The Polynomial Is an Approximation; the Remainder Is the Error
A Taylor polynomial about a center \(a\) uses the derivatives of a function at \(a\) to approximate its values near that center. It is important to distinguish the polynomial itself from the difference between that polynomial and the function. The polynomial is the approximation; the difference is the approximation error.
This distinction matters in applications. A computed polynomial value is not automatically the function's exact value, and the difference can be positive or negative. Its sign tells us whether the polynomial underestimates or overestimates the function at the point in question. Its magnitude tells us how far the approximation is from the true value.
The definition fixes the sign convention: remainder equals function minus polynomial. If \(R_n(x)>0\), then \(f(x)>T_n(x)\), so the polynomial underestimates the function there. If \(R_n(x)<0\), the polynomial overestimates it. The absolute error is \(\lvert R_n(x)\rvert\); it gives the size of the discrepancy without recording its direction.
The remainder is a function of \(x\), not a single constant attached to the polynomial. Even for one fixed Taylor polynomial, it can have different signs and magnitudes at different points. At the center, however, the polynomial agrees with the function: \(R_n(a)=0\).
Why the Remainder and Its Derivatives Vanish at the Center
The Taylor polynomial is constructed so that its derivatives at \(a\) match those of \(f\), through order \(n\). This explains the close local matching encoded by the polynomial. The following proof checks that fact directly from the definition.
For any \(j\) with \(0\leq j\leq n\), differentiating a term of degree less than \(j\) gives zero. A term of degree at least \(j\) gives $$ \frac{d^j}{dx^j}\left(\frac{f^{(k)}(a)}{k!}(x-a)^k\right) = \frac{f^{(k)}(a)}{(k-j)!}(x-a)^{k-j}. $$
At \(x=a\), every term with \(k>j\) contains a positive power of \(x-a\), so it is zero. The term with \(k=j\) equals \(f^{(j)}(a)\). Therefore \(T_n^{(j)}(a)=f^{(j)}(a)\).
Since \(R_n=f-T_n\), it follows that $$ R_n^{(j)}(a)=f^{(j)}(a)-T_n^{(j)}(a)=0 \qquad (0\leq j\leq n). $$
This result says that the remainder has a zero of order at least \(n+1\) at the center, provided the derivatives in question exist. In particular, it is not merely the function value that matches at \(a\): the derivatives through order \(n\) match as well. This local agreement does not, by itself, guarantee that a Taylor polynomial is accurate at an arbitrary distance from the center.
Worked Examples: Finding the Remainder and Reading Its Sign
Worked Example: A Rational Function About Zero
Find the degree-2 Taylor polynomial and remainder for \(f(x)=\frac{1}{1+x}\) about \(a=0\), then evaluate both at \(x=\frac15\).
The derivatives needed at zero are \(f(0)=1\), \(f'(x)=-(1+x)^{-2}\), so \(f'(0)=-1\), and \(f''(x)=2(1+x)^{-3}\), so \(f''(0)=2\). Hence $$ T_2(x)=1-x+\frac{2}{2!}x^2=1-x+x^2. $$
Subtracting this polynomial from the function gives an exact expression: $$ R_2(x)=\frac{1}{1+x}-(1-x+x^2) =\frac{1-(1+x)(1-x+x^2)}{1+x}. $$ Since \((1+x)(1-x+x^2)=1+x^3\), this simplifies to $$ R_2(x)=-\frac{x^3}{1+x}. $$
At \(x=\frac15\), the polynomial value is $$ T_2\left(\frac15\right)=1-\frac15+\frac1{25}=\frac{21}{25}. $$ The true function value is \(f\left(\frac15\right)=\frac{1}{6/5}=\frac56\), and the remainder is $$ R_2\left(\frac15\right)=-\frac{(1/5)^3}{1+1/5} =-\frac{1/125}{6/5}=-\frac1{150}. $$
Check by subtraction: $$ f\left(\frac15\right)-T_2\left(\frac15\right) =\frac56-\frac{21}{25} =\frac{125-126}{150} =-\frac1{150}. $$ The negative remainder means the polynomial overestimates the function at this point. Its absolute error is \(\frac1{150}\), approximately \(0.006667\).
Worked Example: A Logarithm About Zero
For \(f(x)=\ln(1+x)\), find the degree-1 Taylor polynomial about zero and the exact remainder at \(x=\frac14\).
Since \(f(0)=0\) and \(f'(x)=\frac{1}{1+x}\), we have \(f'(0)=1\). Thus $$ T_1(x)=x. $$ To find the remainder, use the integral definition of the natural logarithm: $$ R_1(x)=\ln(1+x)-x =\int_0^x\left(\frac{1}{1+t}-1\right)\,dt =-\int_0^x\frac{t}{1+t}\,dt. $$
For \(x=\frac14\), evaluating the exact difference gives $$ R_1\left(\frac14\right) =\ln\left(\frac54\right)-\frac14 \approx 0.223144-0.250000 =-0.026856. $$ The displayed decimals are rounded to six places. The integral also verifies the sign: for \(0<t\leq\frac14\), \(t/(1+t)>0\), so the negative integral is strictly negative.
