# Tag Info

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 $n=2^{20}$

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

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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

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

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 $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} ... 1 No. It is impossible. Initial values (x_i,f(x_i)) are experimental data. You should measure them from real world phenomena or they should be given. Although you can approximate f(x_2) by Newton forward difference formula using one of the pairs \{(x_0,f(x_0)),(x_3,f(x_3))\}, \{(x_0,f(x_0)),(x_4,f(x_4))\}, \{(x_1,f(x_1)),(x_3,f(x_3))\} or ... 1 Note that if: g_1(x)=x\Longrightarrow e^x/e^a=x\Longrightarrow e^x=e^ax\Longrightarrow e^x-e^ax=0\equiv f(x)=0 how f have only two roots then g_1 have exactly two fixed points. g_2(x)=x\Longrightarrow a + \ln{x}=x\Longrightarrow \ln{e^a}+\ln{x}=\ln{e^x}\Longrightarrow \ln{e^ax}=\ln{e^x}\Longrightarrow e^ax=x \Longrightarrow x-e^ax=0\equiv ... 1 Notice that$$f(P)=0\Rightarrow e^P-e^aP=0\Rightarrow e^P=e^aP\Rightarrow P=\frac{e^P}{e^a}=g_1(P)$$In the same way for Q. So g_1(x) has two fixed points. Now$$f(P)=0\Rightarrow e^P-e^aP=0\Rightarrow e^P=e^aP\Rightarrow P=a+\ln(P)=g_2(P) The same technique for $Q$.

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I observe that the left column is obviously exponential, and the right column seemingly so. (Each item in the right column is somewhat more than three times the previous item.) So it makes sense to construct a log-log plot of the data: The $x$-axis is the logarithm of the left-hand column, and the $y$-axis is the logarithm of the right-hand column. At ...

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