Exercises on Differential Operators

Exercise 1

L[i_]:=If[i==1,D[f[x],x]-f[x],D[L[i-1],x]-i L[i-1]]

Do[Print[DSolve[{L[5]==0,{f[0],f'[0],f''[0],f'''[0],f''''[0]} \
                     ==IdentityMatrix[5][[i]]},f[x],x]],{i,5}]

Exercise 2
a)

DSolve[f''[x]-3f'[x]-4f[x]==e^x,f[x],x]

DSolve[f''[x]-3f'[x]-4f[x]==0,f[x],x]

b)

DSolve[f'[x]-4f[x]==0,f[x],x]

DSolve[f'[x]+f[x]==0,f[x],x]

We notice that the solution of the homogeneous equation
in part (a) is a linear combination of the solutions in
part (b).

Exercise 3

Clear[f,L]
F[f_]:=D[f,x]-2f
L[f_]:=F[F[F[f]]]

a)

L[E^(2x)]

L[x E^(2x)]

L[x^2 E^(2x)]

b)

L[f[x]]//Expand

f[x]/.DSolve[L[f[x]]==0,f[x],x]

B=Table[(%[[1]]-(%[[1]]/.{C[i]->0}))/.{C[i]->1},{i,3}]

M=Table[D[B[[j]],{x,i-1}],{i,3},{j,3}]; M//TableForm

W[0]=Det[M]/.{x->0}

We can now conclude that the solutions in part (a) are
a basis for the kernel of L, taking also into account the
fact that the kernel has dimension 3. 

c)
We can conjecture that (x^i)E^(r x) , with 0 <= i < f ,
will be solutions.

L[f_]:=F[F[F[F[F[f]]]]]

Do[Print[L[x^i E^(2x)]//Expand],{i,0,4}]

Exercise 4
a)

Clear[f,L]
F1[f_]:=D[f,x]+2f
F2[f_]:=F1[F1[f]]+3^2f
L[f_]:=F2[F2[F2[f]]]

L[x^i E^(-2x)Cos[3x]]//Expand/.{i->{0,1,2}}

b)

L1[i_]:=L[x^i E^(-2x)Sin[3x]]//Expand
L2[i_]:=L[x^i E^(-2x)Cos[3x]]//Expand
Do[Print[L1[i],",",L2[i]],{i,0,2}]

c)

L[f[x]]//Expand

f[x]/.DSolve[L[f[x]]==0,f[x],x]

B=Table[(%[[1]]-(%[[1]]/.{C[i]->0}))/.{C[i]->1},{i,6}]

M=Table[D[B[[j]],{x,i-1}],{i,3},{j,3}]; M//TableForm

W[0]=Det[M]/.{x->0}

d)
They are the same set of six linearly independent solutions.

Exercise 5

Clear[f,L]
L[f_]:=D[f,{x,3}]-6D[f,{x,2}]+11D[f,x]-6f
L[f[x]]

a)

f[x]/.DSolve[L[f[x]]==0,f[x],x]

b)

B=Table[(%[[1]]-(%[[1]]/.{C[i]->0}))/.{C[i]->1},{i,3}]

c)

f[x]/.DSolve[L[f[x]]==E^(3x),f[x],x]

Exercise 6

p=x^5+x^4+x^3+x^2+x+1

Clear[L,f]
L[f_]:=Sum[Coefficient[p,x^i]D[f,{x,i}],{i,0,5}]
L[f[x]]

a)

p//Factor

b)

f[x]/.DSolve[L[f[x]]==0,f[x],x]

c)

f[x]/.DSolve[{L[f[x]]==0,{f'[0],f''[0],f'''[0],f''''[0]}=={0,0,0,0}},f[x],x]

d)

DSolve[{L[f[x]]==0,{f[0],f'[0],f''[0],f'''[0],f''''[0]} \
                   =={0,0,0,0,0}},f[x],x]

Check:

f[x]=%%/.Solve[(%%/.{x->0})==0,C[5]]

Exercise 7

Clear[f,F,L]
F[f_]:=D[f,{x,3}]+D[f,{x,2}]+f
L[f_]:=F[F[f]]
L[f[x]]

p=Sum[Coefficient[L[f[x]],D[f[x],{x,i}]]x^i,{i,0,6}]

a)

p//Factor

b)

s=x/.Solve[p==0,x];
sR=Re[s]

sI=Im[s]

We will have to use floating point representations here,
otherwise the result gets too complicated for Mathematica.

j=Mod[2i,6]+1;
f=Sum[If[i<4,C[i]E^(sR[[j]]x)(Cos[sI[[j]]x]+I Sin[sI[[j]]x]),       \
             x C[i]E^(sR[[j]]x)(Cos[sI[[j]]x]+I Sin[sI[[j]]x])],{i,6}];
f//N

Solution check (gives a floating point approximation
close enough to 0):

L[f//N]/.{x->5}//Simplify

c)

x/.Solve[p==1,x]

We notice from above that f=E^(-x)is a particular 
solution for the nonhomogeneous equation.

