What Is a Taylor Polynomial?
Let's start with the definition. Given a function f, a specific point x = a (called the center), and a positive integer n, the Taylor polynomial of f at a, of degree n, is the polynomial T of degree n that best fits the curve y = f(x) near the point a, in the sense that T and all its first n derivatives have the same value at x = a as f does.
If the center is 0, then T may be called a Maclaurin polynomial.
Appearing here is the formula to find the Taylor polynomial:
Equivalently, if you expand the sum out, you see something like this one appearing here:
Now we know it may look daunting at first, but there's a step-by-step procedure for creating a Taylor polynomial. As long as you've had plenty of experience with derivatives, and if you know your way around factorials (that's the i! in the formula), then it shouldn't be too hard.
Computing a Taylor Polynomial
Notice that each term of the formula requires:
- A derivative of some order. In fact, the little (i) notation on the f means the derivative of order i (it doesn't mean f to the i power, even though the little (i) seems to look like an exponent).
- Plugging the given value a into that derivative.
- Dividing by the factorial number i!. Recall that i! = i(i - 1)(i - 2). . . (2)(1), if i > 0, and, by convention, 0! = 1.
- Multiplying by (x - a) to the i power. Note, in this formula x is just an unknown variable, and we leave it in the expression as part of the polynomial.
Once each term is built up, then the entire Taylor polynomial may be read off as the sum of those terms. Let's see this by example!
Example: Find the third degree Taylor polynomial for f(x) = 4/x, centered at x = 1.
First, we rewrite 4/x = 4x(-1) to make derivatives easier to find. Notice the table appearing on your screen right now starts with i = 0.
|i||i-th deriv.||Plug in center (a)||Divide by i!||Mult. by (x - a)i|
|0||4x(-1)||4(1)(-1) = 4||4/0! = 4/1 = 4||4|
|1||-4x(-2)||-4(1)(-2) = -4||-4/1! = -4/1 = -4||-4(x - 1)|
|2||8x(-3)||8(1)(-3) = 8||8/2! = 8/2 = 4||4(x - 1)2|
|3||-24x(-4)||-24(1)(-4) = -24||-24/3! = -24/6 = -4||-4(x - 1)3|
Therefore, you should always start with 0. In fact, it's the easiest part of the table, because this term simply boils down to f(a). Here, a = 1, because that's the given center.
So the answer is: T = 4 - 4(x - 1) + 4(x - 1)2 - 4(x - 1)3.
Probably one of the most important functions in calculus is the natural exponent, ex, or as it is often written, exp(x). What makes this function so special is that the only function whose derivative is equal to itself is a constant multiple of exp(x). This fact makes finding Taylor polynomials of exp(x) quite easy!
So here's our example: Find the fifth degree Maclaurin polynomial for exp(x).
As you look over this table, recall that a Maclaurin polynomial is simply a Taylor polynomial centered at a = 0.
|i||i-th deriv.||Plug in center (a)||Divide by i!||Mult. by (x - a)^i|
|0||exp(x)||exp(0) = 1||1/0! = 1||1|
|1||exp(x)||exp(0) = 1||1/1! = 1||x|
|2||exp(x)||exp(0) = 1||1/2! = 1/2||(1/2)x2|
|3||exp(x)||exp(0) = 1||1/3! = 1/6||(1/6)x3|
|4||exp(x)||exp(0) = 1||1/4! = 1/24||(1/24)x4|
|5||exp(x)||exp(0) = 1||1/5! = 1/120||(1/120)x5|
In many situations, we leave the factorial notation as it is - for example, writing 5! instead of 120 in the answer. This equation appearing here is the fifth-degree Maclaurin series for exp(x):
Taylor Polynomials: Functions
The reason we care about Taylor and Maclaurin polynomials is that they typically do a good job approximating functions. Say you have a complicated function that's hard to analyze. Well if you could replace it with a polynomial, which is typically easier to work with, then you'd be all set. Furthermore, the higher the degree n, the better the approximation will be!
But how good is the approximation? Let's see how the graph of exp(x) compares to its Maclaurin polynomials of degrees up to 4. For reference, here are the Maclaurin polynomials appearing on your screen right now:
How about this example? Use a 3rd degree Maclaurin polynomial to estimate exp(1).
Just plug in x = 1 and, as you can see, we get:
Not a bad estimate, since exp(1) = 2.718.
All right, let's briefly review what we've learned. We first learned that a Taylor polynomial is the polynomial T of degree n that best fits the curve y = f(x) near the point a. We also learned that the n-th degree Taylor polynomial for a function f, centered at x = a is found by this formula appearing here:
We also learned that a Maclaurin polynomial is just a Taylor polynomial with center 0. And, finally, before delving into some examples, we learned that the higher the degree, the better a Taylor polynomial will approximate the given function near its center.
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Function Represented by the Sum of a Taylor Series
Tn (x) is the nth degree Taylor polynomial expanded around x = a
If f(x) = Tn (x) + Rn (x) and, in the limit, as n approaches infinity, Rn (x) approaches zero on the interval |x - a| < D, then f(x) can be represented by the sum of the Taylor series on the interval |x-a| < D.
Taylor's Inequality for the remainder:
- M |x-a|n + 1 / (n+1)! ≤ Rn (x) ≤ M |x-a|n + 1 / (n+1)!
Where |f(n + 1) (x)| ≤ M on the interval |x-a| < D.
Find the Maclaurin series for sin(x) and prove that it represents the function for all x.
Computing the Maclaurin series:
f(x)=sin(x) ; f(0) = 0
f'(x) = cos(x) ; f'(0) = 1
f(x) = -sin(x) ; f(0) = 0
f'(x) = -cos(x) ; f'(0) = -1
f""(x) = sin(x) ; f""(0) = 0
The series is:
f(0) + f'(0)/1! + f(0)/2! + f'(0)/3! + ... =
= x - x3 /3! + x5 /5! + ... =
=Σ (-1)n x2n + 1 / (2n + 1)!
Since the absolute value of all derivatives in the Taylor inequality for the remainder is less than one, we can use M =1.
- |x|n + 1 / (n+1)! ≤ Rn (x) ≤ |x|n + 1 / (n+1)!
In the limit, as n becomes infinite, the remainder approaches zero.
The function can be represented by its series expansion for all values of x.
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