# Tag Info

13

This is a "visual proof" of the Archemidean limiting regular polygons. You could implement the recursion and it would progressively approach $\pi$. The proof lies in the "squeezing" argument. $\pi$ is transcendental (no solution" to this recurrence in a closed algebraic expression). Whatever implementation of recursion with approach $\pi$ that is ...

12

Some explanations first The substitution in the question introduces the reduced wave function $u(r)$ by solving the original radial equation in polar coordinates, $$-\frac{1}{2}\left(R''(r)+\frac {1}{r}R'(r)\right) - \frac{1}{r}R(r) + \frac {m^2}{2r^2}R(r) = E R(r)$$ using the ansatz $$R(r)\equiv \frac{1}{\sqrt{r}}u(r)$$ The apparently divergent ...

8

You should use RecurrenceTable: N@RecurrenceTable[{a[n + 1] == 2 a[n] b[n]/(a[n] + b[n]), b[n + 1] == Sqrt[a[n + 1] b[n]], a[0] == 2 Sqrt[3], b[0] == 3}, {a[n], b[n]}, {n, 1, 10}] N@Pi with result: {{3.21539, 3.10583}, {3.15966, 3.13263}, {3.14609, 3.13935}, {3.14271, 3.14103}, {3.14187, 3.14145}, {3.14166, 3.14156}, {3.14161, ...

7

You can use Mathematica to prove the induction step in a proof that $a_n, b_n \rightarrow L$ for some limit $L$ by showing that $a_n$ is decreasing and $b_n$ increasing toward each other and that the difference goes to zero. To show $L = \pi$ you would need a definition of $\pi$ you can relate to this sequence. For instance, if you define $\pi$ to be the ...

7

With[{x = Array[x, Dimensions[mA]]}, Solve[mA .x - x. mB + mC == 0, Flatten@x]] Or With[{x = Array[x, Dimensions[mA]]}, x /. Solve[mA .x - x. mB + mC == 0, Flatten@x]] {{{3, 1}, {0, 3}}}

5

x = FindInstance[{Cot[Sqrt[λ]] == Sqrt[λ], λ >= 0, λ < 100}, λ, 100] You can take the numerical value of the solution using N and use ReplaceAll (/.) to get the values and finally take the square root: λ /. N[x] // Sqrt {0.860334, 3.42562, 6.4373, 9.52933} This should work regardless of the number of solutions.

3

I guess the answer is "no," Solve does not always find solutions in terms of radicals when they exist. Example: The polynomial $x^5 + 20 x^3 + 20 x^2 + 30 x + 10$ has root expressible in terms of radicals (see Wikipedia): poly = x^5 + 20 x^3 + 20 x^2 + 30 x + 10; x1 = 2^(1/5) - 2^(2/5) + 2^(3/5) - 2^(4/5); poly /. x -> x1 // Simplify (* 0 *) But ...

2

Here you are plotting a complex function, as you know, so you need to plot the real and imaginary parts separately. Plot[ReIm[(-1)^x + 2^x - 2 x - 1], {x, -4, 4}, Evaluated -> True] You can get some help on this from Wolfram Alpha, if you start your expression with two equal signs: This shows that you need to plot the real and imaginary parts ...

2

I think in general no one know's how to express roots of polynomials in terms of radicals, or even determine when it's possible. Quintics has been solved and there's a Mathematica package to solve them. Radicals.nb SolveQuintic[x^5 + 20 x + 32 == 0, x] For sextics I believe the most that is known is how to determine which equations can be expressed in ...

2

I'm going to restrict to the case of rational coefficients. There are ways to extend to complex rational coefficients but that's more than I have time or desire to consider right now. I'll illustrate an efficient methodology with this example. Along the way I will say a bit about modest improvements that can be made. We'll start with the polynomial and the ...

2

eqn = ((h*s + l - r*s - r)/(s^2 + s)) (1 + s) - h == (h*s + l - r*s - r)/(s^2 + s) - r // Simplify (* (h - l)/(1 + s) == 0 *) Note that r does not appear in the simplified expression so Mathematica returns "no solution" Solve[eqn, r] (* {} *)

2

I post this as a way to generate integer solutions. The desired result could be extracted. I appreciate letting Mathematica do it with constraints already does this. I am not clear what is being asked. Perhaps this will lead to clarification: func[p_] := Module[{v = {a, b, c, d, e, f}, m}, m = {#1^2, #2^2, #1 #2, #1, #2, 1} & @@@ p; v /. ...

2

List @@ Roots[x^3 - 5 x + 4 == 0, x][[All, -1]] (* {(1/2)*(-1 - Sqrt[17]), (1/2)*(-1 + Sqrt[17]), 1} *) EDIT: To understand a complex expression, decompose it and rebuild it, step-by-step, examining each intermediate step. expr1 = Roots[x^3 - 5 x + 4 == 0, x] (* x == (1/2)*(-1 - Sqrt[17]) || x == (1/2)*(-1 + Sqrt[17]) || x == 1 *) The ...

2

A methode i like to use and a starter. Please see Bisection method and Bisection, in particular: Let $a_n$ and $b_n$ be the endpoints at the nth iteration (with $a_1=a$ and $b_1=b$) and let $r_n$ be the nth approximate solution. Then the number of iterations required to obtain an error smaller than epsilon is found by noting that ...

1

Introduction For many transcendental functions, NSolve can solve for the roots, but not in this case. Since the roots are real we can apply the "Chebyshev-proxy rootfinder" method (CPR) which is based on the "colleague matrix" of a truncated Chebyshev series approximation to the function (see this answer by J.M. and the book by Boyd (2014)). The first ...

1

This is a good example of the "dynamic range" problem in Boyd's CPR method (see also this answer by J.M.). As α moves toward -100, the OP's oscillatory Whittaker function decays exponentially. According to Boyd, where $ε_{mach}$ is the machine epsilon and $f_{max}$ is the maximum of $f(x)$ over an interval $[a,b]$, "If a function $f(x)$ has a magnitude as ...

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