6 concepts16 worked examples27 exercises3 exam-level6 figures
How often this comes up in exams: 12 questions, 8/39 of the years
all years (1988-2026)Final 6 · Midterm 1 5 · Midterm 2 1
last ten yearsFinal 1
also touched by2 questions that are filed under another section but still use this one
The questions in this section name themselves clearly in the archive.
Counted from the department's own exam archive: 106 exams, 599 question items. On an independent sample of 60 questions read by hand, the topic labels were 83% right, so treat these counts as a direction, not a certainty.
What are you here for?
05 Power series representations, Taylor and Maclaurin series, and their approximations
Start with this
One question before you read anything. Getting it wrong is the point: it shows you what this section is for.
§05.0 - geometric warmup
Before we bend the into new shapes, make sure the plain version is automatic.
Find(a) Does it converge, and if so to what number?
Giventhe series $\sum_{n=0}^{\infty}\left(\tfrac{1}{3}\right)^{n}$
Hint 1/4
Read off the first term $a$ and the ratio $r$, then check $|r|<1$.
Hint 2/4
A convergent geometric series sums to $\dfrac{a}{1-r}$.
Hint 3/4
Here $a=1$ and $r=\tfrac{1}{3}$, so the sum is $\dfrac{1}{1-\tfrac13}$.
Hint 4/4
The series converges to $\tfrac{3}{2}$.
Show solution
Apply the formula
$$a=1,\ r=\tfrac13,\ |r|<1$$
the ratio is below one, so the closed form applies
Type $e^{0.3}$ into a calculator and twenty digits come back. There is no exponential curve etched into the chip, only a processor that can add, subtract and multiply. So somewhere inside, the machine is turning $e^{0.3}$ into a short sum of fractions it actually can work out. What sum, and how does it know when to stop?
By the end you can write $e^{x}$, $\sin x$ and $\cos x$ as explicit infinite sums, reuse them to expand things like $e^{-x^{2}}$, and decide how many terms pin a value to a set number of decimals.
In 60 seconds
A power series is a function written as an infinite polynomial; this week is about finding that polynomial and controlling the error when you cut it short.
Geometric representation
$$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}$$
the function can be bent into the shape $\tfrac{1}{1-r}$
Taylor / Maclaurin coefficients
$$c_{n}=\frac{f^{(n)}(a)}{n!}$$
no known series fits and you must build one from derivatives
Remainder decides equality
$$f(x)=\sum c_{n}(x-a)^{n}\iff R_{n}(x)\to 0$$
you must justify that a series really equals the function
Taylor's Inequality
$$|R_{n}(x)|\le\frac{M}{(n+1)!}|x-a|^{n+1}$$
you need a numeric error bound or a term count
Three most common mistakes
Reusing the geometric series for $\tfrac{1}{1+x}$ without reading the ratio as $r=-x$, so every sign comes out wrong.
Building the Maclaurin coefficients as $f^{(n)}(0)$ and forgetting to divide by $n!$.
Using $n!$ instead of $(n+1)!$ in Taylor's Inequality, or bounding $f^{(n+1)}$ only at the center instead of across the whole interval.
On this course the graded weight you can see is quizzes 10% and homework 10%; the rest of the scheme is not stated here, so treat every week as quiz-and-homework relevant and do not read exam emphasis into it.
How much time do you have?
10 minutes
You walk in with the four boxed results and the six library series, enough to expand a function and bound one error.
The 60-second card · Where the coefficients come from · Taylor's Inequality · Formula card
45 minutes
You can build a representation two ways (geometric bending and derivatives), reuse the library, and finish a full error estimate.
The 60-second card · Bending the geometric series into new functions · Reusing the standard series and the binomial series · Where the coefficients come from · Taylor's Inequality · Scaffolding comes off
full read
Every route in, plus why a smooth function can fail to equal its own series, and the term-by-term calculus that links 11.9 to 11.10.
everything in order
By the end of this section
Represent a function as a power series by bending the geometric series through substitution and multiplication, and state the interval.
Differentiate and integrate a power series term by term to build new representations such as those for $\arctan x$ and $\ln(1+x)$.
Compute a Taylor or Maclaurin series directly from a function's derivatives using the coefficient formula $f^{(n)}(a)/n!$.
Reuse the standard Maclaurin library by substitution, multiplication and the binomial series instead of differentiating repeatedly.
Explain, via the Taylor polynomial and remainder, exactly when a function equals its Taylor series on an interval.
Bound the approximation error with Taylor's Inequality and choose the number of terms needed for a target accuracy.
