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infinite sequences and series

Power series represent functions as infinite polynomials, with convergence determining where the representation is valid.

Wed May 06

Infinite Sequences and Series

Objective

  1. Radius and Interval of Convergence
  2. Representations
  3. Differentiation and Integration of Power Series
  4. Taylor Series and Maclaurin Series

1. Power Series

The basic form of a power series is:

∑n=0∞cn(x−a)n\sum_{n=0}^{\infty} c_n(x-a)^n

where:

  • cnc_n are the coefficients;
  • xx is the variable;
  • aa is the center of expansion;
  • (x−a)n(x-a)^n represents powers centered at aa.

If a=0a=0, the power series becomes:

∑n=0∞cnxn\sum_{n=0}^{\infty} c_nx^n

This is called a power series centered at 00.

An ordinary polynomial has finitely many terms, for example:

1+x+x2+x31+x+x^2+x^3

whereas a power series has infinitely many terms:

1+x+x2+x3+⋯1+x+x^2+x^3+\cdots

Many complicated functions can be written in the form of a power series.

For example:

11−x=1+x+x2+x3+⋯\frac{1}{1-x}=1+x+x^2+x^3+\cdots

That is:

11−x=∑n=0∞xn\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n

2. Radius of Convergence

2.1 What is convergence?

Whether an infinite series is meaningful depends on whether it converges.

For example:

1+12+14+18+⋯1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots

The sum of this series gets closer and closer to 22, so it converges.

However, a series such as:

1+1+1+1+⋯1+1+1+1+\cdots

grows without bound, so it diverges.

2.2 Why do we need to discuss convergence for power series?

A power series contains a variable xx. Therefore, for different values of xx, the same power series may sometimes converge and sometimes diverge.

For example:

∑n=0∞xn\sum_{n=0}^{\infty}x^n

If x=12x=\frac{1}{2}:

1+12+14+18+⋯1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots

it converges.

If x=2x=2:

1+2+4+8+⋯1+2+4+8+\cdots

it diverges.

So we need to study:

For which values of xx does this power series converge?

2.3 Radius of convergence RR

For the power series:

∑n=0∞cn(x−a)n\sum_{n=0}^{\infty}c_n(x-a)^n

there usually exists a number RR such that:

  • when ∣x−a∣<R|x-a|<R, the series converges;
  • when ∣x−a∣>R|x-a|>R, the series diverges;
  • aa is the center of convergence.

image

This number RR is called the radius of convergence.

Three cases for RR

Case 1: The series converges only at the center

If R=0R=0, this means the series converges only at:

x=ax=a

Case 2: The series converges for all real numbers

If R=∞R=\infty, this means the series converges for all xx.

That is, the interval of convergence is:

(−∞,∞)(-\infty,\infty)

Case 3: The series converges near the center

If 0<R<∞0<R<\infty, then the series converges when:

∣x−a∣<R|x-a|<R

This inequality can be rewritten as:

a−R<x<a+Ra-R<x<a+R

So the basic interval of convergence is:

(a−R,a+R)(a-R,a+R)

3. Interval of Convergence

3.1 What is the interval of convergence?

The interval of convergence is the interval consisting of all values of xx for which the power series converges.

If the radius of convergence is RR and the center is aa, then the interior of the interval must be:

(a−R,a+R)(a-R,a+R)

However, the endpoints x=a−Rx=a-R and x=a+Rx=a+R must be checked separately.

Why do the endpoints need to be checked separately?

Because the radius of convergence only tells us that:

  • the series definitely converges inside the interval;
  • the series definitely diverges outside the interval.

However, at the endpoints, where ∣x−a∣=R|x-a|=R, the series may converge or diverge. Therefore, the endpoints must be substituted back into the original series and checked separately.

Thus, if the center is aa and the radius is RR, the interval of convergence may be one of the following four possibilities:

(a−R,a+R)(a-R,a+R) (a−R,a+R](a-R,a+R] [a−R,a+R)[a-R,a+R) [a−R,a+R][a-R,a+R]

In other words:

  • the left endpoint may or may not be included;
  • the right endpoint may or may not be included.

