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Instructor's Manual MATHEMATICAL METHODS FOR PHYSICISTS A Comprehensive Guide SEVENTH EDITION George B. Arfken Miami. Instructor's Manual MATHEMATICAL METHODS FOR PHYSICISTS A Comprehensive Guide SEVENTH EDITION George B. Arfken Miami University Oxford, OH. homeranking.info - Ebook download as PDF File .pdf), Text The seventh edition of Mathematical Methods for Physicists is a substantial and.

This is valid for any order of the indices. Use the identity of Exercise 3. Fourier Series If we demand a polynomial solution then n must be a nonnegative integer. The y in the integrand can be dropped because, by symmetry, it makes no net contribution to the integral. Do not further factor into quantities linear in s.

Some may be useful as test questions or additional study material. Complete methods of solution have been provided for all the problems that are new to this seventh edition. This feature is useful to teachers who want to determine, at a glance, features of the various exercises that may not be com- pletely apparent from the problem statement. While many of the problems from the earlier editions had full solutions, some did not, and we were unfortunately not able to undertake the gargantuan task of generating full solutions to nearly problems.

The authors invite users of the text to call attention to errors or ambiguities, and it is intended that corrections be listed in the chapter of this Manual entitled Errata and Revision Status. Errata and comments may be directed to the au- thors at harris at qtp. If users choose to forward additional materials that are of general use to instructors who are teaching from the text, they will be considered for inclusion when this Manual is updated. We particularly want to acknowledge the assis- tance of our Editorial Project Manager, Kathryn Morrissey, whose attention to this project has been extremely valuable and is much appreciated.

Chapter 2 Errata and Revision Status Last changed: Page Figure Page Exercise The answer is then correct. Page Eq. Page After Eq.

For arfken physicists manual pdf mathematical methods solution

Consistency with the duplication formula then determines C2. The text assumes it to be kr. The right-hand side of the second equation should read: The right-hand side of the third equation should read: Disregard it.

Page Table The column of references should, in its entirety, read: Corrections and Additions to Exercise Solutions None as of now. Chapter 3 Exercise Solutions 1. Mathematical Preliminaries 1.

This expression approaches 1 in the limit of large n. The solution is given in the text. Let sn be the absolute value of the nth term of the series.

MATHEMATICAL METHODS FOR PHYSICISTS ARFKEN SOLUTION MANUAL PDF - PDF Drive

Therefore this series converges. Because the sn are larger than corre- sponding terms of the harmonic series, this series is not absolutely con- vergent. With all signs positive, this series is the harmonic series, so it is not aboslutely convergent. We therefore see that the terms of the new series are decreasing, with limit zero, so the original series converges.

With all signs positive, the original series becomes the harmonic series, and is therefore not absolutely convergent. The solutions are given in the text. The upper limit x does not have to be small, but unless it is small the convergence will be slow and the expansion relatively useless. The integrated terms vanish, and the new integral is the negative of that already treated in part a.

Use mathematical induction. Thus, we want to see if we can simplify 1 p! The formula for un p follows directly by inserting the partial fraction decomposition.

After inserting Eq. Using now Eq. Insertion of this expression leads to the recovery of Eq. Applying Eq. Using these in Eq. P and Q are antiparallel; R is perpendicular to both P and Q. Now take real and imaginary parts to get the result. All other identities are shown similarly. Separating this into real and imaginary parts for real z1, z2 proves the addition theorems for real arguments.

Analytic continuation extends them to the complex plane. The nth term of the x expansion will be xn n!

Apply an integration by parts to the integral in Table 1. This integral can also be evaluated using contour integration see Exam- ple The series in parentheses is that discussed in Exercise 1. Integrate by parts, to raise the power of x in the integrand: Note that the integrated terms vanish. The integral can now be recognized see Table 1. Write erf as an integral and interchange the order of integration.

Write E1 as an integral and interchange the order of integration. Now the outer u integration must be broken into two pieces: Integrating over one quadrant and multiplying by four, the range of x is 0, a and, for given x, the range of y is from 0 to the positive y satisfying the equation for the ellipse.

