Taylor Polynomials of Quotients

Why Coefficient Matching Works for Quotients

A quotient can be harder to differentiate repeatedly than either of its parts. When the denominator is nonzero at the center, there is a direct way to find the quotient’s Taylor polynomial: choose its coefficients so that multiplying by the denominator reproduces the numerator’s Taylor polynomial to the required degree.

The method is finite. It does not require expanding an infinite series or differentiating the quotient repeatedly. The key is to account for every cross-term in the product.

Central idea: If $f(x)=u(x)/v(x)$ and $v(a)\ne 0$, find the Taylor polynomial $Q_n$ of the quotient by matching the coefficients of $v(x)Q_n(x)$ with those of $u(x)$ through degree $n$, using powers of $x-a$.

Definitions and the Quotient Condition

For a function $F$ with $n$ derivatives at the center $a$, its degree-$n$ Taylor polynomial centered at $a$ is

T_nF(x;a)=\sum_{k=0}^{n}\frac{F^{(k)}(a)}{k!}(x-a)^k.

Write $h=x-a$ to make the powers around the center easier to see. Suppose $u$ and $v$ have Taylor polynomials through degree $n$, and suppose $v(a)\ne 0$. By continuity, $v(x)$ remains nonzero in some interval around $a$, so $f=u/v$ is defined there.

Write the Taylor polynomials of the numerator and denominator as

T_nu(x;a)=u_0+u_1h+\cdots+u_nh^n,\qquad T_nv(x;a)=v_0+v_1h+\cdots+v_nh^n,

where $u_k=u^{(k)}(a)/k!$ and $v_k=v^{(k)}(a)/k!$. In particular, $v_0=v(a)$, which is nonzero. We seek a polynomial

Q_n(x)=q_0+q_1h+\cdots+q_nh^n

whose coefficients make the product with the denominator agree with the numerator through degree $n$.

Definition: Matching through degree $n$ means that the coefficients of $h^0,h^1,\ldots,h^n$ in the product of the denominator’s Taylor polynomial and $Q_n$ equal the corresponding coefficients in the numerator’s Taylor polynomial. Terms of degree greater than $n$ do not affect this matching.

The Coefficient Equations

When multiplying the denominator polynomial by $Q_n$, the coefficient of $h^k$ is a sum of products. Each pair of indices that adds to $k$ contributes one term:

v_0q_k+v_1q_{k-1}+\cdots+v_kq_0=u_k \qquad (0\le k\le n).

For $k=0$, this gives $v_0q_0=u_0$. Since $v_0\ne0$, each successive equation can be solved for the next unknown coefficient.

Coefficient recurrence: Set $q_0=u_0/v_0$. For $1\le k\le n$, calculate $$ q_k=\frac{u_k-\displaystyle\sum_{j=1}^{k}v_jq_{k-j}}{v_0}. $$ The sum accounts for all the product terms involving previously calculated coefficients.

In particular, the first equations are

v_0q_0=u_0,\qquad v_0q_1+v_1q_0=u_1,\qquad v_0q_2+v_1q_1+v_2q_0=u_2.

This system is triangular: the equation for $q_k$ involves $q_k$ and coefficients already found at lower degrees. The nonzero value of $v_0$ is what permits each step.

Why the Method Gives the Taylor Polynomial

The coefficient equations are not merely a convenient approximation rule. They ensure that the resulting polynomial is the Taylor polynomial of the quotient. Here is the justification.

Match the product coefficients.
Choose $q_0,\ldots,q_n$ so that $T_nv(x;a)Q_n(x)$ and $T_nu(x;a)$ have the same coefficients through degree $n$. Thus their difference has no terms of degree at most $n$.

Relate the polynomial product to the functions.
The Taylor polynomials approximate $u$ and $v$ to degree $n$; multiplying by the fixed polynomial $Q_n$ preserves agreement through degree $n$. Consequently, $u(x)-v(x)Q_n(x)$ has zero Taylor coefficients through degree $n$.

Divide by the denominator.
Because $v(a)\ne0$, the denominator stays nonzero sufficiently near $a$. There, $$ \frac{u(x)}{v(x)}-Q_n(x)=\frac{u(x)-v(x)Q_n(x)}{v(x)}. $$ The numerator has no Taylor terms through degree $n$, so dividing by the nonzero denominator does not introduce any such terms. Therefore $Q_n$ is the degree-$n$ Taylor polynomial of $u/v$.

This reasoning also explains why the denominator condition matters: if $v(a)=0$, division by $v$ near the center may not be possible, and the coefficient equations cannot be solved by dividing by $v_0$.

Worked Example: A Quotient at the Origin

Find the degree-3 Taylor polynomial at $x=0$ for $f(x)=(1+x)/(2-x)$.

Use $h=x$. The numerator coefficients through degree 3 are $u_0=1$, $u_1=1$, $u_2=0$, and $u_3=0$. The denominator coefficients are $v_0=2$, $v_1=-1$, $v_2=0$, and $v_3=0$. Since $v_0=2\ne0$, coefficient matching applies.

For the constant coefficient, $2q_0=1$, so $q_0=1/2$.

For the coefficient of $x$, $2q_1-q_0=1$. Hence $2q_1=1+1/2=3/2$, so $q_1=3/4$.

For the coefficient of $x^2$, $2q_2-q_1=0$. Hence $2q_2=3/4$, so $q_2=3/8$.

For the coefficient of $x^3$, $2q_3-q_2=0$. Hence $2q_3=3/8$, so $q_3=3/16$. Thus

T_3f(x;0)=\frac12+\frac34x+\frac38x^2+\frac{3}{16}x^3.

