** Assignment:** Page 151: Nos. 1,2,3,4,8 Page 161: Nos. 1-8

** Page 151**

**1.**Can a continuous functon on**R**have a periodic point of period 48 and not of of period 56? Why?

**Yes**. Observe that 56=(2^3)*7 precedes 48=(2^4)*3 in the Sarkovskii ordering.

Thus, if a continuous function F has a cycle of period 56, then it also has a cycle of period 48. But, a continuous function F with a periodic point of period 48 may or**may not**have a periodic point of period 56.

**2.**Can a continuous functon on**R**have a periodic point of period 176 and not of of period 96? Why?

**No**. In this case, 176=(2^4)*11 precedes 96=(2^5)*3 in the Sarkovskii ordering, and so, by Sarkovskii's theorem, a continuous function with a periodic point of period 176 must also have a periodic point of period 96.**3.**Give and example of a funciton F:[0,1] --> [0,1] that has a periodic point of period 3 and**no**other periods. Can this happen?

Sarkovskii's theorem applies only to continuous functions and so it follows that such a function must be discontinuous. Consider the function,

This function has lots of three cycles:

0->1/3->2/3->0

1/6->1/2->6/6->1/6

1/12->5/12->9/12->1/12

Indeed, the graph clearly shows that

f[0,1/3]=[1/3,2/3]

f[1/3,2/3]=[2/3,1)

f[2/3,1]=[0,1/3]

and so**every**point is of period 3 (except the endpoint 1 which is eventually period 3)

**4.**The graphs shown each have a cyle of period 4 given by {0,1,2,3}. One of these functions has a cycle of all other periods, and one has only period 1, 2, and 4. Identify which function has each of these properties.

To see the**graphs**under consideration, click here.

**Note:**Each of these functions is contuniuous and so we need only find a 3-cycle to identify which has cycles of all periods.

From the graph of Fig. A {naming this function F}, it's clear that

F[0,1]=[1,2]

F[1,2]=[2,3]

F[2,3]=[0,3]

and so we have [0,1] ( [0,3] = F^3[0,1]. Therefore, F has a period 3 point in [0,1].

Furthermore the equation of the piecewise linar graph in Fig. B {naming this function G} is G(x) = { 3-x if 0<=x<=1, 4-2x if 1<=x<=2, x-2 if 2<=x<=3

Now let 0<=x<=1, Then -1<=-x<=0 and so 2<=3-x<=3. In other words, G[0,1]=[2,3], and similarly, G[2,3]=[0,1]. (These facts are easily verified by inspection fo the graph of G.) This means that no point in [0,1] U [2,3] can have odd period. Now let's see what happens when we take and arbitary x in [0,1] and iterate G.

G(x)=3-x , since x is in [0,1]

G^2(x)=G(3-x)=(3-x)-2=1-x , since 3-x is in [2,3]

G^3(x)=G(1-x)=3-(1-x)=2+x , since 1-x is in [0,1]

G^4(x)=G(2+x)=(2+x)-2=x , since 2+x is in [2,3]

So every point in [0,1] U [2,3] is of period 4. Now G[1,2]=[0,2] and G[0,2]=[0,3]. But as soon as a point leaves the closed interval [1,2], it's locked into a 4-cycle. So the question is: are there points in [1,2] that remain in [1,2] for all time? The answer is**no**! The orbits of points close to the fiexed point x=4/3 oscillate away from the fixed point and eventually enter [0,1], an interval of 4-cycles.

**8)**Consider the function whose graph is displayed in fig. 11.17b on page 152. Prove that this function has cycles of all even periods but no odd periods (except 1).First observe that 2 is fixed by F. (In fact any multiple of a power of 2 in [0,4] is eventually fixed.) Inspection of the graph shows that

[3,4]->[0,1]->[2,4] and [2,3]->[1,2]->[2,4]

But [2,4]->[0,2] and vice versa. In general,

F^(2n+1)[0,1]=F^(2n+1)[1,2]=[2,4] and

F^(2n+1)[2,3]=F^(2n+1)[3,4]=[0,2]

Hence there can be no odd periodic points.On the other hand, all even periodic orbits exist because

F^(2n)[0,1]=F^(2n)[1,2]=[0,2] and

F^(2n)[2,3]=F^(2n)[3,4]=[2,4].

