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I found the following identity. While trying to prove it, I found some things that I don’t quite understand:

$$\frac{\pi}{4}=\sqrt{5} \sum_{n=0}^{\infty} \frac{(-1)^n F_{2n+1}}{(2n+1) \phi^{4n+2}}$$

(where $\phi=\frac{\sqrt{5}+1}{2}$).

What I tried

I first considered the series: $$F(x)=\sum_{n=0}^{\infty}F_{2n+1}x^{n}=\frac{1-x}{x^2-3x+1}$$ (when it converges).

Then I replaced $x$ with $x^2$ and tried integrating to get something like:

$$A(x)=\sum_{n=0}^{\infty}F_{2n+1} \frac{x^{2n+1}}{2n+1}=\int \frac{1-x^2}{x^4-3x^2+1}$$

This is where one question arises:

  1. Is this integration a valid thing to do? The sum on the left has a value, but the integral on the right has some constant added to it. In that case, how should I choose the constant?

Now, making $x=\frac{1}{\phi^2}$ would make the $\phi^{4n+2}$ term appear, but we still need to put a $(-1)^n$ in there, so I thought of putting an $i$ there (because $i^{2n+1}=i \cdot (-1)^n$) this way:

$x=\frac{i}{\phi^2}$, and then $$A(x)=i \sum_{n=0}^{\infty} \frac{(-1)^n F_{2n+1}}{(2n+1) \phi^{4n+2}}.$$ So now I need to prove that the integral for this $x$ is exactly $\frac{i \pi}{4\sqrt{5}}$, but the problem I have is that the integral has logarithms and I don’t know how to find logarithms of complex numbers like $\log(5+2i)$. (I found on Wikipedia the Taylor series for logarithms, but I can’t see how this makes the problem simpler.)

More questions

  1. Does it make sense to plug this complex value into the power series and the integral? If so, then how one would evaluate the integral for this specific $x=\frac{i}{\phi^2}$?

  2. Is there any other path to prove this intriguing identity?

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    I would expect it's easier to use $\sqrt{5} \cdot F_{2n+1} = \phi^{2n+1} - \psi^{2n+1}$, where $\psi = -\frac{1}{\phi}$. – Daniel Fischer Sep 06 '15 at 19:54
  • A very good idea . Thanks a lot , this simplifies everything and I think will work without much trouble (I will try it ) .But I still want answers to the other questions regarding my approach . –  Sep 06 '15 at 19:58
  • @OussamaBoussif But in the series I have $x^{2n+1}$ and not $x^n$ . –  Sep 06 '15 at 20:13
  • Yes you are right I didn't see that – Oussama Boussif Sep 06 '15 at 20:14
  • If you want to find the logarithm of complex numbers just write them in exponential form and you apply the logarithm. – Oussama Boussif Sep 06 '15 at 20:27
  • @OussamaBoussif I didn't thought that it is so easy . But a question : $$e^{\pi i}=e^{3 \pi i}$$ so how should I choose the correct logarithm between $\pi i$ and $3 \pi i$ or any other one of the form $(2k+1) \pi i$ .Thanks –  Sep 06 '15 at 20:34
  • We don't really need to use complex numbers: the claim is just a consequence of the identity $\frac{1}{\varphi}+\frac{1}{\varphi^3}=1-\frac{1}{\varphi^4}$. – Jack D'Aurizio Sep 06 '15 at 20:35
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    You can set a constant of integration simply by knowing the value at any particular point, such as at $x=0$. However, your integral has branch cuts, so things are much more complicated than a mere "constant of integration". Try using a definite integral instead of an indefinite integral! A path integral is probably best, using your favorite path from $x=0$ to $x=\mathbf{i} \varphi^{-2}$. P.S. you can avoid complex numbers by making the substitution $x \mapsto -x^2$ rather than $x \mapsto x^2$. Or even better, don't both substituting at all, and later plug in $x = -\varphi^{-4}$. –  Sep 07 '15 at 01:24

3 Answers3

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By the explicit formula for Fibonacci numbers it follows that: $$\color{red}{S}=\color{blue}{\sqrt{5}}\sum_{n\geq 0}\frac{(-1)^n \color{blue}{F_{2n+1}}}{(2n+1)\,\color{blue}{\varphi^{4n+2}}}=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\left(\color{blue}{\frac{1}{\varphi^{2n+1}}+\frac{1}{\varphi^{6n+3}}}\right),$$ hence by the arctangent Taylor series and the (arc)tangent sum formulas: $$ \color{red}{S} = \arctan\frac{1}{\varphi}+\arctan\frac{1}{\varphi^3}=\arctan\frac{\frac{1}{\varphi}+\frac{1}{\varphi^3}}{1-\frac{1}{\varphi^4}}=\arctan 1=\color{red}{\frac{\pi}{4}}$$ as wanted.

