2. Maclaurin Series
By M. Bourne
In the last section, we learned about Taylor Series, where we found an approximating polynomial for a particular function in the region near some value x = a.
We now take a particular case of Taylor Series, in the region near x = 0. Such a polynomial is called the Maclaurin Series.
The infinite series expansion for f(x) about x = 0 becomes:
f '(0) is the first derivative evaluated at x = 0, f ''(0) is the second derivative evaluated at x = 0, and so on.
[Note: Some textbooks call the series on this page "Taylor Series" (which they are, too), or "series expansion" or "power series".]
Example: Expanding sin x
Find the Maclaurin Series expansion for f(x) = sin x.
We plot our answer
to see if the polynomial is a good approximation to f(x) = sin x.
We observe that our polynomial (in red) is a good approximation to f(x) = sin x (in blue) near x = 0. In fact, it is quite good between -3 ≤ x ≤ 3.
Example: Expanding ex
Find the Maclaurin Series expansion of f(x) = ex.
Exercise
Find the Maclaurin Series expansion of cos x.
Finding Pi Using Infinite Series
In the 17th century, Leibniz used the series expansion of arctan x to find an approximation of π.
We start with the first derivative:
The value of this derivative when x = 0 is 1.
Similarly for the subsequent derivatives:
f ''(0) = 0
f '''(0) = -2
f iv(0) = 0
f v(0) = 24
Now we can substitute into the Maclaurin Series formula:
Considering that (see the 45-45 triangle)
we can substitute x = 1 into the above expression and get the following expansion for π
All very well, but it was not a good way to find the value of π because this expansion converges very slowly.
Even after adding 1000 terms, we don't have 3 decimal place accuracy.
(We know now that π = 3.141 592 653 5...)
I wonder whether Leibniz was aware it was not such a good approximation...?
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