Wemazi

Topic 10.15 · BC

Combine series to represent a new function

Builds on integrating a power series term by term. 8 minutes.

Multiplying two series

The function \(h(x)=\dfrac{e^x}{1-x}\) is a product of two functions whose series you know. The series of \(e^x\) is \(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots\), and the series of \(\frac1{1-x}\) is \(1+x+x^2+x^3+\cdots\).

In the table, the columns are the terms of the first series and the rows are the terms of the second. Each cell is the product of its column and its row.

Press a button to choose a power of \(x\). The orange cells hold that power. For \(x^2\), they are \(1\cdot x^2\), \(x\cdot x\) and \(\frac{x^2}{2}\cdot1\), and their coefficients add to \(1+1+\frac12=\frac52\).

The four powers give the coefficients 1, 2, \(\frac52\) and \(\frac83\), so the series of \(h\) begins \(1+2x+\frac52x^2+\frac83x^3\).

Any two series

The two series are now \(\sum a_jx^j\) and \(\sum b_kx^k\). The cell in column \(j\) and row \(k\) holds \(a_jb_kx^{j+k}\).

Press a button to choose a power \(n\). The orange cells lie on one diagonal, and in every orange cell the two exponents add to \(n\).

The coefficient of \(x^n\) is the sum of the orange cells. A diagonal holds every product with the power \(n\), so no product is missed.

Notation

  1. To add two series, add the coefficients of equal powers. To scale a series, multiply every coefficient.

    \[\begin{aligned} &\sum a_nx^n+\sum b_nx^n \\ &\quad=\sum\left(a_n+b_n\right)x^n \end{aligned}\]

  2. To substitute, replace \(x\) in a known series by an expression. Then simplify each term.

    \[\begin{aligned} f(2x)&=\sum c_n(2x)^n \\ &=\sum c_n2^nx^n \end{aligned}\]

  3. To multiply, find the coefficient of each power \(x^n\). Add every product whose exponents add to \(n\). On the graph, the orange curve is this series of \(h\) through \(x^5\), and it follows the chalk curve \(h\) near 0.

    \[c_n=\cy{a_0b_n+a_1b_{n-1}+\cdots+a_nb_0}\]

  4. The new series converges where both original series converge, so its radius is at least the smaller of the two radii. For \(h\), the teal bar marks \(|x|<1\).

When it fails

The series of \(e^x\) converges for every \(x\). The series of \(\frac1{1-x}\) converges only for \(|x|<1\).

Move \(x\) to 1.5 and raise the degree. The orange curve moves further from the chalk curve, because the series of \(\frac1{1-x}\) diverges at \(x=1.5\), and so does the product series.

Example 1: a sum with a substitution

Find the first two nonzero terms of the series of \(\sin 2x-2\sin x\).

  1. The chalk curve is \(\sin 2x-2\sin x\). Start with the series of \(\sin x\).

    \[\sin x=x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\cdots\]

  2. To find \(\sin 2x\), substitute \(2x\) for \(x\). Each term \(x^n\) becomes \(2^nx^n\).

    \[\begin{aligned} \sin 2x&=2x-\dfrac{8x^3}{6}+\dfrac{32x^5}{120}-\cdots \\ &=2x-\dfrac{4x^3}{3}+\dfrac{4x^5}{15}-\cdots \end{aligned}\]

  3. Multiply the series of \(\sin x\) by 2.

    \[2\sin x=2x-\dfrac{x^3}{3}+\dfrac{x^5}{60}-\cdots\]

  4. Subtract the coefficients of equal powers. The \(x\) terms cancel, so the series starts at \(x^3\). The orange curve shows \(-x^3\) and then \(-x^3+\frac{x^5}{4}\), and both follow the chalk curve near 0.

    \[\begin{aligned} &\sin 2x-2\sin x \\ &\quad=\left(-\dfrac43+\dfrac13\right)x^3 \\ &\qquad+\left(\dfrac4{15}-\dfrac1{60}\right)x^5-\cdots \\ &\quad=-x^3+\dfrac{x^5}{4}-\cdots \end{aligned}\]

  5. As a check, at \(x=0.5\) these two terms give about \(-0.1172\), and the function is about \(-0.1174\). Both series converge for every \(x\), so the result does too.

