Spring 2016, problem 16

Let $ f : \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable, $ 2 \pi$-periodic, even function. Prove that if $$f''(x)+f(x)=\frac{1}{f(x+ \frac{3\pi}{2} )}$$ holds for every $ x$, then $ f$ is actually $ \frac{\pi}{2}$-periodic.


1 year ago

I assume, that the requirement satisfied by $f$ is to be read as:

$f(x + \frac{3 \pi}{2}) ( f''(x) + f(x) ) = 1$ (Eq. 1)

and we can thus infer, that $f(x+ \frac{3 \pi}{2})$ (and therefore $f(x)$) is nonzero everywhere. Since $f$ is in particular assumed continuous, it follows, that either $f >0$ everywhere or or $f \lt 0$ everywhere (otherwise $f$ would have a zero!). We can assume, that the first inequality holds, for if $f \lt 0$ everywhere, then $g := -f > 0$ everywhere and g satisfies the same requirements as $f$.

Since $f$ is even, so is $\frac{1}{f'' + f}$. Also, $f(x + \frac{3 \pi}{2}) = f(x - \frac{\pi}{2})$ from $2\pi$-periodicity. So $f(x - \frac{\pi}{2})$ is even. It is easy to show, that any even function, that stays even when shifted by $\frac{\pi}{2}$, is $\pi$-periodic. And so we have the intermediate result:

$f$ is $\pi$-periodic.

By using the $\pi$ periodicity, it is easily shown, that:

$f(x) = \frac{1}{f''(x - \frac{\pi}{2}) + f(x - \frac{\pi}{2})}$

This equation, taken together with $(Eq. 1)$, yields:

$f(x) f''(x - \frac{\pi}{2}) = f''(x) f(x - \frac{\pi}{2})$.

Define $g(x) := f(x - \frac{\pi}{2})$. Then:

$fg'' = f''g \rightarrow \partial (f'g - f g') = 0 \rightarrow f'g = f g' + C$

with $C$ a constant (the $\partial$ symbol denotes the ordinary derivative operator).

But evaluating the last equation at $y := x + \frac{\pi}{2}$ gives:

$f'(x + \frac{\pi}{2}) f(x) = f(x + \frac{\pi}{2}) f'(x) + C$

And therefore, using $\pi$-periodicity:

$g'(x) f(x) = g(x) f'(x) + C$.

This is only compatible with $f'g = f g' + C$, if $C = 0$. So we have:

$\frac{f'}{f} - \frac{g'}{g} = \partial [ ln(\frac{f}{g}) ] = 0$

So $f = \lambda g$ for some positive constant $\lambda$.

The value of $\lambda$ can't be anything else than 1, for otherwise $f(x)$ would converge to zero for large $x$ (case $ \lambda \lt 1 $ or would be unbounded from above (case $\lambda > 1$). Both cases are excluded for a positive, periodic function. Therefore $\lambda = 1$ and we're finished.

Sir, If We assume that that the fundaamental period of f(x) be T. then it implies that f" has a fundamental period of the the form T/n where n is a Natural number.Now by the LCM RULE of the the periodicity f"/f has a fundaMENTAL period of T.Because if it would not happen then It would be of the form T/m where m is a Natural number.Now T/m must be >= T by the LCM RULE.hence m must be <= 1.Hence m=1. therefore pi/2 must be a multiple of T.Hence T<=pi/2.Hence f has a period of pi/2 . SIR, IS MY LOGIC CORRECT,If wrong please point out my mistake

SRIJIT 1 year ago

Hi, Babin. I can't really point to a mistake here. It's hard to tell if your logic is correct, if you don't expose your arguments more clearly. Please elaborate.

My best guess is, you're reasoning goes somewhere along these lines:

  1. $\frac{f''}{f}$ has the same fundamental period $T$ as $f$.
  2. $\frac{f''}{f}$ is $\frac{\pi}{2}$-periodic.
  3. Therefore $f$ is $\frac{\pi}{2}$-periodic.