Determinants and Matrices 2. Therefore no nontrivial solution exists. This is the general solution for arbitrary values of x. The sum over i collects the quantities that multiply all the aij in column j of the determinant. If a set of forms is linearly dependent, one of them must be a linear combination of others.

The determinant whose value is not changed by the operation will be seen to be zero. Interchanging p and q gives two terms, hence the factor 2. By direct matrix multiplications and additions. By direct matrix multiplication we verify all claims. Then make a cyclic permutation if needed to reach CBA. This is proved in the text. Same answers as Exercise 2. This summation replaces Tij by unity, leaving that the sum over Pj equals the sum over Qi, hence conserving people.

The answer is given in the text. If Jx and Jy are real, so also must be their commutator, so the commuta- tion rule requires that Jz be pure imaginary. The anticommutation can be demonstrated by matrix multiplication.

In block form, Eq. The requirements the gamma matrices must satisfy are Eqs. Use the same process that was illustrated in the solution to Exer- cise 2. Then proceed as in the solution to Exercise 2. Vector Analysis 3. Both vectors are of unit length. If a and b both lie in the xy-plane their cross product is in the z-direction.

Arfken mathematical methods for physicists 7th edition pdf

The cross product of two parallel vectors is zero. The parallelpiped has zero height above the BC plane and therefore zero volume. If an incoming ray strikes the xy plane, the z component of its direction of propagation is reversed.

A strike on the xz plane reverses its y component, and a strike on the yz plane reverses its x component. These properties apply for an arbitrary direction of incidence, and together the reverse the propagation direction to the opposite of its incidence orientation. Because S is orthogonal, its transpose is also its inverse. Therefore, to achieve the same polar orientation, we must place the x1 axis where the x2 axis was using the text rotation. From these we verify that each matrix element of Eq.

Adding these contributions, we get the required result. This corresponds to the rotational transformation given in Eq. The product rule directly implies a and b. Carry out the indicated operations, remembering that derivatives operate on everything to their right in the current expression as well as on the function to which the operator is applied. To determine the direction of the stream lines, pick a convenient point on a circle, e.

Further Properties 3. Hence the four terms. Write the x components of all the terms on the right-hand side of this equation. All terms cancel except those corresponding to the x component of the left-hand side of the equation.

The x component of the left-hand side of Eq. Using Exercise 3. From Eq. Equating to zero the real and imaginary parts of all components of the above vector, we recover two Maxwell equations. By direct matrix multiplication we verify this equation. Here we spend energy. The work is path dependent which is consistent with the physical interpretation that With the branch cuts of Example For all points in the upper half-plane, both the branch cuts of Exam- ple In Example But in Exercise The original value of both terms cannot be recovered simultaneously until the number of circuits around the branch point is the smallest common multiple of 3 and 4, i.

Start by making a partial fraction expansion on the integrand: Break the integral into its two parts and for each part make a binomial expansion of the denominator in a form that will converge for the region of integration: An attempt to use sinz directly in Eq. Applying Eq Therefore, the integral of f. The semicircle does not contribute to the integral, and the contour encloses the pole, so our formula for u s will be equal to the residue, namely unity.

But now the pole is not within the contour, so our expression for u s will evaluate to zero. This contour encloses no singularities and therefore evaluates to zero. Thus, the principal-value integral will be equal to that over the small semicircle traversed counterclockwise.

This is a case of Example Hence the result. In both these cases the integral does not exist. The residue at the pole of order 3 is 1 2! At large [z[, the integrand of the contour integral of Example At small [z[, the denominator of the integrand approaches unity, so the integrand is basically of the form z p. To reconcile Eqs. Next write the integral as shown below, note that its integrand is an even function of x, and break the integral into two parts, in one of which replace bx by x and in the other replace ax by x: This integral, which is the topic of Exercise Note that the integrated term vanishes.

Combining the above, we obtain the answer in the text. For cos kx we rescale: Thus, the contour encloses no singularities, so the contour integral vanishes. Since there are no singularities, the residue theorem gives zero for the loop integral. Contour surrounds branch cut and two poles. Applying Exercise The integrand has a branch point at the origin and we choose a contour of the form shown in Fig. The contour consists of four pieces: The integral on the last line was the topic of Example Use the contour of Fig.

