Peano Remainder

The Error as a Smaller-Order Term

The Lagrange form describes the Taylor remainder using a derivative at an intermediate point. There is another useful description when the target $x$ is close to the center $a$: the remainder may be negligible compared with the last power retained in the Taylor polynomial. This is the Peano form of the Taylor expansion. It describes how the error behaves as $x$ approaches $a$, without requiring the error to equal a particular derivative value at an intermediate point.

Write $h=x-a$, and let $T_{n,f,a}$ be the degree-$n$ Taylor polynomial of $f$ at $a$. Its signed remainder is $R_{n,f,a}(x)=f(x)-T_{n,f,a}(x)$. For $n\geq 1$, the key comparison is between this remainder and $h^n$. Because $h$ can be positive or negative, it is the absolute value of the ratio that determines whether the remainder is negligible.

Definition (Peano Remainder): Let $n\geq 1$, and suppose the Taylor polynomial $T_{n,f,a}$ is defined. The remainder has Peano form of order $n$ at $a$ if $$ R_{n,f,a}(x)=o\bigl((x-a)^n\bigr)\quad\text{as }x\to a. $$ This means that $$ \lim_{\substack{x\to a\\x\ne a}} \frac{R_{n,f,a}(x)}{(x-a)^n}=0. $$ Equivalently, for every $\varepsilon>0$, there is a $\delta>0$ such that $0<|x-a|<\delta$ implies $|R_{n,f,a}(x)|\leq\varepsilon|x-a|^n$.

The inequality in the definition is local: for each chosen $\varepsilon$, it holds sufficiently close to $a$. It is not, in general, a fixed bound of the form $C|x-a|^{n+1}$. Nor does it say the remainder is zero. It says that after dividing by the last retained power, the error tends to zero.

Taylor’s Theorem in Peano Form

The Peano form needs fewer derivatives than the Lagrange form of Taylor’s Theorem. For a positive integer $n$, it is enough that derivatives through order $n$ exist in a neighborhood of the center. In particular, this conclusion does not require $f^{(n+1)}$. For $n=1$, differentiability at $a$ itself gives the result directly from the definition of the derivative.

Theorem (Taylor’s Theorem, Peano Form): Let $I$ be an open interval, let $a\in I$, and let $n\geq1$. Suppose $f:I\to\mathbb{R}$ has derivatives through order $n$ on $I$. Then $$ f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k +o\bigl((x-a)^n\bigr) \quad\text{as }x\to a. $$ In other words, $R_{n,f,a}(x)=o((x-a)^n)$. For $n=0$, the corresponding assertion $f(x)=f(a)+o(1)$ holds whenever $f$ is continuous at $a$.

Proof. We prove the assertion for positive $n$ by induction. When $n=1$, differentiability at $a$ means $$ \lim_{\substack{x\to a\\x\ne a}} \frac{f(x)-f(a)}{x-a}=f'(a). $$ Subtracting $f'(a)$ from the quotient gives $$ \lim_{\substack{x\to a\\x\ne a}} \frac{f(x)-f(a)-f'(a)(x-a)}{x-a}=0. $$ This is precisely the claimed Peano remainder for the degree-one Taylor polynomial.

