{"id":2796,"date":"2026-09-01T05:00:58","date_gmt":"2026-09-01T09:00:58","guid":{"rendered":"https:\/\/mathvoices.ams.org\/featurecolumn\/?p=2796"},"modified":"2026-08-26T09:28:40","modified_gmt":"2026-08-26T13:28:40","slug":"a-golden-formal-power-series","status":"publish","type":"post","link":"https:\/\/mathvoices.ams.org\/featurecolumn\/2026\/09\/01\/a-golden-formal-power-series\/","title":{"rendered":"A Golden Formal Power Series"},"content":{"rendered":"

Luckily, such a formula exists, and there are lots of ways to find it. In this post, we\u2019ll find it using power series. Read on, brave bunny farmer…<\/em><\/span><\/p>\n

A Golden Formal Power Series<\/h1>\n

Courtney Gibbons<\/strong>
\nHamilton College<\/strong><\/p>\n

Some days, I feel a lot like the angel in Dürer’s Melencolia I looks: grumpy, tired, and annoyed. <\/p>\n

\"A<\/a>
A grumpy angel stares at a truncated triangular trapezohedron in Dürer’s Melencolia I. The cross ratio of Dürer’s solid is the Golden Ratio! Who doesn’t love a golden polyhedron?<\/figcaption><\/figure>\n

Unlike the melancholy angel, for me a little mathematical treat is a nice pick-me-up that takes my mind off my troubles and recharge my batteries.<\/p>\n

Sequences and series are important for lots of applied mathematics, but this column is my<\/em> favorite application of these things you learn in your introductory undergraduate calculus sequence: finding a closed form for the $n$-th Fibonacci number! Along the way, we see our friend the Golden Ratio, $\\Phi = \\frac{1+\\sqrt{5}}{2}$ (and its counterpart, the Other Ratio, $\\Psi = \\frac{1-\\sqrt{5}}{2}$) appear as we build a power series, apply convergence tests, use the rearrangement theorem for absolutely convergent series, dust off partial fraction decomposition and the quadratic formula, and do some very careful scratch work.<\/p>\n

The Fibonacci Sequence<\/h2>\n

and a formal power series to hold it all together<\/h3>\n

The ubiquitous Fibonacci sequence, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55…! Maybe you’ve encountered it out in the wild? Around 1200 C.E., Leonardo Pisano Fibonacci introduced it in a thought experiment about the breeding habits of rabbits.<\/p>\n

\"A
An oldie but a goodie from my days as semi-professional math doodler “Brown Sharpie”<\/figcaption><\/figure>\n

The sequence is defined by a recursion relation that tracks the number of pairs of (immortal) rabbits every month under the assumptions that breeding pairs produce two offspring every month, and the offspring need a month of adulthood before they start breeding new bunnies. Mathematically speaking, if $F_n$ is the number of pairs of rabbits in the $n$th month, we have
\n$$F_0 = 1, F_1 = 1, F_n = F_{n-1} + F_{n-2},$$
\nso that $$F_2 = 2, F_3 = 3, F_4 = 5, F_5 = 8, F_6 = 13, F_7 = 21, \\ldots.$$<\/p>\n

Let\u2019s say for some reason, you set your heart on knowing the value of $F_{413}$ (the number of pairs of rabbits in the 413th month of this thought experiment). I hope you have some time on your hands if you\u2019re planning to add all the way up to that because the recursive formula means you need to know $F_{412}$ and $F_{411}$, and for those… well, you’re reading the AMS Feature Column. You get the idea. Instead, wouldn\u2019t it be nice if we had a simple formula that we could use (i.e., a formula that was not<\/em> recursive) to figure out the $n$th Fibonacci number?<\/p>\n

Luckily, such a formula exists, and there are lots of ways to find it. In this post, we\u2019ll find it using power series. Read on, brave bunny farmer!<\/p>\n

\"A<\/a>
Fibonacci Melodie, a short melody by Markus Bautsch (and my favorite ringtone!). Click the image to go to the MP3.<\/figcaption><\/figure>\n

Building a Power Series<\/h3>\n

and ensuring it has a nonzero radius of convergence<\/h4>\n

First thing we\u2019re going to do is build a power series:
\n$$\\begin{array}{r c l} F(x) &=& \\sum_{n=0}^\\infty F_n x^{n} \\\\
\n&=& 1+ 1x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + 13x^6 + 21x^7 + \\cdots.
\n\\end{array}$$<\/p>\n

We need to convince ourselves this power series converges for some nonzero values of $x$, which we can do by comparison to the geometric power series $\\sum_{n=0}^\\infty 2^n x^2$ (which converges for $|x| < \\frac{1}{2}$) and a little induction.<\/p>\n