Now item number 3 is clear. Item number 2 is true, but you use it without any justification. Nevertheless it can be justified easily enough: First you proof that $f$ is actually $\pi$-periodic (you need to use the evenness property for this, as far as I can see. You didn't use this anywhere.). Then number 2 follows from the equation in the problem statement pretty easily: just divide by $f(x)$ and note that $f(x)f(x+\frac{3 \pi}{2})$ is $\frac{\pi}{2}$-periodic for any $\pi$-periodic function.

But I don't see the argument for item number 1. Please state the "LCM rule" rule, you are using. Is there a theorem, that says: If $g$ and $h$ are functions with fundamental period $G$ and $H$, resp. and $h \neq 0$ everywhere and if LCM($G$,$H$) exists, then $\frac{f}{g}$ has fundamental period LCM($G$,$H$) ?? This is not true, by the way, if you don't force additional restrictions on $g$ and $h$, as $g = h = 2 + sin(x)$ would be a counterexample. I can't think of a less, trivial, or less technical counterexample, sorry. But maybe it is true for the particular case of $g = h''$,?I don't know.

Nelix 1 year ago

Yes, Sir due to the various exams in my near future i am getting less time to give line by line logical approach.I apologise for the inconvenience. Yes sir, I have exactly done in the way you have pointed out. Regarding the L.C.M rule, your claim is correct that it is due to f"(x). It can be prooved from the first principle of derivative that the fundamental period of f"(x) is some divisor of T or T itself. Hence the L.C.m rule is valid here. Sir, one Thing I must appreciate is the beauty of the question posted here. Thanks and Regards to The Purdue University and Michael Golomb.

SRIJIT 1 year ago

Hi, Nelix. I love your answer, but there one place where I do not understand what you have done.

"f is π -periodic.

By using the π periodicity, it is easily shown, that:


This equation, taken together with (Eq.1) yields:

f(x)f″(x−π2)=f″(x)f(x−π2)" Could you please show me in detail? I hope you can answer me soon :)

Raw231 1 year ago

Hi Raw231,

sorry, to have kept you waiting for a week. You've probably figured it all out yourself by now, but anyway, here's the details:

We start at the point, where $\pi$-periodicity is already established, that is we have:

$$f(x + z \pi) = f(x) \forall z \in \mathbb{Z}$$

Now in (Eq. 1), we are certainly allowed to substitute $x - \frac{\pi}{2}$ instead of $x$, in all places, where $x$ occurs, so:

$$f(x - \frac{\pi}{2} + \frac{3 \pi}{2}) = \frac{1}{f''(x - \frac{\pi}{2}) + f(x - \frac{\pi}{2})}$$

But the $\pi$ periodicity says, that the left side is equal to $f(x)$. This should make the first equation in your question clear.

Now rearrange this equation:

$$f(x) f(x - \frac{\pi}{2}) = 1 - f(x) f''(x - \frac{\pi}{2})$$

Also rearrange (Eq. 1):

$$f(x) f(x + \frac{3\pi}{2}) = 1 - f''(x) f(x + \frac{3\pi}{2})$$

But, since $f(x + \frac{3 \pi}{2}) = f(x + \frac{3 \pi}{2} - 2 \pi) = f( x - \frac{\pi}{2})$, the last two equations reduce to:

$$f(x) f''(x - \frac{\pi}{2}) = f''(x) f(x - \frac{\pi}{2})$$.

The proof would probably have been clearer, had I simply stated, that $f$ and $f''$ are $\pi$ periodic, and $f(x) f(x - \frac{\pi}{2})$ (and therefore $f''(x) f(x - \frac{\pi}{2})$ by (Eq. 1)) is $\frac{\pi}{2}$-periodic, from which the last equation above is almost obvious.

