5. Taylor Polynomials, Series, and Error Control

Learning goals and preparation

We assume derivatives of all elementary functions, definite integrals, integration by parts, and limits of sequences. This chapter extends tangent-line approximation to higher-degree polynomials. You will construct Taylor polynomials, derive and bound their remainder, distinguish a finite approximation from a convergent infinite representation, and use expansions only within their justified domains.

Writing several terms followed by dots does not prove that an infinite series equals a function. A reliable approximation includes a remainder statement; a reliable series identity additionally proves that the remainder tends to zero.

5.1 Sequences, series, and the geometric model

A sequence $(s_n)$ converges to $S$ when for every $\varepsilon>0$, all sufficiently late terms satisfy $|s_n-S|<\varepsilon$. An infinite series $\sum_{k=0}^\infty a_k$ converges to $S$ when its partial sums $s_n=\sum_{k=0}^n a_k$ converge to $S$.

The finite geometric identity is

$$1+x+\cdots+x^n=\frac{1-x^{n+1}}{1-x}\quad(x\ne1).$$

It follows by subtracting $x$ times the sum from the sum. Therefore

$$\frac1{1-x}=\sum_{k=0}^n x^k+\frac{x^{n+1}}{1-x}.$$

If $|x|<1$, the remainder tends to zero and $\sum_{k=0}^\infty x^k=1/(1-x)$. At $x=1$ the partial sums grow; at $x=-1$ they oscillate. Endpoints cannot be included merely because they lie near the convergence interval.

Worked example 1: a certified geometric approximation. At $x=0.1$, using terms through $x^3$ gives $1.111$. The exact error is $0.1^4/(1-0.1)=1/9000$, about $0.0001111$. The finite identity proves the error, rather than just observing agreement on a calculator.

5.2 Taylor coefficients are forced by derivative matching

A degree-at-most-$n$ polynomial around $a$ is $P(x)=c_0+c_1(x-a)+\cdots+c_n(x-a)^n$. Evaluating its successive derivatives at $a$ gives $P^{(k)}(a)=k!c_k$. Thus the unique such polynomial matching $f$ and its first $n$ derivatives at $a$ is

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

This is the Taylor polynomial of order $n$. When $a=0$ it is called a Maclaurin polynomial. For $n=1$ it is the tangent-line approximation; higher orders match additional derivatives, not merely additional sample values.

Worked example 2: expand about a nonzero point. For $f(x)=\ln x$ about $a=1$, the derivatives at one are $f(1)=0$, $f'(1)=1$, $f''(1)=-1$, and $f'''(1)=2$. Hence

$$T_3(x)=(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}.$$

The centre is part of the approximation. Replacing $(x-1)$ by $x$ would give a different polynomial and would fail even at the matching point.

5.3 Deriving a remainder formula

The order-$n$ Taylor polynomial at $a$ is $T_n(x)=\sum_{k=0}^nf^{(k)}(a)(x-a)^k/k!$, whose degree may be smaller than $n$ if leading derivatives vanish.

Theorem: Taylor's integral remainder

Let $n\ge0$ and let $f$ have continuous derivatives through order $n+1$ on the closed interval between $a$ and $x$. Then

$$f(x)=T_n(x)+\frac1{n!}\int_a^x f^{(n+1)}(t)(x-t)^n\,dt.$$

Consequently, if $|f^{(n+1)}(t)|\le M$ there, then $|f(x)-T_n(x)|\le M|x-a|^{n+1}/(n+1)!$.

Proof. At $n=0$, the identity is the fundamental theorem $f(x)-f(a)=\int_a^xf'(t)dt$. For $n\ge1$, integrate the remainder by parts, taking $u=(x-t)^n/n!$ and $dv=f^{(n+1)}(t)dt$. The boundary at $x$ is zero and at $a$ contributes $-f^{(n)}(a)(x-a)^n/n!$. The remaining integral is the order-$(n-1)$ remainder. Induction therefore yields the identity. This uses oriented integrals, so it also holds when $x<a$. Taking absolute values, reversing integration limits if needed, and integrating $|x-t|^n$ gives the stated bound. Continuity of the highest derivative makes every use of parts and the bound valid.

