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Equalities with a Large Number of Terms in Differential equations

    For the proof of an equality with a large number of terms, all authors up to now omit the proof and its discussion altogether. It might take readers a week to finish the needed calculations. These authors only say that one can prove it if one is patient and careful enough. Maybe they lack feelings and have nothing to say in response to monumental experiences. Maybe they are so kind that they do not want to torture the eyes of a typesetter. Maybe they think that the proof only requires brutal force as if nothing else could be learned. Let us consider the following two questions: How do we find the most effective method to prove the equality? How do we debug errors we might make in the middle of a proof so that we may avoid committing more mistakes? In my opinion, answering these two questions is essential for one to build a solid theoretical background. Furthermore, to avoid any discussion of the proof is irresponsible to the readers who desperately need assistance.
  1. [Kre, p.94, (27.15)].
        Suppose A, B, and C are the first, second and third determinants of [Kre, p.94, l.7] respectively. Let D and E be the first and second determinants of [Kre, p.94, (27.15)] respectively. We want to prove AB - C2 = D - E. Our strategy is to use the following methods to reduce calculations:
    1. (Using symmetry) E is a 3´3 determinant. Each entry has three terms. Therefore, E contains 33´6 terms. Using symmetry we may write
      E = (x12+y12+z12)(x22+y22+z22)(x122+y122+z122)
      +2(x12x1+y12y1+z12z1)(x1x2+y1y2+z1z2)(x2x12+y2y12+z2z12)
      -(x12x1+y12y1+z12z1)2(x22+y22+z22)
      -(x1x2+y1y2+z1z2)2(x122+y122+z122)
      -(x12+y12+z12)(x12x2+y12y2+z12z2)2.
    2. (Using cancellations) Cancellations reduce E to 21 terms. Similarly, cancellations reduce D to 33 terms.
      The expansion of       AB - C2 contains 57 terms.
    3. (Using signs) We divide the 54 (=21+33) terms on the right-hand side into two groups according to the sign of each term. Choose any term from the left-hand side and check to see if the term is also on the right-hand side. If the sign of the term is +, then we need not check the group with minus signs.
    4. (Using factorizations)
          Obviously, the term x2z1y1z2y11y22 does not appear in (x12+y12+z12)(x22+y22+z22)(x11x22+y11y22+z11z22) (a part of the expansion of D).
    Remark. After cancellation, the right-hand side of [Kre, p.91, (27.6)] contains only 54 terms, while the right-hand side of [Lau, p.65, l.-13] contains 9072 terms. Thus, in terms of efficiency of calculation, [Lau, p.65, Theorem 5.5.1] fails to live up to its name (an excellent theorem) even though it is the tensor version of [Kre, p.91, Theorem 27.1].

  2. [Ber, p. 49, (4.1.2)].
    Proof. We preserve intermediate steps in order to reveal cancellations. Let
    x1 = r sinq cosf,
    x2 = r sinq sinf,
    and x3 = r cosq.
    dx1 = dr sinq cosf + r d (sinq cosf),
    dx2 = dr sinq sinf + r d (sinq sinf),
    and dx3 = dr cosq + r d (cosq).
    (dx1)2 = (dr sinq cosf)2 + r2 [d (sinq cosf)]2 + 2rdr sinq cosf (cosq dq cosf - sinq sinf df),
    (dx2)2 = (dr sinq sinf)2 + r2 [d (sinq sinf)]2 + 2rdr sinq sinf (cosq dq sinf + sinq cosf df),
    and (dx3)2 = (dr cosq)2 + r2 [d (cosq)]2 - 2rdr sinq cosq dq.

  3. (d cos a/du)2 + (d cos b/du)2 + (d cos g/du)2 = (1+p2+q2) -2 T -4 [For, p.17, l.10].
    Hint. T'2 = (p'p"+q'q"+p2q'q"-pqq'p"-pqp'q"+q2p'p")2/[p'2+q'2+(pq'-qp')2].
    TT' = pp"+qq"+p2q'q"-pqq'p"-pqp'q"+q2p'p" (get rid of the denominator of T' or consider TT' the derivative of T2/2).
    T4[(d cos a/du)2 + (d cos b/du)2 + (d cos g/du)2]
    = (Tq"-q'T')2 + (-Tp"+p'T')2 + (Tpq"-Tqp"-pq'T'+qp'T')2
    = (p'p"+q'q"+p2q'q"-pqq'p"-pqp'q"+q2p'p")2 (the sum of the coefficients of T'2 cancels the denominator of T'2)
    + T2(p"2+q"2+p2q"2+q2p"2-2pqp"q")
    - 2(TT')2 (recognize the pattern: the sum of the coefficients of TT' is -2TT').

