gugo82 ha scritto:Exercise:
Let \(x,y,z\) be real numbers.
1. Find a closed form for:
\[
\tag{A}
\sum_{n=0}^N (y+nz)\ x^n\; ,
\]
where \(N\in \mathbb{N}\).
Testo nascosto, fai click qui per vederlo
Let:
\[
S_N:=\sum_{n=0}^N (y+nz)\ x^n\; .
\]
First, assume \(x=1\). Then:
\[
\begin{split}
S_N &= \sum_{n=0}^N (y+nz)\\
&= y\ \sum_{n=0}^N 1 + z\ \sum_{n=0}^N n\\
&= y\ (N+1) + z\ \frac{N(N+1)}{2}\\
&= \frac{N+1}{2}\ (y+Nz)\; .
\end{split}
\]
Now, assume \(x\neq 1\). Then:
\[
\begin{split}
S_N &= y &+ (y+z)\ x &+ (y+2z)\ x^2 &+\cdots &+ (y+Nz)\ x^N\\
-x\ S_N &= &-y\ x &- (y+z)\ x^2 &-\cdots &- (y+(N-1)z)\ x^N - (y+Nz)\ x^{N+1}\\
\hline\\
(1-x)\ S_N &= y &+ z\ x &+z\ x^2&+\cdots &+z\ x^N - (y+Nz)\ x^{N+1}
\end{split}
\]
and a little algebra gives:
\[
\begin{split}
S_N &= y\ \frac{1-x^{N+1}}{1-x} +z\ \frac{1}{1-x}\ \left(x\ \sum_{n=0}^{N-1} x^n - N\ x^{N+1}\right) \\
&=y\ \frac{1-x^{N+1}}{1-x} + z\ \frac{1}{1-x}\ \left( x\ \frac{1-x^N}{1-x} - N\ x^{N+1}\right)\\
&=y\ \frac{1-x^{N+1}}{1-x} + z\ \left( \frac{1-x^N}{1-x} - N\ x^N\right)\ \frac{x}{1-x}\\
&= y\ \frac{1-x^{N+1}}{1-x} + z\ \frac{1-(N+1)\ x^N + x^{N+1}}{(1-x)^2}\ x\; .
\end{split}
\]
Therefore the closed form to be determined is:
\[
\tag{1}
S_N = \begin{cases} \frac{N+1}{2}\ (y+Nz) &\text{, if } x=1\\
y\ \frac{1-x^{N+1}}{1-x} + z\ \frac{1-(N+1)\ x^N + x^{N+1}}{(1-x)^2}\ x &\text{, otherwise.}
\end{cases}
\]
*
gugo82 ha scritto:2. Let \(y,z\) be held fixed. Find both the convegence set and the sum of the series:
\[
\tag{B}
\sum_{n=0}^\infty (y+nz)\ x^n
\]
w.r.t. the variable \(x\) alone.
Testo nascosto, fai click qui per vederlo
If \(y=0=z\), then series (B) converges in the whole of \(\mathbb{R}\) to the zero function \(f(x;0,0)=0\).
So, assume \(y\neq 0\) or \(z\neq 0\). Then, passing (1) to the limit, it becomes clear the series (B) converges iff \(|x|<1\) to the function:
\[
\begin{split}
f(x;y,z) &:= \frac{y}{1-x} + \frac{z\ x}{(1-x)^2} \\
&= \frac{y+(z-y)\ x}{(1-x)^2}\; .
\end{split}
\]
*
gugo82 ha scritto:3. Find the convergence set and analyse the mode of convergence of the series (B) w.r.t. the three variables \(x,y,z\).
Testo nascosto, fai click qui per vederlo
Thanks to part 2, series (B) converges in the set:
\[
D:=\underbrace{\{(x,0,0),\ x\in \mathbb{R}\}}_{=:D_0} \cup \underbrace{\{(x,y,z):\ |x|<1\}}_{=: D_1} \subseteq \mathbb{R}^3
\]
(which is the union of the \(x\)-axis \(D_0\) and of the unbounded double layer \(D_1\)) to the three variables function:
\[
s(x,y,z) := f(x;y,z) = \begin{cases}
0 &\text{, if } (x,y,z)\in D_0\\
\frac{y+(z-y)\ x}{(1-x)^2} &\text{, if } (x,y,z)\in D_1\; .
\end{cases}
\]
Let \(M_n\) be the least upper bound of \(|y+nz|\ |x|^n\) in \(D\). Then \(M_n=+\infty\) and Weierstrass test does not apply to (B) over \(D\).
On the other hand, the least upper bound \(M_n(x_0,y_0,z_0)\) of \(|y+nz|\ |x|^n\) over any region of the type:
\[
D(x_0,y_0,z_0):=D_0\cup \big( ]-x_0,x_0[\times ]-y_0,y_0[\times ]-z_0,z_0[ \big)
\]
(with \(0<x_0<1\) and \(y_0,z_0>0\)) is less than \(( y_0+nz_0)\ x_0^n\), therefore (B) converges totally (thus uniformly and absolutely) in \(D(x_0,y_0,z_0)\) by the Weierstrass test.
Thus, the convergence of (B) is total over bounded sets contained in \(D\) for every bounded set \(B\subset D\) is contained into a suitable set of the type \(D(x_0,y_0,z_0)\).
Sono sempre stato, e mi ritengo ancora un dilettante. Cioè una persona che si diletta, che cerca sempre di provare piacere e di regalare il piacere agli altri, che scopre ogni volta quello che fa come se fosse la prima volta. (Freak Antoni)
