Uniqueness of the Taylor Polynomial

Why Uniqueness Matters

A list of derivative values at a point can be used to specify the coefficients of a polynomial of bounded degree. The key question here is not how to calculate those coefficients, but whether two different polynomials could satisfy the same conditions. The answer is no: once the center and degree bound are fixed, matching all derivatives through that degree forces the polynomials to be identical.

This claim is about uniqueness, not existence. It says that there can be at most one polynomial satisfying the conditions. It does not, by itself, show that a polynomial exists for every possible set of derivative values. In the Taylor polynomial setting, the usual formula supplies a polynomial when the function has the required derivatives; the uniqueness result explains why no second polynomial can satisfy the same conditions.

Derivative Data at a Center

Let \(a\) be a real number and \(n\) a nonnegative integer. A polynomial \(P\) of degree at most \(n\) is said to match a function \(f\)'s derivatives through order \(n\) at \(a\) when \(f\) has derivatives through order \(n\) at \(a\) and

P^{(k)}(a)=f^{(k)}(a) \qquad\text{for } k=0,1,\dots,n.

Here the zeroth derivative means the function itself, so \(P^{(0)}(a)=P(a)\). Thus the conditions include matching the function value, the first derivative, the second derivative, and so on through the \(n\)th derivative. The degree bound means that \(P\) has no powers higher than \(n\).

The Taylor polynomial of degree \(n\) for \(f\) at \(a\), when the derivatives in the formula exist, is

T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k.

The uniqueness theorem says that any polynomial of degree at most \(n\) matching those same derivative values must be this polynomial. The center \(a\) is fixed throughout: matching data at a different center is a different set of conditions.

Theorem: Let \(P\) and \(Q\) be polynomials of degree at most \(n\), and let \(a\) be a real number. If \(P^{(k)}(a)=Q^{(k)}(a)\) for every \(k=0,1,\dots,n\), then \(P=Q\) as polynomials.

Why Derivatives Reveal the Coefficients

A polynomial of degree at most \(n\) can be written in powers of \(x-a\):

P(x)=c_0+c_1(x-a)+c_2(x-a)^2+\cdots+c_n(x-a)^n.

This is a polynomial expansion centered at \(a\). The coefficient \(c_j\) is directly recoverable from the \(j\)th derivative at \(a\). To see why, differentiate \(j\) times. A term with exponent less than \(j\) vanishes after \(j\) differentiations. A term with exponent greater than \(j\) still contains a positive power of \(x-a\), so it vanishes when evaluated at \(a\). The term with exponent exactly \(j\) becomes the constant \(j!c_j\).

P^{(j)}(a)=j!c_j \qquad\text{and therefore}\qquad c_j=\frac{P^{(j)}(a)}{j!}.

This identity holds for each \(j=0,1,\dots,n\), including \(j=0\), when it simply says \(P(a)=c_0\). It explains the central logic of uniqueness: if two polynomials have the same derivatives at \(a\), their coefficients in powers of \(x-a\) agree one by one.

Proof of the Uniqueness Theorem

Suppose \(P\) and \(Q\) both have degree at most \(n\), and their derivatives agree at \(a\) through order \(n\). Consider their difference \(R=P-Q\). It is also a polynomial of degree at most \(n\). For every \(k=0,1,\dots,n\), the derivative of the difference at \(a\) is

R^{(k)}(a)=P^{(k)}(a)-Q^{(k)}(a)=0.

Write \(R\) in powers of \(x-a\):

R(x)=c_0+c_1(x-a)+\cdots+c_n(x-a)^n.

By the coefficient identity just established, \(c_j=R^{(j)}(a)/j!\) for each \(j=0,1,\dots,n\). Since every derivative \(R^{(j)}(a)\) is zero, every coefficient is zero. Hence

R(x)=0 \qquad\text{for every }x.

Because \(R=P-Q\), this gives \(P(x)=Q(x)\) for every \(x\). Thus \(P\) and \(Q\) are identical polynomials. This proves the theorem.

Why the proof works: The degree bound ensures that the polynomial has no coefficients beyond \(c_n\). The derivatives through order \(n\) recover every coefficient, leaving no freedom for a second polynomial.

Worked Examples

Worked Example: Two Quadratics at a Nonzero Center

Consider the polynomials

P(x)=2+3(x-1)-4(x-1)^2 \qquad\text{and}\qquad Q(x)=-4x^2+11x-5.

Both have degree at most \(2\). We check whether they have the same derivatives through order \(2\) at \(a=1\). For \(P\), the centered form gives

P(1)=2,\qquad P'(x)=3-8(x-1),\qquad P'(1)=3,\qquad P''(x)=-8,\qquad P''(1)=-8.

For \(Q\), differentiate its expanded form:

Q'(x)=-8x+11,\qquad Q''(x)=-8.

At the center, the values are

Q(1)=-4+11-5=2,\qquad Q'(1)=-8+11=3,\qquad Q''(1)=-8.