This theorem provides a sufficient regularity setting, not the weakest possible hypotheses. For a Taylor series to represent $f$ at a fixed $x$, the remainders must tend to zero as $n\to\infty$. A finite-order error bound alone does not prove that without control of its dependence on $n$.

With $h=x-a$, the substitution $t=a+sh$ gives

$$R_n(x)=\frac{h^{n+1}}{n!}\int_0^1(1-s)^nf^{(n+1)}(a+sh)ds.$$

This form controls either sign of $h$. If the derivative magnitude is at most $M$, then $|R_n|\le M|h|^{n+1}/(n+1)!$. Let $n$ be a nonnegative integer, let $f^{(n+1)}$ be continuous on the closed segment between $a$ and $x$, and suppose the integral remainder above holds. If $h\ne0$, put $w(s)=f^{(n+1)}(a+sh)$. The weight $(1-s)^n$ is positive on $[0,1)$ and has integral $1/(n+1)$. When $n=0$, it is the constant polynomial $1$ on all of $[0,1]$. If $w$ is nonconstant, continuity makes $w-\min w$ and $\max w-w$ strictly positive on some interior subintervals. The weight is positive there, so the corresponding integrals are both strictly positive. Thus the weighted average lies strictly between the minimum and maximum; the intermediate value theorem gives some $s_*\in(0,1)$ attaining it. If $w$ is constant, choose any interior $s_*$. Therefore

$$R_n(x)=\frac{f^{(n+1)}(a+s_*h)}{(n+1)!}h^{n+1}.$$

This also holds for $h<0$, with the sign retained in $h^{n+1}$. If $h=0$, directly $R_n(a)=0$ and no intermediate point is needed.

The unknown $\xi$ is not an input we may choose at will; for computation, a bound on the whole segment is more useful.

Worked example 3: approximate an exponential with a guaranteed error. For $e^{0.2}$, use $T_3=1+0.2+0.2^2/2+0.2^3/6=1.221333\ldots$. On $[0,0.2]$, every derivative equals $e^t<2$, so

$$|R_3|\le\frac{2(0.2)^4}{4!}=\frac1{7500}<0.000134.$$

The derivative bound need not be sharp to be useful. Merely identifying the next coefficient without a remainder theorem would not by itself certify an arbitrary Taylor approximation.

Why a power series permits calculus inside its radius

Let $a\in\mathbb R$ and let $(c_n)_{n\ge0}$ be real coefficients. Set $S=\{r\ge0:\sum_{n=0}^{\infty}|c_n|r^n<\infty\}$, interpreting the constant term at $r=0$ as $|c_0|$, and set $R=\sup S\in[0,\infty]$. For $0\le r<s<R$, choose a finite $M$ with $|c_n|s^n\le M$ for all $n$. Such $s$ has absolute convergence: the definition of the supremum supplies a member of $S$ greater than $s$, and comparison transfers convergence to $s$. Thus on $|x-a|\le r$,

$$|c_n(x-a)^n|\le M(r/s)^n.$$

The tails are bounded by geometric tails independently of $x$. Real completeness gives the pointwise limits, and this same estimate proves uniform convergence on every smaller closed interval. If the original series converged at $h=x-a$ with $|h|>R$, its terms would be bounded, say $|c_nh^n|\le B$. Then every $0\le t<|h|$ would belong to $S$, by comparison with $B(t/|h|)^n$, contradicting $R<|h|$. This proves the radius assertion without deciding either endpoint. If $R=0$, there is no open interval of convergence to which the calculus assertion applies.

For $0<r<s<R$, put $q=r/s<1$. The derivative terms satisfy

$$|nc_n(x-a)^{n-1}|\le (M/s)nq^{n-1}\qquad(n\ge1,\ |x-a|\le r).$$

The numerical majorant is summable: its successive-term ratio tends to $q<1$, so its tail is bounded by a geometric tail. Hence the derivatives $P_N'$ of the polynomial partial sums converge uniformly on $[a-r,a+r]$ to a continuous function $g$. Continuity follows directly from the three-error estimate: approximate $g$ uniformly by one fixed continuous polynomial derivative, then use that polynomial's continuity. Since $P_N(a)=c_0$, the fundamental theorem gives

$$P_N(x)=c_0+\int_a^xP_N'(t)\,dt.$$

The integral error is at most $r\sup|P_N'-g|$, so passage to the limit yields $f(x)=c_0+\int_a^xg(t)dt$. The fundamental theorem now proves $f'=g$ on the interval interior. Every $|x-a|<R$ lies in such an interior. Similarly, uniform convergence of $P_N$ bounds the integration error by $r\sup|P_N-f|$, proving

$$f'(x)=\sum_{n=1}^{\infty}nc_n(x-a)^{n-1},\qquad \int_a^xf(t)dt=\sum_{n=0}^{\infty}\frac{c_n(x-a)^{n+1}}{n+1}.$$

