Let’s extend our bridge between infinite series and closed form formulas for Geometric series.
Geometric Template
Currently, we have an equivalence between a geometric series and a rational function.
This equivalence is valid while \(|f(x)| < 1\). This gives us a domain known as the convergence
interval of the series. The domain is centered around \(0\).
Another way to view this is that the convergence interval is centered around the zero of \(f(x)\), which has been \(0\) up to this point. What if a different domain number is a zero of \(f\)?
Suppose some number, \(c\) is the zero of \(f\). Our interval of convergence would be centered at \(c\) instead.
Create a series equivalent to \(g(x)=\frac {3}{7 - 2x}\) centered at \(5\).
First a quick analysis of \(g(x)=\frac {3}{7 - 2x}\). We can see that \(\frac {7}{2}\) is a singularity. The domain of our infinite series representation of \(g(x)\) cannot contain this singularity. The domain will be centered at \(5\) and come right up to \(\frac {7}{2}\). That gives an interval radius of \(\frac {3}{2}\). Therefore, the interval of convergence should be \(\left ( \frac {7}{2}, \frac {13}{2} \right )\).
Now, let’s create the series representation.
We must rewrite \(\frac {3}{7 - 2x}\) in the form \(\frac {a}{1-something}\) where \(something(5) = 0\)
Almost. We have \(1 + something\) in the denominator. It must be \(1 - something\) to match our template.
The first few few terms look like
This is an example of an alternating series, because the signs of the terms
alternate.
Now, for the radius. We know the interval of convergence for a basic Geometric series is \((-1, 1)\).
Therefore, solve
\begin{align*} \frac {2(x-5)}{3} & = 1 \\ 2(x-5) & = 3 \\ x-5 & = \frac {3}{2} \\ x & = \frac {3}{2} + 5 = \frac {13}{2} \end{align*}
\begin{align*} \frac {2(x-5)}{3} & = -1 \\ 2(x-5) & = -3 \\ x-5 & = -\frac {3}{2} \\ x & = -\frac {3}{2} + 5 = \frac {7}{2} \end{align*}
The interval of convergence is \(\left ( \frac {7}{2}, \frac {13}{2}\right )\), as predicted.
We have two different formulas for the same function.
Polynomial Approximation
We have two different formulas for the same function, at least on \(\left ( \frac {7}{2}, \frac {13}{2}\right )\).
Let’s compare their graphs.
Of course, we cannot type in an infinite sum. We’ll approximate the infinite series with just the first few terms.
Just a fourth degree polynomial is doing a pretty good job approximating \(g(x)\) on a subinterval of our domain. More terms may do a better job.
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more examples can be found by following this link
More Examples of Geometric Series