Complexité algo :
O ( 1 ) {\displaystyle {\mathcal {O}}(1)} O ( log n ) {\displaystyle {\mathcal {O}}(\log {n})} O ( n ) {\displaystyle {\mathcal {O}}(n)} O ( n 2 ) {\displaystyle {\mathcal {O}}(n^{2})} O ( n 3 ) {\displaystyle {\mathcal {O}}(n^{3})} O ( n log n ) {\displaystyle {\mathcal {O}}(n\log {n})} O ( 2 n 1 / 3 ) {\displaystyle {\mathcal {O}}(2^{n^{1/3}})} O ( n ! ) {\displaystyle {\mathcal {O}}(n!)} f ( n ) = O ( g ( n ) ) ⇔ lim n ↦ ∞ sup | f ( n ) g ( n ) | < ∞ ⇔ ∃ x O , ∃ M > 0 / ∀ x > x 0 , | f ( x ) | ≤ M | g ( x ) | {\displaystyle f(n)={\mathcal {O}}(g(n))\Leftrightarrow \lim _{n\mapsto \infty }\sup \left|{\frac {f(n)}{g(n)}}\right|<\infty \Leftrightarrow \exists x_{O},\exists M>0\,/\,\forall x>x_{0},|f(x)|\leq M|g(x)|} 3 n 3 + 2 n 2 {\displaystyle 3n^{3}+2n^{2}} 3 n 3 {\displaystyle 3n^{3}} n 2 {\displaystyle n^{2}} n 3 {\displaystyle n^{3}} O ( n a ) {\displaystyle {\mathcal {O}}(n^{a})} O ( n b ) {\displaystyle {\mathcal {O}}(n^{b})} a < b {\displaystyle a<b} P ⊂ NP {\displaystyle {\mbox{P}}\subset {\mbox{NP}}} P = NP {\displaystyle {\mbox{P}}={\mbox{NP}}} t i {\displaystyle t_{i}} t ( n ) = t 1 + t 2 + n t 3 + n t 4 + t 5 + t 6 = O ( n ) {\displaystyle t(n)=t_{1}+t_{2}+nt_{3}+nt_{4}+t_{5}+t_{6}={\mathcal {O}}(n)} t 5 {\displaystyle t_{5}} 2 t 5 {\displaystyle 2t_{5}} 2 t 4 {\displaystyle 2t_{4}} 3 t 4 {\displaystyle 3t_{4}} t 5 + 2 t 5 + . . . + ( n − 1 ) t 5 + n t 5 = n ( n + 1 ) 2 t 5 {\displaystyle t_{5}+2t_{5}+...+(n-1)t_{5}+nt_{5}={\frac {n(n+1)}{2}}t_{5}} 2 t 4 + 3 t 4 + . . . + n t 4 + ( n + 1 ) t 4 = t 4 ( 1 + 2 + 3 + . . . + n + ( n + 1 ) ) − t 4 = t 4 ( n + 1 ) ( n + 2 ) 2 − t 4 {\displaystyle {\begin{aligned}2t_{4}+3t_{4}+...+nt_{4}+(n+1)t_{4}&=t_{4}\left(1+2+3+...+n+(n+1)\right)-t_{4}\\&=t_{4}{\frac {(n+1)(n+2)}{2}}-t_{4}\\\end{aligned}}} t ( n ) = t 1 + t 2 + n t 3 + t 4 ( n + 1 ) ( n + 2 ) 2 − t 4 + n ( n + 1 ) 2 t 5 + n t 6 + t 7 + t 8 = n 2 ( t 4 + t 5 2 ) + n ( 3 t 4 + t 5 2 + t 3 + t 6 ) + ( t 1 + t 2 + t 7 + t 8 ) = O ( n 2 ) {\displaystyle {\begin{aligned}t(n)&=t_{1}+t_{2}+nt_{3}+t_{4}{\frac {(n+1)(n+2)}{2}}-t_{4}+{\frac {n(n+1)}{2}}t_{5}+nt_{6}+t_{7}+t_{8}\\&=n^{2}\left({\frac {t_{4}+t_{5}}{2}}\right)+n\left({\frac {3t_{4}+t_{5}}{2}}+t_{3}+t_{6}\right)+(t_{1}+t_{2}+t_{7}+t_{8})\\&={\mathcal {O}}(n^{2})\end{aligned}}} L = { [ 1 , b ] , [ 2 , b ] , [ 3 , a ] } {\displaystyle L=\left\{[1,b],[2,b],[3,a]\right\}} t r i 1 ( L ) = { [ 3 , a ] , [ 1 , b ] , [ 2 , b ] } {\displaystyle tri_{1}(L)=\left\{[3,a],[1,b],[2,b]\right\}} t r i 2 ( L ) = { [ 