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Geometric series for values between 0 and 1. 0. Formula for a geometric series. 0. Writing geometric ...
A clever solution to find the expected value of a geometric r.v. is those employed in this video lecture of the MITx course "Introduction to Probability: Part 1 - The Fundamentals" (by the way, an extremely enjoyable course) and based on (a) the memoryless property of the geometric r.v. and (b) the total expectation theorem.
The definition of a geometric series is a series where the ratio of consecutive terms is constant. It doesn't matter how it's indexed or what the first term is or whether you have a constant. That stuff just has to do with how you write the series.
Matrices and Geometric Series. Related. 0. Sum of series with generic term inside it. 3.
So now my question is, why do they give the same result? Is there some relationship between taylor series and geometric series? Also for what type of functions do they give the same result? (i.e. polynomials, trig functions etc.) Because I found computing the taylor series using the geometric series approach a lot quicker.
Generalized geometric series with long range dependence. 1. Geometric Series/Sequence Partial Sums. 0.
I realize this is an old thread, but I wanted to expand on the above answers on how to derive the formula for anyone else that might come along. Starting with the geometric series and taking successive derivatives:
$\begingroup$ So surely you see the answer now, but I'll state it for the record: a power series is a geometric series if its coefficients are constant (i.e. all the same). In particular, not all power series are geometric. For example $\sum x^n$ is geometric, but $\sum \frac{x^n}{n!}$ is not. $\endgroup$ –
Using geometric series for computing Integrals. 1. Finite geometric series of matrices. 0.
In fact, there is a simpler solution to find the sum of this series only with these given variables. By modifying geometric series formula, Sn = a(1-r^n)/1-r is equal to a-ar^n/1-r. And a is the first term and ar^n is the term after the last term, ar^n-1. Both are given by the problem: a=8 and ar^n-1=52488.