Therefore, since both [0,1] and [1,2] are contained in [0,2] and both [2,3] and [3,4] are contained in [2,4] even periodic orbits are guaranteed to exist.

**Page 161**

**1.**Compute the Schwarzian derivative for the following functions and decide if SF(x)<0 for all x.Recall: SF(x)=F'''(x)/F'(x)-(3/2)[F''(x)/F'(x)]^2

**a)**F(x) = x^2

F'(x)=2x; F''(x)=2; and F'''(x)=0

SF(x)=0-(3/2)[2/2x]^2 = -3/2x^2 <0 for all x.**b)**F(x) = x^3

F'(x)=3x^2; F''(x)=6x; and F'''(x)=6

SF(x)=6/3x^2-(3/2)[6x/3x^2]^2 = -4/x^2 <0 for all x.**c)**F(x) = e^3x

F'(x)=3e^3x; F''(x)=9e^3x; and F'''(x)=27e^3x

SF(x)=27e^3x/3e^3x-(3/2)[9e^3x/3e^3x]^2 = -9/2<0 for all x.**d)**F(x) = cos(x^2+1)

F'(x)=-2xsin(x^2+1); F''(x)=-4x^2cos(x^2+1)-2sin(x^2+1);and F'''(x)=8x^3sin(x^2+1)-12xcos(x^2+1)

SF(x)=[8x^3sin(x^2+1)-12xcos(x^2+1)]/[-2xsin(x^2+1)]- (3/2)[-4x^2cos(x^2+1)-2sin(x^2+1)]/[-2xsin(x^2+1)]^2

=6cot(x^2+1)-4x^2-(3/2)[2xcot(x^2+1)+(1/x)]^2

=6cot(x^2+1)-4x^2-(3/2)[4x^2cot^2(x^2+1)+4cot(x^2+1)+(1/x^2)]

=-[4x^2+6x^2cot^2(x^2+1)+(3/2x^2)] <0 for all x.

Note: You could use the chain rule for this problem let C(x)=cos(x) and G(x)=x^2+1....**e)**F(x) = arctan(x)

F'(x)=1/(x^2+1); F''(x)=-2x/(x^2+1)^2; and F'''(x)=2(3x^2-1)/(x^2+1)^3

SF(x)=2(3x^2-1)/(x^2+1)^2-(3/2)[-2x/(1+x^2)]^2

=-2/(1+x^2)^2 <0 for all x.**2.**Is it true that S(F+G)(x)=SF(x) + SG(x)? If so, prove it. If not, give a counter example.

**False.**Unlike ordinary differentiation, the Schwarzian derivative does not distribute over addition. Let F(x)=G(x)=e^x. Then

recalling, SF(x)= F'''(x)/F'(x) - 3/2[F''(x)/F'(x)]^2, we get

SF(x) + SG(x) = 2(e^x/e^x - 3/2[e^x/e^x]^2)

=2(1-3/2)

=-1

But,

S(F+G)(x) = (e^x +e^x)/(e^x +e^x) -3/2[(e^x +e^x)/(e^x +e^x)]^2

=1-3/2

=-1/2

**3.**Is it true that S(F*G)=SF(x)*G(x) + F(x)SG(x)? If so, prove it. If not, give a counter example.

**False.**Unfortunately, there is no product-like rule for Schwarzian derivatives. Let F(x)=e^2x and G(x)=e^3x. Then

SF(x)=8*e^2x/2*e^2x - 3/2[4*e^2x/2*e^2x]^2

=-2

SG(x)=27*e^3x/3*e^3x - 3/2[9*e^3x/3*e^3x]^2

=-9/2

So,

SF(x)*G(x) + F(x)SG(x) = -2*e^3x - (9/2)*e^2x

While,

S(F*G)(x)=-25/2 {since F*G(x)=e^5x}.

