# 
restart;
# Exam 2b
# Problem 1(i)
phi:=unapply(x-f(x)/D(f)(x),x);
f:=x->x+4*cos(x);
phi(x);
# Problem 1(ii)
Digits:=20;
x1:=phi(2.0);
x2:=phi(x1);
x3:=phi(x2);
x4:=phi(x3);
# Problem 1(iii)
x1:=phi(4.0);
x2:=phi(x1);
x3:=phi(x2);
x4:=phi(x3);
# Problem 2(iv)
plot(f(x),x=1..4);
# The solutions converge to different
# limits because f(x)=0 has multiple
# solutions and the solution closest
# to 2 is a different solution than the
# solution closest to 4.
# Problem 2(i)
S:=(n,x)->Sum((-1)^l*x^(2*l+1)/(2*l+1)!,l=0..n);
seq(value(S(n,2)),n=1..4);
Digits:=12;
seq(evalf(S(n,2)),n=1..4);
# Problem 2(ii)
g:=s->2*s*sqrt(1-s^2);
seq(value(g(S(n,1))),n=1..4);
seq(evalf(g(S(n,1))),n=1..4);
# Problem 2(iii)
# This can be done using trigonometry and the
# formula
#     sin(2x)=2sin(x)cos(x)
#            =2sin(x)sqrt(1-sin(x)^2)
T1:=sin(2)-g(sin(1));
simplify(T1);
# Problem 2(iv)
# Error in sequence from part (i)
seq(evalf(S(n,2)-sin(2)),n=1..4);
# Error in sequence from part (ii)
seq(evalf(g(S(n,1))-sin(2)),n=1..4);
# The second sequence converges faster.  Looking
# at the series for the first sequence we see
# there is a factor of 2^(2l+1) in the numerator.
# Thus, the terms in the first sequence are, in
# general larger than for the second sequence,
# in where the factor 1^(2l+1)=1 for every l.
# This implies the remainer Rn will tend to zero
# more slowly than for the second sequence.
# Alternatively, one can estimate Rn directly.
# The integral for the remainder in the first
# sequence has limits x=0..2 which is larger than
# the intergral for the remainder in the second
# sequence with has limits x=0..1.
# Problem 3(i)
f:=x->4*sin(ln(1+x));
plot(f(x),x=1..10);
L:=Int(sqrt(1+D(f)(x)^2),x=1..10);
evalf(L);
# Problem 3(ii)
Vx:=Int(Pi*f(x)^2,x=1..10);
evalf(Vx);
# Problem 3(iii)
Vy:=Int(2*Pi*x*f(x),x=1..10);
evalf(Vy);
# Problem 4(i)
# Let x by x be the dimensions of the base.
# Let y be the height of the holding tank.
S:=x*x+4*x*y;
V:=x*x*y;
s1:=solve(V=2000,{y});
s2:=subs(s1,S);
dS:=diff(s2,x);
s2:=solve(dS=0,{x});
s3:=evalf(s2[1]);
subs(s3,s1);
# Problem 4(ii)
# It was assumed that the weight of the tank
# was proportional to the total surface area.
# Problem 5
# This proof problem was done using only
# pencil and paper.
# Problem 6
eq1:=D(y)(x)-y(x)=x^2;
eq2:=y(1)=3;
dsolve({eq1,eq2},y(x));
# The integrating factor is
u:=exp(-x);
y1:=(1/u)*(C+int(u*x^2,x));
y2:=subs(x=1,y1);
s4:=solve(y2=3,{C});
y3:=subs(s4,y1);
expand(y3);

#