Closed Form

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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 $(-0.75, 0.75)$ ?

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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 of Geometric Series

2025-01-12 23:34:11

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