3 , a ] , [ 2 , b ] , [ 1 , b ] } {\displaystyle tri_{2}(L)=\left\{[3,a],[2,b],[1,b]\right\}} t ( n ) = t 1 + n t 2 + ( n − 1 ) t 3 + ( ∑ i = 2 n i ) t 4 + ( ∑ i = 2 n ( i − 1 ) ) t 5 + ( n − 1 ) t 6 + t 7 + t 8 = t 1 + n t 2 + ( n − 1 ) t 3 + ( n ( n − 1 ) 2 − 1 ) t 4 + ( n ( n − 1 ) 2 ) t 5 + ( n − 1 ) t 6 + t 7 + t 8 = n 2 ( t 4 + t 5 2 ) + n ( 2 t 2 + 2 t 3 − t 4 − t 5 + 2 t 6 2 ) + ( t 1 − t 3 − t 4 − t 6 + t 7 + t 8 ) {\displaystyle {\begin{aligned}t(n)&=t_{1}+nt_{2}+(n-1)t_{3}+\left(\sum _{i=2}^{n}i\right)t_{4}+\left(\sum _{i=2}^{n}(i-1)\right)t_{5}+(n-1)t_{6}+t_{7}+t_{8}\\&=t_{1}+nt_{2}+(n-1)t_{3}+\left({\frac {n(n-1)}{2}}-1\right)t_{4}+\left({\frac {n(n-1)}{2}}\right)t_{5}+(n-1)t_{6}+t_{7}+t_{8}\\&=n^{2}\left({\frac {t_{4}+t_{5}}{2}}\right)+n\left({\frac {2t_{2}+2t_{3}-t_{4}-t_{5}+2t_{6}}{2}}\right)+(t_{1}-t_{3}-t_{4}-t_{6}+t_{7}+t_{8})\end{aligned}}} 2 n 2 {\displaystyle 2n^{2}} 1000 n 2 {\displaystyle 1000n^{2}}
SI :
θ ˙ = − 2 x ( t ) x ˙ 4 l 3 2 ( l 2 2 + l 1 2 ) − ( x 2 − l 1 2 − l 2 2 − l 3 2 ) 2 {\displaystyle {\dot {\theta }}={\frac {-2x(t){\dot {x}}}{\sqrt {4l_{3}^{2}(l_{2}^{2}+l_{1}^{2})-(x^{2}-l_{1}^{2}-l_{2}^{2}-l_{3}^{2})^{2}}}}} θ ˙ β ˙ = − 2 p 2 π ( p 2 π β + l 2 2 + ( l 1 + l 3 ) 2 ) 1 − ( p 2 π β + l 2 2 + ( l 1 + l 3 ) 2 ) 2 − ( l 1 2 + l 2 2 + l 3 2 ) 2 l 3 l 1 2 + l 2 2 {\displaystyle {\frac {\dot {\theta }}{\dot {\beta }}}={\frac {-2{\frac {p}{2\pi }}\left({\frac {p}{2\pi }}\beta +{\sqrt {l_{2}^{2}+(l_{1}+l_{3})^{2}}}\right)}{\sqrt {1-{\frac {\left({\frac {p}{2\pi }}\beta +{\sqrt {l_{2}^{2}+(l_{1}+l_{3})^{2}}}\right)^{2}-(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})}{2l_{3}{\sqrt {l_{1}^{2}+l_{2}^{2}}}}}}}}} C ( t ) = K p ϵ ( t ) + K i ∫ ϵ ( t ) d t + K d d ϵ ( t ) d t {\displaystyle C(t)=K_{p}\epsilon (t)+K_{i}\int \epsilon (t)dt+K_{d}{\frac {d\epsilon (t)}{dt}}} K 1 − e − t τ {\displaystyle {\frac {K}{1-e^{-{\frac {t}{\tau }}}}}} K 1 + 2 ξ p ω 0 + p 2 ω 0 2 {\displaystyle {\frac {K}{1+{\frac {2\xi p}{\omega _{0}}}+{\frac {p^{2}}{\omega _{0}^{2}}}}}}
Cette année, je fais mon TIPE sur les mémoires holographiques.
![{\displaystyle L=\left\{[1,b],[2,b],[3,a]\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ffb9f6599f563218a3328335b168abec23fc7ebd)
![{\displaystyle tri_{1}(L)=\left\{[3,a],[1,b],[2,b]\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39cf7932082d66a81a67d15abeb0d994d02bb228)
![{\displaystyle tri_{2}(L)=\left\{[3,a],[2,b],[1,b]\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f0e8d9dfe6d57b6c5d70e7c40fb39b48a649d481)