Both resulting series have radius exactly $R$. The arguments above give convergence throughout $|x-a|<R$. Conversely, convergence of the derivative series at $h\ne0$ bounds $nc_nh^{n-1}$ and therefore bounds $c_nh^n$; geometric comparison then gives absolute convergence of the original series at every smaller radius. Convergence of the integrated series at $h\ne0$ bounds $c_nh^{n+1}/(n+1)$; for $0\le t<|h|$ this gives $|c_n|t^n\le B(n+1)(t/|h|)^n/|h|$, again summable by a geometric-ratio bound. Neither new series can therefore have a larger radius. This reasoning also covers radius zero and infinite radius. Endpoint convergence is still a separate question.

Theorem: uniform completeness

A sequence of real-valued functions is uniformly convergent if and only if it is uniformly Cauchy.

Proof. Uniform convergence gives the Cauchy estimate by two errors of size $\varepsilon/2$. Conversely, uniform Cauchy implies that for each fixed $x$, $(f_n(x))$ is a real Cauchy sequence, so define $f(x)=\lim_n f_n(x)$. Fix $\varepsilon>0$ and use uniform Cauchy with tolerance $\varepsilon/2$. For every $x$ and $n\ge N$, pass $m\to\infty$ in the estimate to get $|f_n(x)-f(x)|\le\varepsilon/2<\varepsilon$. The same $N$ works everywhere, proving uniform convergence. Completeness of the target $\mathbb R$ supplied the pointwise limits.

Theorem

Uniform limit preserves continuity

If each $f_n$ is continuous at $a\in A$ relative to $A$ and $f_n\to f$ uniformly on $A$, then $f$ is continuous at $a$.

Proof

The three-term estimate

Given $\varepsilon>0$, choose $N$ so $|f_N(x)-f(x)|<\varepsilon/3$ for all $x\in A$. By continuity of the fixed function $f_N$ at $a$, choose $\delta$ so $|f_N(x)-f_N(a)|<\varepsilon/3$ whenever $|x-a|<\delta$. Then

$$|f(x)-f(a)|\le|f(x)-f_N(x)|+|f_N(x)-f_N(a)|+|f_N(a)-f(a)|<\varepsilon.$$

Derivative interchange without $C^1$ hypotheses

Theorem

Let $a<b$. Suppose every $f_n$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f_n(x_0)$ converges at one fixed $x_0\in[a,b]$ and $f_n'$ converges uniformly on $(a,b)$ to $g$, then $f_n$ converges uniformly on $[a,b]$ to a continuous $f$, differentiable on $(a,b)$, and $f'=g$. Continuity of the individual derivatives is not required.

Proof of existence. Uniform convergence of derivatives makes them uniformly Cauchy. For $m,n$ sufficiently large, the mean value theorem applied to $f_n-f_m$ between $x_0$ and any $x$ gives

$$|f_n(x)-f_m(x)|\le|f_n(x_0)-f_m(x_0)|+(b-a)\sup_{(a,b)}|f_n'-f_m'|.$$

The anchor sequence is Cauchy, so the right side is uniformly small. Uniform completeness gives a uniform limit $f$, which is continuous as a uniform limit of continuous functions.

