## Fun integral problem

Aryan
Posts: 12
Joined: Sat Mar 16, 2024 11:46 am

### Fun integral problem

Let $P(x) = (x - 3)(x - 7)(x - 8)$. Determine the value of $$\int_{0}^{12} \underbrace{P(P(\ldots P(x) \ldots))}_{69 \text{ times}} \, dx.$$
Riemann
Posts: 178
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

### Re: Fun integral problem

First of all we note that $\mathrm{P}(x) + \mathrm{P} \left ( 12 - x \right ) = 12$ since

\begin{align*}
\mathrm{P}(x) - 12 &= \left ( x -4 \right ) \left ( x - 5 \right ) \left ( x -9 \right ) \\
&= -\left ( 4 - x \right ) \left ( 5- x \right ) \left ( 9 - x \right ) \\
&= - \left [ \left ( 12 - x \right ) - 8 \right ] \left [ \left ( 12 - x \right ) - 7 \right ] \left [ \left ( 12 - x \right ) - 3 \right ] \\
&= - \mathrm{P} \left ( 12 - x \right )
\end{align*}

because the roots of the polynomial $\mathrm{P}(x) - 12 = (x-3)(x-7)(x-8) - 12$ are $4, \,5,\,9$. Inductively, we will prove that $\mathrm{P}^{(n)}(x) + \mathrm{P}^{(n)} \left ( 12 - x \right ) = 12$ for all $x \in \mathbb{R}$ where $\mathrm{P}^{(n)} (x) = \underbrace{P\left ( P\left ( ... P\left ( P(x) \right ) ... \right ) \right )}_{n \; \text{times}}$. To this end, for $n=1$ the last relation is valid. If we assume that it holds for $n=k$, we'll prove that it also holds for $n=k+1$. Indeed,

\begin{align*}
\mathrm{P}^{(k+1)} (12 - x) &= \mathrm{P} \left ( \mathrm{P}^{(k)}(12 - x) \right ) \\
&= \mathrm{P} \left ( 12 - \mathrm{P}^{(k)}(x) \right ) \\
&= 12 - \mathrm{P} \left ( \mathrm{P}^{(k)}(x) \right ) \\
&= 12 - \mathrm{P}^{(k+1)}(x)
\end{align*}

Finally,

\begin{align*}
\int_{0}^{12} \mathrm{P}^{\left (n \right )}(x) \, \mathrm{d}x & \overset{x \mapsto 12-x}{=\! =\! =\! =\! =\! =\!} \int_{0}^{12} \mathrm{P}^{(n)} \left ( 12 -x \right )\, \mathrm{d}x \\
&= \frac{1}{2} \int_{0}^{12} \left ( \mathrm{P}^{\left ( n \right )}(x) + \mathrm{P}^{(n)} \left ( 12 -x \right ) \right ) \, \mathrm{d}x \\
&= \frac{1}{2} \int_{0}^{12} 12 \, \mathrm{d}x \\
&= \frac{12 \cdot 12}{2} \\
&= 72
\end{align*}
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Aryan
Posts: 12
Joined: Sat Mar 16, 2024 11:46 am

### Re: Fun integral problem

Riemann wrote: Fri Mar 22, 2024 7:16 pm First of all we note that $\mathrm{P}(x) + \mathrm{P} \left ( 12 - x \right ) = 12$ since

\begin{align*}
\mathrm{P}(x) - 12 &= \left ( x -4 \right ) \left ( x - 5 \right ) \left ( x -9 \right ) \\
&= -\left ( 4 - x \right ) \left ( 5- x \right ) \left ( 9 - x \right ) \\
&= - \left [ \left ( 12 - x \right ) - 8 \right ] \left [ \left ( 12 - x \right ) - 7 \right ] \left [ \left ( 12 - x \right ) - 3 \right ] \\
&= - \mathrm{P} \left ( 12 - x \right )
\end{align*}

because the roots of the polynomial $\mathrm{P}(x) - 12 = (x-3)(x-7)(x-8) - 12$ are $4, \,5,\,9$. Inductively, we will prove that $\mathrm{P}^{(n)}(x) + \mathrm{P}^{(n)} \left ( 12 - x \right ) = 12$ for all $x \in \mathbb{R}$ where $\mathrm{P}^{(n)} (x) = \underbrace{P\left ( P\left ( ... P\left ( P(x) \right ) ... \right ) \right )}_{n \; \text{times}}$. To this end, for $n=1$ the last relation is valid. If we assume that it holds for $n=k$, we'll prove that it also holds for $n=k+1$. Indeed,

\begin{align*}
\mathrm{P}^{(k+1)} (12 - x) &= \mathrm{P} \left ( \mathrm{P}^{(k)}(12 - x) \right ) \\
&= \mathrm{P} \left ( 12 - \mathrm{P}^{(k)}(x) \right ) \\
&= 12 - \mathrm{P} \left ( \mathrm{P}^{(k)}(x) \right ) \\
&= 12 - \mathrm{P}^{(k+1)}(x)
\end{align*}

Finally,

\begin{align*}
\int_{0}^{12} \mathrm{P}^{\left (n \right )}(x) \, \mathrm{d}x & \overset{x \mapsto 12-x}{=\! =\! =\! =\! =\! =\!} \int_{0}^{12} \mathrm{P}^{(n)} \left ( 12 -x \right )\, \mathrm{d}x \\
&= \frac{1}{2} \int_{0}^{12} \left ( \mathrm{P}^{\left ( n \right )}(x) + \mathrm{P}^{(n)} \left ( 12 -x \right ) \right ) \, \mathrm{d}x \\
&= \frac{1}{2} \int_{0}^{12} 12 \, \mathrm{d}x \\
&= \frac{12 \cdot 12}{2} \\
&= 72
\end{align*}
Great

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