\section*{Problem 1 - 171241}

If \(f\) and \(g\) are continuous functions in the interval \(-1 \leq x \leq 1\), then \(d_{1}(f, g)\) and \(d_{2}(f, g)\) are defined as follows:

\[
d_{1}(f, g)=\max |f(x)-g(x)|
\]

\(d_{2}(f, g)\) is the area in square units of a Cartesian coordinate system bounded by the images of the functions \(f\) and \(g\) and two segments on the lines \(x=-1\) and \(x=1\).

(The area of each subregion, regardless of its orientation, is considered positive.) Now let \(f_{n}(n=1,2,3, \ldots)\) and \(h\) be the functions defined in the interval \(-1 \leq x \leq 1\) by

\[
f_{n}(x)=\sum_{k=1}^{n}(1-x) x^{k-1} \quad \text { and } \quad h(x)=1
\]

a) Determine \(\lim _{n \rightarrow \infty} d_{1}\left(f_{n}, h\right)\), if this limit exists.

b) Determine \(\lim _{n \rightarrow \infty} d_{2}\left(f_{n}, h\right)\), if this limit exists.
}

The functions \(f_{n}\) and \(h\) in the interval \(-1 \leq x \leq 1\) are continuous. Using the sum formula for finite geometric partial sums, we get, provided \(x \neq 1\):

\[
f_{n}(x)=(1-x) \frac{x^{n}-1}{x-1} \quad \text { so } \quad f_{n}(x)=1-x^{n}
\]

Since \(f_{n}(1)=0\), the relation (1) also holds for \(x=1\). Therefore, \(\left|f_{n}(x)-(x)\right|=\left|1-x^{n}-1\right|=\left|x^{n}\right|\), so \(d_{1}\left(f_{n}, h\right)=\max \left|x^{n}\right|=1\). Furthermore, the definition of \(d_{2}(f, g)\) is equivalent to

\[
d_{2}(f, g)=\int_{-1}^{1}|f(x)-g(x)| d x \quad \text { thus } \quad d_{2}\left(f_{n}, h\right)=\int_{-1}^{1}\left|x^{n}\right| d x
\]

Since the image of \(y=\left|x^{n}\right|\) for \(-1 \leq x \leq 0\) is transformed into the image of \(y=x^{n}\left(=\left|x^{n}\right|\right)\) for \(0 \leq x \leq 1\) by reflection across the y-axis, it follows that

\[
d_{2}\left(f_{n}, h\right)=2 \cdot \int_{0}^{1} x^{n} d x=2 \cdot \frac{1^{n+1}-0^{n+1}}{n+1}=\frac{2}{n+1}
\]

Thus, we obtain
a) \(\lim _{n \rightarrow \infty} d_{1}\left(f_{n}, h\right)=1\)
b) \(\lim _{n \rightarrow \infty} d_{2}\left(f_{n}, h\right)=0\)

Adapted from \([3]\)