Syllabus coverage
covered
11.9 — covered
Representations of functions as power series
Bending the geometric series comes first; the term-by-term differentiation and integration that 11.9 also introduces is the block right after it.
covered
11.10 — covered
Taylor and Maclaurin series
The coefficient formula, the standard library and the binomial series span three blocks; the remainder that decides equality opens the bridge to 11.11.
covered
11.11 — covered
Applications of Taylor polynomials
Taylor's Inequality, term counts for a target accuracy, and a physics-style approximation live here.
Recall first
Geometric series
For $|r|<1$, $\;\sum_{n=0}^{\infty}ar^{n}=\dfrac{a}{1-r}$. For $|r|\ge 1$ the series diverges.
It is the one series we already trust, and 11.9 builds every other representation by bending it.
A power series $\sum c_{n}(x-a)^{n}$ has a radius $R$ so that it converges for $|x-a|<R$ and diverges for $|x-a|>R$; the two endpoints are decided one at a time.
Every representation this week comes with an interval, and term-by-term calculus is stated in terms of $R$.
Ratio test
If $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_{n}}\right|=L$, then $\sum a_{n}$ converges absolutely when $L<1$ and diverges when $L>1$.
It is how we read the radius $R$ off a series, including the new ones we build.
Derivatives of the standard functions
$\dfrac{d}{dx}e^{x}=e^{x}$, $\;\dfrac{d}{dx}\sin x=\cos x$, $\;\dfrac{d}{dx}\cos x=-\sin x$, and $\dfrac{d}{dx}x^{n}=nx^{n-1}$.
Taylor coefficients are built from repeated derivatives, so fluency with these is the whole engine of 11.10.
Try it yourself first (2 questions)
1§05.0 - an endpoint trap
A power series can converge at one endpoint and diverge at the other. This one catches almost everyone.
Find(a) True or false: the series converges at $x=1$.
Given
the series $\sum_{n=1}^{\infty}\dfrac{x^{n}}{n}$
radius $R=1$
Hint 1/4
At $x=1$ the series becomes a specific number series; write it out.
Hint 2/4
Setting $x=1$ gives $\sum \tfrac{1}{n}$, which is the harmonic series.
Hint 3/4
The harmonic series $\sum \tfrac1n$ is the classic divergent series.
Hint 4/4
So at $x=1$ it diverges; it is $x=-1$ that converges, by alternation.
Show solution
Substitute the endpoint
$$x=1:\ \sum_{n=1}^{\infty}\tfrac{1}{n}$$
the endpoint is not covered by |x|<1; test it directly
$$\sum \tfrac1n\ \text{diverges}$$
this is the harmonic series, a known divergence
Answer $$\boxed{\text{diverges at } x=1}$$
Check
At $x=-1$ the same series is the alternating harmonic series, which converges: the two ends really do differ.
2§05.0 - a third derivative at zero
Taylor coefficients are built from repeated derivatives at the center, so warm up the machinery on a function you know.
Find(a) What is $f'''(0)$?
Given$f(x)=e^{x}$
Hint 1/4
Ask what every derivative of $e^{x}$ equals.
Hint 2/4
Each derivative of $e^{x}$ is again $e^{x}$.
Hint 3/4
So $f'''(x)=e^{x}$, and you evaluate at $x=0$.
Hint 4/4
$f'''(0)=e^{0}=1$.
Show solution
Differentiate three times
$$f'(x)=f''(x)=f'''(x)=e^x$$
the exponential is fixed by differentiation
$$f'''(0)=e^0=1$$
evaluate at the center x=0
Answer $$\boxed{1}$$
Check
Every derivative of $e^x$ at 0 is 1, which is exactly why its coefficients are $1/n!$.
Notation
symbol
reads as
means
watch out
$\sum_{n=0}^{\infty} c_n x^n$
the sum over n from zero to infinity of c-n times x to the n
a power series: an infinite polynomial whose coefficients are the numbers $c_n$
The index usually starts at $0$; after differentiating it starts at $1$ because the constant term is gone.
$f^{(n)}(a)$
the n-th derivative of f evaluated at a
differentiate $f$ a total of $n$ times, then plug in the center $a$
$f^{(0)}$ means $f$ itself; evaluate at $a$, never at a running $x$.
$n!$
n factorial
the product $1\cdot 2\cdot 3\cdots n$, with $0!=1$
Factorials grow faster than any power, which is why Taylor tails collapse.
$T_n(x)$
the n-th Taylor polynomial
the series cut off after the degree-$n$ term
$T_n$ is a finite polynomial, not the whole series; do not call it the series.
$R_n(x)$
the n-th remainder
the leftover $f(x)-T_n(x)$ after chopping at degree $n$
$R_n$ is generally nonzero for every finite $n$; only its limit as $n\to\infty$ may vanish.
