# Charming approximation of $\pi$: $2\left(\frac{1}{2}\right)^{\phi/2}+2< \pi$, where $\phi$ is the golden ratio

Prove that :

$$2\left(\frac{1}{2}\right)^{\phi/2}+2< \pi$$ where $$\phi:=\frac12(1+\sqrt{5})=1.618\ldots$$ is the golden ratio.

How I came across this approximation?

Well, I was studying the following function:

$$f(x)=x^{\phi(1-x)}+(1-x)^{\phi x }+2$$

The approximation corresponds to the maximum of $$f(x)$$

I can solve it using power series it's not hard and using some approximation of $$\pi$$

But I would like to know if there is a proof without derivatives using by example inequality like Bernoulli's inequality .

Can you help me ?

Thanks a lot for all your contributions.

• You need to go up to $72/89$ with the Fibonacci-based lower bounds to $\phi/2$ to make that estimate work. Curious :-) – Jyrki Lahtonen Jun 27 '20 at 7:45
• Nice function. What approximation of pi did you use and what did you expand in power series ? Could you write some more details about your derivation ? – Thomas Jun 27 '20 at 8:35
• Equivalently the fixed point of $f(x)=1-2^{-\cos(2x/5)}$ is less than $\pi/2$. – TheSimpliFire Jun 27 '20 at 10:24

If we'll prove that $$2^{\sqrt5}>\frac{212}{45},$$ so it's enough to prove that: $$\frac{2}{\sqrt[4]2\cdot\sqrt[4]{\frac{212}{45}}}+2<\pi$$ or $$\sqrt[4]{\frac{90}{53}}<\pi-2$$ for which it's enough to prove that $$\sqrt[4]{\frac{90}{53}}<\frac{1613}{1413},$$ which is true because $$1413^4\cdot90-1613^4\cdot53=-1802797643<0.$$ I hope it will help.

To prove the main result

$$2\left(\frac{1}{2}\right)^{\frac{\phi}{2}}+2<\pi$$

we shall show that $$\pi <\sqrt{10}$$ and interestingly, we also deduce that $$3<\pi<4$$

Preliminaries

Consider the set $$S=\left\{X_n= \frac{1}{n^3(n+1)^3} : n\in\mathbb {N} \right\}$$ and here we show that set $$S$$ is bounded set with lower and bounds $$0$$ and $$\frac{1}{8}$$ respectively. Note that for $$n\geq 1$$ the \begin{aligned} S_{n+1}-S_n & =\frac{1}{(n+1)^3}\left[\frac{1}{(n+2)^3}-\frac{1}{n^3}\right]\\ & =\frac{1}{(n+1)^3}\left[\frac{n^3-(n+2)^3}{n^3(n+2)^2}\right]\cdots(1)\end{aligned} since for all $$n>1$$, $$n< n+2\implies n^3-(n+2)^3<0$$ from $$(1)$$ it follows that $$S_{n+1}-S_n <0$$ implies the sequence $$X_n$$ contained in set $$S$$ is a decreasing sequence and thus \begin{aligned} \operatorname{sup}\left\{X_n : n\in \mathbb{N}\right\}&=\frac{1}{8}<1\\ \operatorname {inf}\left\{X_n:n\in\mathbb{N} \right\}&=0\end{aligned}Therefore, $$0 < X_n \leq \frac{1}{8} <1$$. Futher, $$n^3(n+1)^3> n(n+1)=Y_n$$ and hence $$0 <\sum_{n\geq 1} X_n < \sum_{n\geq 1} (Y_n)^{-1}=1\cdots(2)$$ since we have telescoping series as $$\displaystyle \sum_{n\geq 1} (Y_n)^{-1} =\sum_{n\geq 1} \left(\frac{1}{n}-\frac{1}{n+1}\right)=1$$ Now \begin{aligned} \sum_{n\geq 1} X_n & =\sum_{n\geq 1} \left(\frac{1}{Y_n}\right)^3\\&=\sum_{n \geq 1}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)\\&-\sum_{n\geq 1}\frac{3}{(Y_n)^2}=\zeta(3)-\zeta(3)+1\\& 1-3\sum_{n\geq 1} \left(\frac{1}{n^2}+\frac{1}{(n+1)^2}\right)\\&-2\sum_{n\geq 1}\frac{1}{(Y_n)}= 1-6\zeta(2)+9 \\&= 10-\pi^2\end{aligned} and thus from $$(2)$$ $$\sum_{n\geq 1}X_n >0\implies \pi <\sqrt{10}$$ since $$9-\pi^2 < 10-\pi^2<16-\pi^2$$ which implies $$3< \pi < 4$$.

Proof of the main result

If the left hand expression of main result has to be less than $$\pi$$ then it should also be less than $$\sqrt {10}$$. To prove the result we suppose that the inequality is true. That is; $$A= 2\left(\frac{1}{2}\right)^{\frac{\phi}{2}}+2 < \pi ,\;\; A< \sqrt {10}$$ Squaring both sides we yield \begin{aligned}\left( \frac{1}{2}\right)^{\phi}+ \left(2\left(\frac{1}{2}\right)^{\frac{\phi}{2}}+2\right)-1<\frac{5}{2}\end{aligned}\\ \left(\frac{1}{2}\right)^{\phi}+A<\frac{7}{2} hence$$\left(\frac{1}{2}\right)^{\phi} <\frac{7-2\sqrt{10}}{2}=\left(1+\frac{5-2\sqrt{10}}{2}\right)^{\frac{1}{\phi}}=(1+y)^{\phi^{-1}}$$ since $$\phi>1\implies \frac{1}{\phi}<1$$ and hence by Bernoulli inequality we have $$(1+y)^{\phi^{-1}}<1+ \frac{y}{\phi}=\\ 1-\frac{15}{(5+2\sqrt{10})(1+\sqrt{5})}<1-\frac{15}{11 \cdot 3} =1-\frac{15}{33}=\frac{18}{33}$$ Since \begin{aligned} \frac{7-2\sqrt {10}}{2}=(1+y)^{\phi{-1}} <\frac{18}{33}\end{aligned}. We claim that $$A<\sqrt {10}$$ which also says $$\frac{1}{2} <(1+y)^{\phi^{-1}}$$ also we have $$(1+y)^{\phi^{-1}} <\frac{18}{33}$$. Therefore we must have $$\frac{1}{2} < \frac{18}{33}$$ which is true since $$\frac{1}{2} -\frac{18}{33} =\frac{33-36}{66} =-\frac{1}{22}<0$$ As we claimed inequality to be true and hence we came up $$-\frac{1}{22}<0$$ to be true and thus,

$$2 \left(\frac{1}{2}\right)^{\frac{\phi}{2}}+2<\pi$$ must be true.

• Assuming the details are correct and you have proved $A < \sqrt{10},$ how does it follow that $A < \pi$? – Calum Gilhooley Jul 2 '20 at 12:12
• @Calum Gilhooley, Can you please check once again, I had made few changes in the solution? – Naren Jul 3 '20 at 18:37
• (I'm sorry about the slow response. I was unable to post for 7 days. It's a long story!) Putting the details of your argument in brackets, for the moment, its overall structure appears to be: if $A < \pi,$ then $A < \sqrt{10},$ and this is true (assuming that the aforementioned details are correct, and also using the proposition $\pi < \sqrt{10},$ which is correct in any case), therefore the hypothesis $A < \pi$ is true. But this is a fallacious argument. (I haven't downvoted it yet, in case I have misunderstood what you're getting at.) – Calum Gilhooley Jul 10 '20 at 15:50