Now let $n\geq2$, and assume the theorem holds for order $n-1$. The function $f'$ has derivatives through order $n-1$ on $I$, so applying the induction hypothesis to $f'$ at $a$ gives $$ f'(t)=\sum_{k=0}^{n-1}\frac{f^{(k+1)}(a)}{k!}(t-a)^k +o\bigl((t-a)^{n-1}\bigr) \quad\text{as }t\to a. $$ Let $P(t)=T_{n,f,a}(t)$, and set $g(t)=f(t)-P(t)$. Since $P'(t)$ is the displayed polynomial part of the expansion for $f'(t)$, it follows that $$ g'(t)=o\bigl((t-a)^{n-1}\bigr)\quad\text{as }t\to a. $$ Also $g(a)=0$. For any $x\ne a$ sufficiently close to $a$, the Mean Value Theorem applies to $g$ on the interval with endpoints $a$ and $x$: $g$ is continuous there and differentiable in its interior. It gives a point $c$ strictly between $a$ and $x$ such that $$ g(x)=g'(c)(x-a). $$ Consequently, $$ \left|\frac{g(x)}{(x-a)^n}\right| = \left|\frac{g'(c)}{(c-a)^{n-1}}\right| \left|\frac{c-a}{x-a}\right|^{n-1}. $$ As $x\to a$, the point $c$ lies between $a$ and $x$, so $c\to a$, and the first factor tends to zero by the estimate for $g'$. The second factor is at most $1$, since $|c-a|\leq|x-a|$. Thus the whole product tends to zero. This proves $g(x)=o((x-a)^n)$, as required. For $n=0$, continuity at $a$ gives $f(x)-f(a)\to0$, which is exactly $o(1)$. $\square$

The induction uses the Mean Value Theorem to pass from a small derivative error to a small function error. The intermediate point $c$ is used only in the proof; the Peano conclusion itself is a limit at $a$. This differs from the Lagrange form, which represents the remainder at a particular target by a derivative value between the center and that target.

Worked Examples

Worked Example: A Degree-One Expansion Without a Second Derivative

Define $f(x)=|x|^{3/2}$ for real $x$, and expand at $a=0$. We have $f(0)=0$. For $h\ne0$, $$ \frac{f(h)-f(0)}{h} = \frac{|h|^{3/2}}{h}. $$ The absolute value of this quotient is $|h|^{1/2}$, which tends to zero as $h\to0$. Therefore $f'(0)=0$, and the degree-one Taylor polynomial is $T_{1,f,0}(x)=0$. Its remainder satisfies $$ \frac{R_{1,f,0}(x)}{x} = \frac{|x|^{3/2}}{x} \longrightarrow 0. $$ Thus $R_{1,f,0}(x)=o(x)$. This example needs only differentiability at the center for the degree-one Peano conclusion; it does not require a second derivative at the center.

Worked Example: A Degree-Two Expansion for a Logarithm

Let $f(x)=\ln(1+x)$, with center $a=0$. This function has derivatives through order two on an open interval containing zero, and $$ f(0)=0,\qquad f'(0)=1,\qquad f''(0)=-1. $$ The degree-two Taylor polynomial and its remainder are $$ T_{2,f,0}(x)=x-\frac{x^2}{2}, \qquad R_{2,f,0}(x)=\ln(1+x)-x+\frac{x^2}{2}. $$ Taylor’s Theorem, Peano Form, gives $$ \lim_{\substack{x\to0\\x\ne0}} \frac{\ln(1+x)-x+x^2/2}{x^2}=0. $$ Hence the error is negligible compared with $x^2$. This is a statement about the limit of the normalized error; the theorem does not claim that the error is identically zero or that it is given by a third-derivative value.

Worked Example: A Peano Error Need Not Be of the Next Power

For $0<|x|<e^{-1}$, define $$ f(x)=x+\frac{x}{\ln(1/|x|)},\qquad f(0)=0. $$ The logarithm in the denominator is positive on this punctured interval. For $x\ne0$, $$ \frac{f(x)-f(0)}{x} = 1+\frac{1}{\ln(1/|x|)} \longrightarrow 1 \quad\text{as }x\to0. $$ Thus $f'(0)=1$, and its degree-one Taylor polynomial at zero is $T_{1,f,0}(x)=x$. The remainder is $R_{1,f,0}(x)=x/\ln(1/|x|)$, so $$ \frac{R_{1,f,0}(x)}{x} = \frac{1}{\ln(1/|x|)} \longrightarrow 0. $$ This is a Peano remainder of order one. But its size is not bounded by a constant times $x^2$ near zero: for $x>0$, $$ \frac{|R_{1,f,0}(x)|}{x^2} = \frac{1}{x\ln(1/x)} \longrightarrow\infty. $$ For the last limit, put $u=1/x$. Then $u\to\infty$ and the quotient becomes $u/\ln u$, which tends to infinity by the earlier result that logarithmic growth is slower than every positive power. So $o(x)$ does not imply $O(x^2)$.

