Dgdrxrt

this is a problem from the entrance exam of the University of Tokyo and unfortunately, the official doesn't offer solutions.

1) Using mathematical induction. Assume $f_n(x)=c_n x^a_n$ holds for $n$, and by $f_n+1(x)=p int_0^xleft(f_n(t)right)^1 / q mathrmd t$ we can get $f_n+1(x)$. Then comparing the coeffient and the exponent will show that it holds for $n+1$ too.

2) 3) 4) 5) 6) I've no idea. I tried to calc the derivatives of $g_n$, but I don't know what to do next.

I can get $a_n+1$ from the recursive formula, but the form of it is kinda complex which makes it hard to get $c_n+1$. So I guess the rest questions could be done without knowing what actually $a_n$ and $c_n$ are. But I don't know how to do that.

Also, $1 / p+1 / q=1$ seems a frequent condition, how is it usually used in solutions?

To get you started, the main notion to know to handle the first two questions is that of an arithmetico-geometric sequence (in the French context which differs from the English one).

Then, you will easily notice how $a_n$ is an arithmetico-geometric sequence.
That gives you both:

• an explicit formula for $a_n = frac1-q^-n+11-q^-1=sum_k=0^n-2q^-k = p(1-q^-n+1)$




• its limit $lim_nto infty a_n=frac11-q^-1=p$

For 2):
$$
g'_n(x_n)=a_nx_n^a_n-1-px_n^p-1=0 Leftrightarrow x_n=left( fraca_np right)^frac1p-a_n
$$
which holds because $lim_nto infty a_n=p$ and $(a_n)_ngeq 0$ is increasing so $p-a_n> 0$ thus $x_n$ exists and is unique.
It is a maximum because of the sign of the difference of $xmapsto a_nx^a_n-1$ and $xmapsto px_n^p-1$ (try to get this result by yourself).

For 3):

Since $ymapsto x^y=e^yln x$ is continuous over $mathbbR$ whatever $xin mathbbR^+*$, $lim_nto inftyx^a_n=x^lim_nto inftya_n=x^p$. Thus, $lim_nto inftyg_n(x)=0$ for $xin ( 0,1 ]$. For $x=0$, the result is trivial.