4. Geometric Series

The most important starting point is:

11−x=1+x+x2+x3+⋯\frac{1}{1-x}=1+x+x^2+x^3+\cdots

It can also be written as:

11−x=∑n=0∞xn\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n

This is valid when:

∣x∣<1|x|<1

Why is this true?

The finite partial sum is:

sn=1+x+x2+⋯+xns_n=1+x+x^2+\cdots+x^n

This is the sum of a finite geometric sequence:

sn=1−xn+11−xs_n=\frac{1-x^{n+1}}{1-x}

When ∣x∣<1|x|<1:

xn+1→0x^{n+1}\to 0

So:

sn→11−xs_n\to \frac{1}{1-x}

Therefore:

1+x+x2+⋯=11−x1+x+x^2+\cdots=\frac{1}{1-x}

5. Representing Functions Using Geometric Series

The core idea of this section is:

Rewrite the target function into the form 11−r\frac{1}{1-r}, and then apply the geometric series formula.

1. Example: Represent 11+x2\frac{1}{1+x^2}

We know:

11−x=∑n=0∞xn\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n

Now we want to represent:

11+x2\frac{1}{1+x^2}

Notice that:

1+x2=1−(−x2)1+x^2=1-(-x^2)

So:

11+x2=11−(−x2)\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}

Let:

r=−x2r=-x^2

Then:

11+x2=∑n=0∞(−x2)n\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-x^2)^n

Expanding:

11+x2=1−x2+x4−x6+x8−⋯\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8-\cdots

It can also be written as:

11+x2=∑n=0∞(−1)nx2n\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^nx^{2n}

The convergence condition is:

∣−x2∣<1|-x^2|<1

That is:

x2<1x^2<1

So:

∣x∣<1|x|<1

The interval of convergence is:

(−1,1)(-1,1)

2. Example: Represent 1x+2\frac{1}{x+2}

We want to rewrite it into the form 11−r\frac{1}{1-r}.

First factor out 22 from the denominator:

x+2=2(1+x2)x+2=2\left(1+\frac{x}{2}\right)

So:

1x+2=12(1+x2)\frac{1}{x+2}=\frac{1}{2\left(1+\frac{x}{2}\right)}

That is:

1x+2=12⋅11+x2\frac{1}{x+2}=\frac{1}{2}\cdot\frac{1}{1+\frac{x}{2}}

Rewrite the plus sign as a minus sign:

1+x2=1−(−x2)1+\frac{x}{2}=1-\left(-\frac{x}{2}\right)

So:

1x+2=12⋅11−(−x2)\frac{1}{x+2}=\frac{1}{2}\cdot \frac{1}{1-\left(-\frac{x}{2}\right)}

Apply the geometric series formula:

11−r=∑n=0∞rn\frac{1}{1-r}=\sum_{n=0}^{\infty}r^n

Let:

r=−x2r=-\frac{x}{2}

Then:

1x+2=12∑n=0∞(−x2)n\frac{1}{x+2}=\frac{1}{2}\sum_{n=0}^{\infty}\left(-\frac{x}{2}\right)^n

Simplifying:

1x+2=∑n=0∞(−1)n2n+1xn\frac{1}{x+2}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}x^n

The convergence condition is:

∣−x2∣<1\left|-\frac{x}{2}\right|<1

So:

∣x∣2<1\frac{|x|}{2}<1

That is:

∣x∣<2|x|<2

The interval of convergence is:

(−2,2)(-2,2)

3. Example: Represent x3x+2\frac{x^3}{x+2}

Since we already know that:

1x+2=∑n=0∞(−1)n2n+1xn\frac{1}{x+2}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}x^n

we have:

x3x+2=x3⋅1x+2\frac{x^3}{x+2}=x^3\cdot \frac{1}{x+2}

Thus:

x3x+2=x3∑n=0∞(−1)n2n+1xn\frac{x^3}{x+2}=x^3\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}x^n

Multiplying by x3x^3:

x3x+2=∑n=0∞(−1)n2n+1xn+3\frac{x^3}{x+2}=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}x^{n+3}

Expanding the first few terms:

x3x+2=12x3−14x4+18x5−116x6+⋯\frac{x^3}{x+2} =\frac{1}{2}x^3-\frac{1}{4}x^4+\frac{1}{8}x^5-\frac{1}{16}x^6+\cdots

The interval of convergence remains unchanged:

(−2,2)(-2,2)

because multiplying by x3x^3 does not change the convergence condition of the geometric series itself.

