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Power Series and Taylor Polynomials · Tutorial 875 of 999

Diagnosing Center, Factorial, and Sign Errors

Use the Taylor-polynomial formula as a checklist to find common errors and verify a corrected polynomial.

Advanced 10 min read

What You'll Learn

  • Write the Taylor polynomial of degree n centered at a in its standard form
  • Distinguish the center a from the displacement x minus a
  • Check that each derivative is evaluated at the center
  • Use the factorial denominator for the matching power
  • Determine signs from derivative values rather than assuming a pattern
  • Verify a polynomial by checking its derivatives at the center

Why Taylor-Polynomial Errors Are Easy to Miss

A Taylor polynomial can look convincing even when one coefficient is wrong. The powers may be in the right order, and the arithmetic may be tidy, while a misplaced center, omitted factorial, or incorrect sign changes the polynomial.

A dependable check begins with the formula: each coefficient comes from a derivative evaluated at the center, divided by the factorial matching that derivative's order. The power uses the displacement from the center. These parts work together, so checking them one at a time is more reliable than judging the finished expression by appearance.

The Formula as a Diagnostic Checklist

Definition: If \(f\) has derivatives through order \(n\) at \(x=a\), its Taylor polynomial of degree \(n\) centered at \(a\) is $$ P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k. $$ Here \(f^{(0)}(a)=f(a)\), and \(0!=1\).

Read each term as a matched set: the derivative order \(k\), the value \(f^{(k)}(a)\), the denominator \(k!\), and the displacement power \((x-a)^k\) must all have the same index \(k\). The center \(a\) is where derivatives are evaluated; the variable \(x\) appears in the displacement.

For example, the term of degree two is \(f''(a)(x-a)^2/2!\). Replacing \(f''(a)\) by \(f''(x)\) makes the coefficient depend on \(x\); replacing \((x-a)^2\) with \(x^2\) changes the center; omitting \(2!\) doubles that term.

1
Identify the center and degree.
Write down \(a\) and \(n\) before calculating coefficients.
2
Make a derivative table.
Find \(f(a), f'(a), \ldots, f^{(n)}(a)\), evaluating each at the same center.
3
Build terms in order.
For each \(k\), use \(f^{(k)}(a)/k!\) as the coefficient of \((x-a)^k\).
4
Check the result.
Differentiate the polynomial and verify that its derivatives through order \(n\), at \(a\), match those of \(f\).

Why the Coefficients Have This Form

The formula is not an arbitrary recipe. Its coefficients are exactly the ones that make the polynomial agree with the function's derivatives at the center. To see why, write a degree-\(n\) polynomial in powers of the displacement as \(P_n(x)=\sum_{k=0}^{n}c_k(x-a)^k\).

Consider its \(j\)th derivative evaluated at \(x=a\). A term with \(k<j\) disappears when differentiated \(j\) times. A term with \(k>j\) still contains a positive power of \(x-a\), so it is zero at \(a\). Only the term with \(k=j\) contributes, giving \(P_n^{(j)}(a)=j!c_j\). Therefore, requiring \(P_n^{(j)}(a)=f^{(j)}(a)\) forces \(c_j=f^{(j)}(a)/j!\). This applies to each \(j\) from zero through \(n\), which gives the stated formula.

Diagnostic principle: A correct degree-\(n\) Taylor polynomial centered at \(a\) matches \(f\) and its derivatives through order \(n\) at \(a\). This checks the coefficients; it does not, by itself, establish how accurate the polynomial is away from the center.

Worked Example: Finding a Missing Factorial

Worked Example: Finding a Missing Factorial

Find the degree-three Taylor polynomial for \(f(x)=\ln x\) centered at \(a=2\), and diagnose the error in using \(f'''(2)\) directly as the coefficient of \((x-2)^3\).

Differentiate and evaluate each derivative at \(2\):

$$ \begin{aligned} f(2)&=\ln 2,\\ f'(x)&=\frac{1}{x}, & f'(2)&=\frac{1}{2},\\ f''(x)&=-\frac{1}{x^2}, & f''(2)&=-\frac{1}{4},\\ f'''(x)&=\frac{2}{x^3}, & f'''(2)&=\frac{2}{8}=\frac{1}{4}. \end{aligned} $$

Apply the formula, keeping each derivative with its matching factorial and displacement power:

$$ \begin{aligned} P_3(x) &=f(2)+f'(2)(x-2)+\frac{f''(2)}{2!}(x-2)^2+\frac{f'''(2)}{3!}(x-2)^3\\ &=\ln 2+\frac{1}{2}(x-2)+\frac{-1/4}{2}(x-2)^2+\frac{1/4}{6}(x-2)^3\\ &=\ln 2+\frac{1}{2}(x-2)-\frac{1}{8}(x-2)^2+\frac{1}{24}(x-2)^3. \end{aligned} $$

The cubic coefficient is \(1/24\), not \(1/4\): the third derivative value must be divided by \(3!=6\). To check, differentiate the polynomial:

$$ \begin{aligned} P_3'(x)&=\frac{1}{2}-\frac{1}{4}(x-2)+\frac{1}{8}(x-2)^2,\\ P_3''(x)&=-\frac{1}{4}+\frac{1}{4}(x-2),\\ P_3'''(x)&=\frac{1}{4}. \end{aligned} $$

At \(x=2\), these give \(P_3(2)=\ln 2\), \(P_3'(2)=1/2\), \(P_3''(2)=-1/4\), and \(P_3'''(2)=1/4\), matching the function's values and derivatives there. If the cubic coefficient were incorrectly set to \(1/4\), its third derivative would be \(6(1/4)=3/2\), not \(f'''(2)=1/4\).

