Find $\sum f_n(x)$

Question

Problem

For non-negative integer $n$, define

$$f_n\colon [0, 1]\to\mathbb{R}, \quad f_n(x)=\int_0^x f_{n-1}(t)dt\quad (n>0),\quad f_0(x) = e^x$$

Find $g(x)=\displaystyle\sum_{n=0}^\infty f_n(x)$

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My attempt

Compute $f_n(x)$ for some $n$ :

$f_0=e^x$

$f_1=e^x-1$

$f_2=e^x-x-1$

$f_3=e^x-\frac{x^2}{2}-x-1$

so, $f_n$ has series expansion until $(n-1)$-th degree of $e^x$.

My conjecture is :

$$g(x)= \sum_{n=0}^\infty f_n(x)=e^x + \sum_{n=1}^\infty f_n(x) = e^x$$

But I don't know how to justify this one.

Answer 1

A theorem states

If $(f_{n})$ is a sequence of differentiable functions on $[a,b]$ such that $\lim\limits_{n\to \infty }f_{n}(x_{0})$ exists (and is finite) for some $x_{0}\in [a,b]$ and the sequence $(f^\prime_{n})$ converges uniformly on $[a,b]$, then $(f_{n})$ converges uniformly to a function $f$ on $[a,b]$, and $f^\prime(x)=\lim\limits_{n\to \infty }f^\prime_{n}(x)$ for $x\in [a,b]$.

Now using Taylor's Theorem, for $x \in [0,1]$ and $n \ge 0$, it exists $\xi \in (0,x)$ such that

$$f_{n+1}(x) = e^x - P_n(x) = \frac{e^\xi}{(n+1)!} x^{n+1}$$ where

$$P_n(x) = \sum_{k=0}^n \frac{x^k}{k!}.$$ Therefore $\vert f_{n+1}(x) \vert \le \frac{e}{(n+1)!}$. Which implies that $\sum f^\prime_n$ converges uniformly according to Weierstrass M-test. As $f_n(0) = 0$ for $n \ge 1$, we can apply the theorem mentioned above.

And conclude that

$$g(x)=(x+1) e^x$$ as stated in JustANoob answer.

Answer 2

If you find a way to justify that we can differentiate $g$ termwise, we get:

$$g'(x) = \sum_{n\ge 0} f_n'(x) = f'_0(x) + \sum_{n\ge 1} f_n'(x) = f'_0(x) + \sum_{n\ge 1} f_{n-1}(x) \implies$$

$$g'(x) = f'_0(x) + \sum_{n\ge 0} f_{n}(x) = e^x + g(x)$$

We get $g(x) = (x+1) e^x$