The large semicircle does not contribute to the integral. Sector contour. If L denotes the triangular path of Fig. This problem is similar to Exercise This change causes the last line of the solution of Exercise Use the contour in Fig. Take the contour to be that shown in Fig. This contour consists of four line segments: We now evaluate the contour integral using the residue theorem.

Contours for Exercise Note that together these curves enclose a region in which the inte- grand is analytic, so the line integrals on the two curves are equal and CHAPTER 3. Referring to Fig. Arguments and moduli of singular factors. Contour avoiding branch cut. The limiting behavior for I N that is needed for the contour-integral eval- uation of sums is produced by the behavior of f z for large [z[, which in applicable cases becomes small rapidly enough that the integrand of I N becomes negligible on its entire contour and therefore evaluates to zero.

Because each of these exponentials occurs in both the numerator and denominator, this expression will remain of order unity except where the denominator approaches zero i. This residue is 1 2! This summation is most easily done by decomposing the summand into partial fractions: Invoking the periodicity of the cotangent, these are seen to add to zero. Thus, if z meets the conditions of Exercise That is why mappings by an analytic function are termed conformal.

MATHEMATICAL METHODS FOR PHYSICISTS ARFKEN SOLUTION MANUAL PDF

Further Topics in Analysis We now interchange the summation and integration and evaluate the sum: The three terms on the left-hand side of Eq.

To reach Eq. The unnumbered identity following Eq. Equation Using the Cauchy integral for the power series for the generating functions yields the results. Multiply numerator and denominator of the fraction on the right-hand side by e t and then put both right-hand terms over a common denominator. Multiply numerator and denominator of the right-hand side by e t. The B n can be read out of Table Putting everything together, we get n.

This problem can be approached in a way similar to the solution of Exer- cise Substitution of these quantities leads to the expected result. Since this ratio increases without limit, the series can only be asymptotic.

A similar analysis applies to the function Q. Applying a binomial expansion to the denominator, the integral is asymp- totically represented by the series N. Gamma Function The denominator of Eq. From Fig. See also Exercise Our formula is then entirely equivalent to that in the exercise. Using Exercise Referring to the solution of Exercise Using the formula of Exercise The y in the integrand can be dropped because, by symmetry, it makes no net contribution to the integral.

The remainder of the expression now contains an integral of the form treated in Exercise All together, these contributions to Eq. Substituting this expression for the Bernoulli sum, we reach the answer in the text, in which there is a remaining summation of the zeta functions of CHAPTER 3.

We therefore also subtract the right-hand side of this equation from the formula of part a. But, from Eq. First form, using Eq. Inserting the expansion for E 1 from Eq. One way to obtain the argument of a complex quantity is to identify it as the imaginary part of its logarithm.

From Eqs.

Expanding all the beta functions, and using Eq. This is symmetric in a, b, and c, so the presumed relation must be correct. This is a case of Eq. Make a change of the variable of integration to make the integration limits zero and one: Therefore, using Eq. The integrals at issue here are cases of Eq.

For n even, a similar process can be used to convert the gamma function of half-integer argument to the more convenient form given as the listed answer. This can be brought to the form in the text using Eq. These integrals are cases of Eq. A reduction similar to that of part a leads to the listed answer. This reduces in a way similar to part a. The integrals occurring here are cases of Eq. The integrals of this exercise are cases of Eq.

Use Eq. Compute the logarithm, then exponentiate. The result is 8. Since ln n! Hence the series diverges. It is convenient to work with logarithms of the factors. Note that the individual terms of scaling greater than N have combined in a way that makes S an extensive quantity.

For details, see Section These steps proceed as follows: Here the integrated term vanishes and the second term contains the inte- gral of Eq. The denominator of the integrand with a factor t in the numerator is the generating function for the Bernoulli numbers, so we can introduce that expansion and integrate termwise.

The integral in this expression is a case of Eq. Make a binomial expansion of the denominator of the integrand as given in the exercise: Insert this into the integral of this exercise and integrate termwise: Referring to Eq. This integration is a case of Exercise Use the binomial theorem to expand the denominator of the integrand and then integrate termwise.

