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The seventh edition of Mathematical Methods for Physicists is a substantial and detailed revision of its predecessor. The changes extend not only to the topics and their presentation, but also to the exercises that are an important part of the student experience. The new edition contains 271 exercises that were not in previous editions, and there has been a wide-spread reorganization of the previously existing exercises to optimize their placement relative to the material in the text. Since many instructors who have used previous editions of this text have favorite problems they wish to continue to use, we are providing detailed tables showing where the old problems can be found in the new edition, and conversely, where the problems in the new edition came from. We have included the full text of every problem from the sixth edition that was not used in the new seventh edition. Many of these unused exercises are excellent but had to be left out to keep the book within its size limit. 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 1400 problems. Not part of this Instructor’s Manual but available from Elsevier’s on-line web site are three chapters that were not included in the printed text but which may be important to some instructors. These include
In addition, also on-line but external to this Manual, is a chapter (designated
Page 665 Exercise 14.2.4 Change Eq. (11.49) to Eq. (14.44).
Page 686 Exercise 14.5.5 In part (b), change l to h in the formulas for amn and bmn (denominator and integration limit).
Page 687 Exercise 14.5.14 The index n is assumed to be an integer.
Page 695 Exercise 14.6.3 The index n is assumed to be an integer.
Page 696 Exercise 14.6.7(b) Change N to Y (two occurrences).
Page 709 Exercise 14.7.3 In the summation preceded by the cosine function, change (2z)^2 s^ to (2z)^2 s+1.
Page 710 Exercise 14.7.7 Replace nn(x) by yn(x).
Page 723 Exercise 15.1.12 The last formula of the answer should read P 2 s(0)/(2s + 2) = (−1)s(2s − 1)!!/(2s + 2)!!.
Page 754 Exercise 15.4.10 Insert minus sign before P (^1) n(cos θ).
Page 877 Exercise 18.1.6 In both (a) and (b), change 2π to
2 π.
Page 888 Exercise 18.2.7 Change the second of the four members of the
first display equation to
x + ip √ 2
ψn(x), and
change the corresponding member of the second display equation to
x − ip √ 2
ψn(x).
Page 888 Exercise 18.2.8 Change x + ip to x − ip.
Page 909 Exercise 18.4.14 All instances of x should be primed.
Page 910 Exercise 18.4.24 The text does not state that the T 0 term (if present) has an additional factor 1/2.
Page 911 Exercise 18.4.26(b) The ratio approaches (πs)−^1 /^2 , not (πs)−^1.
Page 915 Exercise 18.5.5 The hypergeometric function should read
2 F 1
( (^) ν 2 +^
1 2 ,^
ν 2 + 1;^ ν^ +^
3 2 ;^ z
Page 916 Exercise 18.5.10 Change (n − 12 )! to Γ(n + 12 ).
Page 916 Exercise 18.5.12 Here n must be an integer.
Page 917 Eq. (18.142) In the last term change Γ(−c) to Γ(2 − c).
Page 921 Exercise 18.6.9 Change b to c (two occurrences).
Page 931 Exercise 18.8.3 The arguments of K and E are m.
Page 932 Exercise 18.8.6 All arguments of K and E are k^2 ; In the integrand of the hint, change k to k^2.
Page 978 Exercise 20.2.9 The formula as given assumes that Γ > 0.
Page 978 Exercise 20.2.10(a) This exercise would have been easier if the book had mentioned the integral
representation J 0 (x) =
π
0
cos xt √ 1 − t^2
dt.
Page 978 Exercise 20.2.10(b) Change the argument of the square root to x^2 − a^2.
Page 978 Exercise 20.2.11 The l.h.s. quantities are the transforms of their r.h.s. counterparts, but the r.h.s. quantities are (−1)n^ times the transforms of the l.h.s. expressions.
Page 978 Exercise 20.2.12 The properly scaled transform of f (μ) is (2/π)^1 /^2 injn(ω), where ω is the transform variable. The text assumes it to be kr.
Page 980 Exercise 20.2.16 Change d^3 x to d^3 r and remove the limits from the first integral (it is assumed to be over all space).
Page 980 Eq. (20.54) Replace dk by d^3 k (occurs three times)
Page 997 Exercise 20.4.10 This exercise assumes that the units and scaling of the momentum wave function correspond to the formula ϕ(p) =
(2πℏ)^3 /^2
ψ(r) e−ir·p/ℏ^ d^3 r.