6. Term-by-Term Differentiation of Power Series

6.1 Core idea

If a function can be written as a power series:

f(x)=∑n=0∞cn(x−a)nf(x)=\sum_{n=0}^{\infty}c_n(x-a)^n

then inside the interval of convergence, we can differentiate it term by term just like an ordinary polynomial:

f′(x)=∑n=1∞ncn(x−a)n−1f'(x)=\sum_{n=1}^{\infty}n c_n(x-a)^{n-1}

Notice that after differentiation, the sum starts from n=1n=1, because the constant term corresponding to n=0n=0 differentiates to 00.

6.2 Why can we differentiate term by term?

Finite polynomials can certainly be differentiated term by term:

ddx(1+x+x2)=0+1+2x\frac{d}{dx}(1+x+x^2)=0+1+2x

A power series is like an infinite polynomial.

The theorem tells us that inside the radius of convergence, a power series can also be differentiated term by term.

6.3 Does the radius of convergence change?

Important

After term-by-term differentiation of a power series, the radius of convergence RR remains unchanged, but the behavior at the endpoints may change.

In other words:

  • the original radius is still RR;
  • but whether the endpoints converge needs to be checked again.

Example: Represent 1(1−x)2\frac{1}{(1-x)^2}

Start from the basic formula:

11−x=1+x+x2+x3+⋯\frac{1}{1-x}=1+x+x^2+x^3+\cdots

That is:

11−x=∑n=0∞xn\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n

Differentiate both sides.

Left-hand side:

ddx(11−x)=1(1−x)2\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{1}{(1-x)^2}

Right-hand side:

ddx(1+x+x2+x3+⋯ )=1+2x+3x2+4x3+⋯\frac{d}{dx}(1+x+x^2+x^3+\cdots)=1+2x+3x^2+4x^3+\cdots

Therefore:

1(1−x)2=1+2x+3x2+4x3+⋯\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+\cdots

This can be written as:

1(1−x)2=∑n=1∞nxn−1\frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}n x^{n-1}

If we want to write it starting from n=0n=0, we can re-index:

1(1−x)2=∑n=0∞(n+1)xn\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}(n+1)x^n

The radius of convergence is still:

R=1R=1

7. Term-by-Term Integration of Power Series

7.1 Core formula

If:

f(x)=∑n=0∞cn(x−a)nf(x)=\sum_{n=0}^{\infty}c_n(x-a)^n

then:

∫f(x) dx=C+∑n=0∞cn(x−a)n+1n+1\int f(x)\,dx=C+\sum_{n=0}^{\infty}c_n\frac{(x-a)^{n+1}}{n+1}

That is, each term can be integrated separately.

Just like differentiation:

After term-by-term integration, the radius of convergence RR remains unchanged. However, the endpoints still need to be checked again.

Example: Represent ln⁡(1−x)\ln(1-x)

We know:

11−x=1+x+x2+x3+⋯\frac{1}{1-x}=1+x+x^2+x^3+\cdots

and:

ddxln⁡(1−x)=−11−x\frac{d}{dx}\ln(1-x)=-\frac{1}{1-x}

So:

−ln⁡(1−x)=∫11−x dx-\ln(1-x)=\int \frac{1}{1-x}\,dx

Use the series to integrate the right-hand side:

−ln⁡(1−x)=∫(1+x+x2+x3+⋯ ) dx-\ln(1-x)=\int (1+x+x^2+x^3+\cdots)\,dx

Integrating term by term:

−ln⁡(1−x)=x+x22+x33+x44+⋯+C-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots+C

When x=0x=0:

−ln⁡(1−0)=0-\ln(1-0)=0

The right-hand side is:

0+C0+C

So:

C=0C=0

Therefore:

−ln⁡(1−x)=x+x22+x33+x44+⋯-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots

Multiplying both sides by −1-1:

ln⁡(1−x)=−x−x22−x33−x44−⋯\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\cdots

In summation notation:

ln⁡(1−x)=−∑n=1∞xnn\ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}

Valid when:

∣x∣<1|x|<1

Radius of convergence:

R=1R=1

Example: Represent tan⁡−1x\tan^{-1}x

We know:

ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}

and we have already obtained:

11+x2=1−x2+x4−x6+⋯\frac{1}{1+x^2}=1-x^2+x^4-x^6+\cdots

So we integrate:

tan⁡−1x=∫11+x2 dx\tan^{-1}x=\int \frac{1}{1+x^2}\,dx

Substitute in the series:

tan⁡−1x=∫(1−x2+x4−x6+⋯ ) dx\tan^{-1}x=\int (1-x^2+x^4-x^6+\cdots)\,dx

Integrating term by term:

tan⁡−1x=x−x33+x55−x77+⋯\tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots

In summation notation:

tan⁡−1x=∑n=0∞(−1)nx2n+12n+1\tan^{-1}x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}

Valid when:

∣x∣<1|x|<1

Radius of convergence:

R=1R=1

8. Using Power Series to Approximate Integrals

8.1 Why use power series to approximate integrals?

Some functions do not have simple elementary antiderivatives.

For example:

∫11+x7 dx\int \frac{1}{1+x^7}\,dx

This integral is relatively difficult to compute directly.

However, we can write the integrand as a power series and then integrate term by term.

Example: Compute ∫11+x7 dx\int \frac{1}{1+x^7}\,dx

Rewrite:

11+x7\frac{1}{1+x^7}

as:

11−(−x7)\frac{1}{1-(-x^7)}

Applying the geometric series:

11+x7=∑n=0∞(−x7)n\frac{1}{1+x^7}=\sum_{n=0}^{\infty}(-x^7)^n

That is:

11+x7=1−x7+x14−x21+⋯\frac{1}{1+x^7}=1-x^7+x^{14}-x^{21}+\cdots

Integrate term by term:

∫11+x7 dx=∫(1−x7+x14−x21+⋯ ) dx\int \frac{1}{1+x^7}\,dx=\int(1-x^7+x^{14}-x^{21}+\cdots)\,dx

We get:

∫11+x7 dx=C+x−x88+x1515−x2222+⋯\int \frac{1}{1+x^7}\,dx=C+x-\frac{x^8}{8}+\frac{x^{15}}{15}-\frac{x^{22}}{22}+\cdots

In summation notation:

∫11+x7 dx=C+∑n=0∞(−1)nx7n+17n+1\int \frac{1}{1+x^7}\,dx=C+\sum_{n=0}^{\infty}(-1)^n\frac{x^{7n+1}}{7n+1}

The convergence condition is:

∣−x7∣<1|-x^7|<1

That is:

∣x∣<1|x|<1

If we want to compute:

∫01/211+x7 dx\int_0^{1/2}\frac{1}{1+x^7}\,dx

using the antiderivative above and taking C=0C=0:

∫01/211+x7 dx=[x−x88+x1515−x2222+⋯ ]01/2\int_0^{1/2}\frac{1}{1+x^7}\,dx=\left[ x-\frac{x^8}{8}+\frac{x^{15}}{15}-\frac{x^{22}}{22}+\cdots \right]_0^{1/2}

Substituting x=12x=\frac{1}{2}:

=12−18⋅28+115⋅215−122⋅222+⋯=\frac{1}{2}-\frac{1}{8\cdot 2^8}+\frac{1}{15\cdot 2^{15}} -\frac{1}{22\cdot 2^{22}}+\cdots

This is an alternating series.