  4. (Systematic reduction due to the intrinsic structure)
    Prove ds ds sin c = V(dpdq - dpdq) [For, p.35, l.-9].
    Proof. cos2 c +sin2 c
    = E[E(dp/ds)2+2F(dp/ds)(dq/ds)+G(dq/ds)2](dp/ds)2
    + 2F[E(dp/ds)2+2F(dp/ds)(dq/ds)+G(dq/ds)2](dp/ds)(dq/ds)
    + G[E(dp/ds)2+2F(dp/ds)(dq/ds)+G(dq/ds)2](dq/ds)2
    = (ds2/ds2)[E(dp/ds)2+2F(dp/ds)(dq/ds)+G(dq/ds)2] = 1.

  5. High-level cancellations versus low-level cancellations.
        Low-level cancellations are term by term cancellations. High-level cancellations are cancellations of groups of terms. It would be too much work to perform cancellations at the lowest level even though success would be guaranteed. If possible, we would like to use higher-level cancellations because these cancellations are more efficient.
    Example 1. Prove V(2 V/ x12) = GV2 [For, p.37, l.1].
    Proof. V(2 V/ x12) = y22+z22-y22Z2+2y2z2YZ-z22Y2.
    GX2 = x22+y22+z22-x22Y2-y22Y2-z22Y2-x22Z2-y22Z2-z22Z2.
    Up to this stage, we retain Y and Z as they are. We need not use [For, p.36, l.15] to expand them. Due to these high-level cancellations, it is enough to prove x22X2 = (y2Y+z2Z)2. Only for this step is it necessary to expand X, Y and Z using [For, p.36, l.15].
    Example 2. Prove V(2 V/ x1 y1) = GXY [For, p.37, l.4].
        In this case, unlike Example 1, if we want to compare the terms on the two sides for possible cancellations, we must expand X, Y and Z using [For, p.36, l.15]. We have no way to retain high-level quantities X, Y and Z and enjoy cancellations at the same time.

  6. Using the hierarchy of mathematical structures as the guide to find the right direction for cancellations.
     Example. Prove the coefficient of a1 is 0 [For, p.53, l.9].
        A good start is half of the battle. If we try to perform cancellations using the high level relations in [For, p.48, l.-8-p.49. l.6], it will lead nowhere. Consequently, we follow the hierarchy to go to a lower level.
    (/q)(s/V2) = (1/V2)(s/q)-2(s/V3)(V/q)
    = (F2M+FM2-G2L-GL2)/V2-2s(G'+D")/V2.
    Then we use [For, p.44, l.-1; p.45, l.8] to expand G2 and use [For, p.44, l.-4; p.45, l.7-l.8] to expand F2.

  7. Do not open Pandora's box. Otherwise, you will unnecessarily divide a proof into numerous cases.
    Example. Prove x1x2+y1y2+z1z2 = 0 [Wea1, vol. 1, p.125, (5); p.131, (16)].
    Proof. The trick is to use the identity x1x2 = [(xx1)(xx2)]/(x2),
    where xx1 and xx2 can be easily derived from [Wea1, vol. 1, p.130, (15)].
    Remark. Do not try to find x1 from x12 because you will have a difficult time determining the sign of x1. Even if you are able to figure out the sign of x1 case by case, you still unnecessarily complicate the proof by dividing it into numerous cases.

