In this post, I show how to find a Taylor series for a function and how to find its convergence.
Example
Solution
Instead of applying the general formula of Taylor series, let's say that the Taylor series is of the form
We already know a =1. All we have to do is to determine fⁿ(a). In order to do that we determine first f'(1), f"(1). f'''(1) then f⁴(n):
The order number derivative is equal to the factorial of the number with the derivative changing sign alternatively. We can generalize by saying fⁿ(1) = (-1)ⁿ n!.
The Taylor series of f at x = 1 then is by substituting fⁿ(1):
To determine the interval convergence of the series, let's determine its convergence. Let's use the ratio test by starting to determine the ratio:
The limit of this ratio when n approaches infinity is equal to the limit of ❘x-1❘. This limit is equal to ❘x-1❘
The series is convergent if ❘x-1❘< 0 i.e to 0 <x<2. Let's check the convergence of the series at the endpoints i.e at x = 0 and x = 2.
At x= 0 we have
Let's use the divergence test by calculating the limit of 1 when n approaches infinity. This limit is equal to 1. The series diverges since the limit is different of 0.
At x = 2, we have:
The limit of (-1)ⁿ is probably unknown, therefore different of 0.
The series diverges when x = 2. The interval of convergence is (0,2).
Practice
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