The value and both derivatives agree. The theorem therefore says the polynomials are identical. Direct expansion checks the conclusion:

\begin{aligned} P(x) &=2+3x-3-4(x^2-2x+1)\\ &=2+3x-3-4x^2+8x-4\\ &=-4x^2+11x-5\\ &=Q(x). \end{aligned}

Thus the derivative data at \(1\) determine at most one polynomial of degree at most \(2\), and these two expressions are simply different forms of the same polynomial.

Worked Example: Cubics at a Negative Center

Let

P(x)=1+2(x+1)+3(x+1)^2-(x+1)^3 \qquad\text{and}\qquad Q(x)=-x^3+5x+5.

Both have degree at most \(3\). At \(a=-1\), the centered expression for \(P\) gives

P(-1)=1,\qquad P'(-1)=2,\qquad P''(-1)=6,\qquad P'''(-1)=-6.

For \(Q\), compute the derivatives and evaluate them at \(-1\):

\begin{aligned} Q(-1)&=-(-1)^3+5(-1)+5=1-5+5=1,\\ Q'(x)&=-3x^2+5, &Q'(-1)&=-3+5=2,\\ Q''(x)&=-6x, &Q''(-1)&=6,\\ Q'''(x)&=-6, &Q'''(-1)&=-6. \end{aligned}

The four derivative values, from order zero through order three, agree. The theorem implies \(P=Q\). To verify the identity by expansion, use \((x+1)^2=x^2+2x+1\) and \((x+1)^3=x^3+3x^2+3x+1\):

\begin{aligned} P(x) &=1+2x+2+3x^2+6x+3-x^3-3x^2-3x-1\\ &=-x^3+5x+5\\ &=Q(x). \end{aligned}

This example also illustrates the role of the center: expressing a polynomial in powers of \(x+1\) makes the derivative values at \(-1\) apparent.

Worked Example: Why the Degree Bound Cannot Be Dropped

Take \(n=2\), center \(a=0\), and compare the zero polynomial \(P(x)=0\) with \(Q(x)=x^3\). The zero polynomial has degree at most \(2\), but \(Q\) has degree \(3\), so the theorem's degree condition does not hold for this pair. Check the derivatives through order \(2\):

\begin{aligned} P(0)&=0, &P'(x)&=0, &P'(0)&=0, &P''(x)&=0, &P''(0)&=0,\\ Q(0)&=0, &Q'(x)&=3x^2, &Q'(0)&=0, &Q''(x)&=6x, &Q''(0)&=0. \end{aligned}

These derivative values agree, but the polynomials are not identical: for example, \(P(1)=0\) and \(Q(1)=1\). The example does not contradict the theorem, because \(Q\) is not of degree at most \(2\). It shows that derivatives only through order \(n\) cannot, without the degree bound, rule out additional terms of higher degree.

Uniqueness Is Not the Same as Existence

Suppose a function \(f\) has derivatives through order \(n\) at \(a\). The formula for \(T_n\) gives a polynomial of degree at most \(n\), and differentiating its terms shows that \(T_n^{(j)}(a)=f^{(j)}(a)\) for \(j=0,1,\dots,n\). So in this setting a matching polynomial exists. The uniqueness theorem proves that any other polynomial of degree at most \(n\) satisfying the same conditions must equal \(T_n\).

Keep the two logical claims distinct. Existence means at least one polynomial satisfies the conditions. Uniqueness means no more than one does. The uniqueness proof itself begins by assuming two candidates exist and showing they coincide; it does not need to construct either candidate.

A common mistake is to infer uniqueness from the derivative conditions while forgetting the degree restriction. For example, matching derivatives through order \(2\) only constrains the constant, linear, and quadratic terms in a centered expansion. A cubic term such as \((x-a)^3\) has zero value and zero first and second derivatives at \(a\), so it could be added without changing those data. The degree bound excludes exactly this possibility.

Why the Result Matters

The Taylor polynomial is specified by local derivative information: a function value and a finite list of derivatives at one center. Uniqueness gives that specification a precise meaning. There is not a choice among multiple degree-at-most-\(n\) polynomials that fit the same data; the data determine a single polynomial. This is the foundation for treating a Taylor polynomial as the polynomial associated with those derivatives, rather than merely one possible fit.

The result is also a useful check on algebra. If two proposed degree-at-most-\(n\) expressions have the same derivatives through order \(n\) at the same center, they must simplify to the same polynomial. If they do not, then either a derivative calculation, an expansion, or an assumption has been mishandled. Conversely, if the degree bound is absent, matching derivatives alone does not justify that conclusion.

Check Your Understanding

Use the uniqueness theorem and its conditions to answer the following questions.

State the degree and derivative-matching conditions needed for two polynomials to be guaranteed identical.
In the proof, why is it useful to consider the difference \(R=P-Q\)?
For a polynomial written in powers of \(x-a\), what is the coefficient of \((x-a)^j\) in terms of its \(j\)th derivative at \(a\)?
Does uniqueness alone prove that a matching polynomial exists? Explain the distinction.
Why do \(0\) and \((x-a)^3\) have the same derivatives through order \(2\) at \(a\) without being identical?