Proof of the derivative identity. Fix an interior point $x$ and $\varepsilon>0$. Choose $N$ with $|f_n'(t)-f_N'(t)|\le\varepsilon$ for all interior $t$ and all $n\ge N$, and also $|g(x)-f_N'(x)|\le\varepsilon$. For sufficiently small nonzero $h$, apply the mean value theorem to $f_n-f_N$ on the segment from $x$ to $x+h$:

$$\left|\frac{f_n(x+h)-f_n(x)}h-\frac{f_N(x+h)-f_N(x)}h\right|\le\varepsilon.$$

Pass $n\to\infty$ at these two fixed inputs. The same inequality holds with $f$ replacing $f_n$. Since the fixed $f_N$ is differentiable at $x$, its difference quotient is within $\varepsilon$ of $f_N'(x)$ for small enough $h$. The triangle inequality bounds the difference quotient of $f$ from $g(x)$ by $3\varepsilon$. Arbitrary $\varepsilon$ proves differentiability and equality. This argument does not integrate $g$, so it does not silently assume $g$ continuous or Riemann integrable.

Learning goals and allowed earlier results

Construct rather than assume an elementary function, prove its algebraic laws, and justify the logarithm as an inverse. We use the previously proved ratio test, uniform convergence of power and derivative series on compact intervals, the mean value theorem, and the intermediate value theorem.

Define

$$E(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!},\qquad x\in\mathbb R.$$

For fixed $x\ne0$, the ratio of successive absolute terms is $|x|/(n+1)\to0$, so the series converges absolutely; at $x=0$ it equals one directly. On any fixed interval $|x|\le M$, both its series and derivative series have summable factorial majorants. The uniform derivative theorem therefore applies to polynomial partial sums and gives $E'(x)=E(x)$ everywhere, with $E(0)=1$.

Theorem: positivity and exponential law

For every real $x,y$, $E(x)>0$, $E(-x)=1/E(x)$, and $E(x+y)=E(x)E(y)$.

Proof. Differentiate $H(x)=E(x)E(-x)$ to obtain $H'(x)=E(x)E(-x)-E(x)E(-x)=0$. The mean value theorem makes $H$ constant; at zero it equals one. Thus $E$ never vanishes. Since it is continuous and $E(0)>0$, an intermediate-value argument prevents it from becoming negative anywhere, proving positivity and the reciprocal identity. For fixed $y$, differentiate $K(x)=E(x+y)E(-x)$; the same cancellation gives zero derivative. At zero $K(0)=E(y)$, so $E(x+y)E(-x)=E(y)$. Multiplication by $E(x)$ yields the addition law. No rearrangement of a double infinite series was needed.

Since $E'=E>0$, it is strictly increasing. For $x\ge0$, all terms are nonnegative and $E(x)\ge1+x$, so $E(x)\to\infty$ at positive infinity. The reciprocal identity gives $E(x)\to0$ at negative infinity. Continuity and monotonicity therefore make $E:\mathbb R\to(0,\infty)$ a bijection.

Define $\exp=E$, $e=E(1)$, and $\ln:(0,\infty)\to\mathbb R$ as its inverse. The inverse is continuous: if $x_0=E(t_0)$, then for any $\varepsilon>0$, $E(t_0-\varepsilon)<x_0<E(t_0+\varepsilon)$; taking $x$ within the smaller output gap confines $\ln x$ between those inputs.

Theorem: logarithmic derivative and laws

For $x>0$, $\ln'(x)=1/x$. For positive $x,y$ and real $r$, $\ln(xy)=\ln x+\ln y$ and $\ln(x^r)=r\ln x$, where $x^r$ is defined as $E(r\ln x)$.

Proof. Write $x_0=E(t_0)$, $x=E(t)$ and use inverse continuity to ensure $t\to t_0$ as $x\to x_0$. Then

$$\frac{\ln x-\ln x_0}{x-x_0}=\frac{t-t_0}{E(t)-E(t_0)}\longrightarrow\frac1{E'(t_0)}=\frac1{x_0}.$$

The exponential law gives $E(\ln x+\ln y)=xy$, and uniqueness of inverse values proves the product law. The real-power definition and inverse identity give the other formula. Differentiating that definition yields $(x^r)'=rx^{r-1}$ for $x>0$. This derivation did not first assume the real-power derivative to define the logarithm.

Worked examples and quantitative estimates

Approximating the constructed constant. The partial sum through $n=4$ gives $1+1+1/2+1/6+1/24=65/24$. For the tail starting at $n=5$, write $n=5+k$. Since $(5+k)!\ge5!6^k$, the tail is at most $(1/120)\sum_{k\ge0}6^{-k}=1/100$. Thus $65/24<e\le65/24+1/100$. This is a legitimate simple bound, though sharper factorial estimates are available.

