23. Taylor Series

e. Taylor Remainder

When we approximate a power series, S=n=0an\displaystyle S=\sum_{n=0}^\infty a_n, by one of its partial sums Sk=n=0kan\displaystyle S_k=\sum_{n=0}^k a_n, the error in the approximation, Ek=SSk=n=k+1an\displaystyle E_k=S-S_k=\sum_{n=k+1}^\infty a_n, may be bounded by an integral bound, a comparison bound or an alternating series bound, as appropriate. When we approximate a function by a Taylor polynomial, the error in the approximation is called the Taylor remainder.

Taylor Remainder
The Taylor remainder of degree kk centered at x=ax=a for the function f(x)f(x) is Rkf(x)=f(x)Tkf(x) R_k f(x)=f(x)-T_k f(x) where the Taylor polynomial of degree kk is Tkf(x)=n=0kf(n)(a)n!(xa)n T_k f(x)=\sum_{n=0}^k \dfrac{\,f^{(n)}(a)}{n!}(x-a)^n

For values of xx for which the Alternating Series Test, the Integral Test or the Simple Comparison test are sufficient to prove convergence, the corresponding bounds may be applied to the remainder. (This was done in the example on the previous page.) However, when these techniques fail, the Taylor Remainder Inequality (given below) will give a bound on the error in the Taylor polynomial approximation. (See the example and exercise on the next page.) Further, the Taylor Remainder Inequality will finally allow us to determine whether the Taylor series for a function f(x)f(x) actually converges to the function f(x)f(x). (See the example and exercise on the following page.) This inequality is based on the Taylor Remainder Formula given in the next theorem.

1. Formula and Inequality

The above definition says what the remainder is, but it does not give a formula for it or even put an upper bound on it. The following theorem gives a formula and the next one says how to put an upper bound on it.

Taylor Remainder Formula (Extended Mean Value Theorem)
If the (k+1)st(k+1)^\text{st} derivative exists on an open interval containing aa and xx, then there is a number cc between aa and xx such that the Taylor remainder of degree kk is Rkf(x)=f(k+1)(c)(k+1)!(xa)k+1 R_k f(x)=\dfrac{f^{(k+1)}(c)}{(k+1)!}(x-a)^{k+1}   The proof is similar to that for the Mean Value Theorem .

The Taylor remainder may be combined with the Taylor polynomial to say: f(x)=Tkf(x)+Rkf(x)=f(a)+f(a)(xa)+f(a)2(xa)2++f(k)(a)k!(xa)k+f(k+1)(c)(k+1)!(xa)k+1\begin{aligned} f(x) &=T_k f(x)+R_k f(x) \\ &=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2}(x-a)^2 \\ &\quad+\cdots+\dfrac{\,f^{(k)}(a)}{k!}(x-a)^k +\dfrac{f^{(k+1)}(c)}{(k+1)!}(x-a)^{k+1} \end{aligned} So the remainder is just the next term in the Taylor series except that the (k+1)st(k+1)^\text{st} derivative is evaluated at cc instead of aa.

For k=0k=0, the remainder is R0f(x)=f(x)f(a)R_0 f(x)=f(x)-f(a), and the Taylor Remainder Formula says R0f(x)=f(c)(xa)R_0 f(x)=f'(c)(x-a). This is just the Mean Value Theorem for Derivatives:

Mean Value Theorem for Derivatives
If f(x)f'(x) exists on an open interval containing aa and xx, then there is a number cc between aa and xx such that f(x)=f(a)+f(c)(xa) f(x)=f(a)+f'(c)(x-a) or equivalently, f(c)=f(x)f(a)xa f'(c)=\dfrac{f(x)-f(a)}{x-a}

Notice that the first version looks like the equation of the tangent line except that the derivative f(c)f'(c) is evaluated at cc instead of aa. Consequently, Taylor's Remainder Theorem is often called the Extended Mean Value Theorem.

The problem with these theorems is that there is no way to determine the value of cc guaranteed by the theorem. So instead, we find a bound on f(k+1)(c)f^{(k+1)}(c) and use it to put a bound on the remainder Rkf(x)R_k f(x):

Taylor Remainder Inequality
If aa is in an open interval II and Mf(k+1)(c)M \ge |f^{(k+1)}(c)| for all cc in the interval II, then the Taylor remainder of degree kk centered at x=ax=a satisfies Rkf(x)M(k+1)!xak+1 |R_k f(x)| \le \dfrac{M}{(k+1)!}|x-a|^{k+1} for all xx in the interval II. The quantity on the right is called the Taylor bound on the remainder.

On the next page we will see examples in which the Taylor Remainder Inequality is used to bound the error in a Taylor polynomial approximation. Then on the following page we will see how it is used to show the Taylor series for a function f(x)f(x) actually converges to the function f(x)f(x).

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