Here are a few base cases to get us warmed up before using the induction hypothesis that $F_k < 2^k$ for all $1 \\le k < n$.
\n$$\\begin{array}{r c l} F_0 &= 1 =& 2^0 \\\\
\nF_1 &= 1 \\lt & 2 = 2^1 \\\\
\nF_2 &= 2 \\lt & 4 = 2^2 \\\\
\nF_3 = F_2 + F_1 & \\lt 2^1 + 2^2 \\lt & 2^3 \\\\
\n& \\vdots & \\\\
\nF_n = F_{n-1} + F_{n-2} & \\lt 2^{n-1} + 2^{n-2} & \\lt & 2^n.
\n\\end{array}$$<\/p>\n

Since we have a series where $0 < F_n x^n < 2^n x^n$ and our comparison series converges for $0 < x < \\frac{1}{2}$, our Fibonacci series converges for those values too (and possibly others).<\/p>\n

Rearranging a Power Series<\/h3>\n

but carefully, and under the right conditions<\/h4>\n

From here out, let’s assume $0 < x < \\frac{1}{2}$ to make sure the terms of our series are positive and $x$ is within the radius of convergence of the Fibonacci series. For these values of $x$, we can use any theorem about absolutely convergent series, like The Rearrangement Theorem, which says that absolutely convergent series satisfy the commutative property of addition (so we can rearrange the terms and add them up and still get the same value, which seems like it’s obvious until someone blows your mind by rearranging the alternating harmonic series).<\/p>\n

Now we’ll use the recursion (for $n \\geq 2$) to rewrite our power series (for values of $x$ where it converges absolutely):
\n$$\\begin{array}{r c l} F(x) &=& \\displaystyle\\sum_{n=0}^\\infty F_n x^n \\\\
\n&=& 1 + x + \\displaystyle\\sum_{n=2}^\\infty(F_{n-1} + F_{n-2})x^n \\\\
\n&=& 1 + x + \\displaystyle \\sum_{n=2}^\\infty(F_{n-1}x^n + F_{n-2}x^n) \\\\
\n&=& 1 + x + \\displaystyle \\sum_{n=2}^\\infty F_{n-1}x^n + \\displaystyle \\sum_{n=2}^\\infty F_{n-2}x^n.
\n\\end{array}$$<\/p>\n

(That last step is where we need the series to converge absolutely in order to use the Rearrangement Theorem!)<\/p>\n

Now we can adjust the index $n$ to obtain the equalities:
\n$$\\begin{array}{r c l}
\n\\displaystyle \\sum_{n=2}^\\infty F_{n-1}x^n &=& \\displaystyle \\sum_{n=1}^\\infty F_n x^{n+1} \\quad =\\quad x(F(x) – 1),\\\\
\n\\displaystyle \\sum_{n=2}^\\infty F_{n-2}x^n &=& \\displaystyle \\sum_{n=0}^\\infty F_n x^{n+2} \\quad = \\quad x^2 F(x).
\n\\end{array}
\n$$
\nBy substitution,
\n$$\\begin{array}{r c l} F(x) & = & \u00a01 + x F(x) + x^2 F(x).
\n\\end{array}$$
\nFinally, if we solve all of that for $F(x)$, we get the marvelous rational function $$F(x) = \\frac{1}{x^2 – x – 1}.$$<\/p>\n

\"A
Talk about the halo effect!<\/figcaption><\/figure>\n

Whenever I encounter a rational function, the partial fraction decomposition neuron in my brain fires up. We need the irreducible (over $\\mathbb{R}$) factors of our denominator, and we can use the quadratic formula to find the roots of $x^2-x-1$ or evaluate at $\\Phi$ and $\\Psi$ to believe that $x^2 – x – 1 = (x – \\Phi)(x- \\Psi)$. (Golden tip: from this equality, we learn $\\Phi \\Psi = -1$ and $\\Phi + \\Psi = 1$, which are useful for simplifying many computations involving the Golden Ratio and Other Ratio.<\/em>)\u00a0 Then, cranking out the partial fraction decomposition, I found that
\n$$\\frac{1}{x^2 – x – 1} = \\frac{1}{\\sqrt{5}(x-\\Phi)} + \\frac{-1}{\\sqrt{5}(x-\\Psi)}.$$<\/p>\n

We turn the summands on the right into geometric series, so that
\n$$
\n\\begin{array}{r c l c r c l}
\n\\displaystyle{\\frac{1}{\\sqrt{5}(x-\\Phi)}} &=& \\displaystyle{\\frac{1}{\\sqrt{5}} \\sum_{n = 0}^\\infty \\Phi^n x^n} & \\quad & \\displaystyle{\\frac{-1}{\\sqrt{5}(x-\\Psi)}} &=& \\displaystyle{\\frac{-1}{\\sqrt{5}} \\sum_{n=0}^\\infty \\Psi^n x^n}.
\n\\end{array}
\n$$<\/p>\n