An integral definition emerges afterwards. Since $(\ln x)'=1/x$ and $\ln1=0$, the fundamental theorem gives $\ln x=\int_1^xdt/t$ for every $x>0$. One could instead begin with that integral and construct its inverse, but using both definitions without proving their equivalence would be circular. Here the equality follows from already established derivatives.

5.4 Common series and their justified domains

For every real $x$,

$$e^x=\sum_{k=0}^\infty\frac{x^k}{k!},\qquad \sin x=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!},\qquad \cos x=\sum_{k=0}^\infty(-1)^k\frac{x^{2k}}{(2k)!}.$$

The coefficient patterns come from repeated derivatives. To prove equality, fix a finite $x$. For the exponential, derivatives on the segment are bounded by $e^{|x|}$, and $e^{|x|}|x|^{n+1}/(n+1)!\to0$. For sine and cosine all derivative magnitudes are at most one, giving the same factorial control. The factorial dominates a fixed power because successive ratios eventually have magnitude below, for example, $1/2$.

The logarithmic series is

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\qquad(-1<x\le1).$$

One derivation integrates the finite geometric identity for $1/(1+t)$ from zero to $x$, retaining an integral remainder. For $|x|<1$ that remainder tends to zero by a geometric bound. At $x=1$, its magnitude is at most $1/(n+1)$ after a suitable truncation, since $1/(1+t)\le1$ on $[0,1]$. At $x=-1$, the logarithm is undefined and the negative harmonic terms do not give a finite sum. Endpoints are checked separately rather than inherited from the interior argument.

A power series $\sum c_k(x-a)^k$ has an interval of absolute convergence centered at $a$, with endpoints requiring separate analysis. For the series of $1/(1-2x)$, the geometric condition is $|2x|<1$, or $|x|<1/2$, not $|x|<1$. Substituting an expression into a known series also substitutes into its convergence condition.

3. Dyadic condensation: the monotonicity is doing work

Theorem 2. If $a_n\ge0$ and $a_n$ is nonincreasing, then

$$\sum_{n=1}^{\infty}a_n\quad\hbox{converges if and only if}\quad\sum_{k=0}^{\infty}2^k a_{2^k}\quad\hbox{converges}.$$

Proof. The block $B_k=\sum_{n=2^k}^{2^{k+1}-1}a_n$ contains exactly $2^k$ terms. Every term is at most $a_{2^k}$ and at least $a_{2^{k+1}}$, so

$$\frac12\bigl(2^{k+1}a_{2^{k+1}}\bigr)\le B_k\le2^k a_{2^k}.$$

If the condensed series converges, the upper bound makes the partial sums of the block series bounded. If the block series converges, the lower bound makes the condensed series from $k=1$ onward bounded. Adding its one missing term does not change convergence. Finally, the block endpoints are a subsequence of the original nonnegative partial sums. Bounded block-endpoint sums bound every intermediate partial sum, because it lies between neighboring endpoints. Hence the block series converges precisely when the original series does. $\square$

The full p-series threshold by dyadic grouping

For real $p\le0$, $n^{-p}\ge1$, so the terms do not tend to zero and the series diverges. For $p>0$, $a_n=n^{-p}$ is nonnegative and nonincreasing. The proved dyadic test applies, and its condensed terms are $2^k a_{2^k}=2^{k(1-p)}$. Their geometric series converges exactly when $2^{1-p}<1$, equivalently $p>1$. When $0<p\le1$, the condensed terms are at least one and their series diverges, so the original series diverges too. This proves the threshold for every real $p$ without invoking an unproved improper-integral test.