Page 1007 Exercise 20.6.1 The second and third orthogonality equa- tions are incorrect. The right-hand side of the second equation should read: N , p = q = (0 or N/2); N/2, (p + q = N ) or p = q but not both; 0, otherwise. The right-hand side of the third equation should read: N/2, p = q and p + q 6 = (0 or N ); −N/2, p 6 = q and p + q = N ; 0, otherwise.
Page 1007 Exercise 20.6.2 The exponentials should be e^2 πipk/N^ and e−^2 πipk/N^.
Page 1014 Exercise 20.7.2 This exercise is ill-defined. Disregard it.
Page 1015 Exercise 20.7.6 Replace (ν − 1)! by Γ(ν) (two occurrences).
Page 1015 Exercise 20.7.8 Change M (a, c; x) to M (a, c, x) (two
1.1.1. (a) If un < A/np^ the integral test shows
n un^ converges for^ p >^1. (b) If un > A/n,
n un^ diverges because the harmonic series diverges.
1.1.2. This is valid because a multiplicative constant does not affect the conver- gence or divergence of a series.
1.1.3. (a) The Raabe test P can be written 1 +
(n + 1) ln(1 + n−^1 ) ln n
This expression approaches 1 in the limit of large n. But, applying the Cauchy integral test, (^) ∫ dx x ln x
= ln ln x,
indicating divergence. (b) Here the Raabe test P can be written
n + 1 ln n
ln
n
ln^2 (1 + n−^1 ) ln^2 n
which also approaches 1 as a large-n limit. But the Cauchy integral test yields (^) ∫ dx x ln^2 x
ln x
indicating convergence.
1.1.4. Convergent for a 1 − b 1 > 1. Divergent for a 1 − b 1 ≤ 1.
1.1.5. (a) Divergent, comparison with harmonic series.
(b) Divergent, by Cauchy ratio test. (c) Convergent, comparison with ζ(2). (d) Divergent, comparison with (n + 1)−^1. (e) Divergent, comparison with 12 (n + 1)−^1 or by Maclaurin integral test.
1.1.6. (a) Convergent, comparison with ζ(2). (b) Divergent, by Maclaurin integral test. (c) Convergent, by Cauchy ratio test.
(d) Divergent, by ln
n
n
(e) Divergent, majorant is 1/(n ln n).
1.1.7. The solution is given in the text.
1.1.8. The solution is given in the text.
1.1.10. In the limit of large n, un+1/un = 1 +
n
Applying Gauss’ test, this indicates divergence.
1.1.11. Let sn be the absolute value of the nth term of the series.
(a) Because ln n increases less rapidly than n, sn+1 < sn and limn→∞ sn =
1 2 n + 1
2 n + 2
2 n + 3
2 n + 4
this series converges. With all signs positive, this series is the harmonic series, so it is not aboslutely convergent. (c) Combining adjacent terms of the same sign, the terms of the new series satisfy
, etc.
The general form of these relations is
2 n n^2 − n + 2
sn >
n + 1
1.1.16. (a) Write
ζ(3) = 1 +
n=
n^3
n=
(n − 1)n(n + 1)
n=
n^3
n(n^2 − 1)
n=
n^3 (n^2 − 1)
(b) Now use α′ 2 and α′ 4 =
n=
n(n^2 − 1)(n^2 − 4)
ζ(3) = 1 +
n=
n^3
n=
n(n^2 − 1)
α′ 2 −
n=
n(n^2 − 1)(n^2 − 4)
n=
n^3
n(n^2 − 1)
n(n^2 − 1)(n^2 − 4)
n=
4 − (1 + B)n^2 n(n^2 − 1)(n^2 − 4)
The convergence of the series is optimized if we set B = −1, leading to the final result
ζ(3) =
n=
n(n^2 − 1)(n^2 − 4)
(c) Number of terms required for error less than 5 × 10 −^7 : ζ(3) alone, 999; combined as in part (a), 27; combined as in part (b), 11.
1.2.1. (a) Applying Leibniz’ test the series converges uniformly for ε ≤ x < ∞ no matter how small ε > 0 is. (b) The Weierstrass M and the integral tests give uniform convergence for 1 + ε ≤ x < ∞ no matter how small ε > 0 is chosen.
1.2.2. The solution is given in the text.
1.2.3. (a) Convergent for 1 < x < ∞. (b) Uniformly convergent for 1 < s ≤ x < ∞.