And when we add these together, we end up with $$F(x) = \\frac{1}{\\sqrt{5}} \\sum_{n=0}^\\infty (\\Phi^n – \\Psi^n) x^n.$$<\/p>\n

And now, another helpful neuron fires to remind us that $x^n$ in $F(x)$ is actually $F_n$, and therefore (drumroll, please!)<\/p>\n

$$F_n = \\frac{1}{\\sqrt{5}}(\\Phi^n – \\Psi^n).$$<\/p>\n

And so, to calculate $F_{413}$, just pop $$\\frac{1}{\\sqrt{5}} \\left( \\left(\\frac{1+\\sqrt{5}}{2} \\right)^{413} – \\left(\\frac{1-\\sqrt{5}}{2}\\right)^{413}\\right)$$ into your into your favorite calculator, which will tell you that this Fibonacci number is merely<\/p>\n

$$\\begin{array}{l l}
\n91,708,675,472,160,090,711,036,102,616,287,632,089,318, & \\\\
\n\\, \\, 911,882,123,915,231,537,729,562,617,689,923,506,638,302,133 &
\n\\end{array}$$
\nor 91 septenvigintillion, 708 sexvigintillion, 675 quinvigintillion, 472 quattuorvigintillion, 160 trevigintillion, 90 duovigintillion, 711 unvigintillion, 36 vigintillion, 102 novemdecillion, 616 octodecillion, 287 septendecillion, 632 sexdecillion, 89 quindecillion, 318 quattuordecillion, 911 tredecillion, 882 duodecillion, 123 undecillion, 915 decillion, 231 nonillion, 537 octillion, 729 septillion, 562 sextillion, 617 quintillion, 689 quadrillion, 923 trillion, 506 billion, 638 million, 302 thousand, 133 (roughly a million times the number of atoms in the universe).<\/p>\n

Further Reading<\/h3>\n

[1] The man himself, Leonardo Pisano Fibonacci, listed the first several terms in his manuscript, which is archived here: https:\/\/archive.org\/details\/conventi-soppressi-c.-i.-2616\/page\/n127\/mode\/1up<\/a>
\nI went down a bit of a, ahem, rabbit hole at this point. My medieval Latin isn’t great, but I think a credulous reader (like me) could believe there are words related to cuniculus<\/em>, which is Latin for rabbit, in the text. The translation into English<\/a> bears this out.<\/p>\n

\"A
Coniculox? Cunicules?<\/figcaption><\/figure>\n

Editor’s note: Courtney was on the right track, but missed a scribal abbreviation. The screenshot begins “Quot paria coniculo\ua75d”, a scribal abbreviation (with some creative spelling) for “Quot paria cuniculorum”, which translates as “How many pairs of rabbits…”<\/em><\/p>\n

[2] Here’s a nice short paper about The Rearrangement Theorem from the Pi Mu Epsilon journal: https:\/\/www.jstor.org\/stable\/24340038<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Luckily, such a formula exists, and there are lots of ways to find it. In this post, we\u2019ll find it using power series. Read on, brave bunny farmer… A Golden Formal Power Series Courtney Gibbons Hamilton College Some days, I feel a lot like the angel in Dürer’s Melencolia IRead More →<\/a><\/span><\/p>\n","protected":false},"author":5,"featured_media":1599,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"advanced_seo_description":"","jetpack_seo_html_title":"","jetpack_seo_noindex":false,"jetpack_seo_schema_type":"","_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_post_was_ever_published":false},"categories":[213,10,26],"tags":[238,239,240],"class_list":["entry","author-cgibbons","post-2796","post","type-post","status-publish","format-standard","has-post-thumbnail","category-213","category-courtney-gibbons","category-real-and-complex-analysis","tag-fibonacci-sequence","tag-golden-ratio","tag-sequences-and-series"],"jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"jetpack_featured_media_url":"https:\/\/i0.wp.com\/mathvoices.ams.org\/featurecolumn\/wp-content\/uploads\/sites\/2\/2023\/03\/mathvoices-banner-feat-col.png?fit=2760%2C580&ssl=1","_links":{"self":[{"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/posts\/2796","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/comments?post=2796"}],"version-history":[{"count":19,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/posts\/2796\/revisions"}],"predecessor-version":[{"id":2842,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/posts\/2796\/revisions\/2842"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/media\/1599"}],"wp:attachment":[{"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/media?parent=2796"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/categories?post=2796"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathvoices.ams.org\/featurecolumn\/wp-json\/wp\/v2\/tags?post=2796"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}