To identify this summation as a polygamma function we need to change the indexing to move the lower summation limit from 0 to 1. When a is an integer, this reduces to the answer in the text. Applying this result m times in succession yields the formula to be proved. Substitute the formula of Exercise In Eq. The integral evaluates to n! This is the answer to part c. This result corresponds to the answer we require.

When these expressions are substituted into the form for V r we recover the answer in the text. This is shown in Eqs. Rearranging, we reach the desired expression: Now introduce the Maclaurin series for sint and integrate termwise. The result is the answer in the text. Insert the relation connecting the incomplete gamma functions to the iden- tity of part a of Exercise A similar approach can be used to verify that part b of Exercise The formula indicates that n must be a nonnegative integer.

Start by expanding the denominator in the integrand: The u integral evaluates to j! Bessel Functions For real x the inequalities follow. Combining them yields the answer to this part of the exercise. Collecting terms and dividing through by 2 yields the desired result.

In the generating function for the J n as given in Eq. Proceed by applying Eq. Other cases can be treated by obvious extensions of the method to be used here.

There must be at least one zero of J. Then note that by Eq. The integration that remains can also be rewritten using Eq. To do so it will be convenient to have a formula similar to that of Eq. Now we form x 2 C. In this way we reach x 2 C.

This is the linear ODE we seek. Evaluating the derivative in Eq. The contour in s is the same as the contour for t.

The contour consists of three parts: Range 1: The second integral can be expanded into real and imaginary parts. Write J 0 bx as its series expansion, and integrate termwise, recognizing the integrals as factorials. This result can now be analytically continued to the entire region for which the integral representation converges. Expand the integrand: See Table This, therefore, is the fraction of the intensity in the central maximum. Combining these forms of the two integrals, the J 1 terms cancel, leaving the result given for part b of the exercise.

The exercise asks for the allowable values of k, to which the foregoing provides an answer. We may then use Eq. This expression reduces to the value given for the second Lommel integral. In the value given for that integral, the J. This result is proved in Exercise The equation referenced in this exercise should have been Eq. Pure imaginary roots can be excluded because when z is pure imaginary, all terms of the power-series expansion have the same sign and therefore cannot sum to zero.

The normalization integral in the denominator has the value given in Eq. The normalization integral in the denominator has the value given in the solution to Exercise Substitute the Bessel series for f x into the integral for the Parseval relation and invoke orthogonality of the Bessel functions: Now invoking Eq. Use the recursion relations, Eqs. The second recursion is proved similarly.

From Exercise Using this result, the second formula in Exercise The left-hand side of the formula of this exercise is the Wronskian W of the two solutions. For an ODE in the form y. The result is the answer given in the text. Let y denote the integral of this exercise. The integrand is identically zero. Parts a through e of this exercise are easily proved using the Wronskian formula, Eq. Similar processes prove parts b through e.

For parts f and g we need the relationship in Eq. As an example, write the formula of part g as follows: The formula of part f is proved similarly. Thus, the s and t contours are identical except that they are traversed in opposite directions, which can be compensated by introducing a minus sign.

Methods pdf arfken manual solution physicists mathematical for

Substituting into Eq. This substitution leads to the integral of part b. After this divi- sion, we get the answer shown in the text after making a typographical correction to change l into h. Here use the recurrence relations for I v , Eq. These additions change only the relative signs of the two sides of the equations. Agreement with Eq. Note the factors i that accompany the derivatives. Because notations like J.

Start from the integral representation, Eq. In this exercise n is assumed to be an integer. Starting from the generating function for I n given in Exercise These steps lead to the claimed result.

The integrated terms vanish, and the resultant integration cancels the z 2 sinh 2 t term of the original integral. We thereby attain the desired zero result. Saddle points are where w. Inserting these data into the steepest-descents formula, Eq. This problem can be solved by deforming the integration contour as needed to pass through a saddle point and use the steepest descents method. In- serting these data into the steepest-descents formula, Eq.

In writing these equations we have used the fact that the integrated end- point terms all vanish. Here is where it is necessary that n be an integer. In the limit of small z, the integral becomes 1! The term of any given r will therefore contain in its denominator 4 r which combines with the remaining factors of the summation in Eq.

Inserting the form given for y z , we get initially z 2 y. If this term is zero i. We only need the initial term of each asymptotic expansion. In Part b , N should be replaced by Y.