Uniqueness of the Coefficients

A Peano expansion does more than provide one approximation: among polynomials of degree at most $n$, its coefficients are forced. This matters when expansions are found by different methods. If both satisfy the same error condition, their coefficients cannot disagree.

Theorem (Uniqueness of Peano Polynomial Coefficients): Fix $a\in\mathbb{R}$ and $n\geq0$. Suppose polynomials $P$ and $Q$, each of degree at most $n$, both satisfy $$ f(x)-P(x)=o\bigl((x-a)^n\bigr), \qquad f(x)-Q(x)=o\bigl((x-a)^n\bigr) $$ as $x\to a$, for $n\geq1$. Then $P=Q$. For $n=0$, the analogous uniqueness holds if the two differences tend to zero.

Proof. Subtracting the two error relations gives $P(x)-Q(x)=o((x-a)^n)$. Write $h=x-a$ and let $D(h)=P(a+h)-Q(a+h)$, a polynomial of degree at most $n$. If $D$ were not the zero polynomial, it would have a least index $j$, with $0\leq j\leq n$, whose coefficient $d_j$ is nonzero. Then $$ \frac{D(h)}{h^j}\longrightarrow d_j\ne0 \quad\text{as }h\to0. $$ But $D(h)=o(h^n)$, so $$ \frac{D(h)}{h^j} = \frac{D(h)}{h^n}h^{n-j}\longrightarrow0, $$ since $D(h)/h^n\to0$ and $h^{n-j}$ remains bounded and tends to zero if $j<n$ (and equals $1$ if $j=n$). This contradicts the nonzero limit $d_j$. Hence $D$ is the zero polynomial and $P=Q$. For $n=0$, two constant polynomials whose differences from $f(x)$ both tend to zero must have equal constants by subtracting the limits. $\square$

What the Peano Form Does—and Does Not—Say

The Peano form is useful when the question concerns local asymptotic accuracy: it specifies that the error is smaller than the last power included in the polynomial. Its hypotheses and conclusion should not be confused with the Lagrange form. For a degree-$n$ Peano expansion, derivatives through order $n$ suffice under the theorem’s hypotheses; a Lagrange representation for the degree-$n$ remainder uses the next derivative. In return, Peano form gives a limit rather than an intermediate-derivative formula.

A common pitfall is to treat “smaller than $|x-a|^n$” as though it meant “bounded by a constant times $|x-a|^{n+1}$.” The logarithmic example shows the distinction: its remainder divided by $x$ tends to zero, but its remainder divided by $x^2$ is unbounded. A more precise bound needs additional information. Conversely, if a later argument establishes a bound by a higher power, that stronger statement automatically implies the corresponding Peano estimate.

When using a Peano expansion, keep the order in the little-o term explicit. For example, an error $o((x-a)^n)$ can be divided by $(x-a)^n$ and sent to zero, but it cannot automatically be divided by a smaller power and assigned a particular limit, nor does it imply a fixed numerical error estimate. The ratio definition is the safest guide to exactly what has been proved.

Check Your Understanding

Use the definition, theorem, and examples to check your understanding of Peano remainders.

What limit expresses that $R_{n,f,a}(x)=o((x-a)^n)$?
What differentiability is required in Taylor’s Theorem, Peano Form, for a positive integer $n$?
Why does the Mean Value Theorem step in the induction proof give a factor bounded by $1$?
Does an error of $o((x-a)^n)$ necessarily have an $O((x-a)^{n+1})$ bound? Explain using the logarithmic example.
Why must two degree-at-most-$n$ polynomials giving Peano expansions of order $n$ have the same coefficients?

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