  8. Once you recognize a pattern, you will see the light at the end of the tunnel.
    Example. Prove [Wea1, vol. 1, p.185, (7)].
        Keep (b+b') as one unit. If it is a factor, do not try to expand the product by breaking the unit apart. After you substitute [Wea1, vol. 1, p.185, (5)] into the left-hand side of [Wea1, vol. 1, p.185, (7)], you try to find the numerator by calculating its coefficients of (du)4, (dv)4, (du)3(dv), (du)(dv)3, and (du)2(dv)2. You hope every coefficient is zero, but none of them are. The calculations for the first pair are easiest. The calculations for the second pair are more difficult. The calculations for the term (du)2(dv)2 are most difficult. After you calculate the coefficients for the first pair, you will find their coefficients are the corresponding coefficients of
    (-1/4)[Wea1, vol. 1, p.185, (6)]2.
    Then the rest calculations become routine and any error that you commit can be easily debugged.

  9. (Creating checkpoints) The proof of [Gon1, p.429, (5.17-20)] requires long calculations. It is frustrating if one discovers that one's final answer does not match the answer given in the book. One might miss a term; one might copy a term incorrectly. One does not know where one has made a mistake. If one has made a mistake, then all one's work beyond that point is simply a waste. Consequently, it is important to create some checkpoints along the way to help one detect one's mistakes at an early stage and to establish the reliability of one's work by passing some tests at the checkpoint. If one can pass these tests, then it is less likely that one's work before the checkpoint contains any mistakes. For example, one may want to prove the simpler formula [Gon1, p.429, (5.17-21)] first (the first checkpoint) because its highest power is 6 instead of 7. If one can prove [Gon1, p.429, (5.17-21], the proof at least ensures that one's Taylor series expansion of Tan2 (z/2) is correct. Then one may use this correct expansion to prove the more complicated formula [Gon1, p.429, (5.17-20)]. After one expresses the coefficient f(l) of z7 in terms of l, one may want to check if f(1) = - (7!)-1 (the second checkpoint) before one actually substitutes l = 1 - 2k2 into f(l). When one substitutes l = 1 - 2k2 into f(l), one should check the coefficient of k4 first (the third checkpoint). This will give one another chance to ensure that the coefficients of the powers of l in f(l) are correct before one calculates the more complicated coefficient of k2.

  10. (Reducing gamma functions to sine functions)
        [Guo, p.160, (3)] can be proved as follows:
    By [Guo, p.99, (2)], it suffices to prove
    [sin (pa)]-1 [sin (pb)]-1([sin (p[g-a])]-1 [sin (p[g-b])]-1 - [sin (pg)]-1 [sin (p[g-a-b])]-1)
    = ([sin (pg)][sin (p[g-a])][sin (p[g-b])][sin (p[a+b-g])])-1.