Nelix 1 year ago
1 year ago

If $f$ is even $f''$ is even and $x \mapsto f\left( {x + 3\frac{\pi }{2}} \right)$ is even. Then $f\left( {x + 3\frac{\pi }{2}} \right) = f\left( { - x + 3\frac{\pi }{2}} \right) = f\left( {x - 3\frac{\pi }{2}} \right)$ and $f$ is $3\pi $ periodic and therefore $\pi $ periodic.

$f''\left( x \right) + f\left( x \right)\left( {1 - \frac{1}{{f\left( x \right)f\left( {x + \frac{\pi }{2}} \right)}}} \right) = 0 $

$f''\left( {x + \frac{\pi }{2}} \right) + f\left( {x + \frac{\pi }{2}} \right)\left( {1 - \frac{1}{{f\left( x \right)f\left( {x + \frac{\pi }{2}} \right)}}} \right) = 0$

$f$ and $g: x \mapsto f\left( {x + \frac{\pi }{2}} \right)$ are two even solutions of the linear équation

$y'' + \left( {1 - \frac{1}{{f\left( x \right)f\left( {x + \frac{\pi }{2}} \right)}}} \right)y = 0$ (1)

$f\left( 0 \right) = a$, $f'\left( 0 \right) = 0$, $a \ne 0$ otherwhise $f=0$

$g\left( 0 \right) = b$, $g'\left( 0 \right) = 0$, $b \ne 0$ otherwhise $g=0$

$h = bf - ag$ is a solution of (1). $h\left( 0 \right) = h'\left( 0 \right) = 0$. Then $h = 0$

$f\left( {x + \frac{\pi }{2}} \right) = \lambda f\left( x \right)$ where $\lambda = \frac{b}{a}$

$f\left( x \right) = f\left( {x + \pi } \right) = {\lambda ^2}f\left( x \right)$

For all $x$, $f\left( x \right) \ne 0$. Then ${\lambda ^2} = 1$

But if $f\left( {x + \frac{\pi }{2}} \right) = - f\left( x \right)$, for some $x \in \mathbb{R}$, $f\left( x \right) = 0$.

$\lambda = 1$ and $f$ is $\frac{\pi }{2}$ periodic.


All the solutions of $y'' + y = \frac{1}{y}$ seem to be periodic.

They verify $2y'y'' + 2y'y = 2\frac{{y'}}{y}$

and an equation of this type: ${y'^2} + {y^2} = 2\ln y + C$ , $C \ge 1$

The period of this solution is $T = 2\int_{{y_0}}^{{y_1}} {\frac{{dy}}{{\sqrt {C - {y^2} + 2\ln y} }}} $ , ${y_0} \le {y_1}$

where ${y_0}$ and ${y_1}$ are the solutions of ${y^2} - 2\ln y = C$

With $\ln y = \frac{x}{2}$

$T = \int\limits_{{x_0}}^{{x_1}} {\frac{{{e^{\frac{x}{2}}}{\rm{dx}}}}{{\sqrt {C - {e^x} + x} }}} $

where ${x_0}$ and ${x_1}$ are the solutions of ${e^x} = x + C$

With $x = t - \ln \frac{{{\mathop{\rm sh}\nolimits} t}}{t}$

Let $\varphi \left( t \right) = {e^x} - x = \frac{{t{e^t}}}{{{\mathop{\rm sh}\nolimits} t}} - t + \ln \left( {\frac{{{\mathop{\rm sh}\nolimits} t}}{t}} \right) = t\coth t + \ln \left( {\frac{{{\mathop{\rm sh}\nolimits} t}}{t}} \right)$

$\varphi $ is even.

The two solutions of the equation $\varphi \left( t \right) = C$ are opposite.

