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3

Here are a few books as stated in this answer: The first two books reference each other, so I recommend having both on hand. The third is in my reading queue. All are by Jonathan Borwein and David Bailey. Experimentation in Mathematics Mathematics by Experiment Experimental Mathematics in Action These books cover Mathematica, Maple, and many other tools. ...


3

Let $a$ be a positive real number. Then $$\lim_{n\to\infty}(a^\frac1n-1)\cdot n=\lim_{n\to\infty}\frac{a^{\frac1n}-1}{\frac1n}=\lim_{n\to\infty} \frac{a^\frac1n\cdot\ln a\cdot\frac{-1}{n^2}}{\frac{-1}{n^2}}=\ln a$$ Your result is a special case of this with $a=5$, and $n=2^{20}$


3

The point is that when $x$ is small, $\cos(x) \approx 1$ and so you can expect loss of precision in $y$. One remedy is this:


2

I checked this in Mathematica. It really does seem to converge to Log(5) accurately after many iterations: a = N[Sqrt[5], 30]; Do[ a = Sqrt[a], {n, 1, 10000 - 1}] (a - 1)*2^10000 Output: 1.60943791243410037460075933323 compared to: N[Log[5], 30] 1.60943791243410037460075933323


2

With steplength $h$, the implicit Euler method for this equation would say $$y_{n+1}-y_n=hy_{n+1}^2.$$ Seeing this as a quadratic equation to solve for $y_{n+1}$, we get $$y_{n+1}=\frac{1\pm\sqrt{1-4hy_n}}{2h},$$ but you need to pick the correct sign. For concistency, you absolutely need to have $y_{n+1}$ converging to $y_n$ if you let $h\to0$, and that ...


2

The identity is equivalent to $$\sum_{n\ge 1} \frac{1-\cos(2\pi nx)}{n^2} = \pi^2 x(1-x).$$ The sum term $$S(x) = \sum_{n\ge 1} \frac{1-\cos(2\pi nx)}{n^2}$$ is harmonic and may be evaluated by inverting its Mellin transform. Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} ...


2

This is called a fixed-point iteration.


1

Let's start with $$ \begin{align} \sum_{n=1}^\infty\frac{\sin(2\pi nx)}{n} &=\mathrm{Im}\left(\sum_{n=1}^\infty\frac{e^{2\pi inx}}n\right)\\ &=\mathrm{Im}\left(-\log\left(1-e^{2\pi ix}\right)\right)\\ &=-\arctan\left(\frac{-\sin(2\pi x)}{1-\cos(2\pi x)}\right)\\ &=\arctan\left(\frac{2\sin(\pi x)\cos(\pi x)}{2\sin^2(\pi x)}\right)\\[9pt] ...


1

One of the oldest ISC (Inverse Symbolic Computer) is the Plouffe's inverter, cited here for memory because it is no longer available at http://pi.lacim.uqam.ca/ Others were already quoted in the peceeding answers : http://oldweb.cecm.sfu.ca/ http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html Also, WolframAlpha provides such tool : ...


1

Referring to differential equations: 1)the analytical approach is about trying to prove the existence, the uniqueness and find an explicit form for the solution. As finding an explicit form is almost always very difficult (or impossible) one can go through two different ways 2)the numerical approach is about trying to find an approximate solution using ...


1

No. It is impossible. Initial values $(x_i,f(x_i))$ are experimental data. You should measure them from real world phenomena or they should be given. Although you can approximate $f(x_2)$ by Newton forward difference formula using one of the pairs $\{(x_0,f(x_0)),(x_3,f(x_3))\}$, $\{(x_0,f(x_0)),(x_4,f(x_4))\}$, $\{(x_1,f(x_1)),(x_3,f(x_3))\}$ or ...



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