Composition of Functions and Sequence

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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.

Accepted Answer

rifat
rifat on 1 Sep 2013
Edited: rifat on 1 Sep 2013
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 Comments
Roger Stafford
Roger Stafford on 1 Sep 2013
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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More Answers (2)

Matt J
Matt J on 1 Sep 2013
Edited: Matt J on 1 Sep 2013
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 on 1 Sep 2013
Edited: Roger Stafford on 1 Sep 2013
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 Comment
Kondiik
Kondiik on 2 Sep 2013
Edited: Kondiik on 2 Sep 2013
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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