Wemazi

Topic 10.14 · BC

Build a series by substitution or scaling

Builds on recognizing the standard series. 7 minutes.

Replace u by 3x

The chalk curve is \(\sin 3x\). To build its series, start with the sine series \(u-\frac{u^3}{3!}+\frac{u^5}{5!}-\cdots\) and replace every \(u\) by \(3x\).

The readouts show both series. Each power of \(x\) matches the power of \(u\), and its coefficient is multiplied by \(3^n\), so \(\frac{u^3}{3!}\) becomes \(\frac{27x^3}{3!}=\frac92x^3\).

Move the slider to add terms. The orange curve is the sum of the terms so far, and each new term makes it follow \(\sin 3x\) over a wider range of \(x\).

Every term changes together

The top line of the table is the \(n\)-th term \(c_nu^n\) of a known series. Each column replaces \(u\) by one expression: \(2x\), \(x^2\) or \(-x^3\).

Press a button to select a column, and its new term appears below the table. The factor row shows what multiplies \(c_n\), and the next row shows the new power of \(x\).

In every column, the \(n\)-th term of the new series comes from the \(n\)-th term of the old one and keeps \(c_n\). Replacing \(u\) by \(kx^m\) multiplies \(c_n\) by \(k^n\) and changes \(x^n\) to \(x^{mn}\).

Notation

  1. Choose a known series in \(u\) and write its \(n\)-th term. For the sine series, \(c_n\) is 0 for even \(n\) and \(\pm\frac{1}{n!}\) for odd \(n\).

    \[c_n\,u^n\]

  2. Choose the replacement \(u=kx^m\) and use it in every term. The coefficient and the power of \(x\) both change, because \((kx^m)^n=k^nx^{mn}\).

    \[c_n\,(kx^m)^n=c_n\,\cy{k^n}\,x^{\cy{mn}}\]

    For example, \((-2x)^n=(-2)^nx^n\) and \((x^2)^n=x^{2n}\). In the figure, \(k=3\) and \(m=1\).

  3. If the function does not match a standard series, rewrite it first. For example, factor 4 out of \(4-x\), so that \(\frac{1}{1-u}\) appears with \(u=\frac{x}{4}\).

    \[\dfrac{1}{4-x}=\dfrac14\cdot\dfrac{1}{1-\frac{x}{4}}\]

  4. To multiply the series by \(x^p\), add \(p\) to every power of \(x\) and keep the coefficients, because \(x^p\cdot x^n=x^{n+p}\). In the figure, \(p=1\).

    \[x^p\sum c_n x^n=\sum c_n x^{n+p}\]

  5. The new series equals its function only where \(u\) is in the range of the known series. The last skill of this topic shows how to find that range.

Example 1: a negative scale

Write the first four nonzero terms of the series for \(\ln(1-2x)\).

  1. Write \(\ln(1-2x)\) as \(\ln(1+u)\) with \(u=-2x\), and start with the series for \(\ln(1+u)\). The function is defined only for \(x\lt\frac12\), left of the dashed line.

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

  2. Replace \(u\) by \(-2x\) in every term.

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

  3. Find each power.

    \[\begin{aligned} (-2x)^2&=4x^2,\quad (-2x)^3=-8x^3, \\ (-2x)^4&=16x^4 \end{aligned}\]

  4. Divide each power by its denominator. Every term is negative, because the sign \((-1)^{n+1}\) from the series times the sign \((-1)^n\) of \((-2)^n\) is \(-1\).

    \[-2x-2x^2-\dfrac{8x^3}{3}-4x^4-\cdots\]

Example 2: a power of x

Write the first three nonzero terms of the series for \(\cos(x^2)\).

  1. Write \(\cos(x^2)\) as \(\cos u\) with \(u=x^2\), and start with the cosine series.

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

  2. A tempting wrong move: The powers of \(x\) become \(x^4\) and \(x^8\), so change the factorials to \(4!\) and \(8!\) to match.

    \[1-\dfrac{x^4}{4!}+\dfrac{x^8}{8!}\]

    At \(x=1\), \(\cos(x^2)\) is \(\cos 1\), about 0.540. The red curve is this polynomial, and its dot at about 0.958 lies far above the chalk dot.

  3. The factorial belongs to the power of \(u\), not to the power of \(x\). Replace \(u\) by \(x^2\) and keep the factorials. At \(x=1\), this polynomial gives about 0.542, close to \(\cos 1\).

    \[\begin{aligned} &1-\dfrac{(x^2)^2}{2!}+\dfrac{(x^2)^4}{4!} \\ &\quad=1-\dfrac{x^4}{2}+\dfrac{x^8}{24} \end{aligned}\]

Practice

Write the first four terms of the series for \(x\,e^{2x}\).

First write the series for \(e^{2x}\). Then multiply every term by \(x\).

Solution

  1. Replace \(u\) by \(2x\) in the exponential series.

    \[\begin{aligned} e^{2x}&=1+2x+\dfrac{(2x)^2}{2!} \\ &\quad+\dfrac{(2x)^3}{3!}+\cdots \\ &=1+2x+2x^2+\dfrac43x^3+\cdots \end{aligned}\]

  2. Multiply every term by \(x\). Each power of \(x\) goes up by 1, so the series starts at \(x\).

    \[\begin{aligned} x\,e^{2x}&=x+2x^2+2x^3 \\ &\quad+\dfrac43x^4+\cdots \end{aligned}\]

Questions

The chalk curve is a function, and the orange curve is the first terms of its series, which is built from a standard series. The readout below the figure shows these terms. Press a button to show the curves for question 1 or question 2.

  1. Which replacement of u was made in the cosine series?

    • \(u=2x\)
    • \(u=x^2\)
    • \(u=\dfrac{x}{2}\)
    Answer

    \(u=2x\)

  2. Which function has the series in the figure for question 2?

    • \(x\sin 3x\)
    • \(\sin(3x^2)\)
    • \(3\sin(x^2)\)
    Answer

    \(x\sin 3x\)

  3. What is the coefficient of \(x^4\) in the series for \(e^{-2x^2}\)?

    • 2
    • 4
    • −1
    • \(\dfrac12\)
    Answer

    2

Summary

You replace u by the new input in every term of a known series, so both the coefficient and the power of x change.

One replacement builds the whole series, because the n-th term of the new series comes from the n-th term of the old one.

You use this to build the series for sin 3x, ln(1 − 2x), cos x² or x eˣ without taking derivatives.

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