**4.**Is it true that S(cF)(x)=cS(F(x)) where c is a constant? If so, prove it. If not, give a counter example.

**False.**For any function F, (cF)^n=c*F^n for all n, so we have that

S(cF)(x) = cF'''(x)/cF''(x) - 3/2[cF''(x)/cF'(x)]^2

=F'''(x)/F''(x)- 3/2[F''(x)/F'(x)]^2

=SF(x).

**5.**Give and example of a function that has SF(x)>0 for at least some x-values.

According to the Schwarzian Min-Max Principle, we need to find a function, F, whose derivative, F',**has**a positive local minimum or a negative local maximum.

Let F(x) = -x - x^3. Then F'(x)=-1 -3x^2 which has a negative local maximum at (0,-1).

And so,

SF(x) = -6/(-1-3x^2) - 3/2[-6x/(-1-3x^2)]^2

=(6-36x^2)/(1+3x^2)^2

which is positive when 6-36x^2>0 or when |x| < sqrt(6)/6.

**6.**Prove that S(1/x)=0 and S(ax+b) =0. Conclude that SF(x)=0 where F(x) = 1/(ax+b).

Let R(x) =1/x and L(x)=ax+b. Then R'(x)=-1/x^2, R''(x)=2/x^3, and R'''(x)=-6/x^4. Also, L'(x)=a and L''(x)=L'''(x)=0 for all n>1.

Therefore,

SR(x)=-6x^(-4)/-x^(-2) - 3/2[2x^(-3)/-x^(-2)]^2

=6/x^2 - 6/x^2 = 0

and,

SL(x)=0/a - 3/2[0/a]^2 = 0

Applying the chain rule for Schwarzian derivatives,

S(R(L(x)) = S(R(L(x))*[L'(x)]^2 + SL(x) = 0*a^2 + 0 = 0

**7.**Compute SM(x) where M(x)=(ax+b)/(cx+d).Note: This solution makes use of problem #6.

We will show that the linear fractional transformation, or Mobius transformation , has**zero**Schwarzian derivative. Our goal will be to find constants k1, k2 and k3 such that

(ax+b)/(cx+d)=k1 + k2/(x+k3)

Then the Mobius transformation is the composition of functions that we have already shown to have zero Schwarzian derivative (exercise 6). Hence, the Mobius tranformatin itself has zero Schwarzian derivative by repeated appliations of the chain rule.

Proceding, note that if c does not equal zero then we may write

(ax+b)/(cx+d)=[(a/c)x+(b/c)]/[x+(c/d)]**{1}**

also note

k1 + k2/(x+k3)=[k1(x+k3)+k2]/[x+k3]

=[k1x+(k1k3+k2)]/[x+k3]**{2}**

Equating coefficients in**{1}**and**{2}**we see immediately that k1=a/c and k3=d/c also,

b/c=k1k3+k2=ad/cc+k2 which implies that

k2=(b-ad)/c thus,

(ax+b)/(cx+d)=[a/c+((b-ad)/c)]/(x+d/c)

and we are done since we have shown that the Mobius transformation is the composition of both linear and inverse transformations.**8.**Let M be as in the previous exercise. Prove that S(MoF)=SF.

Note: F is the function 1/(ax+b) defined in problem 6.According to the chain rule for Schwarzian derivatives

S(MoF)(x)=(SF(M(x)))(M'(x)^2) + SM(x)

by problem 6 we know that SF(x)=0 and by problem 7 we know that SM(x)=0 therefore

S(MoF)(x)=0(M'(x)^2) + 0 = 0 = SF(x).

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