We have $f:RR->RR,f(x)=e^(x-1),$ and the string $(X_n)_(n>=1)$ with ${(x_1=2),(x_(n+1)=f(x_n)):}$ .How to demonstrate that ( $X_n$ ) is strictly monotone?

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Answer 1 · 2017-04-18T14:00:28

See below.

From

we know that for $x_k ne 1$

$x_k < e^(x_k -1)$

so calling $x_(k+1)=e^(x_k -1)$ we will have a monotonic strictly increasing sequence

$x_1 < x_2 < cdots < x_k < x_(k+1) < cdots$

We have f:RR->RR,f(x)=e^(x-1),f:RR->RR,f(x)=e^(x-1),and the string (X_n)_(n>=1) (X_n)_(n>=1) with {(x_1=2),(x_(n+1)=f(x_n)):}{(x_1=2),(x_(n+1)=f(x_n)):}.How to demonstrate that (X_nX_n) is strictly monotone?