$\binom{k}{n}$
k choose n, the
for a real $k$, the product $\dfrac{k(k-1)\cdots(k-n+1)}{n!}$
For non-integer $k$ this never becomes zero, so the binomial series does not terminate.
Conventions used here
Interval versus radius
Radius of convergence names how far from the center the series converges; the interval is that radius together with a separate verdict at each of the two endpoints. Term-by-term calculus preserves the radius but not the endpoint verdicts.
Index of a differentiated series
When we differentiate $\sum_{n=0}^{\infty}c_{n}x^{n}$ the constant term dies, so the result starts at $n=1$: $\sum_{n=1}^{\infty}nc_{n}x^{n-1}$. Keeping the lower limit at $0$ is a silent error because it adds a phantom zero term.
Taylor versus Maclaurin
A Taylor series is centered at a general point $a$; a Maclaurin series is the special case $a=0$. When a problem says nothing about a center, it means $a=0$.
Where M lives in the bound
In Taylor's Inequality $M$ must bound $|f^{(n+1)}|$ over the entire interval from the center $a$ out to $x$, not merely at $a$. Taking $M=|f^{(n+1)}(a)|$ is the most common way the bound goes wrong.
Equality is not a picture
A partial sum lying on top of the curve on a plotted window is evidence, not proof. The function equals its series only where the remainder $R_{n}(x)$ tends to zero, which is an inequality argument, not a graph.
5.1Bending the geometric series into new functions
Turn a function into a power series by reshaping the one series we already trust.
Last week closed with a test that answers a backward question: given a series, does it converge? Now we ask the forward one. Handed a function, where does its series come from? We start with the only series we can already write down.
Solvable with what we have
decide whether a given power series converges, using the ratio and root tests
sum a geometric series once we can see the first term and the ratio
Not solvable yet
produce a power series for a function such as $\dfrac{1}{1+x^{2}}$ that arrives with no series attached
We could try to guess the coefficients $c_0,c_1,c_2,\dots$ one at a time by matching values: force the series to agree with $f$ at $x=0$, then at a second point, then a third.
Why it fails
Matching $k$ points pins down only $k$ coefficients, and a power series has infinitely many. The guessing never closes. We need a rule that delivers every coefficient at once, and the geometric series is the first such rule.
One over one minus x equals the sum of every whole-number power of x, and the equality holds precisely when x lies strictly between minus one and one.
Proof
The partial sum satisfies $s_N=1+x+\cdots+x^{N}=\dfrac{1-x^{N+1}}{1-x}$.
When $|x|<1$, $x^{N+1}\to 0$, so $s_N\to\dfrac{1}{1-x}$; when $|x|\ge 1$ the terms do not even shrink.
The blue curve is $\dfrac{1}{1-x}$; the orange dashed curve is the degree-four partial sum. They are one line thick near $x=0$ and separate as $x$ nears the wall at $x=1$, where the true function shoots to infinity.
Looks like this, but is not
It is tempting to reuse the box on $\dfrac{1}{1+x}$ by just flipping the signs to $1-x+x^{2}-\cdots$ and moving on.
That answer is right, but only because you must first read the ratio correctly as $r=-x$: the template is $\dfrac{1}{1-r}$, so $\dfrac{1}{1+x}=\dfrac{1}{1-(-x)}$. Skip that reading and you will get the signs wrong the moment the denominator is $2+x$ or $1+x^{2}$.
A power series for 1 over 1 plus x squared
Represent $f(x)=\dfrac{1}{1+x^{2}}$ as a power series in $x$ and give its interval.
Finda power series for $f$ together with the interval on which it holds
Given
$f(x)=\dfrac{1}{1+x^{2}}$
the only tool so far is $\dfrac{1}{1-r}=\sum r^{n}$ for $|r|<1$
Solution
Match the target to the template
$$\frac{1}{1+x^{2}}=\frac{1}{1-(-x^{2})}$$
the template is 1/(1-r); to get a plus sign in the denominator the ratio must be r=-x^2, not +x^2
pull 4 out so the denominator starts with 1; the plus sign forces the minus into r
$$r=-x^{2}/4$$
the quantity subtracted from 1 is the ratio
Answer $$\boxed{r=-x^{2}/4}$$
Check
Check $|r|<1$ gives $|x|<2$, the expected radius for a $4$ under the square.
⚠ Reusing the plus-sign series without flipping the ratio
the geometric box has a minus sign, so a plus in the denominator is read past too fast
wrong$$\frac{1}{1+x}=\sum x^{n}$$
right$$\frac{1}{1+x}=\sum(-1)^{n}x^{n}$$
⚠ Forgetting to factor the constant
the eye jumps to the variable and treats the leading constant as decoration
wrong$$\frac{1}{2-x}=\sum x^{n}$$
right$$\frac{1}{2-x}=\sum \frac{x^{n}}{2^{n+1}}$$
5.2Differentiating and integrating a power series term by term
Inside its interval a power series can be differentiated and integrated one term at a time.
