Closed Form

$\newenvironment {prompt}{}{} \newcommand {\accordion }[0]{} \newcommand {\ungraded }[0]{} \newcommand {\xmDefaultPreamble }[0]{xmPreamble.tex} \newcommand {\xmDefaultPrintstyle }[0]{xmPrintstyle.sty} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{}$

center

Let’s walk the bridge backwards

Closed Form

Gemetric Series

Given

\[ K(t) = \sum _{n=0}^{\infty } \left ( \frac {t}{2} \right )^2\]

find the closed form formula and the interval of convergence.

This is a Geometric series. The interval of convergence is $(-2, 2)$, since this is when the inside of the general term is $-1$ and $1$.

The closed form is

\[ \frac {1}{1 - \frac {t}{2}} = \frac {2}{2-t} \]

The infinite geometric series will equal $\frac {2}{2-t}$ on $(-2, 2)$. How many terms of the series to we need for a good approximation on just $(-0.75, 0.75)$?

This browser does not support embedded elements.

Gemetric Series

Given

\[ K(t) = \sum _{n=2}^{\infty } \left ( \frac {t}{2} \right )^2 \]

find the closed form formula and the interval of convergence.

This is the same Geometric series, except missing the first two terms. The interval of convergence is $(-2, 2)$, since this is when the inside of the general term is $-1$ and $1$.

The closed form is the same, except subtract off the first two terms.

\[ \frac {1}{1 - \frac {t}{2}} - 1 - \frac {t}{2} = \frac {2}{2-t} - 1 - \frac {t}{2} \]

need a common denominator

\[= \frac {4}{2(2-t)} - \frac {2(2-t)}{2(2-t)} - \frac {t(2-t)}{2(2-t)} = \frac {4-2(2-t)-t(2-t)}{2(2-t)} = \frac {t^2}{2(2-t)} \]

\[ K(t) = \frac {t^2}{2(2-t)} \]

on $(-2, 2)$.

On the other hand, we could have factored out $\left ( \frac {t}{2} \right )$ in order to get the index back to $0$.

\[ K(t) = \sum _{n=2}^{\infty } \left ( \frac {t}{2} \right )^n = \left ( \frac {t}{2} \right )^2 \sum _{n=0}^{\infty } \left ( \frac {t}{2} \right )^n \]

Then the closed form would be

\[ \left ( \frac {t}{2} \right )^2 \cdot \frac {1}{1 - \frac {t}{2}} \]

\[ \frac {t^2}{2(2 - t)} \]

Same formula.

Gemetric Series

Given

\[ K(t) = \sum _{n=0}^{\infty } \left ( \frac {t-3}{2} \right )^2\]

find the closed form formula and the interval of convergence.

This is a Geometric series. The interval of convergence is $(1, 5)$, since this is when the inside of the general term is $-1$ and $1$. Our interval is centered at $3$, which is where the terms equal $0$.

The closed form is

\[ \frac {1}{1 - \frac {t-3}{2}} = \frac {2}{2-(t-3)} = \frac {2}{5-t} \]

Our closed form formula has a singularity at $5$. Therefore, our series equivalence cannot go beyond $5$, which it does not.

This browser does not support embedded elements.

ooooo-=-=-=-ooOoo-=-=-=-ooooo
more examples can be found by following this link
More Examples of Geometric Series

2025-01-12 23:34:11