{"id":936,"date":"2025-06-20T17:23:37","date_gmt":"2025-06-20T17:23:37","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=936"},"modified":"2025-08-18T16:07:25","modified_gmt":"2025-08-18T16:07:25","slug":"operations-with-power-series-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/operations-with-power-series-learn-it-4\/","title":{"raw":"Operations with Power Series: Learn It 4","rendered":"Operations with Power Series: Learn It 4"},"content":{"raw":"

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

Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique?<\/p>\r\n

In other words, if you find a power series for function [latex]f[\/latex] at point [latex]a[\/latex] using one method, could a different technique give you a completely different power series for the same function at the same point?<\/p>\r\n

The answer is no\u2014power series representations are unique. This shouldn't be too surprising if you think of power series as \"infinite polynomials.\"<\/p>\r\n

Intuitively, if two power series are equal:<\/p>\r\n\r\n

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

for all values [latex]x[\/latex] in some open interval [latex]I[\/latex] around zero, then the coefficients must match: [latex]c_n = d_n[\/latex] for all [latex]n \\geq 0[\/latex].<\/p>\r\n

We now state this result formally in Uniqueness of Power Series.<\/p>\r\n\r\n

\r\n

theorem: uniqueness of power series<\/h3>\r\n

If two power series [latex]\\sum_{n=0}^{\\infty} c_n (x-a)^n[\/latex] and [latex]\\sum_{n=0}^{\\infty} d_n (x-a)^n[\/latex] are both convergent and satisfy:<\/p>\r\n

[latex]\\sum_{n=0}^{\\infty} c_n (x-a)^n = \\sum_{n=0}^{\\infty} d_n (x-a)^n[\/latex]<\/p>\r\n

for all [latex]x[\/latex] in an open interval containing [latex]a[\/latex], then the coefficients are identical:<\/p>\r\n

[latex]c_n = d_n[\/latex] for all [latex]n \\geq 0[\/latex]<\/p>\r\n\r\n<\/section>

\r\n

Proof<\/strong><\/p>\r\n\r\n\r\n

\r\n

Let<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)& ={c}_{0}+{c}_{1}\\left(x-a\\right)+{c}_{2}{\\left(x-a\\right)}^{2}+{c}_{3}{\\left(x-a\\right)}^{3}+\\cdots \\hfill \\\\ & ={d}_{0}+{d}_{1}\\left(x-a\\right)+{d}_{2}{\\left(x-a\\right)}^{2}+{d}_{3}{\\left(x-a\\right)}^{3}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

Then [latex]f\\left(a\\right)={c}_{0}={d}_{0}[\/latex]. By Term-by-Term Differentiation and Integration for Power Series, we can differentiate both series term-by-term. Therefore,<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill {f}^{\\prime }\\left(x\\right)& ={c}_{1}+2{c}_{2}\\left(x-a\\right)+3{c}_{3}{\\left(x-a\\right)}^{2}+\\cdots \\hfill \\\\ & ={d}_{1}+2{d}_{2}\\left(x-a\\right)+3{d}_{3}{\\left(x-a\\right)}^{2}+\\cdots ,\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

and thus, [latex]{f}^{\\prime }\\left(a\\right)={c}_{1}={d}_{1}[\/latex]. Similarly,<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill f^{\\prime\\prime}\\left(x\\right)& =2{c}_{2}+3\\cdot 2{c}_{3}\\left(x-a\\right)+\\cdots \\hfill \\\\ & =2{d}_{2}+3\\cdot 2{d}_{3}\\left(x-a\\right)+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

implies that [latex]f^{\\prime\\prime} \\left(a\\right)=2{c}_{2}=2{d}_{2}[\/latex], and therefore, [latex]{c}_{2}={d}_{2}[\/latex]. More generally, for any integer [latex]n\\ge 0,{f}^{\\left(n\\right)}\\left(a\\right)=n\\text{!}{c}_{n}=n\\text{!}{d}_{n}[\/latex], and consequently, [latex]{c}_{n}={d}_{n}[\/latex] for all [latex]n\\ge 0[\/latex].<\/p>\r\n

[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<\/section>\r\n

In this section, we've explored how to find power series representations using algebraic operations, differentiation, and integration. However, we're still limited in the types of functions we can represent this way.<\/p>\r\n

Next, we'll dramatically expand our toolkit by introducing Taylor series<\/strong>\u2014a method that allows us to find power series representations for many more functions.<\/p>","rendered":"

