Au sujet de Maple - Université de Bourgogne

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5/6
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0.8333333333
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1/2−1/2√540
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228826127
2−102334155 2√5
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0.0
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0.000000004370130127
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10
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20
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1/2−1/2√540
>
228826127
2−102334155 2√5
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0.00000000437
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10
1/2−1/2√540
>
228826127
2−102334155 2√5
>
228826127
2+102334155 2√5
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228826127
2−102334155 2√5228826127 2+102334155 2√5
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1
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0.0
0.000000004370130339
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>
>
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−1,1/2+1/2𝑖√3,1/2−1/2𝑖√3
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exprseq
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[−1,1/2+1/2𝑖√33,1/2−1/2𝑖√33]
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[−1,−1,−1]
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1/2+1/2𝑖√3
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𝑥2−𝑥+ 1
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>
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𝑥2+ 2 𝑥+ 1
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𝑛
𝑖=1
𝑖𝑝
𝑛 𝑝
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>
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(𝑝𝑛)𝑠, 𝑘;𝑠:= 0; 𝑘 𝑛 𝑠 := 𝑠+𝑘ˆ𝑝; ; 𝑠;
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385
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10
𝑖=1 𝑖2= 385
>
𝑛
𝑢2𝑢1𝑢2> 𝑢1𝑢3𝑢2
𝑚:= 𝑢1, 𝑗 := 1
𝑖2𝑛
𝑢𝑖> 𝑚 𝑚 := 𝑢𝑖𝑗:= 𝑖
𝑗
𝑧∈𝑧2
2𝑧+ 3𝑖𝑧𝑧2 2𝑧+ 3𝑖
𝑓𝑛(𝑥) = 𝑛𝛼𝑥exp(−𝑛𝑥) [0,+∞]𝛼
(𝑥𝑛) (𝑥𝑛)𝑛exp(𝑥𝑛)=1 𝑥𝑛>0
𝑥𝑛
𝑛= 10
𝑥𝑛

𝑢𝑛+1 =𝑢𝑛+ 4,
𝑣𝑛+2 =𝑣𝑛+1 + 2𝑣𝑛, 𝑣0= 4, 𝑣1=−2.
𝑠𝑘+1 =9
4𝑠𝑘−1 2𝑠𝑘−1, 𝑠1=1 3, 𝑠2=1 12.
𝑘lg[2](∣𝑠𝑘∣)
𝑠𝑘
𝑢𝑛+2 =𝑢𝑛+1 +𝑢𝑛∀𝑛≥0, 𝑢0=𝑢1= 1.
𝛼𝑛𝑛 𝛼 × ⋅⋅⋅ × 𝛼 𝛼 𝑛
𝛼𝑛=𝛼𝑛
2×𝛼𝑛 2𝑛
𝛼×𝛼𝑛−1
2×𝛼𝑛−1 2𝑛.
𝑛2
∂𝑢
∂𝑡 =∂2𝑢 ∂𝑥2, 𝑢(0, 𝑡) = 0 = 𝑢(𝜋, 𝑡)∀𝑡≥0, 𝑢(𝑥, 0) = 1 ∀𝑥∈]0, 𝜋[.
𝑢(𝑥, 𝑡) = Φ(𝑥)𝑇(𝑡)
𝐴𝑛𝑠𝑖𝑛(𝑛𝑥)𝑒−𝑛2𝑡
𝐴𝑛