The geometric trick only reaches functions you can wrestle into $\dfrac{1}{1-r}$. Differentiating and integrating a series term by term breaks out of that cage and reaches $\arctan x$ and $\ln(1+x)$.
TheoremTerm-by-term differentiation and integration
Conditions
$f(x)=\sum_{n=0}^{\infty}c_{n}(x-a)^{n}$ has radius $R>0$
Inside its interval a power series may be differentiated or integrated one term at a time, and the new series keeps the same radius, though the verdict at each endpoint has to be checked again.
Blue is $\arctan x$; orange dashed is the degree-five partial sum $x-\tfrac{x^{3}}{3}+\tfrac{x^{5}}{5}$. Integrating the geometric series builds the arctangent, and at $x=1$ the value is $\tfrac{\pi}{4}$.
Looks like this, but is not
Term-by-term calculus looks like it should carry the whole interval across, endpoints and all.
It preserves the radius, not the endpoints. The geometric series for $\dfrac{1}{1+x^{2}}$ diverges at $x=1$, yet integrating it gives $\arctan x$, whose series converges at $x=1$ to $\tfrac{\pi}{4}$. Integration added an endpoint the original never had.
Building the arctangent series by integrating
Find a Maclaurin series for $\arctan x$ by integrating the series for $\dfrac{1}{1+x^{2}}$.
Finda power series for $\arctan x$ and the value it gives at $x=1$
Given
$\dfrac{1}{1+x^{2}}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}$ for $|x|<1$
5.3Where the coefficients come from: Taylor and Maclaurin series
If a function is a power series at a, its coefficients are forced to be its own derivatives over n factorial.
Substitution and calculus reshaped one seed series. But $e^{x}$, $\sin x$ and $\cos x$ are not disguised geometric series: there is no ratio $r$ to find. For these we go back to first principles and let the function's own derivatives hand us the coefficients. This is the sum the calculator was adding up.
TheoremTaylor and Maclaurin coefficients
Conditions
$f$ has derivatives of every order at $a$
the series so built need not equal $f$; that is settled later by the remainder
If a function is a power series centered at a, then its n-th coefficient must equal the n-th derivative at a divided by n factorial, so the only candidate series is the one built from the function's own derivatives.
Proof
Differentiate $f(x)=\sum c_{n}(x-a)^{n}$ exactly $k$ times term by term.
Every term with $n<k$ vanishes and every term with $n>k$ still carries a factor $(x-a)$.
Set $x=a$: only the $n=k$ term survives, equal to $k!\,c_{k}$, so $c_{k}=f^{(k)}(a)/k!$.
Blue is $\sin x$. The grey line is $T_{1}=x$, orange dashed is $T_{3}=x-\tfrac{x^{3}}{6}$, green is $T_{5}$. Each extra term, built from one more derivative at $0$, widens the stretch where the polynomial hugs the wave.
Looks like this, but is not
Surely any function smooth enough to have all its derivatives at $0$ must equal the series those derivatives build.
No. Take $f(x)=e^{-1/x^{2}}$ for $x\ne 0$ and $f(0)=0$. Every derivative at $0$ works out to $0$, so its Maclaurin series is $0+0\cdot x+\cdots$, the zero function. Yet $f(x)>0$ for all $x\ne 0$. The series exists and equals $f$ at exactly one point. Having the coefficients is not the same as equalling the function.
The Maclaurin series of e to the x
Build the Maclaurin series of $f(x)=e^{x}$ straight from its derivatives.
Findthe Maclaurin series of $e^{x}$
Given
$f(x)=e^{x}$
$c_{n}=f^{(n)}(0)/n!$
Solution
Find every derivative at 0
$$f^{(n)}(x)=e^{x}\ \Rightarrow\ f^{(n)}(0)=1\ \text{for all } n$$
the exponential is fixed by differentiation, so every derivative at 0 equals 1
Assemble the coefficients
$$c_{n}=\frac{f^{(n)}(0)}{n!}=\frac{1}{n!}$$
divide each derivative value by n! as the formula demands
At $x=0.5$: $1-0.125+0.0026-\cdots=0.87758$, matching $\cos 0.5=0.87758$.
Four derivatives reveal the whole repeating pattern; you never differentiate a fifth time.
Checkpoint
§05.3 - the coefficient of x cubed
The whole coefficient formula lives or dies on the factorial. Test it on a single coefficient.