As a check, the polynomial value is \(T_1(\frac14)=\frac14\), and adding the exact remainder gives $$ T_1\left(\frac14\right)+R_1\left(\frac14\right) =\frac14+\ln\left(\frac54\right)-\frac14 =\ln\left(\frac54\right) =f\left(\frac14\right). $$ The polynomial overestimates the logarithm by about \(0.026856\).
Worked Example: A Cosine Approximation About Zero
Use the degree-2 Taylor polynomial for \(f(x)=\cos x\) about zero to approximate \(\cos(\frac{\pi}{6})\), and calculate the signed error.
The values at zero are \(f(0)=1\), \(f'(0)=0\), and \(f''(0)=-1\). Therefore $$ T_2(x)=1-\frac{x^2}{2}. $$ At \(x=\frac{\pi}{6}\), $$ T_2\left(\frac{\pi}{6}\right) =1-\frac{(\pi/6)^2}{2} =1-\frac{\pi^2}{72} \approx 0.862922. $$
The exact function value is \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\approx0.866025\). Subtracting the approximation from the function gives $$ R_2\left(\frac{\pi}{6}\right) =\frac{\sqrt{3}}{2}-1+\frac{\pi^2}{72} \approx0.003103. $$ The decimals are rounded to six places. Thus the polynomial underestimates the cosine at this point, and its absolute error is approximately \(0.003103\).
The sign can also be checked without relying on the rounded values. For \(x>0\), $$ 1-\cos x=\int_0^x (x-t)\cos t\,dt, $$ as follows by integrating the right-hand side by parts. Consequently, $$ \cos x-\left(1-\frac{x^2}{2}\right) =\int_0^x (x-t)(1-\cos t)\,dt. $$ For \(x=\frac{\pi}{6}\), the integrand is positive for \(0<t<x\), so this remainder is positive. Finally, adding the exact remainder back to the polynomial gives $$ 1-\frac{\pi^2}{72} +\left(\frac{\sqrt{3}}{2}-1+\frac{\pi^2}{72}\right) =\frac{\sqrt{3}}{2}, $$ which is the stated function value.
What Sign and Magnitude Tell You
The sign and magnitude answer different questions. The sign records the direction of the error; the magnitude records how large the error is. For an approximation \(T_n(x)\), these meanings follow immediately from \(R_n(x)=f(x)-T_n(x)\).
| Remainder at \(x\) | Interpretation |
|---|---|
| \(R_n(x)>0\) | \(T_n(x)<f(x)\): the polynomial underestimates the function. |
| \(R_n(x)<0\) | \(T_n(x)>f(x)\): the polynomial overestimates the function. |
| \(\lvert R_n(x)\rvert\) | The absolute error, or distance between the approximation and the function value. |
When reporting an approximation, state what value was used and whether an error is signed or absolute. For instance, saying that the error is \(-0.026856\) gives both its size and direction; saying that the absolute error is \(0.026856\) gives only its size. Confusing these quantities can reverse the conclusion about whether an approximation lies above or below the function.
The remainder is also not simply “the next term” of a Taylor series. A finite Taylor polynomial is defined using derivatives at the center, and its remainder is the exact difference \(f(x)-T_n(x)\). A series may help express that difference when it converges to the function, but the definition of the remainder does not require such a series representation.
Using the Remainder Carefully
For a fixed center, the remainder is zero at the center and its derivatives through order \(n\) vanish there when the required derivatives exist. This explains why the Taylor polynomial is designed to match the function locally. But the definition alone does not say how large the remainder is at a particular nonzero distance from the center. Its sign and magnitude must be found from additional information about the function.
Also, a small signed remainder is not the same as a guaranteed small absolute error: the latter concerns \(\lvert R_n(x)\rvert\). If errors from multiple approximations are being considered, their signs may differ, so they should not be combined as though every error had the same direction. Finally, having derivatives at the center is enough to form the polynomial, but it does not by itself ensure that the polynomial accurately represents the function throughout a larger interval.
Check Your Understanding
Use the definition of the remainder and the examples above to answer these questions.
- If \(R_n(x)=-0.04\), is the Taylor polynomial above or below the function value, and what is the absolute error?
- Write the degree-2 Taylor polynomial for \(f(x)=e^x\) about \(a=0\), using the derivatives at the center.
- For \(f(x)=\frac{1}{1+x}\), the degree-2 remainder about zero is \(-\frac{x^3}{1+x}\). What does its sign imply for \(0<x<1\)?
- Why do the derivatives of \(R_n\) through order \(n\) vanish at the center, when they exist?
- Is the signed remainder the same quantity as the absolute error? Explain the difference.