f=f+E^(-x);
f//N

Solve[Table[D[f//N,{x,i-1}]/.{x->0},{i,6}]=={0,0,1,0,0,0}, \
                             {C[1],C[2],C[3],C[4],C[5],C[6]}]

(f//N)/.%

Problems on Differential Operators

Problem 1

DSolve[{17/32s''[t]-(3-5)s[t]==0,s[0]==-12,s'[0]==0},s[t],t]

Plot[-12Cos[8t/Sqrt[17]],{t,0,6}]

Problem 2

DSolve[{17/32s''[t]-(3-5)s[t]+s'[t]==0,s[0]==-12,s'[0]==0},s[t],t]

Plot[E^(-16t/17)(-12Cos[8Sqrt[13]t/17]-24/Sqrt[13]Sin[8Sqrt[13]t/17]), \
                                                               {t,0,6}]

Exercises on Recurrence Relations

<<DiscreteMath`RSolve`

Exercise 1

L=Product[d-i,{i,5}]//Expand

Clear[f]
Do[f[i][n_]:=i^n,{i,5}]
LC[n_]:=Sum[k[i]f[i][n],{i,5}]
LC[n]

Table[LC[n]/.Solve[{Array[LC[#-1]&,5]==IdentityMatrix[5][[i]]}, \
                   {k[1],k[2],k[3],k[4],k[5]}],{i,5}]

Check:

Table[%[[i]]/.{n->j-1},{i,5},{j,5}]//MatrixForm

Exercise 2
a)

s[n]/.RSolve[{s[n+1]-s[n]==(n+1)^2,s[0]==0},s[n],n]//Factor

b)

s[n]/.RSolve[{s[n+1]-s[n]==(n+1)^3,s[0]==0},s[n],n]//Factor

c)

s[n]/.RSolve[{s[n+1]-s[n]==(n+1)^4,s[0]==0},s[n],n]//Factor

d)

s[n]/.RSolve[{s[n+1]-s[n]==(n+1)^5,s[0]==0},s[n],n]//Factor

Exercise 3

We can conjecture that sums for even n's contain the factors 
n(1+n), whereis sums for odd n's contain (n^2)(1+n)^2.
Verification:

s[n]/.RSolve[{s[n+1]-s[n]==(n+1)^10,s[0]==0},s[n],n]//Factor

s[n]/.RSolve[{s[n+1]-s[n]==(n+1)^15,s[0]==0},s[n],n]//Factor

Exercise 4

Clear[n,s,L,F,d]
d[s_]:=s/.{n->n+1}
F[s_]:=d[s]-2s
L[s_]:=F[F[F[s]]]

a)

L[2^n]//Expand

L[n 2^n]//Expand

L[n^2 2^n]//Expand

b)

K=Table[{2^(n-1),(n-1)2^(n-1),(n-1)^2 2^(n-1)},{n,3}];
K//MatrixForm

Det[K]

We can now conclude that the solutions in part (a) are
a basis for the kernel of L, taking also into account the
fact that the kernel has dimension 3. 

c)
We conjecture that (n^i)(r^n) , with 0 <= i < f , will be
solutions.

L[s_]:=F[F[F[F[s]]]]

Do[Print[L[n^i 2^n]//Expand],{i,0,3}]

Exercise 5

F[s_]:=d[s]+2s
L[s_]:=F[F[s]]+(3^2+2^2)s

RSolve[L[s[n]]==0,s[n],n]

Problems on Recurrence Relations

Problem 1
a)

One can imagine a recurrent scheme by which, in order to move
i+1 disks once we need M[i+1]=2M[i]+1 moves, since we have to
move the top n, then move the disk # n+1, then move back the
top n.

M[i]/.RSolve[{M[i+1]==2M[i]+1,M[0]==0},M[i],i]

To achieve our goal, we'll need:

S=Sum[1+%[[1]],{i,0,n-1}]

i.e.,

S=-1+2^n

b)

(S/.{n->64})/((365+1/4)*24*60*60)

%//N

Problem 2
a)

In this case, by the same type of reasoning as before,
if Q[n] is the number of moves required to move n disks from
an end pole to the center and P[n] from the center to an end
pole, we have: P[n+1]=2P[n]+1+Q[n] and Q[n+1]=2Q[n]+P[n]+1.

RSolve[{P[n+1]==2P[n]+1+Q[n],Q[n+1]==2Q[n]+P[n]+1,Q[0]==0,P[0]==0}, \
       {P[n],Q[n]},n]

i.e.,

Q=(-1+3^n)/2

b)

(Q/.{n->64})/((365+1/4)*24*60*60)

%//N

Problem 3
a)

If we denote with "r" the one-month old pairs of rabbits and 
with "R" the pairs of grown-ups, we have: R[n+1]=R[n]+r[n] 
and r[n]=R[n-1], and therefore R[n+1]=R[n]+R[n-1].

2R[n]/.RSolve[{R[n+1]==R[n]+R[n-1],R[0]==n0,R[1]==n0},R[n],n] \
    //Factor

b)

%[[1]]/.{n0->1}/.{n->12*10}//Expand