Wemazi

Topic 10.14 · BC

Recognize standard series

Builds on finding where a power series converges. 7 minutes.

A series with a pattern

The chalk curve shows \(\sin x\). The orange curve is a polynomial made of the first terms of the series \(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\). The symbol \(n!\) means \(1\cdot2\cdots n\), so \(3!=6\).

With one term, the orange curve is the line \(y=x\). Move the slider to add terms. The slider sets the degree, the highest power of \(x\). Each step of 2 adds one term, because the sine series has only odd powers.

Each new term makes the orange curve follow the chalk curve over a wider range of \(x\).

The series has a pattern: only odd powers of \(x\) appear, the signs alternate, and each power is divided by the factorial of that power. This pattern identifies the sine, and four other functions have their own patterns.

Match the pattern

Each row of the table is one standard series. The three columns name the features that tell the series apart: which powers of x appear, how the signs go, and what divides each power.

Press a button to light up a row in orange. The first terms of that series appear below the table.

When the three features of a series match one row, the series is that standard series.

Notation

  1. In the geometric series each term is \(x\) times the term before it, so every power of \(x\) appears and every coefficient is 1. It equals \(\frac{1}{1-x}\) only for \(|x|<1\).

    \[\begin{aligned} \dfrac{1}{1-x}&=1+x+x^2+x^3+\cdots \\ &=\sum_{n=0}^{\infty}x^n \end{aligned}\]

  2. In the series of a function \(f\), the coefficient of \(x^n\) is \(\frac{f^{(n)}(0)}{n!}\). Every derivative of \(e^x\) is 1 at 0, so every power of \(x\) appears with the coefficient \(\frac{1}{n!}\). The series equals \(e^x\) for all \(x\).

    \[\begin{aligned} e^{x}&=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots \\ &=\sum_{n=0}^{\infty}\dfrac{x^n}{n!} \end{aligned}\]

  3. The derivatives of the sine at 0 repeat 0, 1, 0, −1, so the even powers are missing and the signs alternate. Each power is divided by its factorial. The series equals \(\sin x\) for all \(x\).

    \[\begin{aligned} \sin x&=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots \\ &=\sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n+1}}{(2n+1)!} \end{aligned}\]

  4. The derivatives of the cosine at 0 repeat 1, 0, −1, 0, so only even powers appear and the signs alternate. The series equals \(\cos x\) for all \(x\).

    \[\begin{aligned} \cos x&=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots \\ &=\sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n}}{(2n)!} \end{aligned}\]

  5. The logarithm series has no constant term because \(\ln(1+0)=0\). Every power from \(x\) on appears, the signs alternate, and each power is divided by the power itself, not by a factorial. It equals \(\ln(1+x)\) for \(-1\lt x\le 1\).

    \[\begin{aligned} \ln(1+x)&=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots \\ &=\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}x^n}{n} \end{aligned}\]

    It comes from integrating \(\frac{1}{1+t}=1-t+t^2-\cdots\) term by term from 0 to \(x\).

When it fails

At \(x=1.5\), the function \(\frac{1}{1-x}\) has the value −2. The red segment shows the gap between it and the sum of the first terms of the series.

Every term of the series is positive, so these sums, 1, 2.5, 4.75 and so on, cannot reach −2. The geometric series equals its function only for \(|x|<1\), the teal band on the x-axis.

Example 1: a sum of numbers

Find the sum \(1-\dfrac{1}{2!}+\dfrac{1}{4!}-\dfrac{1}{6!}+\cdots\).

  1. The denominators are the factorials of the even numbers 0, 2, 4 and 6, and the signs alternate. Each numerator 1 can be read as a power of 1.

  2. Even powers, alternating signs and the factorial of the power match the cosine row of the table, so this is the cosine series with \(x=1\).

    \[\cos 1=1-\dfrac{1^2}{2!}+\dfrac{1^4}{4!}-\dfrac{1^6}{6!}+\cdots\]

  3. The sum is \(\cos 1\). As a check, the first three terms give 0.5417, and \(\cos 1\approx 0.5403\).

    \[\cy{\cos 1}\]

Example 2: which logarithm

Find the sum \(\dfrac12-\dfrac18+\dfrac1{24}-\dfrac1{64}+\cdots\).

  1. The terms are \((1/2)^n/n\) with alternating signs, starting at \(n=1\). The denominator is the power, so this is the logarithm series with \(x=\frac12\).

    \[\begin{aligned} &\dfrac12-\dfrac{(1/2)^2}{2}+\dfrac{(1/2)^3}{3} \\ &\quad-\dfrac{(1/2)^4}{4}+\cdots \end{aligned}\]

  2. A tempting wrong move: This is the logarithm series, so the sum is \(\ln\frac12\).

    \[\ln\dfrac12\]

    The sums of the first terms, 0.5, 0.375, 0.417 and 0.401, are positive, but \(\ln\frac12\) is about −0.693. The red dot lies far below the orange dot.

  3. The series is for \(\ln(1+x)\), not for \(\ln x\). With \(x=\frac12\) the sum is \(\ln\frac32\), which is about 0.405.

    \[\cy{\ln\dfrac32}\]

Practice

Find the sum \(1+2+\dfrac{4}{2!}+\dfrac{8}{3!}+\dfrac{16}{4!}+\cdots\).

Each power of 2 is divided by the factorial of the same number, and the signs do not alternate. Compare these features with the table.

Solution

  1. The terms are \(2^n/n!\) for \(n=0,1,2,\dots\), and all the signs are positive. This is the exponential series with \(x=2\).

    \[\sum_{n=0}^{\infty}\dfrac{2^n}{n!}\]

  2. The sum is \(e^2\). As a check, the first five terms give 7, and \(e^2\approx 7.389\).

    \[e^{2}\]

Questions

The chalk curve is a function, and the orange curve is a polynomial made of the first terms of its standard series. For questions 1 and 2, press the button for that question.

  1. Which standard series gives the orange curve in the figure for question 1?

    • \(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots\)
    • \(1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots\)
    • \(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots\)
    Answer

    \(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots\)

  2. Which standard series gives the orange curve in the figure for question 2?

    • \(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots\)
    • \(1+x+x^2+x^3+\cdots\)
    • \(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots\)
    Answer

    \(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots\)

  3. What is the sum \(\displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n}{n!}\)?

    • \(e^{-1}\)
    • \(e\)
    • \(\cos 1\)
    • \(\dfrac12\)
    Answer

    \(e^{-1}\)

Summary

You compare the powers of x, the signs and the denominators of a series with the five standard series.

The match names the function, because each standard series is built from the derivatives of one function at 0.

You use the match to name a function from its series or to find the sum of a series of numbers.

This is one of four skills in topic 10.14 of Which & Why: AP Calculus.