Finite logarithm remainders including the endpoint

For every integer $m\ge1$ and $t\ne-1$, finite division gives

$$\frac1{1+t}=\sum_{k=0}^{m-1}(-t)^k+\frac{(-t)^m}{1+t}.$$

For fixed $-1<x\le1$, integrate over the oriented segment from zero to $x$. The fundamental theorem and $\ln1=0$ give

$$\ln(1+x)=\sum_{k=1}^m\frac{(-1)^{k-1}x^k}{k}+\int_0^x\frac{(-t)^m}{1+t}\,dt.$$

If $0\le x\le1$, the absolute remainder is at most $x^{m+1}/(m+1)\le1/(m+1)$. This includes $x=1$, proving convergence to $\ln2$ there. If $-1<x<0$, write $r=|x|<1$. The denominator is at least $1-r$ on the integration segment, so the absolute remainder is at most $r^{m+1}/[(m+1)(1-r)]$, which tends to zero. At $x=-1$, each series term is $-1/k$, giving divergence by the harmonic-series proof, and the real logarithm is undefined there. For $|x|>1$, choose $1<q<|x|$. The ratio of successive term magnitudes $|x|^k/k$ is $|x|k/(k+1)>q$ eventually. Iteration bounds those magnitudes below by one fixed positive magnitude times growing powers of $q$, so the terms do not tend to zero and the series diverges. Thus its real convergence set is exactly $-1<x\le1$, where its sum equals $\ln(1+x)$.

5.5 Smoothness does not guarantee equality with a Taylor series

Define $f(0)=0$ and $f(x)=e^{-1/x^2}$ for $x\ne0$. Every derivative at zero is zero. To understand the mechanism, differentiating at nonzero inputs gives a polynomial in $1/x$ times $e^{-1/x^2}$, and every term tends to zero faster than any needed reciprocal power grows. Indeed, with $t=1/x^2$, the elementary bound $e^t\ge t^N/N!$ for positive $t$ shows $t^me^{-t}\to0$ by choosing an integer $N>m$. Induction through the derivative definitions then gives all zero derivatives at zero.

Its Taylor series at zero is consequently the zero series, but $f(x)>0$ at every nonzero input. This example separates “has derivatives of every order” from “equals the sum of its Taylor series.” Remainder convergence, not visual smoothness or many successful derivatives, is the deciding condition.

Full flat-function differentiation at zero

Use the previously proved real exponential series, positivity, reciprocal law and derivative identity $\exp'=\exp$. Define $F(0)=0$ and $F(x)=\exp(-1/x^2)$ for $x\ne0$. For every real $r\ge0$, set $u=1/x^2$. Choose an integer $k>r/2$. The nonnegative exponential series gives $\exp u\ge u^k/k!$ for $u>0$, hence

$$|x|^{-r}\exp(-1/x^2)=u^{r/2}\exp(-u)\le k!u^{r/2-k}\longrightarrow0.$$

At nonzero inputs, induction gives $F^{(n)}(x)=P_n(1/x)\exp(-1/x^2)$ for a polynomial $P_n$, starting with $P_0=1$ and using $P_{n+1}(z)=-z^2P_n'(z)+2z^3P_n(z)$. Suppose the order-$n$ derivative has this formula off zero and value zero at zero. Every polynomial has finite degree, so the estimate above first shows $F^{(n)}(x)\to0$ and then $F^{(n)}(h)/h\to0$. Thus $F^{(n)}$ is continuous at zero, its derivative at zero exists and equals zero, and the recurrence gives the next off-zero formula. Induction proves $F$ is smooth with every derivative zero at zero. Its Taylor series there is identically zero, whereas $F(x)>0$ for each $x\ne0$. This completes the claimed counterexample without treating smoothness as analyticity.

5.6 Using expansions in calculations

Worked example 4: a limit with cancellation. Since

$$\sin x=x-\frac{x^3}{6}+O(x^5),$$

we obtain $\lim_{x\to0}(\sin x-x)/x^3=-1/6$. The remainder after division is $O(x^2)$ and tends to zero. Simply writing $\sin x\approx x$ would lose the first nonzero term and give no usable information for this quotient.

Worked example 5: a controlled integral approximation. For $0\le x\le0.2$, expand

$$e^{-x^2}=1-x^2+\frac{x^4}{2}+R(x).$$

Taylor's theorem for $e^{-u}$ at $u=0$ gives $|R(x)|\le x^6/6$ because $0\le u=x^2\le0.04$ and the relevant derivative magnitude is at most one. Integrating yields

$$\int_0^{0.2}e^{-x^2}dx\approx0.2-\frac{0.2^3}{3}+\frac{0.2^5}{10},$$

with error at most $0.2^7/42$. This calculation does not require a nonelementary antiderivative or unjustified interchange of an arbitrary infinite sum and an integral: it integrates a finite approximation plus a bounded remainder.