Uniqueness of Power Series<\/h2>\n

Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique?<\/p>\n

In other words, if you find a power series for function [latex]f[\/latex] at point [latex]a[\/latex] using one method, could a different technique give you a completely different power series for the same function at the same point?<\/p>\n

The answer is no\u2014power series representations are unique. This shouldn’t be too surprising if you think of power series as “infinite polynomials.”<\/p>\n

Intuitively, if two power series are equal:<\/p>\n

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

for all values [latex]x[\/latex] in some open interval [latex]I[\/latex] around zero, then the coefficients must match: [latex]c_n = d_n[\/latex] for all [latex]n \\geq 0[\/latex].<\/p>\n

We now state this result formally in Uniqueness of Power Series.<\/p>\n

\n

theorem: uniqueness of power series<\/h3>\n

If two power series [latex]\\sum_{n=0}^{\\infty} c_n (x-a)^n[\/latex] and [latex]\\sum_{n=0}^{\\infty} d_n (x-a)^n[\/latex] are both convergent and satisfy:<\/p>\n

[latex]\\sum_{n=0}^{\\infty} c_n (x-a)^n = \\sum_{n=0}^{\\infty} d_n (x-a)^n[\/latex]<\/p>\n

for all [latex]x[\/latex] in an open interval containing [latex]a[\/latex], then the coefficients are identical:<\/p>\n

[latex]c_n = d_n[\/latex] for all [latex]n \\geq 0[\/latex]<\/p>\n<\/section>\n

\n

Proof<\/strong><\/p>\n

\n

Let<\/p>\n

[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)& ={c}_{0}+{c}_{1}\\left(x-a\\right)+{c}_{2}{\\left(x-a\\right)}^{2}+{c}_{3}{\\left(x-a\\right)}^{3}+\\cdots \\hfill \\\\ & ={d}_{0}+{d}_{1}\\left(x-a\\right)+{d}_{2}{\\left(x-a\\right)}^{2}+{d}_{3}{\\left(x-a\\right)}^{3}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\n

 <\/p>\n

Then [latex]f\\left(a\\right)={c}_{0}={d}_{0}[\/latex]. By Term-by-Term Differentiation and Integration for Power Series, we can differentiate both series term-by-term. Therefore,<\/p>\n

[latex]\\begin{array}{cc}\\hfill {f}^{\\prime }\\left(x\\right)& ={c}_{1}+2{c}_{2}\\left(x-a\\right)+3{c}_{3}{\\left(x-a\\right)}^{2}+\\cdots \\hfill \\\\ & ={d}_{1}+2{d}_{2}\\left(x-a\\right)+3{d}_{3}{\\left(x-a\\right)}^{2}+\\cdots ,\\hfill \\end{array}[\/latex]<\/div>\n

 <\/p>\n

and thus, [latex]{f}^{\\prime }\\left(a\\right)={c}_{1}={d}_{1}[\/latex]. Similarly,<\/p>\n

[latex]\\begin{array}{cc}\\hfill f^{\\prime\\prime}\\left(x\\right)& =2{c}_{2}+3\\cdot 2{c}_{3}\\left(x-a\\right)+\\cdots \\hfill \\\\ & =2{d}_{2}+3\\cdot 2{d}_{3}\\left(x-a\\right)+\\cdots \\hfill \\end{array}[\/latex]<\/div>\n

 <\/p>\n

implies that [latex]f^{\\prime\\prime} \\left(a\\right)=2{c}_{2}=2{d}_{2}[\/latex], and therefore, [latex]{c}_{2}={d}_{2}[\/latex]. More generally, for any integer [latex]n\\ge 0,{f}^{\\left(n\\right)}\\left(a\\right)=n\\text{!}{c}_{n}=n\\text{!}{d}_{n}[\/latex], and consequently, [latex]{c}_{n}={d}_{n}[\/latex] for all [latex]n\\ge 0[\/latex].<\/p>\n

[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n

In this section, we’ve explored how to find power series representations using algebraic operations, differentiation, and integration. However, we’re still limited in the types of functions we can represent this way.<\/p>\n

Next, we’ll dramatically expand our toolkit by introducing Taylor series<\/strong>\u2014a method that allows us to find power series representations for many more functions.<\/p>\n","protected":false},"author":15,"menu_order":14,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":673,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/936"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/936\/revisions"}],"predecessor-version":[{"id":1913,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/936\/revisions\/1913"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/673"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/936\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=936"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=936"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=936"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=936"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}