We also need the derivatives I. Here we essentially reverse the process that was used in solving Exercise The sign alternation in P and Q and the presence of the i multiplying Q are both accounted for by the factor i s in Eq.

Start from Eq. Application to j n z is as follows: The second line of the above equation is reached by writing the double factorials 2p!! The second recurrence formula is a bit less trivial, since j. To complete the proof by math- ematical induction, we need a starting value. For small x, the limiting behavior of j n and y n is given by Eqs.

Then our Wronskian takes the form j n x y. This is the correct power-series expansion of j n z. The resultant integrals are equal in magnitude but opposite in sign; all that remains are the endpoint terms: This equation rearranges to the answer in the text. Consider the result in Exercise Here m can be any nonnegative integer and v is the velocity of sound.

Note that the smallest zero of j. Inserting these into the Wronskian, i 0 x k. Legendre Functions After remov- ing canceling terms, what remains is the Legendre ODE. Start by writing P. It is convenient to change the summation index to k. Formally the k. We are now ready to write all the terms in the Legendre ODE.

Only the term we have already processed causes a change in the indicated powers of x. This formula is easily proved by mathematical induction, using the recur- rence formula. Now assuming P. Since this last equation has to agree with Eq. This completes the proof. Derivation of this formula is presented in the footnote referenced just after Eq. There is a misprint in the answer: Introduce the Rodrigues formula for the Legendre polynomial and inte- grate by parts 2n times to remove the derivatives from the Rodrigues formula.

Using integration by parts, orthogonality and mathematical induction as in Exercise This follows from Eq. See Exercise We now use the result of Exercise Because P. Choose n to be the larger of the two indices. Then use Eq. In the integrand of this expression the term arising from P. The value of a 0 is irrelevant; we set it to zero. The derivative of P 2s leads to associated Legendre functions, Section The second form of the answer is obtained using Eq.

The answer is given in AMS Starting from the solution to Exercise This equation rearranges into the required answer. Using the gradient in polar coordinates from Section 3. A charge q and its image charge q. The charge q is at a distance a from the center of the sphere, its image q. Our task is to show that the potentials produced at P from the two charges add to zero. Now use the law of cosines and the geometry of Fig. These two expressions are clearly equal, completing our proof.

We start from the recurrence formula, Eq. Figure Image charge geometry. We now apply a similar procedure to P m l. The functions P 1 2l 0 vanish because they have odd parity. There are many ways to prove this formula.

Mathematical Methods for Physicists 5th Ed - Arfken - Solution

The formula given here needs a minus sign to be consistent with the sign conventions used throughout the text.

Substituting the delta-function formula: Next, note that Q 1 x is even. Thus, in general Q n has a parity opposite to that of n. First verify by explicit computation that the formulas give correct results for Q 0 and Q 1. We next check that the formulas of parts a and b are consistent with the recurrence formula. Each check involves three parts: It may be useful to organize the terms in a fashion similar to that illustrated in the solution to Exercise To make the exponents correspond for the same index values we replace s by s.

Angular Momentum From these equations we proceed as in Exercise The formula of this exercise is that for angular momentum coupling, so the result must be a spherical tensor of rank J. These states are not yet normalized. Normalizing all these states and using the conventional labeling: The p states are designated as in Exercise Coupling the triplet nuclear state with the electron spin produces a quartet state and a doublet state.

The singlet nuclear state has no spin angular momentum, so coupling it with the electron spin produces the doublet which we mark with a prime 1 2 , 1 2. All these states can now be expanded into weighted sums of the m p , m n , m e states.

After these changes, one can expand the resulting states. The quartet states are the same in both coupling schemes, but the doublets found here, marked with multiple primes, 1 2 , 1 2. One way to show that these doublets span the same space is to write those from part c as linear combinations of those from part b.

The set of Y m l for given l is closed under rotation. This must be the case, because L 2 , a scalar, has a value that is independent of the orientation of the coordinate system.

This problem is a special case of Eq. The quantity A is shown in Eq. This formula is derived at Eq. To solve the problem, use the Laplace expansion, Eq.

