{"id":934,"date":"2025-06-20T17:23:31","date_gmt":"2025-06-20T17:23:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=934"},"modified":"2025-09-11T16:50:30","modified_gmt":"2025-09-11T16:50:30","slug":"operations-with-power-series-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/operations-with-power-series-learn-it-2\/","title":{"raw":"Operations with Power Series: Learn It 2","rendered":"Operations with Power Series: Learn It 2"},"content":{"raw":"

Multiplication of Power Series<\/h2>\r\n

You can also create new power series by multiplying two existing power series together. This multiplication technique gives us another powerful method for finding power series representations of functions.<\/p>\r\n

The process works similarly to multiplying polynomials. When you multiply two power series, you collect like terms to form the coefficients of the resulting series.<\/p>\r\n

For example, suppose you want to multiply:<\/p>\r\n\r\n

[latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots [\/latex]<\/div>\r\n

and<\/p>\r\n\r\n

[latex]\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}={d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\\cdots [\/latex].<\/div>\r\n

The product becomes:<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill \\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}\\right)\\left(\\displaystyle\\sum _{n=-0}^{\\infty }{d}_{n}{x}^{n}\\right)& =\\left({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\\cdots \\right)\\cdot \\left({d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\\cdots \\right)\\hfill \\\\ & ={c}_{0}{d}_{0}+\\left({c}_{1}{d}_{0}+{c}_{0}{d}_{1}\\right)x+\\left({c}_{2}{d}_{0}+{c}_{1}{d}_{1}+{c}_{0}{d}_{2}\\right){x}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n
Notice how each coefficient in the product collects all terms that multiply to give the same power of [latex]x[\/latex]. For [latex]x^n[\/latex], you get contributions from [latex]c_0 d_n + c_1 d_{n-1} + c_2 d_{n-2} + \\cdots + c_n d_0[\/latex].<\/section>
\r\n

theorem: multiplying power series<\/h3>\r\n

If [latex]\\sum_{n=0}^{\\infty} c_n x^n[\/latex] and [latex]\\sum_{n=0}^{\\infty} d_n x^n[\/latex] converge to functions [latex]f[\/latex] and [latex]g[\/latex] respectively on a common interval [latex]I[\/latex], then:<\/p>\r\nDefine the new coefficients as:\r\n

[latex]\\begin{array}{cc}\\hfill {e}_{n}& ={c}_{0}{d}_{n}+{c}_{1}{d}_{n - 1}+{c}_{2}{d}_{n - 2}+\\cdots +{c}_{n - 1}{d}_{1}+{c}_{n}{d}_{0}\\hfill \\\\ & ={\\displaystyle\\sum _{k=0}^{n}{c}_{k}{d}_{n-k}}.\\hfill \\end{array}[\/latex]<\/div>\r\n

\u00a0The product series is:<\/p>\r\n\r\n

[latex]\\left({\\displaystyle\\sum _{n=0}^{\\infty}}{c}_{n}{x}^{n}\\right)\\left(\\displaystyle\\sum _{n=0}^{\\infty }{d}_{n}{x}^{n}\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{e}_{n}{x}^{n}[\/latex]<\/div>\r\nThis new series converges to [latex]f(x) \\cdot g(x)[\/latex] on interval [latex]I[\/latex].\r\n\r\n \r\n

This resulting series is called the Cauchy product<\/strong>.<\/p>\r\n\r\n<\/section>\r\n

We omit the proof of this theorem, as it is beyond the level of this text and is typically covered in a more advanced course.<\/em><\/p>\r\n

We can use this multiplication technique to find power series for functions like: [latex]f(x) = \\frac{1}{(1-x)(1-x^2)}[\/latex] by multiplying the known power series for [latex]y = \\frac{1}{1-x}[\/latex] and [latex]y = \\frac{1}{1-x^2}[\/latex].<\/p>\r\n\r\n

\r\n
\r\n

Multiply the power series representation<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill \\frac{1}{1-x}& ={\\displaystyle\\sum _{n=0}^{\\infty}}{x}^{n}\\hfill \\\\ & =1+x+{x}^{2}+{x}^{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

for [latex]|x|<1[\/latex] with the power series representation<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill \\frac{1}{1-{x}^{2}}& ={\\displaystyle\\sum _{n=0}^{\\infty}}{\\left({x}^{2}\\right)}^{n}\\hfill \\\\ & =1+{x}^{2}+{x}^{4}+{x}^{6}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

for [latex]|x|<1[\/latex] to construct a power series for [latex]f\\left(x\\right)=\\frac{1}{\\left(1-x\\right)\\left(1-{x}^{2}\\right)}[\/latex] on the interval [latex]\\left(-1,1\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n

\r\n

We need to multiply<\/p>\r\n\r\n

[latex]\\left(1+x+{x}^{2}+{x}^{3}+\\cdots \\right)\\left(1+{x}^{2}+{x}^{4}+{x}^{6}+\\cdots \\right)[\/latex].<\/div>\r\n \r\n

Writing out the first several terms, we see that the product is given by<\/p>\r\n\r\n

[latex]\\begin{array}{c}\\left(1+{x}^{2}+{x}^{4}+{x}^{6}+\\cdots \\right)+\\left(x+{x}^{3}+{x}^{5}+{x}^{7}+\\cdots \\right)+\\left({x}^{2}+{x}^{4}+{x}^{6}+{x}^{8}+\\cdots \\right)+\\left({x}^{3}+{x}^{5}+{x}^{7}+{x}^{9}+\\cdots \\right)\\hfill \\\\ =1+x+\\left(1+1\\right){x}^{2}+\\left(1+1\\right){x}^{3}+\\left(1+1+1\\right){x}^{4}+\\left(1+1+1\\right){x}^{5}+\\cdots \\hfill \\\\ =1+x+2{x}^{2}+2{x}^{3}+3{x}^{4}+3{x}^{5}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

Since the series for [latex]y=\\frac{1}{1-x}[\/latex] and [latex]y=\\frac{1}{1-{x}^{2}}[\/latex] both converge on the interval [latex]\\left(-1,1\\right)[\/latex], the series for the product also converges on the interval [latex]\\left(-1,1\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Watch the following video to see the worked solution to the above example.