State where the representation is valid
Where the series equals the function
The chalk curve shows \(\ln(1+x)\). The orange curve is the polynomial of the first terms of its series, \(x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\). The teal band on the axis marks \(-1\lt x\le 1\).
Move the degree slider to add terms. Inside the band, each new term brings the orange curve closer to the chalk curve. To the right of \(x=1\), the orange curve leaves it.
Drag the orange point to \(x=2\), where the function value is \(\ln 3\approx 1.099\). As you add terms, the orange value is above and below \(\ln 3\) in turn, farther from it each time.
The series equals the function only inside the band.
A range for each series
Each button selects one standard series. Its teal band shows its range, the values of x where the series equals its function.
The exponential, sine and cosine series equal their functions for every \(x\). The geometric series equals \(\frac{1}{1-x}\) for \(-1\lt x<1\), and the logarithm series equals \(\ln(1+x)\) for \(-1\lt x\le 1\).
A filled ring means that the end belongs to the range, and an open ring means that it does not.
Notation
Each standard series equals its function only on a range of its input \(u\). The lower line shows the range \(-1\lt u<1\) of the geometric series.
\[\begin{aligned} e^u,\ \sin u,\ \cos u&:\ (-\infty,\infty) \\ \dfrac{1}{1-u}&:\ (-1,1) \\ \ln(1+u)&:\ (-1,1] \end{aligned}\]A new series is built with \(u=g(x)\). Here \(\frac{1}{1+2x}\) is the geometric series with \(u=-2x\). Put \(g(x)\) in place of \(u\) in the range, and the range of \(u\) becomes a condition on \(x\).
\[-1<-2x<1\]Divide every part by \(-2\), which reverses both inequalities. The band for \(x\) is half as wide as the band for \(u\). If a standard series holds for every \(u\), the new series holds for every \(x\).
\[-\dfrac12\lt x<\dfrac12\]At each end of the range of \(x\), the new series is the standard series at an end of the range of \(u\). This means that an end of \(x\) belongs to the range only if its end of \(u\) does. Here both ends of \(u\) are open, so \(x=\pm\frac12\) are left out.
Write the series together with its range of x.
\[\begin{aligned} \dfrac{1}{1+2x}&=\sum_{n=0}^{\infty}(-2x)^n, \\ |x|&<\dfrac12 \end{aligned}\]
When it fails
Suppose you keep the range \(|x|<1\) of the geometric series for \(\frac{1}{1+2x}\).
At \(x=0.8\) the value of \(u\) is \(-1.6\). The partial sums are 1, −0.6, 1.96 and −2.14, and each one is farther from the function value, about 0.385, than the one before. The correct range is \(|2x|<1\), which is \(|x|<\frac12\).
Example 1: a series about 3
The series \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\left(\frac{x-3}{3}\right)^n\) equals \(\ln\dfrac{x}{3}\) for which \(x\)?
This is the logarithm series with \(u=\frac{x-3}{3}\). Then \(1+u=\frac{x}{3}\), so the sum is \(\ln\frac{x}{3}\).
\[\ln(1+u)=u-\dfrac{u^2}{2}+\cdots\]The logarithm series equals \(\ln(1+u)\) for \(-1\lt u\le 1\). Put \(u=\frac{x-3}{3}\) in this range.
\[-1<\dfrac{x-3}{3}\le 1\]Multiply every part by 3. Then add 3 to every part.
\[-3\lt x-3\le 3\ \Rightarrow\ 0\lt x\le 6\]At \(x=6\) the value of \(u\) is 1, where the series converges, so 6 is in the range. At \(x=0\) the value of \(u\) is \(-1\), which the range leaves out.
\[(0,\ 6]\]
Example 2: a square
For which \(x\) does the series for \(\dfrac{1}{9+x^2}\) equal the function?
Factor out 9 so that a geometric form appears. The function is \(\frac19\) times \(\frac{1}{1-u}\), with \(u=-\frac{x^2}{9}\).
\[\dfrac{1}{9+x^2}=\dfrac19\cdot\dfrac{1}{1-\left(-\frac{x^2}{9}\right)}\]The geometric series equals \(\frac{1}{1-u}\) for \(|u|<1\). Put \(u=-\frac{x^2}{9}\) in this range.
\[\left|\dfrac{x^2}{9}\right|<1\iff x^2<9\]A tempting wrong move: From \(x^2<9\), the range is \(x<3\).
\[x<3\]The range \(x<3\) includes \(x=-4\), where \(u=-\frac{16}{9}\) is beyond \(-1\). The series diverges there, so the orange curve is far from the chalk curve.
The condition \(x^2<9\) means that \(x\) is less than 3 away from 0 on either side.
\[-3\lt x<3\]
Practice
For which \(x\) does the series for \(\ln(1+x^2)\) equal the function?
The logarithm series has \(u=x^2\). Ask which \(x\) give \(u=1\), and whether \(x^2\) can reach \(-1\).
Solution
The series is the logarithm series with \(u=x^2\). It equals \(\ln(1+u)\) for \(-1\lt u\le 1\).
\[-1\lt x^2\le 1\]The square \(x^2\) is never negative, so \(x^2>-1\) holds for every \(x\). Only \(x^2\le 1\) remains.
\[x^2\le 1\iff -1\le x\le 1\]At \(x=1\) and at \(x=-1\) the value of \(u\) is 1, where the series converges, so both ends are in the range.
\[[-1,\ 1]\]
Questions
The chalk curve is \(\frac{1}{1-2x}\). The orange curve is the polynomial of eight terms of the geometric series with \(u=2x\).
For which x does the series equal the function?
- \(-\dfrac12\lt x<\dfrac12\)
- \(-1\lt x<1\)
- \(x<\dfrac12\)
- \(-\dfrac12\le x<\dfrac12\)
Answer
\(-\dfrac12\lt x<\dfrac12\)
At \(x=-0.8\) the orange curve is far from the chalk curve. Why?
- The point \(x=-0.8\) is outside the range, so the series diverges.
- The function is not defined at \(x=-0.8\).
- The polynomial has too few terms.
Answer
The point \(x=-0.8\) is outside the range, so the series diverges.
For which \(x\) does the series for \(\ln(1-4x)\) equal the function?
- \(-\dfrac14\le x<\dfrac14\)
- \(-\dfrac14\lt x\le\dfrac14\)
- \(-1\lt x\le 1\)
- \(-\dfrac14\lt x<\dfrac14\)
Answer
\(-\dfrac14\le x<\dfrac14\)
Summary
To find the range of a new series, you substitute \(g(x)\) for \(u\) in the range of the standard series and solve for \(x\).
This works because the new series is the standard series at \(u=g(x)\), so it equals its function only where \(u\) lies in that range.
You state this range each time you build a series by a scale, a shift or a substitution.
This is one of four skills in topic 10.14 of Which & Why: AP Calculus.