Why Reciprocal Functions Make Useful Taylor Models
Reciprocal functions are simple enough for their derivatives to be computed exactly, yet their derivative patterns make sign errors easy to spot. They provide a clear setting for seeing how repeated derivatives determine the coefficients of a Taylor polynomial.
For the function \(f(x)=1/x\), the center must be nonzero: the function and its derivatives are undefined at zero. For the function \(g(x)=1/(1-x)\), the zero center is allowed, while the function is undefined at \(x=1\). In both cases, once the center is in the domain, the Taylor polynomial is found from the function’s derivatives there.
The coefficient of \((x-a)^k\) is the \(k\)th derivative at the center divided by \(k!\). Thus the task is to calculate the derivatives correctly, evaluate them at the center, and keep track of the signs.
Repeated Derivatives of One Over \(x\)
Write \(f(x)=x^{-1}\), for \(x\ne 0\). Differentiating successively gives $$ f'(x)=-x^{-2},\qquad f''(x)=2x^{-3},\qquad f'''(x)=-6x^{-4},\qquad f^{(4)}(x)=24x^{-5}. $$
The signs alternate, while the numerical factors are successive factorials. For every integer \(k\ge 0\), the pattern is $$ f^{(k)}(x)=(-1)^k k!x^{-(k+1)} =\frac{(-1)^k k!}{x^{k+1}}. $$
This formula can be justified by induction, so it is not merely an observed pattern. For \(k=0\), it gives \(f(x)=1/x\). Suppose it holds for some \(k\). Differentiating the proposed expression gives $$ \frac{d}{dx}\left((-1)^k k!x^{-(k+1)}\right) =(-1)^k k!(-(k+1))x^{-(k+2)} =(-1)^{k+1}(k+1)!x^{-(k+2)}. $$ This is the stated formula with \(k+1\) in place of \(k\). Therefore it holds for every \(k\ge 0\) on the domain \(x\ne0\).
At a nonzero center \(a\), substitute \(x=a\) into the derivative formula and then into the Taylor polynomial definition:
The center \(a\) may be positive or negative, but it cannot be zero. When \(a\) is negative, the powers of \(a\) also affect the signs of the coefficients, so they must be evaluated rather than ignored.
Worked Example: A Degree-Four Polynomial for One Over \(x\) at Two
Find the degree-four Taylor polynomial for \(f(x)=1/x\) centered at \(a=2\). The derivative formula gives, for \(k=0,1,2,3,4\), $$ f^{(k)}(2)=\frac{(-1)^k k!}{2^{k+1}}. $$ The Taylor coefficient for each power is \(f^{(k)}(2)/k!\), so the coefficients are $$ \frac12,\qquad -\frac14,\qquad \frac18,\qquad -\frac1{16},\qquad \frac1{32}. $$ Therefore, $$ T_4(x)=\frac12-\frac14(x-2)+\frac18(x-2)^2-\frac1{16}(x-2)^3+\frac1{32}(x-2)^4. $$
Check the result by differentiating the polynomial and evaluating each derivative at \(2\). Its successive derivative values are $$ T_4(2)=\frac12,\quad T_4'(2)=-\frac14,\quad T_4''(2)=\frac14,\quad T_4'''(2)=-\frac38,\quad T_4^{(4)}(2)=\frac34. $$ For the function, the corresponding values from \(f^{(k)}(2)=(-1)^k k!/2^{k+1}\) are $$ \frac12,\quad -\frac14,\quad \frac{2}{8}=\frac14,\quad -\frac{6}{16}=-\frac38,\quad \frac{24}{32}=\frac34. $$ They agree through order four, as required.
Repeated Derivatives of One Over One Minus \(x\)
Now let \(g(x)=1/(1-x)=(1-x)^{-1}\), whose domain excludes \(x=1\). The first derivative is positive: $$ g'(x)=(-1)(1-x)^{-2}(-1)=(1-x)^{-2}. $$ Differentiating again gives \(g''(x)=2(1-x)^{-3}\). Each further differentiation contributes a positive factor: the power rule contributes a negative sign, and differentiating the inner expression \(1-x\) contributes another negative sign. In general, $$ g^{(k)}(x)=\frac{k!}{(1-x)^{k+1}}. $$
For completeness, this pattern also follows by induction. It holds at \(k=0\). If \(g^{(k)}(x)=k!(1-x)^{-(k+1)}\), then the chain rule gives $$ \frac{d}{dx}\left(k!(1-x)^{-(k+1)}\right) =k!(-(k+1))(1-x)^{-(k+2)}(-1) =(k+1)!(1-x)^{-(k+2)}. $$ That is the formula for order \(k+1\).
At the center \(a=0\), every derivative is \(g^{(k)}(0)=k!\). Dividing by \(k!\) in the Taylor polynomial definition gives coefficient \(1\) for each power. Consequently, the degree-\(n\) Maclaurin polynomial is $$ T_n(x)=1+x+x^2+\cdots+x^n. $$
Worked Example: A Degree-Five Maclaurin Polynomial
Find the degree-five Taylor polynomial at zero for \(g(x)=1/(1-x)\). Its derivatives are \(g^{(k)}(x)=k!/(1-x)^{k+1}\), so evaluating at zero gives $$ g(0)=1,\quad g'(0)=1,\quad g''(0)=2,\quad g'''(0)=6,\quad g^{(4)}(0)=24,\quad g^{(5)}(0)=120. $$ The coefficient of \(x^k\) is \(g^{(k)}(0)/k!\). In order, these coefficients are \(1,1,2/2,6/6,24/24,120/120\), all equal to \(1\). Thus $$ T_5(x)=1+x+x^2+x^3+x^4+x^5. $$
To check, the \(k\)th derivative of \(T_5\), evaluated at zero, is \(k!\) for each \(k=0,1,2,3,4,5\): terms of degree below \(k\) have vanished after \(k\) differentiations, the \(x^k\) term contributes \(k!\), and terms of higher degree still contain a positive power of \(x\). These values match the six derivative values of \(g\) listed above.