  11. (Choice will become clear after cancellation)
    Prove the equality given in [Wat1, p.291, l.-14-l.-13].
    Proof. Let A satisfy
    (d2I/dz2)+{(1+a-a')(z-a)-1+(1+b-b')(z-b)-1+(1+g-g')(z-c)-1}(dI/dz)+{A[(z-a)(z-b)(z-c)]-1}I = 0.
    It suffices to prove A = (a+b+g){(a+b+g+1)z + Sa(a+b'+g'-1)}.
    (1). u = (z-a)a(z-b)b(z-c)g.
    du/dz = [a(z-a)a-1(z-b)b(z-c)g+b(z-a)a(z-b)b-1(z-c)g+g(z-a)a(z-b)b(z-c)g-1]I+(z-a)a(z-b)b(z-c)g(dI/dz).
    d2I/dz2 = {a(a-1)(z-a)a-2(z-b)b(z-c)g+b(b-1)(z-a)a(z-b)b-2(z-c)g+g(g-1)(z-a)a(z-b)b(z-c)g-1
    +2[ab(z-a)a-1(z-b)b-1(z-c)g+ag(z-a)a-1(z-b)b(z-c)g-1+bg(z-a)a(z-b)b-1(z-c)g-1]}I
    +2[a(z-a)a-1(z-b)b(z-c)g+b(z-a)a(z-b)b-1(z-c)g+g(z-a)a(z-b)b(z-c)g-1](dI/dz)
    +(z-a)a(z-b)b(z-c)g(d2I/dz2).
    (2). The part of A that the I term of {(1-a-a')(z-a)-1+(1-b-b')(z-b)-1+(1-g-g')(z-c)-1}du/dz contributes is
    {(1-a-a')(z-a)-1+(1-b-b')(z-b)-1+(1-g-g')(z-c)-1}{a(z-a)-1+b(z-b)-1+g(z-c)-1}(z-a)(z-b)(z-c)
    = (1-a-a')(z-a)-1a(z-b)(z-c)+(1-b-b')(z-b)-1b(z-a)(z-c)+(1-g-g')(z-c)-1g(z-a)(z-b)
    +(1-a-a')(bz-bc+gz-gb)+(1-b-b')(az-ac+gz-ga)+(1-g-g')(az-ab+bz-ba).
    Let B = (1-a-a')(z-a)-1a(z-b)(z-c)+(1-b-b')(z-b)-1b(z-a)(z-c)+(1-g-g')(z-c)-1g(z-a)(z-b)
    and C = (1-a-a')(bz-bc+gz-gb)+(1-b-b')(az-ac+gz-ga)+(1-g-g')(az-ab+bz-ba).
    (3). The part of A that the I term of d2I/dz2 contributes is
    [a(a-1)(z-a)-2+b(b-1)(z-b)-2+g(g-1)(z-c)-2
    +2{ab[(z-a)(z-b)]-1+ag[(z-a)(z-c)]-1+ag[(z-a)(z-c)]-1}](z-a)(z-b)(z-c)
    = a(a-1)(z-a)-1(z-b)(z-c)+b(b-1)(z-b)-1(z-a)(z-c)+g(g-1)(z-c)-1(z-a)(z-b)
    +[2abz-2abc+2agz-2agb+2bgz-2bga].
    Let D = a(a-1)(z-a)-1(z-b)(z-c)+b(b-1)(z-b)-1(z-a)(z-c)+g(g-1)(z-c)-1(z-a)(z-b),
    E = 2abz-2abc+2agz-2agb+2bgz-2bga,
    and F = aa'(a-b)(a-c)(z-a)-1+bb'(b-c)(b-a)(z-b)-1+gg'(c-a)(c-b)(z-c)-1.
    Then G=B+D+F=aa'(-z-a+b+c)+bb'(-z-b+a+c)+gg'(-z-c+a+b).
    (4). J = a(a+b'+g'-1)(a+b+g)+ b(b+g'+a'-1)(a+b+g)+c(g+a'+b'-1)(a+b+g)
    = aa(a+b'+g'-1+b+g)+bb(a+b+g'+a'-1+g)+cg(a+b+g+a'+b'-1)+H
    = (-aa'a-bb'b-gg'c)+H.
    A=E+G+C
    =[-aa'-bb'-gg'-a'g-b'a-b'g-ag'-g'b+2a+2b+2g]z+(-aa'a-bb'b-gg'c)+H (after cancellation)
    = (a+b+g)(2-a'-b'-g')z+(-aa'a-bb'b-gg'c)+H
    = (a+b+g)(a+b+g+1)z+J.
    Remark. Before cancellation, we do not know whether we should use 2-a'-b'-g' or a+b+g+1. After cancellation, the answer is clear.

  12. Suppose we want to use [Wat1, p.308, (II) and (III)] to prove [Wat1, p.308, (I)]. See [Wat1, p.309, l.20]. Compare terms in [Wat1, p.308, (III)] with those in [Wat1, p.308, (I)]. All the terms in [Wat1, p.308, (III)] appear in [Wat1, p.308, (I)] except P'n-1(z). After differentiating [Wat1, p.308, (II)], we try to use the resulting equality and [Wat1, p.308, (III)] to cancel the term P'n-1(z).


  13. Prove the formula given in [Wat1, p.317, l.13].
    Proof. I. Express the integrals given in [Wat1, p.317, l.10] in terms of the values of the gamma function [Wat1, p.253, l.-6; p.254, l.-9].
    II. Find the power series expansion in z-2 for F(2-1(n+1), (n/2)+1; n+(3/2); z-2). Use [Guo, p.94, (6)] to express the coefficients of the series in terms of the values of the gamma function.
    III. Use the duplication formula [Wat1, p.240, l.-3].
    Remark 1. The gamma function is a useful tool to replace any quantity that contains many factors such as a factorial or a notation that can be expressed in terms of the gamma function [Guo, p.94, (6)].
    Remark 2. In order to expand the domain of Qn with respect to n, it is unnecessary to repeat the same procedure as in [Wat1, p.317, l.-12-l.-4]
    14. . All we have to do is find its analytic continuation with respect to n using [Wat1, p.288, l.3, l.6, & l.10].