Find(a) What is $c_{3}$ in terms of the derivatives of $f$ at $0$?
Given$f(x)=\sum_{n=0}^{\infty}c_{n}x^{n}$ is the Maclaurin series of $f$
Hint 1/4
Ask which derivative and which factorial attach to the degree-three term.
Hint 2/4
The Maclaurin coefficient is $c_{n}=\dfrac{f^{(n)}(0)}{n!}$.
Hint 3/4
Set $n=3$, so the factorial is $3!=6$ and the derivative is the third.
Hint 4/4
Therefore $c_{3}=\dfrac{f'''(0)}{6}$.
Show solution
Substitute n equals 3
$$c_{3}=\frac{f'''(0)}{3!}=\frac{f'''(0)}{6}$$
3! = 6; the third derivative pairs with the third factorial
Answer $$\boxed{f'''(0)/6}$$
Check
For $f=e^{x}$ this gives $1/6$, matching the $x^{3}/3!$ term of its series.
⚠ Forgetting to divide by n factorial
the derivative feels like the substance and the factorial like bookkeeping
wrong$$c_{n}=f^{(n)}(0)$$
right$$c_{n}=\frac{f^{(n)}(0)}{n!}$$
⚠ Evaluating the derivatives at a running x instead of the center
the derivative is written as a function, so the substitution x equals a is skipped
wrong$$c_{n}=\frac{f^{(n)}(x)}{n!}$$
right$$c_{n}=\frac{f^{(n)}(a)}{n!}$$
5.4Reusing the standard series and the binomial series
Almost every Maclaurin series is a known one after a substitution, a multiplication, or the binomial rule.
Computing $f^{(n)}(0)$ for every $n$ is brutal for anything but the cleanest functions. Almost always the fast route is to bend a series you already have. One new series, the binomial, completes the toolkit.
TheoremThe binomial series
Conditions
$k$ any real number
valid for $\lvert x\rvert<1$
if $k$ is a nonnegative integer the sum stops and holds for all $x$
One plus x, raised to any real power k, equals the sum over n of the generalised binomial coefficient times x to the n, and for a non-integer k this sum runs forever and converges when x is between minus one and one.
Blue is $\sqrt{1+x}$; orange dashed is the binomial quadratic $1+\tfrac{x}{2}-\tfrac{x^{2}}{8}$. They meet exactly at $x=0$ and stay close for small $x$, drifting apart only toward the right edge.
Looks like this, but is not
The binomial series ought to stop, the way $(1+x)^{3}$ stops after four terms.
It terminates only when $k$ is a nonnegative integer, because then some factor $(k-n+1)$ hits zero and kills every later coefficient. For $k=\tfrac12$ or $k=-1$ no factor is ever zero, so the series runs forever.
Function
Maclaurin series
Holds for
$e^{x}$
$\sum_{n\ge 0}\dfrac{x^{n}}{n!}$
all $x$
$\sin x$
$\sum_{n\ge 0}(-1)^{n}\dfrac{x^{2n+1}}{(2n+1)!}$
all $x$
$\cos x$
$\sum_{n\ge 0}(-1)^{n}\dfrac{x^{2n}}{(2n)!}$
all $x$
$\dfrac{1}{1-x}$
$\sum_{n\ge 0}x^{n}$
$\lvert x\rvert<1$
$\ln(1+x)$
$\sum_{n\ge 1}(-1)^{n-1}\dfrac{x^{n}}{n}$
$-1<x\le 1$
$\arctan x$
$\sum_{n\ge 0}(-1)^{n}\dfrac{x^{2n+1}}{2n+1}$
$\lvert x\rvert\le 1$
These six cover almost every 11.10 problem once you allow a substitution or a multiplication; memorise them and the work becomes bookkeeping.
A series for e to the minus x squared
Find the Maclaurin series of $e^{-x^{2}}$ by reusing the series for $e^{x}$.
At $x=0.5$: $1-0.25+0.03125-\cdots=0.77880$, matching $e^{-0.25}=0.77880$.
Differentiating $e^{-x^{2}}$ many times is a nightmare; substitution hands the same series in one line, which is why the library beats raw derivatives.
The binomial series for the square root of 1 plus x
Use the binomial series with $k=\tfrac12$ to expand $\sqrt{1+x}$, then estimate $\sqrt{1.1}$.
Findthe first terms of $\sqrt{1+x}$ and an estimate of $\sqrt{1.1}$
5.5The Taylor polynomial, the remainder, and when a function equals its series
A function equals its Taylor series exactly where the leftover remainder shrinks to zero.
We keep saying the series equals the function, and Concept 3 showed that can fail. Here is the exact condition, and the object that measures the failure: the left after cutting the series short.
