Composition of Functions and Sequence

16 vues (au cours des 30 derniers jours)
Kondiik
Kondiik le 1 Sep 2013
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 le 1 Sep 2013
Modifié(e) : rifat le 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 commentaires
Kondiik
Kondiik le 1 Sep 2013
Modifié(e) : Kondiik le 1 Sep 2013
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 le 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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Plus de réponses (2)

Matt J
Matt J le 1 Sep 2013
Modifié(e) : Matt J le 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 le 1 Sep 2013
Modifié(e) : Roger Stafford le 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 commentaire
Kondiik
Kondiik le 2 Sep 2013
Modifié(e) : Kondiik le 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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