If $C = \varphi \left( \alpha \right)$ with $\alpha > 0$

With $x = t - \ln \frac{{{\mathop{\rm sh}\nolimits} t}}{t}$

$T = \int\limits_{ - \alpha }^\alpha {\frac{{\sqrt {\frac{{t{e^t}}}{{{\mathop{\rm sh}\nolimits} t}}} \left( {1 + \frac{1}{t} - \coth t} \right)}}{{\sqrt {\varphi \left( \alpha \right) - \varphi \left( t \right)} }}} dt$

$T=\int\limits_{ - \alpha }^\alpha {\frac{{\sqrt {\frac{t}{{{\mathop{\rm sh}\nolimits} t}}} \left( {{e^{\frac{t}{2}}} + \frac{{{e^{\frac{t}{2}}}}}{t} - {e^{\frac{t}{2}}}\coth t} \right)}}{{\sqrt {\varphi \left( \alpha \right) - \varphi \left( t \right)} }}dt}$

By parity,

$T=\int\limits_{ - \alpha }^\alpha {\frac{{\sqrt {\frac{t}{{{\mathop{\rm sh}\nolimits} t}}} \left( {{\mathop{\rm ch}\nolimits} \frac{t}{2} + \frac{{{\mathop{\rm sh}\nolimits} \frac{t}{2}}}{t} - {\mathop{\rm sh}\nolimits} \frac{t}{2}\coth t} \right)}}{{\sqrt {\varphi \left( \alpha \right) - \varphi \left( t \right)} }}dt} $

and finally

$T = 2\int\limits_0^\alpha {\sqrt {\frac{t}{{{\mathop{\rm sh}\nolimits} t}}} \left( {\frac{1}{{2{\mathop{\rm ch}\nolimits} \frac{t}{2}}} + \frac{{{\mathop{\rm sh}\nolimits} \frac{t}{2}}}{t}} \right)\frac{{{\mathop{\rm dt}\nolimits} }}{{\sqrt {\varphi \left( \alpha \right) - \varphi \left( t \right)} }}} $

If you want to find the period of the even solution of $y'' + y = \frac{1}{y}$ which verify $y\left( 0 \right) = a$ and $y'\left( 0 \right) = 0$

Calculate $C = {a^2} - 2\ln a$, $C \ge 1$

Find $\alpha > 0$ verifying $\varphi \left( \alpha \right) = C$

$t = \alpha \cos u$

Then $T = 2\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\frac{{\alpha \cos u}}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos u} \right)}}} \left( {\frac{1}{{2{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos u}}{2}}} + \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos u}}{2}}}{{\alpha \cos u}}} \right)}}{{\sqrt {\frac{{\varphi \left( \alpha \right) - \varphi \left( {\alpha \cos u} \right)}}{{{\alpha ^2}{{\sin }^2}u}}} }}{\mathop{\rm du}\nolimits} } $

I think that

1) as $\alpha $ approaches 0, T approaches $\pi \sqrt 2 $

Near from 0, $\varphi \left( \alpha \right) = 1 + \frac{{{\alpha ^2}}}{2} + o\left( {{\alpha ^2}} \right)$

$\mathop {\lim }\limits_{\alpha \to 0} 2\frac{{\sqrt {\frac{{\alpha \cos u}}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos u} \right)}}} \left( {\frac{1}{{2{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos u}}{2}}} + \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos u}}{2}}}{{\alpha \cos u}}} \right)}}{{\sqrt {\frac{{\varphi \left( \alpha \right) - \varphi \left( {\alpha \cos u} \right)}}{{{\alpha ^2}{{\sin }^2}u}}} }} = \frac{2}{{\sqrt {\frac{1}{2}\left( {\frac{{1 - {{\cos }^2}u}}{{{\rm{si}}{{\rm{n}}^2}u}}} \right)} }} = 2\sqrt 2 $

$\mathop {\lim }\limits_{\alpha \to 0} T = \int\limits_0^{\frac{\pi }{2}} {2\sqrt 2 {\mathop{\rm dt}\nolimits} } = \pi \sqrt 2 $

(dominated convergence?)

2) T is a decreasing fonction of $\alpha $

??? No proof...


