Alex Nelson
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 Jan24 comment Metric space not a vector space @SanathDevalapurkar: well, yes and no. Yes, because tangential thought is always helpful in speculative reasoning. Ostensibly, if you were interested in this direction, you would then ask about inverses and automorphisms, then about the diffeomorphism group's structure. That's a good approach, since the "distinct" Riemannian metrics would be $\mathbf{Riem}/\mathbf{Diff}$ (the space of all Riemann metrics modulo the diffeomorphism group). But it won't help coerce $\mathbf{Riem}$ into a vector space :( Jan24 comment Metric space not a vector space @SanathDevalapurkar: You're talking about matrix multiplication? That's kind of tricky, there's a lot of nuances to that question. But remember the flat Riemannian manifold has its metric be (in suitable coordinates) the identity matrix...but that won't help considering the space of metrics as-a vector space (or as-a metric space --- it may be a tangential thought that made you ask, but I felt compelled to note it won't help). Jan24 comment Metric space not a vector space @SanathDevalapurkar The space of all (Semi|Pseudo|vanilla) Riemannian metrics couldn't possibly be a vector space: what's the additive identity? They form a convex cone, though...see, e.g., Arthur Besse's Einstein Manifolds, Ch 4, section B for details and references. Jan24 comment Metric space not a vector space It sounds like you're asking about the Moduli space of Riemannian metrics, which is a "manifold-like" space (not necessarily a vector space). Jan24 comment Metric space not a vector space A relevant discussion: math.stackexchange.com/questions/98179/… Jan1 comment Closed form for this continued fraction Note, though, for $x=i$, the convergents just fluctuates, cycling through $1$, $1+i$, $(1+i)/2$, without settling. So here's a puzzle: for which values of $x\in i(-2,2)$ does the partial fraction experience a divide-by-zero problem? (I mean, aside from the obvious $x=0$ solution!) Dec31 comment Evaluating the precision in the calculation of $\mathrm{e}$ @user21820 I think there is some ambiguity surrounding the term "arbitrary precision arithmetic". Since the OP was interested in having a rational approximation good to $n$ digits, I assumed the OP would be using something like MIT-Scheme which has "Bignum" arbitrary precision rationals. (Hence my concern about the number of division operations: they are the most costly for Bignum arithmetic.) Your use of the phrase "arbitrary precision" seems non-technical...but we have gone far from the ranch, and it seems completely irrelevant to the OP. Dec31 comment Evaluating the precision in the calculation of $\mathrm{e}$ @user21820 err, you seem to be forgetting the topic of discussion is computing $e$ to some desired precision. (I am not worried about $\exp(x)$ for general $x$, just $x=1$.) The 20th continuant of the continued fraction cited is good beyond 100 digits, requires 40 addition operations, 80 multiplication ops, and 1 division op. The Taylor series, OTOH, requires 38 terms, in Horner form that's 38 multiplication ops + 38 addition ops + the killer 38 division ops. That number of division operations makes it practically unacceptable... Dec30 comment Evaluating the precision in the calculation of $\mathrm{e}$ @user21820 A division operation is costlier than a multiplication operation. (About 5 times costlier, in fact, for x86 floating point...and rational arithmetic would be worse, computing the gcd and then performing 2 division operations...) The continued fraction expansion any sane person would have in mind could be googled in a second. Remember $e=\exp(1)$... Dec30 comment Evaluating the precision in the calculation of $\mathrm{e}$ @user21820 Well, look at (e.g.) Wikipedia's page for the general formulas for the numerator and denominator. You can compute them recursively, requiring (for each iteration) 2 additions and 4 multiplications. So $N$ iterations costs 1 division operation (the final division) + $2N$ addition + $4N$ multiplication, far better than the naive Taylor series. Dec30 comment Evaluating the precision in the calculation of $\mathrm{e}$ Unrelated but potentially useful to the OP: using the continued fraction expansion for $\exp(z)$ is more accurate, and if you compute the convergents $A_{n}$, $B_{n}$ recursively...it is faster since it uses fewer division operations... Dec29 comment Identifying a power series This power series does have a real root at $x\approx-1.403761051217752$. Needless to say, all real roots are necessarily negative...but I think this is the only real root. Dec27 comment The difference between $\Delta x$, $\delta x$ and $dx$ Well, $\delta x$ means different things depending on the context. For example, it has a particular meaning in variational calculus, and a completely different one in functional calculus... Dec24 comment Learning math: still paper and pen @AndreySokolov A pop review and a more thorough answer (with many references) at the personal productivity stackexchange. I think from there, you can find additional references to studies done... Dec19 comment Handbook of mathematical drawing? Honestly, learning how to draw in general helped me with technical diagrams and mathematical figures. Perhaps it's the only way... Nov23 comment How to find this limit: $A=\lim_{n\to \infty}\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots+\sqrt{\frac{1}{n}}}}}$ I'm interested: how did you derive your approximation? (I ask because I wouldn't have a clue where to begin with deriving it...) Nov6 comment Derive the equation of motion from Lagrangian of a particle moving in an electromagnetic field Uh, perhaps plug the Lagrangian into the Euler-Lagrange equations...? Oct21 comment General Linear Group of a vector space Well, to be technical, you only need to pick a basis to determine the components of the matrix representing the isomorphism. Oct20 comment Maclaurin series for $e^z /\cos z$. Perhaps some words or explanations would help this jumble of equations make sense... Oct20 comment Converging or diverging series? One quick trick for (a) is to note if $a_{n}=\sqrt{n}\cos^{2}(n)/(n^{2}-2)$, then $0\leq a_{n}\leq \sqrt{n}/(n^{2}-2)$, since $0\leq\cos^{2}(n)\leq1$.