When we use a Taylor polynomial [latex]p_n(x)[/latex] to approximate a function [latex]f(x)[/latex], we need to know how good our approximation is. The remainder [latex]R_n(x)[/latex] tells us exactly that—it’s the difference between the actual function value and our polynomial approximation:
[latex]{R}_{n}(x)=f(x)-{p}_{n}(x)[/latex].
Think of it this way: if you’re using a Taylor polynomial to estimate [latex]f(x)[/latex], the remainder is your error. The smaller the remainder, the better your approximation.
For a Taylor series to converge to [latex]f[/latex], we need two things:
the sequence of Taylor polynomials [latex]{p_n}[/latex] must converge
the remainder [latex]R_n[/latex] must approach zero as [latex]n[/latex] increases.
If [latex]R_n[/latex] doesn’t go to zero, your Taylor series might converge—but not to the function you want! To determine if [latex]R_{n}[/latex] converges to zero, we introduce Taylor’s theorem with remainder. Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the [latex]n[/latex]th Taylor polynomial approximates the function.
Let’s start with the simplest case: the [latex]0[/latex]th-degree Taylor polynomial. Here, [latex]p_0(x) = f(a)[/latex], so:
[latex]R_0(x) = f(x) - f(a)[/latex]
If [latex]f[/latex] is differentiable on an interval containing both [latex]a[/latex] and [latex]x[/latex], the Mean Value Theorem tells us there’s some value [latex]c[/latex] between [latex]a[/latex] and [latex]x[/latex] where:
[latex]R_0(x) = f'(c)(x - a)[/latex]
Using similar reasoning with higher derivatives, we can show that for an [latex]n[/latex]-times differentiable function:
where [latex]c[/latex] is some unknown value between [latex]a[/latex] and [latex]x[/latex].
The [latex]c[/latex] in the remainder formula is NOT the center [latex]a[/latex]—it’s an unknown value somewhere between [latex]a[/latex] and [latex]x[/latex]. We usually can’t find its exact value, but we can still use this formula to bound our error.
Since we don’t know the exact value of [latex]c[/latex], we can’t calculate [latex]R_n(x)[/latex] exactly. However, if we know that [latex]|f^{(n+1)}(x)| \leq M[/latex] for all [latex]x[/latex] in our interval, then:
This bound is incredibly useful—it tells us the maximum possible error when using a Taylor polynomial.
We now state Taylor’s theorem, which provides the formal relationship between a function [latex]f[/latex] and its [latex]n[/latex]th degree Taylor polynomial [latex]{p}_{n}\left(x\right)[/latex]. This theorem allows us to bound the error when using a Taylor polynomial to approximate a function value, and will be important in proving that a Taylor series for [latex]f[/latex] converges to [latex]f[/latex].
theorem: Taylor’s theorem with remainder
Let [latex]f[/latex] be a function that can be differentiated [latex]n+1[/latex] times on an interval [latex]I[/latex] containing the real number [latex]a[/latex]. Let [latex]p_{n}[/latex] be the [latex]n[/latex]th Taylor polynomial of [latex]f[/latex] at [latex]a[/latex] and let
be the [latex]n[/latex]th remainder. Then for each [latex]x[/latex] in the interval [latex]I[/latex], there exists a real number [latex]c[/latex] between [latex]a[/latex] and [latex]x[/latex] such that
If there exists a real number [latex]M[/latex] such that [latex]|{f}^{\left(n+1\right)}\left(x\right)|\le M[/latex] for all [latex]x\in I[/latex], then
We claim that g satisfies the criteria of Rolle’s theorem. Since g is a polynomial function (in t), it is a differentiable function. Also, g is zero at [latex]t=a[/latex] and [latex]t=x[/latex] because
Therefore, g satisfies Rolle’s theorem, and consequently, there exists c between a and x such that [latex]{g}^{\prime }\left(c\right)=0[/latex]. We now calculate [latex]{g}^{\prime }[/latex]. Using the product rule, we note that
Adding the first term on the left-hand side to both sides of the equation and dividing both sides of the equation by [latex]n+1[/latex], we conclude that
as desired. From this fact, it follows that if there exists M such that [latex]|{f}^{\left(n+1\right)}\left(x\right)|\le M[/latex] for all x in I, then
A way to prove that a Taylor series converges to [latex]f[/latex]
A method to bound the error when approximating [latex]f[/latex] with a Taylor polynomial
Notice how the factorial in the denominator grows very quickly as [latex]n[/latex] increases. This rapid growth often causes the remainder to approach zero, which is exactly what we need for convergence. Not only does Taylor’s theorem allow us to prove that a Taylor series converges to a function, but it also allows us to estimate the accuracy of Taylor polynomials in approximating function values. We begin by looking at linear and quadratic approximations of [latex]f\left(x\right)=\sqrt[3]{x}[/latex] at [latex]x=8[/latex] and determine how accurate these approximations are at estimating [latex]\sqrt[3]{11}[/latex].