$\frac{1}{{2{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos \theta }}{2}}} \le \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos \theta }}{2}}}{{\alpha \cos \theta }}$, $\left( {{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right) \ge \alpha \cos \theta } \right)$

$T \ge 2\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\frac{{\varphi \left( \alpha \right) - \varphi \left( {\alpha \cos \theta } \right)}}{{{\alpha ^2}{{\sin }^2}\theta }}} }}\sqrt {\frac{{\alpha \cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right)}}} \left( {\frac{1}{{{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos \theta }}{2}}}} \right){\rm{d\theta }}} $

If $\delta \left( x \right) = \varphi \left( x \right) - 1 - \frac{{{x^2}}}{2}$

$\begin{array}{l}\delta '\left( x \right) & = \varphi '\left( x \right) - x = 2\coth x - \frac{x}{{{{{\mathop{\rm sh}\nolimits} }^2}x}} - \frac{1}{x} - x\\ & = \frac{1}{x}\left( {2x\coth x - {x^2}{{\coth }^2}x - 1} \right)\\ & = - \frac{1}{x}{\left( {1 - x\coth x} \right)^2} \le 0\end{array}$

$\delta$ is a decreasing fonction of $x$ on ${\mathbb{R}^ + }$

and for $x \ge y \ge 0$, $0 \le \varphi \left( x \right) - \varphi \left( y \right) \le \frac{{{x^2} - {y^2}}}{2}$

So $T \ge 2\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\frac{1}{2}\frac{{{\alpha ^2} - {\alpha ^2}{{\cos }^2}\theta }}{{{\alpha ^2}{{\sin }^2}\theta }}} }}\sqrt {\frac{{\alpha \cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right)}}} \left( {\frac{1}{{{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos \theta }}{2}}}} \right){\rm{d\theta }}} $

$T \ge 2\sqrt 2 \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{{\alpha \cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right)}}} \left( {\frac{1}{{{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos \theta }}{2}}}} \right){\rm{d\theta }}} $ which is a decreasing function of $\alpha$

for $\alpha \le 3$, $T \ge 2\sqrt 2 \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{{3\cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {3\cos \theta } \right)}}} \left( {\frac{1}{{{\mathop{\rm ch}\nolimits} \frac{{3\cos \theta }}{2}}}} \right){\rm{d\theta }}} \approx {\rm{2}}{\rm{.341298620}}$

For $x \ge 0$,

$\begin{array}{l}\varphi ''\left( x \right) & = \frac{{2x{\rm{ch}}x}}{{{{{\mathop{\rm sh}\nolimits} }^3}x}} - \frac{2}{{{{{\mathop{\rm sh}\nolimits} }^2}x}} + \frac{1}{{{x^2}}} - \frac{1}{{{{{\mathop{\rm sh}\nolimits} }^2}x}}\\ & = \frac{{2{\mathop{\rm ch}\nolimits} (x)}}{{{{{\mathop{\rm sh}\nolimits} }^3}x}}\left( {x - {\rm{th}}x} \right) + \frac{1}{{{x^2}}} - \frac{1}{{{{{\mathop{\rm sh}\nolimits} }^2}x}} \ge 0\end{array}$

and $\varphi '\left( x \right) \le 2 = \mathop {\lim }\limits_{ + \infty } \varphi '$

For $x \ge y \ge 0$ , $0 \le \varphi \left( x \right) - \varphi \left( y \right)\mathop = \limits_{c \in \left[ {x,y} \right]} \left( {x - y} \right)\varphi '\left( c \right) \le 2\left( {x - y} \right)$

$T \ge 2\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\frac{{\varphi \left( \alpha \right) - \varphi \left( {\alpha \cos \theta } \right)}}{{{\alpha ^2}{{\sin }^2}\theta }}} }}\sqrt {\frac{{\alpha \cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right)}}} \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos \theta }}{2}}}{{\alpha \cos \theta }}{\rm{d\theta }}} $