If we want to approximate it within a certain error bound, we can use the Alternating Series Estimation Theorem:

For an alternating series, if the absolute values of the terms decrease and tend to 00, then the truncation error is less than the absolute value of the next term.

9. Taylor Series

The core question of Taylor Series

Previously, we obtained power series representations of functions by transforming geometric series.

Now we want to ask a more general question:

Can any function f(x)f(x) be written as a power series?

That is, can it be written as:

f(x)=c0+c1(x−a)+c2(x−a)2+c3(x−a)3+⋯f(x)=c_0+c_1(x-a)+c_2(x-a)^2+c_3(x-a)^3+\cdots

If it can, how should we find the coefficients c0,c1,c2,…c_0,c_1,c_2,\ldots?

Suppose:

f(x)=∑n=0∞cn(x−a)nf(x)=\sum_{n=0}^{\infty}c_n(x-a)^n

That is:

f(x)=c0+c1(x−a)+c2(x−a)2+c3(x−a)3+⋯f(x)=c_0+c_1(x-a)+c_2(x-a)^2+c_3(x-a)^3+\cdots

Finding c0c_0

Let:

x=ax=a

Then:

f(a)=c0+c1(a−a)+c2(a−a)2+⋯f(a)=c_0+c_1(a-a)+c_2(a-a)^2+\cdots

Since:

a−a=0a-a=0

all terms except the first one become 00.

Therefore:

f(a)=c0f(a)=c_0

So:

c0=f(a)c_0=f(a)

Finding c1c_1

Differentiate the original expression:

f′(x)=c1+2c2(x−a)+3c3(x−a)2+4c4(x−a)3+⋯f'(x)=c_1+2c_2(x-a)+3c_3(x-a)^2+4c_4(x-a)^3+\cdots

Let:

x=ax=a

Then:

f′(a)=c1f'(a)=c_1

So:

c1=f′(a)c_1=f'(a)

Finding c2c_2

Differentiate again:

f′′(x)=2c2+3⋅2c3(x−a)+4⋅3c4(x−a)2+⋯f''(x)=2c_2+3\cdot 2c_3(x-a)+4\cdot 3c_4(x-a)^2+\cdots

Let x=ax=a:

f′′(a)=2c2f''(a)=2c_2

So:

c2=f′′(a)2c_2=\frac{f''(a)}{2}

That is:

c2=f′′(a)2!c_2=\frac{f''(a)}{2!}

Finding c3c_3

Differentiate once more:

f′′′(a)=3⋅2⋅1c3f'''(a)=3\cdot 2\cdot 1 c_3

So:

f′′′(a)=3!c3f'''(a)=3!c_3

Therefore:

c3=f′′′(a)3!c_3=\frac{f'''(a)}{3!}

Continuing in this way, we obtain:

f(n)(a)=n!cnf^{(n)}(a)=n!c_n

So:

cn=f(n)(a)n!c_n=\frac{f^{(n)}(a)}{n!}

This is where the coefficients of the Taylor Series come from.

Taylor Series Formula

Substitute cnc_n back into the power series:

f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

This is called the Taylor Series of ff at aa.

Written out, it is:

f(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+⋯f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots

10. Maclaurin Series

10.1 What is a Maclaurin Series?

A Maclaurin Series is a special case of a Taylor Series.

When the center of expansion is:

a=0a=0

the Taylor Series becomes the Maclaurin Series.

10.2 Maclaurin Series Formula

Taylor Series:

f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n

Let a=0a=0:

f(x)=∑n=0∞f(n)(0)n!xnf(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n

Expanding:

f(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+⋯f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots

This is the Maclaurin Series.

Difference Between Taylor Series and Maclaurin Series

NameCenter of ExpansionFormula
Taylor SeriesAny point aa∑n=0∞f(n)(a)n!(x−a)n\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n
Maclaurin Series00∑n=0∞f(n)(0)n!xn\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n
Tip

Taylor Series is an expansion around any point aa.
Maclaurin Series is an expansion around 00.

Originally published at mathnest.top by Penny Zhao.

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