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\[
I_3 = \displaystyle\int_{0}^{10}\left\lfloor\left(\dfrac{1+\sqrt{5}}{2}\right)^{\left\lfloor x\right\rfloor}\right\rfloor dx
\]
\[
I_3 = \sum_{k=0}^{9}
\displaystyle\int_{k}^{k+1}\left\lfloor\left(\dfrac{1+\sqrt{5}}{2}\right)^{\left\lfloor x\right\rfloor}\right\rfloor dx
= \sum_{k=0}^{9}
\displaystyle\int_{k}^{k+1}\left\lfloor\left(\dfrac{1+\sqrt{5}}{2}\right)^{k}\right\rfloor dx
= \sum_{k=0}^{9}
\left\lfloor\left(\dfrac{1+\sqrt{5}}{2}\right)^{k}\right\rfloor
= \sum_{k=0}^{9}
\left\lfloor\varphi^{k}\right\rfloor
\]
\[
I_3 = \sum_{k=0}^{9}\left\lfloor\varphi^{k}\right\rfloor
= \left\lfloor\varphi^{0}\right\rfloor
+ \left\lfloor\varphi^{1}\right\rfloor
+ \left\lfloor\varphi^{2}\right\rfloor
+ \left\lfloor\varphi^{3}\right\rfloor
+ \left\lfloor\varphi^{4}\right\rfloor
+ \left\lfloor\varphi^{5}\right\rfloor
+ \left\lfloor\varphi^{6}\right\rfloor
+ \left\lfloor\varphi^{7}\right\rfloor
+ \left\lfloor\varphi^{8}\right\rfloor
+ \left\lfloor\varphi^{9}\right\rfloor
\]
Podem suavitzar un xic el càlcul utilitzant recurrencia amb la golden ratio
\[\varphi=\dfrac{1+\sqrt{5}}{2}=1,6180339887498948482045868343656...\]
\[
\varphi^2=\left(\dfrac{1+\sqrt{5}}{2}\right)^2
=\dfrac{1+5+2\sqrt{5}}{4}
=\dfrac{6+2\sqrt{5}}{4}
=\dfrac{3+\sqrt{5}}{2}
= 1 + \varphi
\]
Amb la relació de recurrència
\[
\begin{array}{rl}
\varphi^{0} &= 1\\
\varphi^{1} &= \varphi\\
\varphi^{2} &= \varphi+1\\
\varphi^{3} &= \varphi(\varphi+1)= \varphi^2+\varphi=2\varphi+1\\
\varphi^{4} &= \varphi(2\varphi+1)= 2\varphi^2+\varphi=3\varphi+2\\
\varphi^{5} &= \varphi(3\varphi+2)= 3\varphi^2+2\varphi=5\varphi+3\\
\varphi^{6} &= \varphi(5\varphi+3)= 5\varphi^2+3\varphi=8\varphi+5\\
\varphi^{7} &= \varphi(8\varphi+5)= 8\varphi^2+5\varphi=13\varphi+8\\
\varphi^{8} &= \varphi(13\varphi+8)= 13\varphi^2+8\varphi=21\varphi+13\\
\varphi^{9} &= \varphi(21\varphi+13)= 21\varphi^2+13\varphi=34\varphi+21
\end{array}
\]
Hem d'entendre que
\[
\left\lfloor z+n\right\rfloor =\left\lfloor z\right\rfloor+n
\hspace{36pt}
\left\lfloor n\cdot z\right\rfloor \neq n\cdot \left\lfloor z\right\rfloor
\]
llavors
\[
\begin{array}{rllcr}
\lfloor\varphi^{0}\rfloor &= 1 &= 1 & = & 1 \\
\lfloor\varphi^{1}\rfloor &= \lfloor\varphi\rfloor &= \lfloor 1.618...\rfloor & = & 1 \\
\lfloor\varphi^{2}\rfloor &= \lfloor\varphi\rfloor+1 &= \lfloor 1.618...\rfloor +1 & = & 2 \\
\lfloor\varphi^{3}\rfloor &= \lfloor2\varphi\rfloor+1 &= \lfloor 3.236...\rfloor +1 & = & 4 \\
\lfloor\varphi^{4}\rfloor &= \lfloor3\varphi\rfloor+2 &= \lfloor 4.854...\rfloor +2 & = & 6 \\
\lfloor\varphi^{5}\rfloor &= \lfloor5\varphi\rfloor+3 &= \lfloor 8.090...\rfloor +3 & = & 11 \\
\lfloor\varphi^{6}\rfloor &= \lfloor8\varphi\rfloor+5 &= \lfloor 12.944...\rfloor +5 & = & 17 \\
\lfloor\varphi^{7}\rfloor &= \lfloor13\varphi\rfloor+8 &= \lfloor 21.034...\rfloor +8 & = & 29 \\
\lfloor\varphi^{8}\rfloor &= \lfloor21\varphi\rfloor+13 &= \lfloor 33.978...\rfloor +13 & = & 46 \\
\lfloor\varphi^{9}\rfloor &= \lfloor34\varphi\rfloor+21 &= \lfloor 55.013...\rfloor +21 & = & 76 \\
\end{array}
\]
Llavors
\[
I_3 = \sum_{k=0}^{9}\left\lfloor\varphi^{k}\right\rfloor
= 1
+ 1
+ 2
+ 4
+ 6
+ 11
+ 17
+ 29
+ 46
+ 76 = 193
\]
\[ I_3 = \displaystyle\int_{0}^{10}\left\lfloor\left(\dfrac{1+\sqrt{5}}{2}\right)^{\left\lfloor x\right\rfloor}\right\rfloor dx = 193 \]
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