Check by multiplying by the denominator:

(2-x)\left(\frac12+\frac34x+\frac38x^2+\frac{3}{16}x^3\right) =1+x-\frac{3}{16}x^4.

The product agrees with the numerator $1+x$ in every degree from 0 through 3. In fact, subtracting the polynomial from the quotient gives $3x^4/[16(2-x)]$, which has no terms of degree at most 3 near $x=0$. This verifies the result.

Worked Example: A Shifted Center

Find the degree-2 Taylor polynomial of $f(x)=x^2/(1+x)$ centered at $x=1$.

Set $h=x-1$, so $x=1+h$. Then the numerator and denominator become

x^2=(1+h)^2=1+2h+h^2,\qquad 1+x=2+h.

Thus $u_0=1$, $u_1=2$, $u_2=1$, while $v_0=2$, $v_1=1$, $v_2=0$. The denominator at the center is 2, so it is nonzero.

The constant equation is $2q_0=1$, giving $q_0=1/2$.

The linear equation is $2q_1+q_0=2$. Therefore $2q_1=2-1/2=3/2$, and $q_1=3/4$.

The quadratic equation is $2q_2+q_1=1$. Therefore $2q_2=1-3/4=1/4$, and $q_2=1/8$. The answer, written in powers of $x-1$, is

T_2f(x;1)=\frac12+\frac34(x-1)+\frac18(x-1)^2.

To check, multiply by the denominator $2+h$:

(2+h)\left(\frac12+\frac34h+\frac18h^2\right) =1+2h+h^2+\frac18h^3.

The numerator is $1+2h+h^2$. The product agrees through degree 2, and the remaining difference is a degree-3 term. Equivalently, the quotient minus the proposed polynomial is $-h^3/[8(2+h)]$. This confirms the degree-2 Taylor polynomial.

Worked Example: A Trigonometric Numerator

Find the degree-3 Taylor polynomial at 0 for $f(x)=\sin x/(1+x)$.

The numerator’s degree-3 Taylor polynomial is $x-x^3/6$, so its coefficients are $u_0=0$, $u_1=1$, $u_2=0$, and $u_3=-1/6$. The denominator has coefficients $v_0=1$, $v_1=1$, $v_2=0$, and $v_3=0$. Since the denominator equals 1 at the center, the quotient is defined near 0.

The constant equation gives $q_0=0$. The linear equation $q_1+q_0=1$ gives $q_1=1$. The quadratic equation $q_2+q_1=0$ gives $q_2=-1$. Finally, the cubic equation $q_3+q_2=-1/6$ gives $q_3-1=-1/6$, so $q_3=5/6$. Therefore

T_3f(x;0)=x-x^2+\frac56x^3.

Check the product:

(1+x)\left(x-x^2+\frac56x^3\right) =x-\frac16x^3+\frac56x^4.

The numerator’s Taylor polynomial is $x-x^3/6$, so the coefficients agree through degree 3. The extra term has degree 4, confirming the result to the requested degree.

A Common Mistake: Ignoring Cross-Terms

A tempting but incorrect shortcut is to divide each numerator coefficient by the denominator’s constant coefficient and use those as the quotient coefficients. For the first example, that shortcut would give $q_1=u_1/v_0=1/2$. But the coefficient of $x$ in the product is not just $2q_1$: it is $2q_1-q_0$, because the denominator contains the term $-x$.

2\left(\frac12\right)-\frac12=\frac12\ne 1.

So the proposed coefficient does not even reproduce the numerator’s linear term. The correct equation includes the cross-term:

2q_1-q_0=1,\qquad 2q_1-\frac12=1,\qquad q_1=\frac34.

Always form the full coefficient equation before solving. Each coefficient of the product collects several pairs of terms; omitting one changes the answer.

A Reliable Calculation Routine

For a quotient $u/v$, the following sequence keeps the calculation organized:

Choose the center $a$, and use $h=x-a$.
Write the numerator and denominator Taylor coefficients through the requested degree.
Check that $v_0=v(a)\ne0$.
Solve the product equations in order, starting with $q_0$.
Write the answer in powers of $x-a$, then multiply by the denominator to check the matching coefficients.

Multiplication is an especially useful final check: it tests the coefficient equations directly without requiring repeated differentiation of the quotient. This method is also a practical application of the Taylor-polynomial product rule: it reverses the product calculation to recover the polynomial for a quotient.

Check Your Understanding

Use coefficient matching to answer these questions. In each case, state the denominator’s value at the center.

Find the degree-2 Taylor polynomial of $(1+x)/(3+x)$ at $x=0$.
For a quotient $u/v$, write the coefficient equation that determines $q_2$ in terms of $u_2,v_0,v_1,v_2,q_0,$ and $q_1$.
Explain why the coefficient method cannot be applied by dividing by $v_0$ when $v(a)=0$.
For $\sin x/(1+x)$, verify by multiplication that the degree-3 polynomial found above matches through degree 3.
Identify the cross-term that must be included when computing $q_1$, and explain what goes wrong if it is omitted.

Taylor Polynomials of Quotients

What You'll Learn

Why Coefficient Matching Works for Quotients

Definitions and the Quotient Condition

The Coefficient Equations

Why the Method Gives the Taylor Polynomial

Worked Example: A Quotient at the Origin

Worked Example: A Quotient at the Origin

Worked Example: A Shifted Center

Worked Example: A Shifted Center

Worked Example: A Trigonometric Numerator

Worked Example: A Trigonometric Numerator

A Common Mistake: Ignoring Cross-Terms

A Reliable Calculation Routine

Check Your Understanding

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