  14. Prove the equality given in [Wat1, p.319, l.-14].
    Proof. 1 - t2 = 4eq(z+1)1/2 (z-1)1/2[eq(z-1)1/2 + (z+1)1/2]-2.
    z - t = [zeq(z+1)1/2 +z(z-1)1/2 - eq(z+1)1/2 +(z-1)1/2]/[eq(z+1)1/2(z-1)1/2].
    dt = 2eq(z+1)1/2(z-1)1/2[eq(z-1)1/2 + (z+1)1/2]-2.
    [eq(z-1)1/2 + (z+1)1/2] [zeq(z+1)1/2 +z(z-1)1/2 - eq(z+1)1/2 +(z-1)1/2]
    = e2q(z2-1)+2eq(z+1)1/2(z-1)1/2 + z2 - 1
    = (2eq(z+1)1/2(z-1)1/2)[(eq+e-q)/2].
    Remark. In order to fully utilize cancellation, we should not expand [eq(z-1)1/2 + (z+1)1/2]2 except when calculating 1 - t2. When a textbook omits long calculations, it can blur the train of thought or the key to solution.


  15. In order to fully take advantage of cancellation, we must group factors properly.
    Example. Prove the formula given in [Wat1, p.329, l.6].
    Proof. Let A = ei(f-w) cosh (x/2) + sinh (x/2),
    B = ei(f-w) sinh (x/2) + cosh (x/2),
    C = eif cosh (x'/2) + sinh (x'/2),
    D = cosh (x'/2)+ eif sinh (x'/2),
    E = sinh (x/2) sinh (x/2) - eiw cosh (x/2) sinh (x'/2),
    F = cosh (x/2) sinh (x/2) - eiw sinh (x/2) sinh (x'/2), and
    G = eiw sinh x sinh2 (x'/2) + e-iw sinh x cosh2 (x'/2) - cosh x sinh x'
    Then dt/df = ieifGD-2; EF = 2-1eiw G;
    (t-1)(t+1)(t-z)-1 = 2eiwAB(CD)-1
    = 2[x+(x2-1)1/2cos (w-f)][x'+(x'2-1)1/2cos (w-f)]-1;
    (t-z)-1 dt = [Di eif G df][CGD2]-1
    =[i eif df][eif (x'+(x'2-1)1/2cos f)]-1.
    Remark. Analytic continuation is a process of generalization. The process is ineffective: there are many ways to select the contour of the integral given in [Wat1, p.328, l.-8]; some of them are poor choices for calculation. Appropriate specification is the key to obtaining the most effective analytic continuation. We should parameterize t with a parameter f which can reduce calculation by a tremendous amount of cancellation. The choice given in [Wat1, p.328, l.-6] meets this requirement.¬


  16. If one can ensure that the essential formulas are correct, then the remaining task will be easy.
        In proving the equality given in [Wat1, p.340, l.2-l.3], if one can obtain the formula for d2v/dz2 correctly, the remaining task will be easy. It is difficult to make serious mistakes except for proving the following two formulas:
    dv/dt =(k-1/2+m)
    17. ò¥(0+)(-t)-k-1/2+m (1+(t/z))k-1/2+m [(-t)/(z2+zt)]e-t dt.
    d2v/dz2 = (k-1/2+m) ò¥(0+)(-t)-k-1/2+m (1+(t/z))k-1/2+m e-t{[2zt+(k+1/2+m)t2]/(z2+zt)2} dt.
    Remark. Watson leaves out a factor, (-1)-k-1/2+m, on the right-hand-side of the equality given in [Wat1, p.340, l.2-l.3].

