By Mats Gyllenberg, Lars-Erik Persson

ISBN-10: 0585328285

ISBN-13: 9780585328287

ISBN-10: 0824792173

ISBN-13: 9780824792176

Offering the court cases of the twenty-first Nordic Congress of Mathematicians at Luleå college of know-how, Sweden, this awesome reference discusses contemporary advances in research, algebra, stochastic approaches, and using desktops in mathematical study.

**Read or Download Analysis, algebra, and computers in mathematical research: proceedings of the Twenty-first Nordic Congress of Mathematicians PDF**

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**Extra resources for Analysis, algebra, and computers in mathematical research: proceedings of the Twenty-first Nordic Congress of Mathematicians**

**Sample text**

When x = 0 is a singular point, the simplest case to consider is when P (x) = P1 x + P2 x2 + · · · . We can then ask what form Q(x) and R(x) must take to ensure that a series solution exists. When x = 0 is an ordinary point, the indicial equation was formed from the y term in the equation alone. Let’s now try to include the y and y terms as well, by making the assumption that R−1 + R0 + · · · . x Then, after substitution of the Frobenius series into the equation, the coeﬃcient of xc−1 gives the indicial equation as Q(x) = Q0 + Q1 x + · · · , R(x) = P1 c(c − 1) + Q0 c + R−1 = 0.

24). We have already seen that x = 0 is a regular singular point. The radii of convergence of xQ(x)/P (x) and x2 R(x)/P (x) ∞ are both unity, and hence the series solution n=0 an xn+c exists, is unique, and will converge for |x| < 1. Example Consider the diﬀerential equation d 2y dy + y = 0. − dx2 dx For this equation, x = 0 is a singular point but it is not regular. The series solution ∞ n+c is not guaranteed to exist, since xQ/P = −1/x3 , which cannot be n=0 an x expanded about x = 0. 3 holds.

13). Since the equation and boundary conditions do not depend upon the azimuthal angle, φ, neither does the solution, and hence Laplace’s equation takes the form ∂T ∂ 1 ∂T 1 ∂ r2 + 2 sin θ = 0. r2 ∂r ∂r r sin θ ∂θ ∂θ Let’s look for a separable solution, T (r, θ) = R(r)Θ(θ). This gives Θ d dr r2 dR dr + R d sin θ dθ sin θ dΘ dθ = 0, 45 46 LEGENDRE FUNCTIONS and hence 1 d R dr r2 dR dr =− d 1 Θ sin θ dθ sin θ dΘ dθ . Since the left hand side is a function of r only and the right hand side is a function of θ only, this equality can only be valid if both sides are equal to some constant, with 1 d R dr r2 dR dr =− d 1 Θ sin θ dθ sin θ dΘ dθ = constant = n(n + 1).

### Analysis, algebra, and computers in mathematical research: proceedings of the Twenty-first Nordic Congress of Mathematicians by Mats Gyllenberg, Lars-Erik Persson

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