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(x)~~f(0)+f’(0)x` `+(f’’(0))/(2!)x^2` `+(f’’’(0))/(3!)x^3` `+(f^("iv")(0))/(4!)x^4` `+...`
`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 1: Expanding sin x
Find the Maclaurin Series expansion for `f(x) = sin x`.
We plot our answer
`f(x)=x-1/6x^3` `+1/120x^5` `-1/5040x^7+...`
to see if the polynomial is a good approximation to `f(x) = sin x`.
We observe that our polynomial (in grey) is a good approximation to `f(x) = sin x` (in green) near `x = 0`. In fact, it is quite good between -π ≤ x ≤ π.
Example 2: Expanding `e^x`
Find the Maclaurin Series expansion of `f(x) = e^x`.
Here's a graph of the answer we just obtained, and the original `f(x) = e^x` graph:
We can see the approximation is very close to the original `f(x)=e^x` in the region `-2 < x < 2.5`.
Find the Maclaurin Series expansion of `cos x`.
Here's a graph showing the answer we just obtained, and the original `f(x) = cos(x)` graph:
Once again, we observe that our polynomial (in grey) is a good approximation to `f(x) = cos x` (in green) between -π ≤ x ≤ π.
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`. That is, `f’(0)=1`.
Similarly for the subsequent derivatives:
Now we can substitute into the Maclaurin Series formula:
`arctan\ x ~~0+(1)x+0/2x^2-2/(3!)x^3+0/(4!)x^4+24/(5!)x^5+...`
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..., and we know many other more efficient ways to find `pi`.)
Here's the graph of y = arctan x (in green) compared to the polynomial we just found (in grey).