Home > Analysis, Olympiad, Uncategorized > Reccurent function sequence SEEMOUS 2010

Reccurent function sequence SEEMOUS 2010


Suppose f_0:[0,1]\to \mathbb{R} is a continuous function, and define the sequence (f_n)_n, f_n:[0,1]\to \mathbb{R} in the following way:

f_n(x)=\int_0^x f_{n-1}(t)dt,\ \forall x \in [o,1].

a) Prove that the series \sum_{n\geq 0}f_n(x) converges for any x \in [0,1].

b) Find an explicit formula in terms of x for the above series.

Seemous 2010, Problem 1

a) It is obvious that f_n'=f_{n-1}. The function f_0 is continuous defined on a compact set, therefore is bounded. This means we can find M>0 such that |f_0(x)|\leq M on [0,1]. But then

\displaystyle |f_1(x)|=|\int_0^x f_0(t)dt|\leq \int_0^x |f_0(t)|dt \leq Mx\leq M on [0,1].

Continuing we get

\displaystyle |f_2(x)|=|\int_0^x f_1(t)dt|\leq \int_0^x |f_1(t)|dt \leq M\frac{x^2}{2} \leq M/2 on [0,1].

By induction we have

\displaystyle |f_n(x)| \leq M\frac{x^n}{n!}\leq \frac{M}{n!} on [0,1].

We are needed to prove that the series \sum_{n=0}^\infty f_n(x) is convergent. For this it is enough to prove that it is absolutely convergent, and for that we have

\displaystyle \sum_{n=0}^\infty |f_n(x)|\leq \sum_{n=0}^\infty \frac{1}{n!}<\infty.

This proves by using the Weierstrass test that the series converges (moreover it converges uniformly).

b) In the second part denote by F(x)=\displaystyle \sum_{n=1}^\infty f_n(x), and note that by differentiating term by term (this is possible due to the uniform convergence) we obtain

F'(x)=f_0+F(x),

a first order linear differential equation which can be immediatley solved.

 

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