TheoremRemainder and the equality criterion
Conditions
$T_{n}(x)=\sum_{i=0}^{n}\dfrac{f^{(i)}(a)}{i!}(x-a)^{i}$ is the degree-$n$ Taylor polynomial
$R_{n}(x)=f(x)-T_{n}(x)$ is the remainder
the statement is pointwise: it may hold at some $x$ and fail at others
The function equals its Taylor series at a point exactly when the remainder, the leftover after cutting the series at degree n, tends to zero as n grows.
Proof
By definition $f(x)-T_{n}(x)=R_{n}(x)$, so $T_{n}(x)\to f(x)$ means precisely $R_{n}(x)\to 0$.
The whole difficulty is therefore pushed into showing $R_{n}\to 0$, which the next block's inequality does.
Blue is $\cos x$; orange dashed is $T_{4}=1-\tfrac{x^{2}}{2}+\tfrac{x^{4}}{24}$. The green segment near $x=\pi$ is the remainder $R_{4}$: the vertical gap between function and polynomial. Equality holds where that gap collapses to zero.
Looks like this, but is not
Since $T_{4}$ and $\cos x$ are indistinguishable on the plot near $0$, one might declare $\cos x$ equals its series and be done.
A picture on $[-1,1]$ says nothing about $x=10$, where $T_{4}$ is enormous and negative while $\cos x$ stays between $-1$ and $1$. Equality everywhere needs $R_{n}(x)\to 0$ for every $x$, an inequality argument, not a graph.
The degree three Taylor polynomial of the logarithm at 1
Write $T_{3}$ for $f(x)=\ln x$ at $a=1$ and use it to estimate $\ln 1.2$.
Findthe polynomial $T_{3}$ and an estimate of $\ln 1.2$
Given
$f(x)=\ln x$
$a=1$
Solution
Collect the derivatives at the center
$$f(1)=0,\ f'(1)=1,\ f''(1)=-1,\ f'''(1)=2$$
differentiate ln x to 1/x, -1/x^2, 2/x^3 and evaluate each at a=1
The gaps fall by more than a factor of $10$ each step, matching the factorial growth in the denominators; the next block proves this happens for every $x$ at once.
Four partial sums are enough to see the collapse; proving it for all x is the next block's job.
Checkpoint
§05.5 - what the Taylor polynomial matches
The remainder measures a gap that starts perfectly closed at the center. Say exactly how closed.
Find(a) At the center $a$, in what sense do $T_{n}$ and $f$ agree?
Given$T_{n}$ is the degree-$n$ Taylor polynomial of $f$ at $a$
Hint 1/4
Think about what the coefficients were built to reproduce at $x=a$.
Hint 2/4
Each coefficient $f^{(i)}(a)/i!$ was chosen to match one derivative of $f$ at $a$.
Hint 3/4
Since $i$ runs from $0$ to $n$, the value and the first $n$ derivatives are forced to agree.
Hint 4/4
So $T_{n}$ matches $f$ in value and in its first $n$ derivatives at $a$.
a good approximation feels exact, so R_n gets treated as nothing
wrong$$R_{n}(x)=0$$
right$$R_{n}(x)=f(x)-T_{n}(x)\neq 0 \text{ in general}$$
Raising the degree does not tilt the polynomial, it extends the stretch over which it still matches: the green gap at x = 2 drops 0.583853, 0.082814, 0.006075, 0.000274. Degree 2 already falls out of the frame near x = 2.4495, which is what a vanishing remainder rules out.
At the edges
degree 8 at x = π R = 0.023978
Further from the centre the same degree leaves a gap almost 90 times larger, so equality is a pointwise claim, not a global one.
degree 0 T = 1
The flat line y = 1 matches only at x = 0; every term after that buys interval, not height.
5.6Taylor's Inequality: turning the remainder into a number
Bound the next derivative by M and the error is at most M over (n+1) factorial times the distance to the power n plus one.
The criterion $R_{n}\to 0$ is a yes-or-no fact. Exams and machines need a number: how big is the error if I stop now? Taylor's Inequality turns the remainder into a bound you can actually compute, and answers the calculator question from the start.
TheoremTaylor's Inequality
Conditions
you must find a single $M$ with $|f^{(n+1)}|\le M$ across the interval
$d$ is the distance from the center out to the farthest $x$ you care about
$$\boxed{\ |f^{(n+1)}(x)|\le M \text{ on } |x-a|\le d\ \Rightarrow\ |R_{n}(x)|\le \frac{M}{(n+1)!}\,|x-a|^{n+1}\ }$$
If the derivative one order past the last term you kept never exceeds M in size on the interval, then the degree-n error is at most M divided by (n+1) factorial, times the distance from the center raised to the n plus one.