Note that we have written the integrations in a way that shows them to be equal. This problem proceeds in a way similar to Exercise This problem is presented in MKS units, with q denoting the electron charge.

As in Exercises Then invoke orthogonality and use the normalization of P N , given in Eq. The answers are given in the text. The parity is the same as that of the spherical harmonic, whose parity is controlled by its value of the lower index L which is the second index of the vector spherical harmonic.

The cross product is orthogonal to both its vectors, so the dot product in the integrand yields zero. Group Theory From Table Moreover, the table shows that for any elements R and R. R, so the group is abelian.

Then build the group multiplication table shown in Table From this and other successive operations, we can identify all elements of the group multiplication table. Since the correspondence is one-to-one, the groups are isomorphic. The Table Thus, all the elements aI, a 2 , ab,. The product of two elements is a group member: The isomorphism further implies that the group of inverses must also be abelian.

Then the positive y-axis of the crystal can be placed applying a rotation in any one of the four directions perpendicular to the direction chosen for the crystal x-axis. Since the matrix of I is a unit matrix, the group operations Ia, aI, Ib, bI, Ic, and cI are consistent with the corresponding matrix products. The transformation does not change I or A. Because all four representation matrices are now block diagonal the two blocks are 1 1 , the elements U 11 of the transformed matrices form a one-dimensional representation, as do the U 22 elements.

Since all these orbitals are of opposite sign above and below this plane, they cannot be used to con- struct a representation that does not change sign when the triangle is turned over. Example This is the representation of D 3 called E and discussed in Example There are therefore four classes.

The only way the dimen- sionality theorem, Eq. There are three distinct ways to assign the minus signs which cannot be assigned to I , leading to the following character table: We check by transforming some of the group elements: All the transformed matrices have the form of a 11 block followed by a 2 2 block.

The block of dimension 1 is the A 1 irreducible representation; that of dimension 2 is the E representation. Comparing with the solution to Exercise The group has eight elements, so the squares of the dimensions of the irreducible representations must add to 8.

The only possibility is to have four repre- sentations of dimension 1 and one of dimension 2. The characters for I must all be equal to the dimension of the representation, and one representation, usually called A 1 , must have all its characters equal to 1. Then, applying Eq. The result is the following character table in which two of the representations are, for reasons we do not discuss, conventionally labeled B 1 and B 2.

The last row is not part of the table but is relevant to Exercise For C. These data are appended to the D 4 character table generated in the solution to Exercise We identify the rows and columns of our representation matrices as corresponding to the basis in the above-given order. The two-dimensional space orthogonal to these basis functions will be spanned by a basis for E.

We assume that the positive lobe of the p x orbital points toward positive x, no matter where it is located. Similar remarks apply to the p y orbitals. Those named A must also be invariant under C 4 , while those named B must change sign under C 4. There are two independent sets of basis functions that satisfy this requirement. The possible cycle structures are: But i, j can be chosen in six ways, so this is a six-member class.

This class therefore has eight members; 5 i, j, k, l ; this cycle structure describes 3! Representations of dimension 3 are customarily labeled T, so our roster of irreducible representations is A 1 , A 2 , E, T 1 , and T 2. It is convenient to use the generators of the SU 2 and SU 3 groups to identify the subgroup structure of the latter.

As pointed out when writing Eq. We now search for SU 2 generators that can be formed from the eight generators of SU 3. To prove that the matrices U n form a group one needs to verify that they satisfy the group postulates: We have a closed subset.

This operator decreases I by 1 and changes u to d. One vector in this subspace is a member of the representation 10; vectors orthogonal to the member of 10 must belong to other repre- sentations.

These functions are not orthonormal, but can be made so: Insertion of the explicit forms for M and M 2 lead directly to U as given in the exercise. Form the transformed electromagnetic tensor as the matrix product F. Then F. The transformed components of B are obtained from B.

Then the generalization of M and M 2 see the solution to Exercise Finally, form the matrix F. More Special Functions It is convenient to use the formula in Eq. However, Eq. The connection of these starting points can, of course, be accomplished in many ways. The answers in the text are incorrect. This formula is equivalent to the corrected form of the answer. The term of the integral containing a linear factor y vanishes due to its odd symmetry.

