Computation for Proposition 14.14 on page 151

For this notebook to be evaluated correctly, the version of Mathematica  must be able to recognize that the coefficient of    (x^(p + q - 1))   in   Expand[ x^(p+q-2)*(1+x+x^2) ]  is  1

Thus, the version of Mathematica must return  True  as the answer to the question

TrueQ[
      Coefficient[Expand[x^(p + q - 2)*(1 + x + x^2) ], 
          x^(p + q - 1) ] == 1]

The version number of Mathematica used to evaluate this notebook:

$VersionNumber

From (14.19), we have

aAA[
        zet_] = f'[zet]^2*
          a[f[zet]] + (f'''[zet]/(2*f'[zet])) - (3/
              4) (f''[zet]/f'[zet])^2

The preceding represents  \(TraditionalForm\`\(\(a\^ ** \)\_\(m, 0, 2\)\)(ζ)\)  and we have chosen  A  to suggest asterisk.

We write

PAA[zet_] = f'[zet]^p*P[f[zet]]

for (14.20).  In particular, we have

D[PAA[zet], zet]

for (14.21) and

Collect[
      D[PAA[zet], {zet, 2}], {P''[f[zet]], P'[f[zet]], 
        P[f[zet]]}, Simplify ]

for (14.22).  Similarly, we introduce

QAA[zet_] = f'[zet]^q*Q[f[zet]]

Now, (14.33) yields

YAA[zet_] = 
        PAA[zet]*D[QAA[zet], {zet, 4}] + 
          y1*D[PAA[zet], zet]*D[QAA[zet], {zet, 3}] + 
          y2*D[PAA[zet], {zet, 2}]*D[QAA[zet], {zet, 2}] + 
          y3*D[PAA[zet], {zet, 3}]*D[QAA[zet], zet] + 
          y4*D[PAA[zet], {zet, 4}]*QAA[zet] + 
          y5*aAA[zet]*PAA[zet]*D[QAA[zet], {zet, 2}] + 
          y6*aAA[zet]*D[PAA[zet], zet]*D[QAA[zet], zet] + 
          y7*aAA[zet]*D[PAA[zet], {zet, 2}]*QAA[zet] + 
          y8*D[aAA[zet], zet]*PAA[zet]*D[QAA[zet], zet] + 
          y9*D[aAA[zet], zet]*D[PAA[zet], zet]*QAA[zet] + 
          y10*D[aAA[zet], {zet, 2}]*PAA[zet]*QAA[zet] + 
          y11*aAA[zet]^2*PAA[zet]*QAA[zet];

and

Y[z_] = 
        P[z]*D[Q[z], {z, 4}] + y1*D[P[z], z]*D[Q[z], {z, 3}] + 
          y2*D[P[z], {z, 2}]*D[Q[z], {z, 2}] + 
          y3*D[P[z], {z, 3}]*D[Q[z], z] + y4*D[P[z], {z, 4}]*Q[z] + 
          y5*a[z]*P[z]*D[Q[z], {z, 2}] + y6*a[z]*D[P[z], z]*D[Q[z], z] + 
          y7*a[z]*D[P[z], {z, 2}]*Q[z] + y8*D[a[z], z]*P[z]*D[Q[z], z] + 
          y9*D[a[z], z]*D[P[z], z]*Q[z] + y10*D[a[z], {z, 2}]*P[z]*Q[z] + 
          y11*a[z]^2*P[z]*Q[z];

For (14.45), we have

differ = f'[zet]^4*( 
              YAA[zet] - f'[zet]^(p + q + 4)*Y[f[zet]] ) // 
          Expand ;

In Version 5.1, the  Coefficient  command functions inefficiently when there are negative numbers in the exponents.  For this reason, we have included the factor  f'[zet]^4 in the preceding expression.

