True or False? The ratio test gives the full interval of convergence.

False. It gives only the open interval and says nothing at the endpoints. Each endpoint has to be substituted in and tested separately.

How is a Taylor polynomial related to a Taylor series?

The polynomial is a partial sum of the series, stopped at a chosen degree. The series is the polynomial carried on without end.

What is the Taylor series of a polynomial function?

The polynomial itself. Its derivatives are all zero beyond a certain order, so the series terminates and reproduces the original function exactly.

How do you find the power series of a product of two functions?

Multiply their individual series together, keeping only terms up to the power you need. Collect like powers at the end.

True or False? Every term produced by multiplying two truncated series is correct.

False. Only the terms up to the power you truncated at are reliable. The higher ones are incomplete, since further terms of the original series would also have contributed to them.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 10: Infinite Sequences and SeriesSeries Representations of Functions

Series Representations of Functions (College Board AP® Calculus BC): Flashcards

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FrontTaylor Polynomial Approximation of a Function
The $n$ th degree Taylor polynomial for $f$ about $x = a$ builds each coefficient from a derivative at $a$ :
$p_{n} (x) = f (a) + f^{'} (a) (x - a) + \frac{f^{''} (a)}{______} {(x - a)}^{2} + \dots$

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The $n$ th degree Taylor polynomial for $f$ about $x = a$ builds each coefficient from a derivative at $a$ :
$p_{n} (x) = f (a) + f^{'} (a) (x - a) + \frac{f^{''} (a)}{______} {(x - a)}^{2} + \dots$
The completed coefficient is $\frac{f^{''} (a)}{2!}$ .
The term in ${(x - a)}^{n}$ has coefficient $\frac{f^{(n)} (a)}{n!}$ , so every coefficient divides a derivative by a factorial.
What is the first-degree Taylor polynomial $p_{1}$ ?
The tangent line to $f$ at $(a, f (a))$ .
It is the linearization already met, so a Taylor polynomial extends that idea to higher degrees.
What is the value of a Taylor polynomial at $x = a$ itself?
Exactly $f (a)$ .
Every term after the first carries a factor of $(x - a)$ , which vanishes there, so the approximation is perfect at the centre.
True or False?
Increasing the degree of a Taylor polynomial improves the approximation.
True.
Each extra term adds a higher power of $(x - a)$ , which improves the fit and extends it further from the centre.
The gain is largest away from $x = a$ .
Find the Taylor polynomial for $g (x) = \sin x$ about $x = \frac{π}{3}$ up to the term in ${(x - \frac{π}{3})}^{3}$ .
The derivatives at $\frac{π}{3}$ are $\frac{\sqrt{3}}{2}$ , $\frac{1}{2}$ , $- \frac{\sqrt{3}}{2}$ and $- \frac{1}{2}$ .
Dividing by the factorials gives $\frac{\sqrt{3}}{2} + \frac{1}{2} (x - \frac{π}{3}) - \frac{\sqrt{3}}{4} {(x - \frac{π}{3})}^{2} - \frac{1}{12} {(x - \frac{π}{3})}^{3}$ .
What is a Maclaurin polynomial?
A Taylor polynomial centred at $x = 0$ .
The formula becomes $f (0) + f^{'} (0) x + \frac{f^{''} (0)}{2!} x^{2} + \dots$ , so for $e^{x}$ it gives $1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3}$ .
Define a power series.
A series of the form $\sum_{n = 0}^{\infty} a_{n} {(x - r)}^{n}$ , in which $x$ is a variable.
Because the terms depend on $x$ , a power series defines a function rather than a single number.
Why does convergence of a power series depend on $x$ ?
Substituting a different $x$ gives a different series of numbers.
It may converge for some values of $x$ and diverge for others.
Is there any value of $x$ at which every power series converges?
Yes, at the centre $x = r$ .
Every term after the first is then zero, so the series collapses to $a_{0}$ .
The ratio test on $\sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \frac{x^{n}}{n \cdot 3^{n}}$ gives a limit of $|\frac{x}{3}|$ , and the series converges where that is below $1$ :
$|x| <______$
The completed condition is $|x| < 3$ .
That is the radius of convergence, half the width of the interval, and it gives the open interval $- 3 < x < 3$ .
True or False?
The ratio test gives the full interval of convergence.
False.
It gives only the open interval and says nothing at the endpoints.
Each endpoint has to be substituted in and tested separately.
A series has radius $3$ about $x = 0$ . At $x = - 3$ it becomes $- \sum \frac{1}{n}$ and at $x = 3$ it becomes $\sum \frac{{(- 1)}^{n + 1}}{n}$ . Give the interval.
It is $- 3 < x \leq 3$ .
The first endpoint gives the negative of the harmonic series, which diverges, and the second gives the alternating harmonic series, which converges.
When applying the ratio test to a power series, how is $x$ treated?
As a constant, since the limit runs over $n$ only.
For $\sum \frac{x^{n}}{n!}$ the ratio $|\frac{x}{n + 1}|$ tends to zero for every fixed $x$ , so that series converges everywhere.
The Taylor series for $f$ about $x = a$ runs on forever, with the general coefficient built from the $n$ th derivative:
$\sum_{n = 0}^{\infty} \frac{f^{(n)} (a)}{______} {(x - a)}^{n}$
The completed series is $\sum_{n = 0}^{\infty} \frac{f^{(n)} (a)}{n!} {(x - a)}^{n}$ .
It is the same coefficient as in a Taylor polynomial, but the sum is never truncated.
How is a Taylor polynomial related to a Taylor series?
The polynomial is a partial sum of the series, stopped at a chosen degree.
The series is the polynomial carried on without end.
True or False?