Within the interior of a power series' convergence interval, termwise differentiation and integration are valid and preserve the radius of convergence. Endpoints must be rechecked afterward. For example, differentiating the geometric series gives $1/(1-x)^2=\sum_{k=1}^\infty kx^{k-1}$ for $|x|<1$; it gives no convergence guarantee at $x=1$.

5.7 Graded exercises

Foundation

1. Find the fourth-order Maclaurin polynomial of $e^x$.
2. Find the second-order Taylor polynomial of $f(x)=1/x$ about $a=2$.
3. Write the geometric series for $1/(1+3x)$ and its open convergence interval.

Core

4. Approximate $\cos(0.1)$ by $1-x^2/2$ and bound the error using the fourth derivative and the zero cubic coefficient.
5. Evaluate $\lim_{x\to0}(e^x-1-x)/x^2$ with a justified remainder order.
6. Find the third-order Taylor polynomial of $\ln(1+x)$ at zero and bound its error for $0\le x\le1/4$ using the fourth derivative.
7. For $|x|<1$, derive a series for $x/(1-x)^2$ from the differentiated geometric series.
8. How many terms through degree $n$ suffice to approximate $1/(1-0.2)$ with error at most $10^{-4}$ using the exact geometric remainder? Find the smallest such $n$.

Synthesis

9. Approximate $\int_0^{0.1}\cos(x^2)dx$ using $1-x^4/2$, and give a rigorous remainder bound.
10. Explain why a list of Taylor coefficients is not a proof that the infinite series represents the function. Use the smooth flat example to identify exactly which required conclusion fails.

5.8 Full solutions

1. Every derivative at zero is one, so $T_4=1+x+x^2/2+x^3/6+x^4/24$.
2. $f(2)=1/2$, $f'(2)=-1/4$, and $f''(2)=1/4$. Thus $T_2=1/2-(x-2)/4+(x-2)^2/8$.
3. Set $r=-3x$ in the geometric identity: $1/(1+3x)=\sum_{k=0}^\infty(-3x)^k$ for $|x|<1/3$.
4. The degree-three Taylor polynomial equals $1-x^2/2$ because the cubic coefficient is zero. The fourth derivative is cosine, bounded by one, so the error at $0.1$ is at most $0.1^4/24=1/240000$.
5. Taylor's theorem gives $e^x=1+x+x^2/2+O(x^3)$ near zero. After subtracting and dividing, the expression is $1/2+O(x)$, so the limit is $1/2$.
6. $T_3=x-x^2/2+x^3/3$. The fourth derivative is $-6/(1+x)^4$, with magnitude at most $6$ on $[0,1/4]$. Thus $|R_3(x)|\le6x^4/24=x^4/4$, at most $1/1024$ at the endpoint.
7. Differentiate $1/(1-x)=\sum_{k=0}^\infty x^k$ inside $|x|<1$, then multiply by $x$. The result is $x/(1-x)^2=\sum_{k=1}^\infty kx^k$, with the same open interval.
8. The error is $0.2^{n+1}/0.8$. For $n=4$ it is $0.0004>0.0001$; for $n=5$ it is $0.00008\le0.0001$. Since the error decreases with $n$, the smallest is $n=5$, meaning six terms from degree zero through degree five.
9. Taylor's theorem for cosine with argument $u=x^2$ gives $\cos u=1-u^2/2+R$, with $|R|\le u^4/24=x^8/24$. Thus the approximation is $0.1-0.1^5/10$, and the integral error is at most $0.1^9/216$.
10. Coefficients determine the polynomials that match derivatives at the centre, but equality with the function requires $R_n(x)=f(x)-T_n(x)\to0$ at the input in question. For the flat example every Taylor polynomial is zero, yet at nonzero $x$ the remainder is the fixed positive value $e^{-1/x^2}$, independent of $n$. It does not tend to zero. Infinite differentiability alone therefore does not establish series representation.

Summary and onward map

Taylor coefficients follow from derivative matching. A finite remainder formula gives usable error bounds. A series representation additionally needs the remainder to vanish with increasing degree. Geometric and factorial remainders establish the common elementary expansions on their stated domains. Endpoint checks, branch restrictions and the first surviving term matter in limits, differentiation and integration.