We introduce

s[i_] := Sum[ c[j]*m[j],  {j, 1, i}]

m[1] = P[f[zet]] Q[f[zet]] a'[f[zet]] ((f'[zet])^(6 + p + q)) f''[zet]

c[1] = Coefficient[differ,  
        m[1]]

dif[1]  =  differ  -  s[1]  // Expand;

m[2] =  a[f[zet]] Q[f[zet]] ((f'[zet])^(6 + p + q)) P'[f[zet]] f''[zet]

c[2] = Coefficient[ differ,  
        m[2] ]

dif[2]  =  differ  -  s[2]  // Expand;

m[3] = a[f[zet]] P[f[zet]] ((f'[zet])^(6 + p + q)) Q'[f[zet]] f''[zet]

c[3] = Coefficient[ differ,  
        m[3] ]

dif[3]  =  differ  -  s[3]  // Expand;

m[4] = a[f[zet]] P[f[zet]] Q[f[zet]] ((f'[zet])^(p + q + 4)) ((f''[zet])^2)

c[4] = Coefficient[ differ, 
        m[4] ]

dif[4]  =  differ  -  s[4]  // Expand;

m[5] = ((f'[zet])^(p + q + 4)) P'[f[zet]] Q'[f[zet]] ((f''[zet])^2)

c[5] = Coefficient[ differ,  
        m[5] ]

dif[5]  =  differ  -  s[5]  // Expand;

m[6] = Q[f[zet]] ((f'[zet])^(2 + p + q)) P'[f[zet]] ((f''[zet])^3)

c[6] = Coefficient[ differ,  
        m[6] ]

dif[6]  =  differ  -  s[6]  // Expand;

m[7] = P[f[zet]] ((f'[zet])^(2 + p + q)) Q'[f[zet]] ((f''[zet])^3)

c[7] = Coefficient[ differ,  
        m[7] ]

dif[7]  =  differ  -  s[7]  // Expand;

m[8] = P[f[zet]] Q[f[zet]] ((f'[zet])^(p + q)) ((f''[zet])^4)

c[8] = Coefficient[ differ,  
        m[8] ]

dif[8]  =  differ  -  s[8] // Expand;

m[9] = ((f'[zet])^(6 + p + q)) Q'[f[zet]] f''[zet] P''[f[zet]]

c[9] = Coefficient[ differ,  
        m[9] ]

dif[9]  =  differ  -  s[9]  // Expand;

m[10] = Q[f[zet]] ((f'[zet])^(p + q + 4)) ((f''[zet])^2) P''[f[zet]]

c[10] = Coefficient[ differ,  
        m[10] ]

dif[10]  =  differ  -  s[10]  // Expand;

m[11] = ((f'[zet])^(6 + p + q)) P'[f[zet]] f''[zet] Q''[f[zet]]

c[11] = Coefficient[ differ,  
        m[11] ]

dif[11]  =  differ  -  s[11]  // Expand;

m[12] = P[f[zet]] ((f'[zet])^(p + q + 4)) ((f''[zet])^2) Q''[f[zet]]

c[12] = Coefficient[ differ,  
        m[12] ]

dif[12]  =  differ  - s[12]  // Expand;

m[13] = a[f[zet]] P[f[zet]] Q[f[zet]] ((f'[zet])^(5 + p + q)) Derivative[3][f][zet]

c[13] = Coefficient[ differ,  
        m[13] ]

dif[13]  =  differ  - s[13]  // Expand;

m[14] =  ((f'[zet])^(5 + p + q)) P'[f[zet]] Q'[f[zet]] Derivative[3][f][zet]

c[14] = Coefficient[ differ,  
        m[14] ]

dif[14]  =  differ  -  s[14]  // Expand;

m[15] = Q[f[zet]] ((f'[zet])^(3 + p + q)) P'[f[zet]] f''[zet] Derivative[3][f][zet]

c[15] = Coefficient[ differ,  
        m[15] ]

dif[15]  =  differ  -  s[15]  // Expand;

m[16] =  P[f[zet]] ((f'[zet])^(3 + p + q)) Q'[f[zet]] f''[zet] Derivative[3][f][zet]

c[16] = Coefficient[ differ,  
        m[16] ]

