Closed Form

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Let’s walk the bridge backwards

1 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 $(-0.75, 0.75)$?

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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.

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more examples can be found by following this link
More Examples of Geometric Series

2025-01-07 04:15:46