  17. If one tries to catch a snake, everything will become easy once one holds its neck. Similarly, if one tries to solve a problem, everything will become easy once one knows how to start.
    Example. Prove [Guo, p.351, (5) & (6)].
    Proof. Even if [Guo, p.351, l.3] provides a hint, it still requires some manipulation of [Guo, p.350, (3); p.351, (4)] to get start.
    [d/(zdz)]n {(sin z)/z} = (-1)n (nz/2)1/2 z-n-1 Jn+1/2(z). ¬
    Remark. We may prove [Guo, p.351, (6)] in a similar manner.

  18. Prove the equality given in [Wat1, p.410, l.-6-l.-3].
    Proof. In this proof, all we need to do is find a general formula to express
    cos 2m z as Sn=0m amn cos 2nz.
    We can write down the recursion formula for amn using mathematical induction: cos 2m z = [(cos 2z +1)/2] cos 2m-2 z.
    Remark. This proof requires the application of the above formulas many times. One should use the formulas rather than repeat the calculations to produce the same formulas many times.
    19. Avoiding unnecessary calculations greatly reduces the trouble for checking errors in a mammoth calculation. One may prove the statements given in [Wat1, p.410, l.-2-l.-1] using §32.2 Fourier Expansions; Functions with q<0 of the following webpage:
    http://www.elsevierdirect.com/companions/9780123846549/Chap_Mathieu.pdf
    In the beginning of this webpage discusses the origins and applications of Mathieu's equation, but they are not the main advantages of this webpage. Its main advantage consists in providing various ideas for solving Mathieu's equation. Without doing exercises, one has no motive to study these new things and cannot sharpen one's problem solving abilities. Mathieu had a deep understanding in theoretic physics. Between 1873 and 1890, he wrote six volumes of mathematical physics (differential equations, capillarity, electrostatics, magnetostatics, electrodynamics and elasticity). He solved differential equations from the viewpoint of physics. Before quantum mechanics was born from laboratory experiments, his mathematical work had already laid the important part of its theoretic foundation. [Coh, vol. 2, chap. XI, Stationary Perturbation Theory] is basically an abstraction and a simplification of the solution to Mathieu's equation. Mathieu is less well-known in the field of quantum mechanics probably because his mathematical work is too difficult to understand for most physicists.

  19. Prove the formula given in [Wat1, p.411, Example 1, (i)].
    Proof. All one needs to do is generalize the method of proving the equality given in [Wat1, p.410, l.-6-l.-3]. There are many aspects that need to be taken care of.
    I. cos 2r z = 2-2r (eiz + e-iz)2r = 2-2r C(2r, r) + 21-2r Sk=0r-1 C(2r, k) cos [(2r - 2k) z].
    II. The qr-term of the coefficient of cos 2rz in the expansion of ce0(z, q) is the coefficient of cos 2rz in br(z).
    (2p)-1ò[-p,p] Sn=0r 32n[(2n)!]-1 cos 2n z cos 2n q br-n (q) dq = Sn=0r an br-n (z)           (*).
    Hence the coefficient of cos 2rz in br(z) is 2r+1[r!r!]-1.
    III. Now consider the coefficient of the qr+2-term of the coefficient of cos 2rz in the expansion of ce0(z, q).
    Since br+1 cannot contain the term cos 2rz, it suffices to consider the coefficient of cos 2rz in br+2(z).
    Replace r with r+2 in (*) and consider only the coefficient of cos 2rz in the expansion of each of the following terms:
    (2p)-1ò[-p,p] [32r+2[[2(r+2)]!]-1 cos 2(r+2) z cos 2(r+2) + 32r+1[[2(r+1)]!]-1 cos 2(r+1) z cos 2(r+1) q b1 (q)
    + 32r[(2r)!]-1 cos 2r z cos 2r q b2 (q) +…] dq = …+ a2br(z) + … + br+2(z).¬