Worked Example: Let the Derivatives Determine the Signs

Worked Example: Let the Derivatives Determine the Signs

Find the degree-four Taylor polynomial for \(f(x)=\cos x\) centered at \(0\). Then check a common sign error: writing a positive \(x^2/2\) term.

Differentiate and evaluate at the center:

$$ \begin{aligned} f(x)&=\cos x, & f(0)&=1,\\ f'(x)&=-\sin x, & f'(0)&=0,\\ f''(x)&=-\cos x, & f''(0)&=-1,\\ f'''(x)&=\sin x, & f'''(0)&=0,\\ f''''(x)&=\cos x, & f''''(0)&=1. \end{aligned} $$

Substitute the derivative values, including their signs, into the formula:

$$ \begin{aligned} P_4(x) &=1+0x+\frac{-1}{2!}x^2+\frac{0}{3!}x^3+\frac{1}{4!}x^4\\ &=1-\frac{x^2}{2}+\frac{x^4}{24}. \end{aligned} $$

Check by differentiating and evaluating at zero:

$$ \begin{aligned} P_4'(x)&=-x+\frac{x^3}{6}, & P_4'(0)&=0,\\ P_4''(x)&=-1+\frac{x^2}{2}, & P_4''(0)&=-1,\\ P_4'''(x)&=x, & P_4'''(0)&=0,\\ P_4''''(x)&=1, & P_4''''(0)&=1. \end{aligned} $$

Also \(P_4(0)=1\), so all five values match those of \(f\). If the quadratic term were mistakenly \(+x^2/2\), the candidate's second derivative at zero would be \(+1\). But \(f''(0)=-1\), so the sign error is exposed immediately. Signs come from the evaluated derivatives; they are not assigned by a rule that every Taylor polynomial must alternate.

Worked Example: Keep the Center in Both Places

Worked Example: Keep the Center in Both Places

Find the degree-two Taylor polynomial for \(f(x)=e^x\) centered at \(a=1\). Compare it with two errors: evaluating derivatives at \(0\) instead of \(1\), and using powers of \(x\) instead of powers of \(x-1\).

Because the function and its first two derivatives are all \(e^x\), the required values at the center are \(f(1)=e\), \(f'(1)=e\), and \(f''(1)=e\). Thus

$$ \begin{aligned} P_2(x) &=e+e(x-1)+\frac{e}{2!}(x-1)^2\\ &=e\left(1+(x-1)+\frac{(x-1)^2}{2}\right). \end{aligned} $$

Check the value and derivatives at the center:

$$ \begin{aligned} P_2(1)&=e\left(1+0+\frac{0}{2}\right)=e,\\ P_2'(x)&=e\bigl(1+(x-1)\bigr), & P_2'(1)&=e,\\ P_2''(x)&=e, & P_2''(1)&=e. \end{aligned} $$

All three match \(f(1)\), \(f'(1)\), and \(f''(1)\). If the derivatives were evaluated at \(0\), the values would each be \(1\), giving the wrong candidate \(1+(x-1)+(x-1)^2/2\); its value at \(1\) is \(1\), not \(e\). If instead the correct coefficients were paired with powers of \(x\), the candidate would be \(e(1+x+x^2/2)\). At \(x=1\) it equals \(e(1+1+1/2)=5e/2\), not \(f(1)=e\). Both errors fail the simplest check: the polynomial must have the same value as the function at its center.

A Common Mistake: Mixing Up the Center and the Displacement

A frequent error is to treat \(a\) itself as the quantity that gets raised to a power, or to substitute the evaluation point into the displacement. The power is always \((x-a)^k\). At the center, \(x=a\), so every positive-degree displacement power is zero. Away from the center, the displacement can be positive or negative; its sign belongs to the expression \(x-a\).

For instance, when the center is \(2\) and the point of evaluation is \(1\), the displacement is \(1-2=-1\), not \(2\) and not \(1\). The quadratic displacement is \((-1)^2=1\), while the cubic displacement is \((-1)^3=-1\). Losing the minus sign in an odd power changes that term's sign.

Takeaway: When a proposed polynomial is in doubt, evaluate it and its derivatives at the center. A mismatch at any order is enough to locate an error in the corresponding coefficient, sign, factorial, derivative value, or displacement power.

Check Your Understanding

Use the Taylor-polynomial formula and the derivative-at-the-center check to answer these questions.

  1. For a polynomial centered at \(a\), what expression should appear in the power of its degree-\(k\) term?
  2. What factorial divides \(f^{(4)}(a)\) in a Taylor polynomial?
  3. If a proposed quadratic Taylor polynomial has second derivative \(3\) at the center, what value must \(f''(a)\) have for the polynomial to be correct?
  4. Why does a term with power greater than \(j\) contribute zero to the \(j\)th derivative evaluated at the center?
  5. Does a correct Taylor polynomial have alternating signs in every case? Explain what determines the signs.

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