1.2.4. From | cos nx| ≤ 1 , | sin nx| ≤ 1 absolute and uniform convergence follow for −s < x < s for any s > 0.
1.2.5. Since | uj+ uj
| ∼ |x|^2 , |x| < 1 is needed for convergence.
1.2.6. The solution is given in the text.
1.2.7. The solution is given in the text.
1.2.8. (a) For n = 0, 1 , 2 ,... we find
d^4 n+1^ sin x dx^4 n+
0
= cos x| 0 = 1,
d^4 n+2^ sin x dx^4 n+
0
= − sin x| 0 = 0,
d^4 n+3^ sin x dx^4 n+
0
= − cos x| 0 = − 1 ,
d^4 n^ sin x dx^4 n
0
= sin x| 0 = 0.
Taylor’s theorem gives the absolutely convergent series
sin x =
n=
(−1)n^
x^2 n+ (2n + 1)!
(b) Similar derivatives for cos x give the absolutely convergent series
cos x =
n=
(−1)n^
x^2 n (2n)!
1.2.9. cot x =
x
x 3
x^3 45
2 x^5 945
− · · · , −π < x < π.
1.2.10. From coth y = η 0 =
ey^ + e−y ey^ − e−y^
e^2 y^ + 1 e^2 y^ − 1 we extract y =
ln
η 0 + 1 η 0 − 1
To check this we substitute this into the first relation, giving
η 0 + 1 η 0 − 1
η 0 + 1 η 0 − 1
= η 0.
The series coth−^1 η 0 =
n=
(η 0 )−^2 n−^1 2 n + 1
follows from Exercise 1.6.1.
1.3.3. The solution is given in the text. Convergent for 0 ≤ x < ∞. 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.
1.3.4. sinh−^1 x = x −
x^3 3
x^5 5 − · · · , − 1 ≤ x ≤ 1.
1.3.5. The expansion of the integral has the form ∫ (^1)
0
dx 1 + x^2
0
1 − x^2 + x^4 − x^6 + · · ·
dx = 1 −
1.3.6. For m = 1, 2 ,... the binomial expansion gives (1+x)−m/^2 =
n=
−m/ 2 n
xn.
By mathematical induction we show that
−m/ 2 n
= (−1)n^
(m + 2n − 2)!! 2 n(m − 2)!!n!
1.3.7. (a) ν′^ = ν
1 ± νc + ν
2 c^2 +^ · · ·
(b) ν′^ = ν
1 ± νc
(c) ν′^ = ν
1 ± νc + 12 ν
2 c^2 +^ · · ·
1.3.8. (a) ν 1 c
= δ + 1/ 2 δ^2.
(b) ν 2 c
= δ − 3 / 2 δ^2 + · · ·.
(c)
ν 3 c
= δ − 1 / 2 δ^2 + · · ·.
w c
α^2 2
α^3 2
1.3.10. x =
gt^2 −
g^3 t^4 c^2
g^5 t^6 c^4
1.3.11. E = mc^2
γ^2 2 n^2
γ^4 2 n^4
n |k|
1.3.12. The solution is given in the text.
1.3.13. The two series have different, nonoverlapping convergence intervals.
1.3.14. (a) Differentiating the geometric series
n=
xn^ =
1 − x
for x = exp(−ε 0 /kT )
yields
x (1 − x)^2
n=
nxn. Therefore, 〈ε〉 =
ε 0 x 1 − x
ε 0 eε^0 /kT^ − 1
(b) Expanding
y ey^ − 1
y 2
〈ε〉 = kT (1 +
ε 0 2 kT
1.3.15. (a) tan−^1 x =
n=
(−1)n
∫ (^) x
0
t^2 ndt =
n=
(−1)n 2 n + 1
x^2 n+1, |x| ≤ 1.
(b) Writing x = tan y as ix = e^2 iy^ + 1 e^2 iy^ − 1
we extract y = − i 2
ln 1 + ix 1 − ix
1.3.16. Start by obtaining the first few terms of the power-series expansion of the expression within the square brackets. Write
2 + 2ε 1 + 2ε
1 + 2ε
= 2 − 2 ε + (2ε)^2 − · · · ,
ln(1 + 2ε ε
ε
2 ε − (2ε)^2 2
(2ε)^3 3
2 + 2ε 1 + 2ε
ln(1 + 2ε ε
ε^2 + O(ε^3 ).
Inserting this into the complete expression for f (ε), the limit is seen to be 4 /3.