Example 2: a product

Find the series of \(\ln(1+x)\cos x\) through the \(x^4\) term.

  1. The columns are the terms of \(\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\). The rows are the terms of \(\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots\). Each cell is a product.

  2. A tempting wrong move: Multiply the coefficients of equal powers.

    \[\dfrac{x^2}{4}-\dfrac{x^4}{96}+\cdots\]

    The product of the two \(x^2\) terms is the orange cell \(\frac{x^4}{4}\), not \(\frac{x^2}{4}\), because exponents add. This series also has no \(x\) term, but the orange cell \(x\cdot1\) gives one. At \(x=0.5\), the function is about 0.356, but this series gives about 0.062.

  3. Multiply each term of one series by each term of the other. Add the products that have the same power. The powers \(x\) and \(x^2\) have one product each, so their coefficients are 1 and \(-\frac12\).

  4. For \(x^3\), the products are \(\frac{x^3}{3}\cdot1\) and \(x\cdot\left(-\frac{x^2}{2}\right)\), and they add to \(-\frac16\). For \(x^4\), the products are \(-\frac{x^4}{4}\cdot1\) and \(\left(-\frac{x^2}{2}\right)\left(-\frac{x^2}{2}\right)\), and they add to 0.

    \[\begin{aligned} &\ln(1+x)\cos x \\ &\quad=x-\dfrac{x^2}{2}-\dfrac{x^3}{6} \\ &\qquad+0\cdot x^4+\cdots \end{aligned}\]

  5. As a check, at \(x=0.5\) the first three terms give about 0.354, and the function is about 0.356. The series converges for \(|x|<1\), the smaller of the two radii.

Practice

Find the first three nonzero terms of the series of \(g(x)=e^{-x^2}-\dfrac1{1+x^2}\).

On the graph, \(g\) is 0 at \(x=0\) and stays close to 0 near it. This means that the lowest terms of its series cancel.

Solution

  1. Substitute \(-x^2\) into the series of \(e^u\).

    \[\begin{aligned} e^{-x^2}&=1-x^2+\dfrac{x^4}{2} \\ &\quad-\dfrac{x^6}{6}+\dfrac{x^8}{24}-\cdots \end{aligned}\]

  2. Substitute \(-x^2\) into the geometric series of \(\dfrac1{1-u}\).

    \[\begin{aligned} \dfrac1{1+x^2}&=1-x^2+x^4 \\ &\quad-x^6+x^8-\cdots \end{aligned}\]

  3. Subtract the coefficients of equal powers. The terms 1 and \(-x^2\) cancel, so the orange curve stays flat until the degree reaches 4.

    \[\begin{aligned} g(x)&=-\dfrac{x^4}{2}+\dfrac{5x^6}{6} \\ &\quad-\dfrac{23x^8}{24}+\cdots \end{aligned}\]

  4. The first series converges for every \(x\). The second converges for \(|x|<1\), so the result converges at least for \(|x|<1\), which the teal bar marks.

Questions

The table multiplies the series of \(\cos x\) (the columns) by the series of \(\dfrac1{1+x}\) (the rows). Each cell is a product. Some cells are 0, because the series of \(\cos x\) has no odd powers.

  1. What is the coefficient of \(x^2\) in the series of \(\dfrac{\cos x}{1+x}\)?

    • \(\dfrac12\)
    • 1
    • \(-\dfrac12\)
    Answer

    \(\dfrac12\)

  2. What is the coefficient of \(x\) in the series of \(\dfrac{\cos x}{1+x}\)?

    • −1
    • 0
    • 1
    Answer

    −1

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

    • \(\dfrac{47}{24}\)
    • \(\dfrac{49}{24}\)
    • 2
    • \(\dfrac1{12}\)
    Answer

    \(\dfrac{47}{24}\)

Summary

You build a series for a new function by adding, substituting into or multiplying known series, and you collect the terms by power.

This works because exponents add when you multiply powers, so every product with the power \(n\) belongs to the coefficient of \(x^n\).

You use it when a function is built from functions with known series, and you keep only the terms that you need.

This is one of five skills in topic 10.15 of Which & Why: AP Calculus.