To show N(2n+1)(x)=(-1)n+1J-2n+1(x).

Short Answer

Expert verified

It is proved thatN(2n+1)(x)=(-1)n+1J-(2n+1)(x)

Step by step solution

01

Concept of Neumann Function:

The Neumann function:

Np(x)=cos(πp)Jp(x)-J-p(x)sin(πp)

02

Calculation of the function N(n+12)(x) :

Consider the Neumann function as follows:

Np(x)=cos(ττp)Jp(x)-J-pττpsinττpNn+12(x)=cos(n+12)ττpJn+12x-J-n+12sinττpn+12

=cosnττ+ττ2Jn+12(x)-J-n+12sinττn+12=-J-n+12x-1n

Simplify further as follows:

Nn+12(x)=-1n+12Jn+12(x)

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Most popular questions from this chapter

Prove the least squares approximation property of Legendre polynomials [see (9.5) and (9.6)] as follows. Let f(x) be the given function to be approximated. Let the functions pl(x)be the normalized Legendre polynomials, that is, pl(x) = √(2l+1)/2 Pl(x) , so that

-11[pl(x)"]"2dx=1.

Show thatLegendre series for f(x)as far as the p2(x)term is

f(x)=c0p0(x)+c1p1(x) +c3p3(x) with c1 =∫-11f(x)pl(x) dx

Write the quadratic polynomial satisfying the least squares condition as b0p0(x)+b1p1(x)+b0p0(x)by Problem 5.14 any quadratic polynomial can be written in this form). The problem is to find b0, b1, b2so that I=∫-11[f2(x)+(b0-c0)2+(b1-c1)2+(b2-c2)2 -c02 -c12 -c22] dx

Now determine the values of the b's to make I as small as possible. (Hint: The smallest value the square of a real number can have is zero.) Generalize the proof to polynomials of degree n.

To show that, ττx/2J1/2(x)=sinx.

Plot

  1. l0(x),l1(x),l2(x), from x = 0 to 2 .
  2. K0(x),K1(x),K2(x), from x = 0.1 to 2 .
  3. role="math" localid="1659270507695" Ai(x)from x = -10 to 10 .
  4. role="math" localid="1659270512971" Bi(x) from x = -10 to 1 .

Solve the differential equations in Problems 5 to 10 by the Frobenius method; observe that you get only one solution. (Note, also, that the two values of are equal or differ by an integer, and in the latter case the larger gives the one solution.) Show that the conditions of Fuchs's theorem are satisfied. Knowing that the second solution is X times the solution you have, plus another Frobenius series, find the second solution.

x(x+1)y''-(x-1)y'+y=0

Expand the following functions in Legendre series.

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