiar with the technique), but I don't immediately see how you integrated this (or what you even integrated with respect to). Did you integrate with respect to $y$? How did you integrate $y''$ with respect to $y$...?

@SteveHuntsman: Ah, what I was having trouble reconciling it with was the Table-Maker's Dilemma... could you explain why that isn't a problem here? As you can see, the article says, "Nobody knows how much it would cost to compute $y^w$ correctly rounded for every two floating-point arguments at which it does not over/underflow."...

@SteveHuntsman: Is this for integer exponentials or does it work for anything? The reason I asked was that I could swear I'd come across a statement that said it is unknown how many bits of precision you'd need to compute an exponential to $n$ bits of accuracy, but I can't find it right now...

What's the time complexity of the scalar exponential?

@DimaPasechnik: Ah okay I see, thanks!

Question -- you say "there are examples of SDPs for which every solution needs exponential space", but earlier you say P = NP would imply that convex optimization can be solved in polynomial time. How does this make sense? Is convex optimization even in NP? And if so, how can an SDP then require exponential space?

I mean I don't understand how you can define a Taylor series when the function doesn't have derivatives. That's why I asked how you found the series, because it couldn't have been by taking derivatives...

But I don't understand, isn't the 3rd term in the 2nd order Taylor series based on the 2nd derivative of the function? How can you say it has a Taylor series when the definition requires its derivatives?

What is the second-order Taylor series for this function?