  20. (Double indices)
        Suppose we count summands row by row first, and then count them column by column, and we want to prove the two resulting sums are equal. It would be difficult if one tries to manage the double indices. It is better to prove the equality for a simple case; the solution will become clear as one proceeds.
    Example 1. (The resulting sums are the same no matter whether we count summands row by row or column by column) Prove the statement given in [Cod, p.146, l.-8-l.-7].
    Proof. I. Case r = 3.
    (P1~Q0-A0P1~)D2 = (A1P0~-P0~Q1)D2                                                                         (1)
    (P2~Q0-A0P2~)D1 = [(A1P1~-P1~Q1)+(A2P0~-P0~Q2)]D1                                            (2)
    (P3~Q0-A0P3~)D0 = [(A1P2~-P2~Q1)+(A2P1~-P1~Q2)+(A3P0~-P0~Q3)]D0                  (3)
    The left-hand side of [Cod, p.144, (2.7(ii)] counts summands row by row, while the right-hand of [Cod, p.144, (2.7(ii)] counts summands column by column.
    II. PkQ0-A0Pk
    = Si=0k (Pi~A0-A0Pi~)Dk-i
    = Si=0k Sl=0i (Ak-iPi-l~-Pi-l~Qk-i)Dl (The sum by adding columns = The sum by adding rows)
    = Si=0k (Ak-iPi-PiQk-i).
    Example 2. (The number of double indices are the same no matter whether we count the indices row by row or column by column) Prove [Cod, p.146, (2.17)].
    Proof. By counting the lattice points row by row or column by column, we have the same index set. Consequently,
    Sl=1r Sk=1r+1-l = Sk=1r Sl=1r+1-k.


  21. Prove that y = (cf1+f2)(cf3+f4)-1 satisfies a Riccati equation [Inc1, p.24, l.-15-l.-13].
    Proof. Eliminate C between the expression for y and the derived expression for y'. It suffices to note that
    C = (yf4-f2)(f1-yf3)-1 and Cf3+f4 = (f4f1-f2f3)(f1-yf3)-1. The numerator of Cf3+f4 does not contain y.


  22. The general solution of z2(d2u/d z2)+(2a-2bn-1)z(du/dz)+{b2g2z2b+a(a-2bn)}u = 0 is u = zbn-aCn(gzb) [Wat, p.97, l.11-l.14].
    Remark. Let z = gxb and u = yx-a. Then dz = gbxb-1 dx.
    du/dx=(du/dz)(dz/dx).
    When calculating du/dz, we may treat x-a as a function of z.
    du/dz = x-a(dy/dz)-ax-a-1(dx/dz)y
    = x-a(dy/dz)-ax-a-1[(gb)-1x1-b]y.
    We may also express x-a in terms of z first.
    du/dz=(dy/dz)(z/r)-a/b +y(-a/b)(z/g)-a/b-1g-1;
    then express z in terms of x. The calculation in the former approach is simpler.


  23. u = z2bn-a is a solution of z2(d2u/d z2)+(2a-2bn-1)z(du/dz)+a(a-2bn)u = 0 [Wat, p.97, l.15-l.16].
    Proof. du/dz = (2bn-a)z2bn-a-1 and d2u/d z2 = (2bn-a)(2bn-a-1)z2bn-a-2.
    z2(d2u/d z2)+(2a-2bn-1)z(du/dz)+a(a-2bn)u
    = [(2bn-a)(2bn-a-1)+(2a-2bn+1)(2bn-a)-a(2bn-a)]z2bn-a
    = [(2bn-a)2-(2bn-a)-(2bn-a)+a+1-a(2bn-a)]z2bn-a
    = (2bn-a)[(2bn-a)-1-(2bn-a)+a+1-a]=0.
    Remark. If one were to expand the brackets before cancelling like terms, it would take more calculations to obtain the same result.


  24. d2{y/c(z)}/d{y(z)}2 - (2n-1)[y(z)]-1(d{y/c(z)}/dy(z))+y/c(z) =  0 can be reduced to
    d2y/dz2-{y"(z)/y'(z)+(2n-1)y'(z)/y(z)+2c'(z)/c(z)}dy/dz
    +[{y"(z)/y'(z)+(2n-1)y'(z)/y(z)+2c'(z)/c(z)}c'(z)/c(z) - c"(z)/c(z)+{y'(z)}2]y = 0 [Wat, p.97, l.7-l.9].
    Proof. dy(z)/dz = y'(z).
    d{y/c(z)}/dz = [c(z)dy/dz-yy'(z)]/c2(z).
    d{y/c(z)}/dy(z) = [c(z)dy/dz-yy'(z)]/[y'(z)c2(z)].
    It suffices to note that d[d{y/c(z)}/dy(z)]/dz = y'(z)[d2{y/c(z)}/d{y(z)}2].