A Check on Why These Coefficients Work
There is a direct reason that the derivative data determine the coefficients. Suppose a polynomial centered at \(a\) is written in powers of \(x-a\): $$ Q(x)=c_0+c_1(x-a)+c_2(x-a)^2+\cdots+c_n(x-a)^n. $$ At \(x=a\), every positive power vanishes, so \(Q(a)=c_0\). After differentiating once, evaluation at \(a\) leaves \(Q'(a)=c_1\); after differentiating twice, it leaves \(Q''(a)=2!c_2\). In general, after \(j\) differentiations, terms with degree less than \(j\) have disappeared and terms with degree greater than \(j\) still contain a factor of \(x-a\). Hence $$ Q^{(j)}(a)=j!c_j, \qquad\text{so}\qquad c_j=\frac{Q^{(j)}(a)}{j!}. $$
This explains the Taylor coefficient rule: choosing \(c_j=f^{(j)}(a)/j!\) makes the polynomial’s \(j\)th derivative at the center match the function’s \(j\)th derivative there, for every \(j\) through the chosen degree. It is the repeated-derivative pattern, not a guess based on a few terms, that produces the reciprocal-function polynomials.
Worked Example: A Degree-Three Polynomial at a Negative Center
Find the degree-three Taylor polynomial for \(f(x)=1/x\) centered at \(a=-2\). First evaluate the function and its first three derivatives: $$ f(-2)=-\frac12,\qquad f'(-2)=-\frac14,\qquad f''(-2)=\frac{2}{(-2)^3}=-\frac14,\qquad f'''(-2)=\frac{-6}{(-2)^4}=-\frac38. $$ The Taylor coefficients are these values divided by \(0!,1!,2!,3!\), respectively. They are $$ -\frac12,\qquad -\frac14,\qquad -\frac18,\qquad -\frac1{16}. $$ Therefore, $$ T_3(x)=-\frac12-\frac14(x+2)-\frac18(x+2)^2-\frac1{16}(x+2)^3. $$
Check by evaluating derivatives at the center. The polynomial gives $$ T_3(-2)=-\frac12,\qquad T_3'(-2)=-\frac14,\qquad T_3''(-2)=2!\left(-\frac18\right)=-\frac14,\qquad T_3'''(-2)=3!\left(-\frac1{16}\right)=-\frac38. $$ These equal \(f(-2), f'(-2), f''(-2)\), and \(f'''(-2)\) calculated above. Notice that the negative center makes all four coefficients negative; the alternating sign in the derivative formula alone does not determine the final signs without accounting for powers of \(-2\).
A Common Sign Error
A frequent mistake is to give \(1/(1-x)\) alternating derivative signs, as though its derivatives followed the pattern for \(1/x\). For instance, the incorrect calculation $$ \left(\frac{1}{1-x}\right)'=-\frac{1}{(1-x)^2} $$ misses the negative sign from differentiating the inner function \(1-x\). Applying the chain rule correctly gives $$ \left((1-x)^{-1}\right)'=(-1)(1-x)^{-2}(-1)=\frac{1}{(1-x)^2}. $$ In particular, the derivative at zero is \(+1\), not \(-1\). A useful correction is to check the first derivative before writing a general pattern, then verify the proposed formula at \(k=0\) and \(k=1\).
Another potential confusion is to assume that a Taylor polynomial is the same function as the reciprocal function. A Taylor polynomial is a polynomial whose derivatives match those of the function at the selected center through a specified degree. It is a finite local model; its definition does not assert equality with the function at every \(x\).
Using the Formulas Carefully
For \(1/x\), first confirm that the center \(a\) is nonzero, then use the coefficient \((-1)^k/a^{k+1}\) for the power \((x-a)^k\). For \(1/(1-x)\) centered at zero, every coefficient through the selected degree is \(1\). In either case, a reliable check is to multiply the coefficient of the \(k\)th power by \(k!\) and confirm that the result equals the \(k\)th derivative at the center.
The function \(1/(1-x)\) is undefined at \(x=1\), and \(1/x\) is undefined at \(x=0\). Those restrictions matter when selecting a center and interpreting the local model. The formulas here construct finite Taylor polynomials; they do not by themselves give an error bound or justify an infinite series.
Check Your Understanding
Use derivative values at the center to justify each coefficient, and check your work before moving on.
- Find the first four derivatives of \(1/x\), and state the general formula for its \(k\)th derivative.
- Write the degree-three Taylor polynomial for \(1/x\) centered at \(a=1\).
- Write the degree-two Taylor polynomial for \(1/x\) centered at \(a=-1\). Explain how the negative center affects the coefficient signs.
- Find the degree-four Maclaurin polynomial for \(1/(1-x)\), and state the derivative value at zero for each order from zero through four.
- Explain why the derivatives of \(1/(1-x)\) at zero are positive rather than alternating in sign.