1.3.17. Let x = 1/A, and write xi 1 = 1 + (1 − x)^2 2 x
ln 1 − x 1 + x
Expanding the logarithm,
ξ 1 = 1 +
(1 − x)^2 2 x
− 2 x −
2 x^3 3
= 2x −
x^2 +
x^3 − · · ·.
The similar expansion of ξ 2 = 2 x 1 + 2x/ 3
yields
ξ 2 = 2x −
x^2 +
x^3 − · · ·.
Comparing these expansions, we note agreement through x^2 , and the x^3 terms differ by (2/9)x^3 , or 2/ 9 A^3.
1.3.18. (a) Insert the power-series expansion of arctan t and carry out the inte- gration. The series for β(2) is obtained. (b) Integrate by parts, converting ln x into 1/x and 1/(1+x^2 ) into arctan x. The integrated terms vanish, and the new integral is the negative of that already treated in part (a).
with integral ∫ (^) x
−x
dt 1 − t^2
ln(1 + x) − ln(1 − x)
x
−x
ln
1 + x 1 − x
)x
−x
The upper and lower limits give the same result, canceling the factor 1/2.
1.5.2. Start by writing the partial-fraction expansion for p + 1 using the assumed form of that for p multiplied by an additional factor 1/(n + p + 1). Thus, we want to see if we can simplify
1 p!
∑^ p
j=
p j
(−1)j n + j
n + p + 1
to get the expected formula. Our first step is to expand the two factors containing n into partial fractions:
1 (n + j)(n + p + 1)
p + 1 − j
n + j
n + p + 1
Replacing the 1/(n + j) term of our original expansion using this result and adding a new 1/(n + p + 1) term which is the summation of the above result for all j, we reach
∑^ p
j=
(−1)j n + j
p!
p j
p + 1 − j
∑^ p
j=
p!
p j
p + 1 − j
(−1)j−^1 n + p + 1
Using the first formula supplied in the Hint, we replace each square bracket by the quantity 1 (p + 1)!
p + 1 j
thereby identifying the first summation as all but the last term of the partial-fraction expansion for p + 1. The second summation can now be written 1 (p + 1)!
∑^ p
j=
(−1)j−^1
p + 1 j
n + p + 1
Using the second formula supplied in the Hint, we now identify the quan- tity within square brackets as
p∑+
j=
(−1)j−^1
p + 1 j
− (−1)p
p + 1 p + 1
p + 1 0
= 1 + (−1)p+1^ − 1 = (−1)p+1,
so the second summation reduces to (−1)p+ (p + 1)!
n + p + 1
as required. Our proof by mathematical induction is now completed by observing that the partial-fraction formula is correct for the case p = 0.
1.5.3. The formula for un(p) follows directly by inserting the partial fraction decomposition. If this formula is summed for n from 1 to infinity, all terms cancel except that containing u 1 , giving the result ∑^ ∞
n=
un(p) =
u 1 (p − 1) p
The proof is then completed by inserting the value of u 1 (p − 1).
1.5.4. After inserting Eq. (1.88) into Eq. (1.87), make a change of summation variable from n to p = n − j, with the ranges of j and p both from zero to infinity. Placing the p summation outside, and moving quantities not dependent upon j outside the j summation, reach
f (x) =
p=
(−1)pcp
xp (1 + x)p+
j=
p + j j
x 1 + x
)j .
Using now Eq. (1.71), we identify the binomial coefficient in the above equation as (^) ( p + j j
= (−1)j
−p − 1 j
so the j summation reduces to ∑^ ∞
j=
−p − 1 j
x 1 + x
x 1 + x
)−p− 1 = (1 + x)p+1.
Insertion of this expression leads to the recovery of Eq. (1.86).
1.5.5. Applying Eq. (1.88) to the coefficients in the power-series expansion of arctan(x), the first 18 an (a 0 through a 17 ) are: 0, −1, 2, − 8 /3, 8/3, − 28 /15, 8/15, 64/105, − 64 /105,− 368 /15, 1376/315, 1376/315, − 25216 /3465, 25216/3465, − 106048 /45045, − 305792 /45045, 690176/45045, − 690176 /45045, 201472/765765. Using these in Eq. (1.87) for x = 1, the terms through a 17 yield the approximate value arctan(1) ≈ 0 .785286, fairly close to the exact value at this precision, 0.785398. For this value of x, the 18th nonzero term in the power series is − 1 /35, showing that a power series for x = 1 cut off after 18 terms would barely give a result good to two significant figures. The 18-term Euler expansion yields arctan(1/