$T \ge 2\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{\sqrt {\frac{{2\left( {\alpha - \alpha \cos \theta } \right)}}{{{\alpha ^2}{{\sin }^2}\theta }}} }}\sqrt {\frac{{\alpha \cos \theta }}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos \theta } \right)}}} \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos \theta }}{2}}}{{\alpha \cos \theta }}{\rm{d\theta }}} = \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{{1 + \cos \theta }}{{\cos \theta }}} \sqrt {{\rm{th}}\frac{{\alpha \cos \theta }}{2}} {\rm{d\theta }}} $ which is an increasing function of $\alpha$.

for $\alpha \ge 3$, $T \ge \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{{1 + \cos \theta }}{{\cos \theta }}} \sqrt {{\rm{th}}\frac{{{\rm{3}}\cos \theta }}{2}} {\rm{d\theta }}} \approx {\rm{2}}{\rm{.123182323}}$

For all $\alpha > 0$ , $T \ge 2$

So the solutions of the problem are $y = 1$ and $y = -1$.

3) as $\alpha $ approaches $+ \infty $ , T approaches $\pi$

as $t$ approaches $+ \infty $ $\varphi \left( t \right) \sim 2t$

as $\alpha $ approaches $+ \infty $

$2\frac{{\sqrt {\frac{{\alpha \cos t}}{{{\mathop{\rm sh}\nolimits} \left( {\alpha \cos t} \right)}}} \left( {\frac{1}{{2{\mathop{\rm ch}\nolimits} \frac{{\alpha \cos t}}{2}}} + \frac{{{\mathop{\rm sh}\nolimits} \frac{{\alpha \cos t}}{2}}}{{\alpha \cos t}}} \right)}}{{\sqrt {\frac{{\varphi \left( \alpha \right) - \varphi \left( {\alpha \cos t} \right)}}{{{\alpha ^2}{{\sin }^2}t}}} }} \sim \frac{{2\sqrt {2\alpha \cos t} \,{e^{ - \frac{{\alpha \cos t}}{2}}}\frac{{{e^{\frac{{\alpha \cos t}}{2}}}}}{{2\alpha \cos t}}}}{{\sqrt {\frac{2}{\alpha }\left( {\frac{{1 - \cos t}}{{{{\sin }^2}t}}} \right)} }}$

$\frac{{2\sqrt {2\alpha \cos t} \,{e^{ - \frac{{\alpha \cos t}}{2}}}\frac{{{e^{\frac{{\alpha \cos t}}{2}}}}}{{2\alpha \cos t}}}}{{\sqrt {\frac{2}{\alpha }\left( {\frac{{1 - \cos t}}{{{{\sin }^2}t}}} \right)} }} = \frac{{\sqrt {1 + \cos t} }}{{\sqrt {\cos t} }}$

$\mathop {\lim }\limits_{\alpha \to + \infty } T = \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{{1 + \cos t}}{{\cos t}}} {\mathop{\rm dt}\nolimits} } = \int\limits_0^{\frac{\pi }{2}} {\sqrt {\frac{2}{{1 - 2{{\sin }^2}\frac{t}{2}}}} \cos \frac{t}{2}{\mathop{\rm dt}\nolimits} } $

With ${u = \sqrt 2 \sin \frac{t}{2}}$

$\mathop {\lim }\limits_{\alpha \to + \infty } T = \int\limits_0^1 {\sqrt {\frac{2}{{1 - {u^2}}}} \sqrt 2 {\rm{du}}} = \pi $

(dominated convergence?)

I think that exept for the solutions $y = 1$ and $y = -1$ the minimum period is $\pi$

I think that exept for the solutions $y=1$ and $y=−1$ the minimum period is $\pi$.


Nelix 1 year ago

In fact I tried with Maple. I think that T is a decreasing fonction of C from $\pi \sqrt 2$ to $\pi$

francois 1 year ago

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