Composition of Functions and Sequence

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Kondiik
Kondiik 2013-9-1
1.2 I tried to sort it but obviously something is wrong...
syms x y
f = 1/(x+1); h = exp(x); g = 1;
h = compose(f,g)
1.3 I have no idea whats it is about.

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rifat
rifat 2013-9-1
编辑：rifat 2013-9-1
1.2 - Try to use finverse
1.3 - Use loops to solve
N=3; % number of terms
v=sqrt(2);
for i=1:N-1
v=sqrt(2+v);
end
v
To find convergence use a very large value of N
  2 个评论
Kondiik
Kondiik 2013-9-1
编辑：Kondiik 2013-9-1
Thank you. I used this one for 1.2 with small modification.
N=3; % number of terms
v=sqrt(2);
v
for i=1:N-1
v=sqrt(2+v);
v
end
Roger Stafford
Roger Stafford 2013-9-1
It is more elegant to use the equation given to you by Matt J:
L=sqrt(2+L)
and solve for L. It is the value for which there is no change in the iteration and it is also the value your sequence converges to. You don't need to carry out the iteration for large N to find it, though if you do, you will see that these answers agree.

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更多回答（2 个）

Matt J
Matt J 2013-9-1
编辑：Matt J 2013-9-1
1.3 To what value does this sequence converge
Hint - the formula for the recursion is
a(n)=sqrt(2+a(n-1))
If a(n) converges to a limit L, then taking limits on both sides leads to the equation
L=sqrt(2+L)
Roger Stafford
Roger Stafford 2013-9-1
编辑：Roger Stafford 2013-9-1
For problem 1.2 I strongly recommend reading up on the definition of function composition. For example:
http://en.wikipedia.org/wiki/Function_composition
You are looking to define a function y = g(x) which has the property that
f(g(x)) = f(y) = h(x) = exp(x)
where f(y) = 1/(y+1). It's easy to solve for y = g(x) in this equation. Then set x equal to 1.
  1 个评论
Kondiik
Kondiik 2013-9-2
编辑：Kondiik 2013-9-2
How to set x = 1?
syms x y; % creat symbolic variables
f = (1/(x+1));
h = exp(x);
g = y;
x = 1;
y = solve(h == compose(f,g,x,y), y)
|y =
exp(-x) - x|

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