Consider the function [latex]f\left(x\right)=\sqrt[3]{x}[/latex].
Find the first and second Taylor polynomials for [latex]f[/latex] at [latex]x=8[/latex]. Use a graphing utility to compare these polynomials with [latex]f[/latex] near [latex]x=8[/latex].
Use these two polynomials to estimate [latex]\sqrt[3]{11}[/latex].
Use Taylor’s theorem to bound the error.
For [latex]f\left(x\right)=\sqrt[3]{x}[/latex], the values of the function and its first two derivatives at [latex]x=8[/latex] are as follows:
The function and the Taylor polynomials are shown in Figure 5.
Figure 5. The graphs of [latex]f\left(x\right)=\sqrt[3]{x}[/latex] and the linear and quadratic approximations [latex]{p}_{1}\left(x\right)[/latex] and [latex]{p}_{2}\left(x\right)[/latex].
Using the first Taylor polynomial at [latex]x=8[/latex], we can estimate
By the Uniqueness of Taylor Series, there exists a c in the interval [latex]\left(8,11\right)[/latex] such that the remainder when approximating [latex]\sqrt[3]{11}[/latex] by the first Taylor polynomial satisfies
We do not know the exact value of c, so we find an upper bound on [latex]{R}_{1}\left(11\right)[/latex] by determining the maximum value of [latex]f^{\prime\prime}[/latex] on the interval [latex]\left(8,11\right)[/latex]. Since [latex]f^{\prime\prime}\left(x\right)=-\frac{2}{9{x}^{\frac{5}{3}}}[/latex], the largest value for [latex]|f^{\prime\prime}\left(x\right)|[/latex] on that interval occurs at [latex]x=8[/latex]. Using the fact that [latex]f^{\prime\prime}\left(8\right)=-\frac{1}{144}[/latex], we obtain
Since [latex]f^{\prime\prime\prime}\left(x\right)=\frac{10}{27{x}^{\frac{8}{3}}}[/latex], the maximum value of [latex]f^{\prime\prime\prime}[/latex] on the interval [latex]\left(8,11\right)[/latex] is [latex]f^{\prime\prime\prime}\left(8\right)\approx 0.0014468[/latex]. Therefore, we have
Watch the following video to see the worked solution to the example above.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
To estimate the error, use the fact that the sixth Maclaurin polynomial is [latex]{p}_{6}\left(x\right)={p}_{5}\left(x\right)[/latex] and calculate a bound on [latex]{R}_{6}\left(\frac{\pi }{18}\right)[/latex]. By the Uniqueness of Taylor Series, the remainder is
for some c between 0 and [latex]\frac{\pi }{18}[/latex]. Using the fact that [latex]|{f}^{\left(7\right)}\left(x\right)|\le 1[/latex] for all x, we find that the magnitude of the error is at most
Solving this inequality for [latex]x[/latex], we have that the fifth Maclaurin polynomial gives an estimate to within 0.0001 as long as [latex]|x|<0.907[/latex].
Watch the following video to see the worked solution to the example above.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
Now that we are able to bound the remainder [latex]{R}_{n}\left(x\right)[/latex], we can use this bound to prove that a Taylor series for [latex]f[/latex] at [latex]a[/latex] converges to [latex]f[/latex].