dif[16]  =  differ  -  s[16]  // Expand;

m[17] = P[f[zet]] Q[f[zet]] ((f'[zet])^(1 + p + q)) ((f''[zet])^2) Derivative[3][f][zet]

c[17] = Coefficient[ differ,  
        m[17] ]

dif[17]  =  differ  -  s[17]  // Expand;

m[18] = Q[f[zet]] ((f'[zet])^(5 + p + q)) P''[f[zet]] Derivative[3][f][zet]

c[18] = Coefficient[ differ,  
        m[18] ]

dif[18]  =  differ  -  s[18]  // Expand;

m[19] = P[f[zet]] ((f'[zet])^(5 + p + q)) Q''[f[zet]] Derivative[3][f][zet]

c[19] = Coefficient[ differ,  
        m[19] ]

dif[19]  =  differ  -  s[19]  // Expand;

m[20] = P[f[zet]] Q[f[zet]] ((f'[zet])^(2 + p + q)) ((Derivative[3][f][zet])^2)

c[20] = Coefficient[ differ,  
        m[20] ]

dif[20]  =  differ  -  s[20]  // Expand;

m[21] = Q[f[zet]] ((f'[zet])^(4 + p + q)) f''[zet] Derivative[3][P][f[zet]]

c[21] = Coefficient[ differ,  
        m[21] ]

dif[21]  =  differ  -  s[21]  // Expand;

m[22] = P[f[zet]] ((f'[zet])^(6 + p + q)) f''[zet] Derivative[3][Q][f[zet]]

c[22] = Coefficient[ differ,  
        m[22] ]

dif[22]  =  differ  -  s[22]  // Expand;

m[23] = Q[f[zet]] ((f'[zet])^(p + q + 4)) P'[f[zet]] Derivative[4][f][zet]

c[23] = Coefficient[ differ,  
        m[23] ]

dif[23]  =  differ  -  s[23]  // Expand;

m[24] = P[f[zet]] ((f'[zet])^(p + q + 4)) Q'[f[zet]] Derivative[4][f][zet]

c[24] = Coefficient[ differ,  
        m[24] ]

dif[24]  =  differ  -  s[24]  // Expand;

m[25] = P[f[zet]] Q[f[zet]] ((f'[zet])^(2 + p + q)) f''[zet] Derivative[4][f][zet]

c[25] = Coefficient[ differ,  
        m[25] ]

dif[25]  =  differ  -  s[25]  // Expand;

m[26] = P[f[zet]] Q[f[zet]] ((f'[zet])^(3 + p + q)) Derivative[5][f][zet]

c[26] = Coefficient[ differ,  
        m[26] ]

dif[26]  =  differ  -  s[26]  // Expand

Thus, we have

Expand[  differ  -  Sum[ c[i]*m[i],  {i, 1, 26}] ]

eqs  =  Table[ c[i] ==  0,  {i, 1, 26} ];

In order to have  differ = 0, it is necessary and sufficient that   y1, y2, y3, y4, y5, y6, y7, y8, y9, y10, y11   be solutions of the system

sol  = 
      Solve[ eqs,  {y1,  y2,  y3,  y4,  y5,  y6,  y7,  y8,  y9,  
          y10,  y11}]

As a check, we set

y1  =  y1   /.  sol[[1]]

y2  =  y2  /.  sol[[1]]  // Factor

y3  =  y3  /.  sol[[1]]

y4  =  y4  /.  sol[[1]]  // Factor

y5  =  y5  /.  sol[[1]]  // Factor

y6  =  y6  /.  sol[[1]]  // Factor

y7  =  y7  /.  sol[[1]]  // Factor

y8  =  y8  /.  sol[[1]]  // Factor

y9  =  y9  /.  sol[[1]]  // Factor

y10  =  y10  /.  sol[[1]]  // Factor

y11  =  y11  /.  sol[[1]]  // Factor

differ // FullSimplify

and see that  differ = 0.

Y[z]

This gives the coefficients in (14.34)--(14.44) for (14.33).