  25. Let s = 2im[zw]1/2/t and f(w) is an arbitrary function of w. Then
    u = t-1z-n/2 ò0¥ exp [-m(z+w)/t]Cn(s)f(w)dw is a solution of
    z(2u/z2)+(1+n)(u/z)-m(u/t) = 0 [Wat, p.99, l.8-l.14].
    Proof. z(2u/z2)+(1+n)(u/z)-m(u/t)
    = (4zt)-1z-n/2 ò0¥ exp [-m(z+w)/t]{s2(2Cn(s)/s2)+s(Cn(s)/s)+(s2-n2)Cn(s)}f(w)dw = 0.


  26. S(bn-gm)(a2-l)-1(b2-l)-1 = l2(a2-l)-1+m2(b2-l)-1+n2(c2-l)-1 [Bel63, p.180, l.16].
    Proof. By the equation given in [Bel63, p.180, l.14-l.15],
    (l2(a2-l)-1+m2(b2-l)-1+n2(c2-l)-1)(a2(a2-l)-1+b2(b2-l)-1+g2(c2-l)-1)-(l2(a2-l)-1+m2(b2-l)-1+n2(c2-l)-1)
    =((al(a2-l)-1+bm(b2-l)-1+)2+gn(c2-l)-1)2.
    If we subtract the term on the right-hand side of the above equation from the first term on the left-hand side, we obtain the term on the left-hand side of the desired equation.


  27. Au2+Bv2+Cw2+2Fvw+2Gwu+2Huv-dD = -S [Bel63, p.217, l.-2].
    Proof. It is unnecessary to lower the order of the determinant by one at a time. We expand S along the 4th column. In order to find the cofactor of u, we expand the cofactor along the third row. By inspection, we obtain -(uA+vH+wG).


  28. Let A=(x2+y2+z2)(a2x2+b2y2+c2z2)-a2(b2+c2)x2-c2(a2+b2)z2+a2b2c2,
    B=b2(r2-a2)(r2-c2)+(a2-b2)(r2-c2)x2-(b2-c2)(r2-a2)z2, and
    C={b(r2-a2)+xxc-1(a2-c2)}{b(r2-c2)+zza-1(a2-c2)}+(a2-c2)(ac)-1{(r2-c2)xxc-1-(r2-a2)zza-1}{axx+czz-abc}. Then
    A=B=C [Bel63, p.269, l.-16-l.-12].
    Proof. I. A=B.
    Proof. The constant term for A and B: a2b2c2.
    The terms of the second degree for B: -a2b2r2-b2c2r2+(a2-b2)(-c2)x2-(b2-c2)(-a2)z2.
    The terms of degrees higher than 2 for B: r4b2+(a2-b2)r2x2-(b2-c2)r2z2=r2[r2b2+(a2-b2)x2-(b2-c2)z2].
    II. B=C.
    Proof. Let D=b2(r2-a2)(r2-c2). It suffices to prove B-D=C-D.
    C-D=b(r2-c2)(a2-c2)xxc-1-b(r2-a2)(a2-c2)zza-1-xzxz(ac)-1(a2-c2)2
    +(a2-c2)(ac)-1{(r2-c2)xxc-1-(r2-a2)zza-1}{-abc}+(a2-c2)(ac)-1{(r2-c2)xxc-1-(r2-a2)zza-1}{axx+czz}
    =-xzxz(ac)-1(a2-c2)2+(a2-c2)(ac)-1{(r2-c2)xxc-1-(r2-a2)zza-1}{axx+czz} (by cancellation)
    =(a2-c2)c-2x2(r2-c2)x2-(a2-c2)a-2z2(r2-a2)z2.
    The result follows from [Bel63, p.268, l.-S3].


  29. Given jk (k=0,…,n-2), find jn-1 using Sk=0n-1[k!(n-k)!]-1jk [Guo, p.2, (6)].
    Solution. We may reduce the calculations by the following method:
    jn-1=-Sk=0n-2[Ckn/n]jk=-Sk=0n-2[(n-1)(n-2)…(n-(k-1))][k(k-1)…2]-1]jk.


  30. Links {1, 2}.