Asked by lawrence
How to show that the composite function of order n
(fof.....order n f)(x)=4^n.x/(1+(4^n-1)x/3m) where f(x)=4x/(1+x/m)
(fof.....order n f)(x)=4^n.x/(1+(4^n-1)x/3m) where f(x)=4x/(1+x/m)
Answers
Answered by
oobleck
Use induction.
true for k=1?
4x/(1 + x/m) = 4mx/(m+x)
f^1(x) = 4^1*x/(1+(4^1-1)x/3m) = 3m*4^1*x/(3m+3x) = 4mx/(m+x)
so P(1) is true
Now we take P(n+1), assuming P(n)
f^(n+1)(x) = 4m(f^n(x))/(m+f^n(x))
plug in f^n(x) and show that you get 4^(n+1)x/(1+4^(n+1)-1)x/3m)
true for k=1?
4x/(1 + x/m) = 4mx/(m+x)
f^1(x) = 4^1*x/(1+(4^1-1)x/3m) = 3m*4^1*x/(3m+3x) = 4mx/(m+x)
so P(1) is true
Now we take P(n+1), assuming P(n)
f^(n+1)(x) = 4m(f^n(x))/(m+f^n(x))
plug in f^n(x) and show that you get 4^(n+1)x/(1+4^(n+1)-1)x/3m)
Answered by
lawrence
Thanks. wonder how to get the result of given closed form by repeatedly making composite of function f. Thanks
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