Approximating $e$ by cutting its series at degree $n$: blue is the true error, orange dashed is the guaranteed bound $\tfrac{3}{(n+1)!}$, scaled so the first bound is $1$. The bound always sits above the true error and both crash as $n$ grows.
Looks like this, but is not
It is tempting to take $M=|f^{(n+1)}(a)|$, the value right at the center, because that number is easy to read off.
The bound needs $|f^{(n+1)}|$ to stay under $M$ across the entire interval to $x$. A derivative that is small at $a$ but large just next to it makes that choice of $M$ too small, and the resulting bound is simply false.
How many terms approximate e to three decimals
How many terms of the series for $e^{x}$ are needed so that $T_{n}(1)$ estimates $e$ with error below $0.001$?
Findthe smallest degree $n$, and the estimate it gives
for any fixed x the factorial eventually outruns the power, so the bound goes to 0
Answer $$R_{n}(x)\to 0 \text{ for all } x,\ \text{so } e^{x}=\sum_{n\ge 0}\tfrac{x^{n}}{n!} \text{ everywhere}$$
Check
The factor $e^{|x|}$ is a fixed constant once $x$ is chosen; only $\tfrac{|x|^{n+1}}{(n+1)!}$ moves, and it is the tail of the convergent series for $e^{|x|}$, so it must vanish.
The same argument, with $|f^{(n+1)}|\le 1$, proves $\sin x$ and $\cos x$ equal their series everywhere: bounded derivatives are the easy case.
Checkpoint
§05.6 - what M has to bound
Taylor's Inequality is only as good as the number $M$ you feed it, and choosing $M$ is where most errors start.
Find(a) The number $M$ must be an upper bound for which quantity, and where?
At $x=0$ the sum is $\tfrac13$, matching $\tfrac{1}{3+0}$; at $x=1$ it gives $\tfrac14=\tfrac{1}{3+1}$.
2 · you write the reasoning
Represent $g(x)=\dfrac{1}{1-x^{3}}$ as a power series with its interval. This one needs no factoring and no sign flip, so fill in the reason for each step.
reasoning
The denominator already reads $1-r$ with $r=x^{3}$, so no factoring is needed and the ratio is read directly.
reasoning
Each $(x^{3})^{n}$ is just $x^{3n}$; the powers land on multiples of three and every coefficient is $1$.
reasoning
The interval comes from $|r|<1$ with $r=x^{3}$, and $|x^{3}|<1$ is the same as $|x|<1$.
3 · find the buried error
A classmate represents $h(x)=\dfrac{1}{4+x^{2}}$ about $0$. Two of the three steps are wrong. Find them.
Step 1.$\dfrac{1}{4+x^{2}}=\dfrac14\cdot\dfrac{1}{1+\frac{x^{2}}{4}}$, factoring the 4 out.
Step 2. Reading $r=\dfrac{x^{2}}{4}$, the series is $\dfrac14\sum\big(\tfrac{x^{2}}{4}\big)^{n}=\sum\dfrac{x^{2n}}{4^{n+1}}$.
Step 3. Converges when $\big|\tfrac{x^{2}}{4}\big|<1$, that is $|x|<4$.
the two buried errors (2)
⚠ step 2
the ratio was read as $+\tfrac{x^{2}}{4}$, dropping the alternating sign
the denominator has a plus sign, so students copy it into $r$ instead of sending it into a minus
right
The form is $\dfrac{1}{1-(-x^{2}/4)}$, so $r=-\tfrac{x^{2}}{4}$ and the series is $\sum(-1)^{n}\dfrac{x^{2n}}{4^{n+1}}$.
⚠ step 3
the interval was solved as $|x|<4$ instead of $|x|<2$
the 4 in the denominator is mistaken for the radius, skipping the square root of $x^{2}<4$
right
From $\big|\tfrac{x^{2}}{4}\big|<1$ we get $x^{2}<4$, so $|x|<2$, not $4$.
4 · the bare problem
§05.1 - represent 1 over 9 plus x squared
Same skeleton, no scaffolding. A $9$ under a square and a plus sign: watch both the sign and the interval.
Find(a) Represent $k$ as a power series about $0$ and give its interval.
Given$k(x)=\dfrac{1}{9+x^{2}}$
Hint 1/4
Factor the 9 out first so the denominator leads with a 1.
Hint 2/4
The plus sign means $r$ carries a minus: $r=-\tfrac{x^{2}}{9}$.
Hint 3/4
Substitute into $\tfrac19\sum r^{n}$; then solve $\big|\tfrac{x^{2}}{9}\big|<1$ for the interval.
Hint 4/4
The series is $\sum(-1)^{n}\dfrac{x^{2n}}{9^{n+1}}$ on $|x|<3$.