A Taylor polynomial is an approximation to $f$ , while the Taylor series it comes from can represent $f$ exactly.
True.
A polynomial stops at a chosen degree, so it carries an error that the Lagrange bound measures.
Where the series converges to $f$ , no terms are dropped and nothing is approximated.
Can a function have two different power series about the same point?
No.
If a power series about $x = a$ converges to $f$ on an interval, it must be the Taylor series for $f$ .
The Taylor series is unique.
Two of the four standard Maclaurin series to know are the geometric one and the exponential one:
$\frac{1}{1 - x} = 1 + x + x^{2} + \dots$ and $e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{______} + \dots$
The completed series is $e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots$
The exponential series converges for every real $x$ , while the geometric one converges only for $|x| < 1$ .
What are the Maclaurin series for $\sin x$ and $\cos x$ ?
They are $x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots$ and $1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \dots$
Both alternate, both converge for all real $x$ , and $x$ is in radians.
What is the Taylor series of a polynomial function?
The polynomial itself.
Its derivatives are all zero beyond a certain order, so the series terminates and reproduces the original function exactly.
How do you find the power series of a composite function?
Start from the standard series for the outer function, then substitute the inner function, in brackets, everywhere $x$ appears.
Expand and collect the powers.
A substitution changes a power series. What happens to its interval of convergence?
It changes too.
Put the inner function into the original condition and solve.
For $\frac{1}{1 + 4 x^{2}}$ the condition $|- 4 x^{2}| < 1$ gives $- \frac{1}{2} < x < \frac{1}{2}$ .
Find the Maclaurin series for $\frac{1}{1 + 4 x^{2}}$ up to the term in $x^{8}$ .
Write it as $\frac{1}{1 - (- 4 x^{2})}$ and substitute into $1 + x + x^{2} + \dots$
That gives $1 - 4 x^{2} + 16 x^{4} - 64 x^{6} + 256 x^{8}$ .
How do you find the power series of a product of two functions?
Multiply their individual series together, keeping only terms up to the power you need.
Collect like powers at the end.
True or False?
Every term produced by multiplying two truncated series is correct.
False.
Only the terms up to the power you truncated at are reliable.
The higher ones are incomplete, since further terms of the original series would also have contributed to them.
Term-by-term differentiation of a power series gives the series for the function's derivative. Applying it to the series for $\sin x$ produces the series for
$\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{______} - \dots$
The completed series is $\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \dots$
Differentiating $x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots$ term by term gives exactly the standard series for $\cos x$ .
Integrating a power series term by term introduces a constant. How do you find it?
Substitute the centre of the series.
For a Maclaurin series put $x = 0$ and choose $C$ so that the series gives the correct value of $f (0)$ .
Integrate the series for $\frac{1}{1 + x^{2}}$ to find the Maclaurin series for $\arctan x$ up to $x^{7}$ .
The series is $1 - x^{2} + x^{4} - x^{6} + \dots$ , and integrating gives $C + x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7}$ .
Since $\arctan 0 = 0$ , the constant is zero.
Define the remainder of a Taylor series.
The difference $R_{n} (x) = f (x) - p_{n} (x)$ between the function and its $n$ th degree Taylor polynomial.
It is the error made by truncating the series.
The Lagrange error bound for the $n$ th degree Taylor polynomial about $x = a$ is given below, with one index missing:
$| R_{n} (x) | \leq \frac{M}{(n + 1)!} | x - a |^{______}$
The completed bound is $|R_{n} (x)| \leq \frac{M}{(n + 1)!} {|x - a|}^{n + 1}$ .
The same $n + 1$ appears in the factorial and in the power, and $M$ bounds the $(n + 1)$ th derivative.
What is $M$ in the Lagrange error bound?
An upper bound on $|f^{(n + 1)} (x)|$ across the interval concerned.
Any valid bound works, but the smallest one gives the tightest error bound.
You want a Lagrange bound at one particular value $x_{0}$ . Over what interval do you find $M$ ?
Over the interval between the centre $a$ and $x_{0}$ , not some larger one.
A narrower interval gives a smaller $M$ and so a tighter bound.
A third-degree Taylor polynomial about $x = 3$ approximates $f (2.9)$ , and $f^{(4)}$ is increasing with $f^{(4)} (3) = \frac{9}{5}$ . Bound the error.
Take $M = \frac{9}{5}$ , the largest value on $[2.9, 3]$ .
Then $|R_{3}| \leq \frac{9 / 5}{4!} {(0.1)}^{4} = 7.5 \times 10^{- 6}$ .
A function is given only as a power series, so its derivatives are unknown. Which error bound can you still use?
The alternating series error bound, provided the series really alternates and $x$ lies in the interval of convergence.
Lagrange needs a derivative, which is unavailable here.
True or False?
A power series with alternating signs in its formula alternates for every value of $x$ .
False.
The series for $e^{- x}$ alternates for positive $x$ but has every term positive when $x$ is negative.
Check the actual signs before using the alternating bound.
The first three terms of the series for $\sin x$ are used on $- \frac{1}{3} \leq x \leq \frac{1}{4}$ . Bound the error.
The next term has size $\frac{{|x|}^{7}}{5040}$ , largest at $x = - \frac{1}{3}$ , the point furthest from the centre.
That gives a bound of $\frac{1}{11022480}$ .

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