Jack D'Aurizio
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    Beautiful and simple proof . Seeing this proof I think that this sum was constructed exactly to fit this expansion for $\arctan$ .Thanks a lot . Can you tell me if there's something I can do with my attempt or if it's completely wrong ? –  Sep 06 '15 at 20:41
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    @ComplexPhi: your approach works, indeed. You just have to exploit partial fraction decomposition to compute your integrals. You simply get more terms to simplify, but the principle is just the same, $\varphi^2=\varphi+1$ leads to interesting consequences. – Jack D'Aurizio Sep 06 '15 at 20:44
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    Nice proof as always @Jack D'Aurizio . I completely forgot that Arctan expansion. – Oussama Boussif Sep 06 '15 at 20:45
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Start with $$ \begin{align} \sum_{k=0}^\infty F_{2k+1}x^{2k} &=\frac{1-x^2}{1-3x^2+x^4}\\ &=\frac{1-x^2}{(x-\phi)(x+\phi)(x+\frac1\phi)(x-\frac1\phi)}\\ &=\frac1{2\sqrt5}\frac1{x+\phi}-\frac1{2\sqrt5}\frac1{x-\phi}+\frac1{2\sqrt5}\frac1{x+\frac1\phi}-\frac1{2\sqrt5}\frac1{x-\frac1\phi}\tag{1} \end{align} $$ Integration yields $$ \begin{align} \sum_{k=0}^\infty\frac{F_{2k+1}}{2k+1}x^{2k+1} &=\frac1{2\sqrt5}\log\left(\frac{(x+\phi)(x+\frac1\phi)}{(x-\phi)(x-\frac1\phi)}\right)\\ &=\frac1{2\sqrt5}\log\left(\frac{x^2+\sqrt5x+1}{x^2-\sqrt5x+1}\right)\tag{2} \end{align} $$ Substituting $x\mapsto ix$ then multiplying by $-i\sqrt5$ gives $$ \begin{align} \sqrt5\sum_{k=0}^\infty(-1)^k\frac{F_{2k+1}}{2k+1}x^{2k+1} &=\frac1{2i}\log\left(\frac{1-x^2+i\sqrt5x}{1-x^2-i\sqrt5x}\right)\\ &=\frac1{2i}\left[\frac12\log\left(1+3x^2+x^4\right)+i\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\right]\\ &-\frac1{2i}\left[\frac12\log\left(1+3x^2+x^4\right)-i\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\right]\\ &=\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\tag{3} \end{align} $$ since $\log(a+ib)=\tfrac12\log(a^2+b^2)+i\arctan\left(\tfrac ba\right)$ for $a\gt0$; that is $|x|\lt1$.

Evaluating $(3)$ at $x=\frac1{\phi^2}$ yields $$ \begin{align} \sqrt5\sum_{k=0}^\infty(-1)^k\frac{F_{2k+1}}{2k+1}\frac1{\phi^{4k+2}} &=\arctan\left(\frac{\sqrt5}{\phi^2-\frac1{\phi^2}}\right)\\ &=\arctan(1)\\[9pt] &=\frac\pi4\tag{4} \end{align} $$

robjohn
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    It might be worth some comment as to why that particular branch of the logarithm is the right one to use. –  Sep 07 '15 at 01:30
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    The other branches of log differ by $2\pi i$, and since the sum for $x=0$ is $0$, we must have the branch that gives $\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)$ – robjohn Sep 07 '15 at 02:22
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1.Is this integration a valid thing to do? because the sum on the left has a value but the integral on the right has some constant added to it .Then how to choose the constant ?

Answer: Yes, integration is a perfectly valid step in a solution to problems like this, but not indefinite integration. The integration must be definite, otherwise the equality is between a function on the LHS:

$$ \sum_{n=0}^\infty{}F_{2n+1}\frac{x^{2n+1}}{2n+1} $$ and a set of functions on the RHS $$ \left\{c + \int\frac{1-x^2}{x^4-3x^2+1}\text{d}x,\ c\in\Bbb{R}\right\} $$ which doesn't make any sense.

So, we have to choose appropriate limits. To choose these limits, think about what you did to the LHS to go from $\sum{F_{2n+1}x^{2n}}$ to $\sum{F_{2n+1}\frac{x^{2n+1}}{2n+1}}$. We integrated the former sum from $0$ to $x$: $$ \int_0^x\sum_{n=0}^\infty{F_{2n+1}t^{2n}}\text{d}t = \sum_{n=0}^\infty{F_{2n+1}\frac{x^{2n+1}}{2n+1}} $$ but since the sum in the integrand is equal to $\frac{1-x^2}{x^4-3x^2+1}$, then we also have $$ \begin{align} \int_0^x\sum_{n=0}^\infty{F_{2n+1}t^{2n}}\text{d}t & = \int_0^x\frac{1-t^2}{t^4-3t^2+1}\text{d}t \\ & = \frac{1}{2\sqrt5}\log\left(\frac{x^2+\sqrt5x+1}{x^2-\sqrt5x+1}\right) \equiv \text{A}(x) \end{align} $$ From here, robjohn's answer explains the rest of the proof.

user3002473
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