As a simple example, you can create the number 10 from smaller numbers: 1 + 2 + 3 + 4.
In the same way, you can piece together simple polynomials to create a more complicated function.
Why would you want to do this? In many cases, you know what a function looks like on a graph; but you don’t know the equation for it. The Maclaurin series can create a duplicate (a doppelgänger, if you like); it’s so close to the real thing that you won’t be able to tell the difference.
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Taylor polynomials can be used to approximate a function around any value for a differentiable function. In other words, when you use a Taylor series, you assume that you can find derivatives for your function. Taylor polynomials look a little ugly, but if you break them down into small steps, it’s actually a fast way to approximate a function.
Use Taylor polynomials to approximate the function cos(x) around the point x = 2.
Notes on the symbols used in the formula:
- ! is the factorial symbol).
- The “c” in the expansion is the point you’re evaluating the function at. In this example, c = 2.
- f′(x) = the first derivative.
- ′′(x) is the second derivative (i.e. take the derivative of the derivative),
- f′′′(x) is the third derivative (i.e. take the derivative of the second derivative)…and so on.
- f(x) = x3.
- First derivative f′(x) = 3x2;
- Second derivative f′′(x): The derivative of 3x2 is 6x, so f′′(x) = 6x
- Third derivative (i.e. take the derivative of the second derivative): f′′′(x) = 6.
- …and so on.
Step 2: Evaluate the function for the second part of the Taylor polynomial.
The first derivative of cos(2) is -sin(2), giving us:
Which simplifies to -sin(2)(x – 2).
Adding this answer to the first part from Step 1, we get:
p(x) = cos(2) – sin(2)(x-2)
Step 3: Evaluate the function for the third part of the Taylor polynomial (adding it to the first and second parts from Step 2). In this step, you’re taking the second derivative (f”(x)):
The second derivative of cos(x) is −cos(x), so we end up with:
p(x) = cos(2) – sin(2)(x-2) – cos(2) ⁄ 2(x-2)2
Step 4: Evaluate the function for the fourth part of the Taylor polynomial (adding it to the first, second and third parts from Step 3):
p(x) = cos(2) – sin(2)(x-2) – cos(2) ⁄ 2(x-2)2 + sin(2)⁄6 (x-2)3
Step 5: Continue evaluating more pieces of the Taylor polynomial, graphing the function periodically to see how well it represents your polynomial.
Tip: Technically, you could go on forever with iterations of the Taylor polynomial, but usually five or six iterations is sufficient for a good approximation.
Watch the video or read the article below:
A Maclaurin series is a special case of a Taylor series, where “a” is centered around x = 0. The series are named after Scottish mathematician Colin Maclaurin. While you can calculate Maclaurin series using calculus, many series for common functions have already been found. For example, the following table shows the Maclaurin series for five common functions, along with the sigma notation for the expansion.
If you don’t have one of the most common functions, the Maclaurin series is fairly easy (although somewhat tedious) to calculate.
How to Find Maclaurin series: Steps
Example problem: Find the Maclaurin series for e5x.
Step 1: Calculate the first few derivatives for the function until you can see a clear pattern. Usually you’ll only need to calculate four or five:
|f′(x) = 5e5x|
|f′′(x) = 52e5x|
|f′′′(x) = 53e5x|
Step 2: Fill in the derivatives you calculated in step 1 with 0 as the input. This gives you the values you need to insert into the Maclaurin series:
|f(x) = e5(0) = 1|
|f′(x) = 5e5(0) = 5|
|f′′(x) = 52e5(0) = 52|
|f′′′(x) = 53e5(0) = 53|
Step 3: Fill in the Maclaurin formula with the values you calculated in Step 2:
Step 4: (Optional): Rewrite using sigma notation:
That’s how to find Maclaurin series!
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