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**Taylor’s Theorem **is a procedure for estimating the remainder of a Taylor polynomial, which approximates a function value. In other words, it gives bounds for the error in the approximation. The remainder given by the theorem is called the **Lagrange form of the remainder** [1].

## Formula for Taylor’s Theorem

The formula is:

Where:

- R
_{n}(x) = The remainder / error, - f
^{(n+1)}= The nth plus one derivative of f (evaluated at z), - c = the center of the Taylor polynomial.

**The equation can be a bit challenging to evaluate.** In many cases, you’re going to want to find the absolute value of both sides of this equation, because of the challenges involved in evaliating the higher derivative: you normally have to take an educated guess here and there for the value of “z”: Sometimes, graphing the function may help with that step, as in the following example.

## Taylor Theorem Example

**Example Question:**What is the expression for the remainder for the following Maclaurin polynomial approximation of f(x) = cos(x) for cos(0.2)?

Step 1: ** Identify the degree of the polynomial. **This is a 4th degree. We can tell that by the highest degree in the equation (the last term is to the 4th power). It’s also identified as T

_{4}on the left side of the equation.

Step 2: **Add 1 to Step 1.**

4 + 1 = 5.

Step 3: **Fill in the formula with what you know so far:**

- c is 0, because all Maclaurin polynomials are centered at zero.
- (n + 1) is the number from Step 2 (for this example, that’s 5).
- x = 0.2 (this is given in the question: we’re finding the approximation for cos(0.2)).

Step 4: **Evaluate the numerator f ^{5} (z).** In English, that is telling us to “evaluate the fifth derivative of the function at point z”. Our function is cos(x):

**f**

^{5}cos(x) = -sin(x).The problem is we can’t directly evaluate it for z (we don’t know what “z” is!). But we do know (from looking at the graph above, in this example), that the cosine function has function values ranging from -1 to 1.

*This is where taking the absolute value of both sides comes in.* Rather than trying to work with -1 to 1, we can just work with the absolute value of the max function value, which in this case, is 1. We can then rewrite our formula as an inequality:

Evaluating this on a calculator, we get:

|R_{4}cos(0.2)| = 0.0000026666.

This is a tiny error, which is what we would expect from the graph.

**Note**: If we reevaluate for cos(2), you would get 0.26666666666, which is a much larger error. This is an upper bound for the error though; In reality the error here is much smaller (around 0.2).

## Why is Taylor’s Theorem Useful?

In practice, the theorem is quite useful. The idea behind it is that a function’s true value is the Taylor approximation T_{n}(x) plus an error (remainder)— a measure of how far off the mark the approximation is. If we can find bounds (i.e. a min or max) for the size of f^{(n + 1)}(x) on the interval (a, b), then we can use the theorem to get concrete error estimates for any Taylor approximation.[2]

## References

Graph: Desmos.com.

[1] Larson, R. & Edwards, B. (2017). Calculus. 11th Edition. Cengage Learning.

[2] Generalizing the Mean Value Theorem – Taylor’s theorem. Retrieved April 26, 2021 from: https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-c-mean-value-theorem-antiderivatives-and-differential-equations/session-34-introduction-to-the-mean-value-theorem/MIT18_01SCF10_ex34sol.pdf

**CITE THIS AS:**

**Stephanie Glen**. "Taylor’s Theorem; Lagrange Form of Remainder" From

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