Combining the recursion formulas in Eqs. Expanding, discarding terms that vanish due to orthogonality, and using the normalization integral, Eq. Substituting this form into the integral of this exercise, invoking orthogo- nality and using the normalization integral, Eq. To verify an operator identity we must show that the two operators in- volved produce identical results when applied to an arbitrary function.

Then we expand J as a binomial series: When the hint is inserted in the integral, each term can be evaluated using Eq. All the function values on the left-hand side of this equation are given in Eq. Compare with the method of solution to Exercise Because U n and T n satisfy the same recurrence formula, and it is independent of n, so also does V n. For T n x , using the expansion in Eq. When this form of T. Note that the right-hand-side integrals are convergent and their cancella- tion is therefore legitimate.

Note that all instances of x and dx in the orthogonality integral should have been primed, i. The shift compresses the original T n into half the original range; this would decrease the orthogonality integral by a factor of 2. These factors of 2 cancel, so the orthogonality condition given for the shifted T n has the same value as that given for the T n.

Here B is a beta function; the integral was evaluated using Eq. This integral can be evaluated by writing it entirely in terms of complex exponentials, expanding the mth power by the binomial theorem, and noting that nearly all the resulting integrals cancel.

Alternatively, it can be recognized as a case of Formula 3. Academic Press, The result can be brought to the same form as in part a. Evaluating these integrals leads to the results given in Eq.

This integral leads to the formulas in Eq. See the solution to part a. To reach the form given in the text, use the recurrence formula, Eq. To derive the second formula of this exercise, start from Eq. Alternatively, the integral can be evaluated by table lookup. The two steps in the forgoing analysis were the use of the basic Legendre recurrence formula, Eq. This observation leads immediately to the expansion in the text.

These recurrence relations are those in Eqs. It is somewhat easier to work backward from the answers than to derive them. A derivation could use the following notions: Parts b , c , and d are transformed in a similar fashion. The representation of part a has a leading factor that was obtained while solving Exercise The other leading factors are checked in the same way. The series in inverse powers for Q l x is given in a convenient form in Exercise The hypergeometric series with which this expansion is to be compared is, from Eq.

We also note that 2 s s! Inserting these relationships, the two forms for Q l are brought into corre- spondence. Introduce a binomial expansion in the integral for B x and perform the integration in t termwise. Inserting the value of the beta function, we obtain the desired result. Use the integral representation of Exercise A simple approach is to use Eq.

The power-series expansion of the error function, Eq. Starting from the formula for y in the exercise, y.

Adding these equations together, we form the ODE relevant to this exer- cise: Using the formulas from Section Note now that, using Eq. Dividing through by n! Note that Eq. If in Eq. In the integral representation, Eq. The generalization to arbitrary derivatives corresponds to the formula in the text. By choosing these terms, we eliminate all x dependence from the integrand except for the single positive factor e xt.

The result is that the entire quantity multiplying e xt now vanishes, and the endpoint integrated terms vanish as well. Moreover, the representation given for U cannot contain M because it vanishes at large x for all parameter values for which the integral converges. The factor multiplying the integral for U is that needed for correct asymptotic behavior; for a proof see the additional readings.

This procedure was used to solve Exercise This formula was derived as a step in the solution of Exercise When a linear second-order ODE is written in self-adjoint form as [p x y.

The proportionality constant, which may depend upon the parameters a and c, may be determined from the behavior of M and U at any convenient value of x. Using the asymptotic values of M and U from Exercise If a is zero or a negative integer, M does not exist, and the Wronskian evaluates to zero.

The Coulomb wave equation is of the form of Eq. This will cause the y. The demonstration is straightforward. As seen in Exercise Thus, the series is convergent for all z on the unit circle. This means that in using Eq. Rewrite Eq. Expand the integrand in the trigonometric form of E m: Here the arguments of K and E are implicitly assumed to be m.

Here E k 2 and K k 2 need to be expanded in power series, and the answer given in the text suggests that we must keep explicit terms in the expansions though k 4. In this exercise, all instances of E and K without arguments refer respec- tively to E k 2 and K k 2. Simplify by rewriting the numerator of the integrand: Before solving this problem, we follow the hint and establish the equation it provides.

Fourier Series Expand f x in a Fourier series.