At $x=1$: $\tfrac{1}{9+1}=0.1$, and $\tfrac19-\tfrac{1}{81}+\cdots=0.1$.
Full exam-style question
Estimating an integral with no elementary antiderivativeexam format
Estimate $\displaystyle\int_{0}^{0.5}\sin(x^{2})\,dx$ to five decimal places and justify the accuracy. The integrand $\sin(x^{2})$ has no antiderivative in the elementary function table, so a series is the standard route.
Finda five-decimal value of the integral with an error justification
the -1 removes the leading 1, and the c^2 factors cancel
Answer $$\boxed{K\approx \tfrac12 mv^{2}}$$
Check
The next term $\tfrac{3}{8}mv^{4}/c^{2}$ is tiny when $v\ll c$, so Newton's formula is the small-speed shadow of Einstein's.
3§05.4 - multiplying two series
Combining two library series by multiplication is a step up: it needs the product rule intuition from the derivatives unit and careful bookkeeping of powers.
Find(a) Find the first three nonzero terms of the Maclaurin series of $e^{x}\sin x$.
Close the page and write, from memory: the geometric series and its interval; the coefficient formula $f^{(n)}(a)/n!$; the six library series; the equality criterion $R_{n}\to 0$; and Taylor's Inequality. Then represent $\dfrac{1}{3-x}$ two ways.
Represent $\dfrac{1}{5+x}$ as a power series and give its interval?
c-geometric-engine
Build the series for $\arctan x$ by integrating a geometric series?
c-calculus-on-series
Compute the Maclaurin series of $\cos x$ from its derivatives at $0$?
c-taylor-machine
Expand $e^{-x^{2}}$ and $\sqrt{1+x}$ by reusing library series?
c-series-library
State when a function equals its Taylor series, and what $R_{n}$ measures?
c-convergence-remainder
Decide how many terms of $e^{x}$ pin $e$ to within $0.001$ using Taylor's Inequality?
c-taylor-inequality
Glossary (12 terms)
power serieskuvvet serisi
A series $\sum c_n(x-a)^n$ with constant coefficients $c_n$ and a variable $x$, centred at the number $a$. Inside its interval of convergence it defines a function of $x$.
geometric seriesgeometrik seri
A series in which each term is a fixed multiple r of the previous one. On this page we use it in the form $\sum ar^{n}=\dfrac{a}{1-r}$ for $|r|<1$, the seed for every representation here.
radius of convergenceyakınsaklık yarıçapı
The number $R$ such that a power series converges for $|x-a|<R$ and diverges for $|x-a|>R$.
interval of convergenceyakınsaklık aralığı
The full set of $x$ where a power series converges: the radius together with a verdict at each endpoint.
Maclaurin seriesMaclaurin serisi
The Taylor series centered at $a=0$, with coefficients $f^{(n)}(0)/n!$.
Taylor seriesTaylor serisi
The series $\sum \dfrac{f^{(n)}(a)}{n!}(x-a)^{n}$ built from a function's derivatives at a center $a$.
Taylor polynomialTaylor polinomu
The finite piece $T_{n}$ of the Taylor series up to the degree-$n$ term.
Taylor remainderkalan terim
The difference $R_{n}(x)=f(x)-T_{n}(x)$ between a function and its degree-$n$ Taylor polynomial; the function equals its series exactly where $R_{n}\to 0$.
Taylor's InequalityTaylor eşitsizliği
The bound $|R_{n}(x)|\le \dfrac{M}{(n+1)!}|x-a|^{n+1}$ whenever $|f^{(n+1)}|\le M$ on the interval.
binomial seriesbinom serisi
The Maclaurin series of $(1+x)^{k}$, namely $\sum \binom{k}{n}x^{n}$, infinite unless $k$ is a nonnegative integer.
binomial coefficientbinom katsayısı
For real $k$, the number $\binom{k}{n}=\dfrac{k(k-1)\cdots(k-n+1)}{n!}$.
coefficientkatsayı
The constant $c_{n}$ multiplying $(x-a)^{n}$ in a power series.
What comes next
§06 · Vectors, lines and planes in space
With series behind us, the next section leaves the number line for space: vectors as arrows you can add and scale, the dot and cross products, and the lines and planes they describe. The algebra is lighter but the pictures are three-dimensional.
Sources
textbookJames Stewart, Calculus, ninth edition, sections 11.9, 11.10 and 11.11 The three sections named on this week's syllabus line; statements, intervals and the binomial series follow this book's conventions.
standard resultThe standard Maclaurin library and Taylor's Inequality The six library series and the remainder bound are standard results, stated here in the form used across single-variable calculus texts.