Vychislitelny_praktikum
.pdfβ5Mk = |
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b5M k = |
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0.029948 |
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0.029851 |
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b5Mk |
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= 1 |
-0.094859 |
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k = 0 |
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0.210559 |
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0.210267 |
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0.434154 |
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0.433667 |
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-0.353464 |
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0.071981 |
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0.071299 |
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0.274272 |
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-0.41309 |
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0.189251 |
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0.190224 |
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0.186837 |
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-0.292016 |
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0.022924 |
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0.21321 |
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-0.139691 |
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0.6 |
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0.36 |
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β5M k |
0.12 |
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b5M k |
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− 0.12 |
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− 0.36 |
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− 0.60 |
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2. Построить графические зависимости δ от параметра χ; μ = 0 - 5, m = 2 - 6. Определить количество локальных экстремумов, значения параметра χopt и соотвествующие им значения погрешностей.
τ4 (ω) := 2 λ(2 + ω2 ) 4 λ λ2 + ω2
χ := γ
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⌠ |
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− λ τ |
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β5(k ,χ ,ω) := 2χ (k + 1) |
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P7(k ,τ ,χ) e |
cos (ω τ ) dτ |
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2 (−1) |
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− ∑ |
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β5(s ,χ ,ω) |
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b5(k ,χ ,ω) := β5(k ,χ ,ω) + |
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s = 0 |
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(m |
+ 1) (m + 2) |
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1 |
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(b5(k ,χ ,ω))2 |
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δ1(χ ,m,ω) := |
1 − |
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∑ |
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2 χ τ4 (ω) |
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k + 1 |
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k = 0 |
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i := 0 .. 100 |
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δ2V0i := δ1(i 0.1 + 0.01,2,0) |
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δ2V2i := δ1(i 0.1 + 0.01,2,2) |
δ2V4i := δ1(i 0.1 + 0.01,2,4) |
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δ2V1i := δ1(i 0.1 + 0.01,2,1) |
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δ2V3i := δ1(i 0.1 + 0.01,2,3) |
δ2V5i := δ1(i 0.1 + 0.01,2,5) |
δ2M 0 |
:= δ2V0 |
δ2M 2 := δ2V2 |
δ2M 5 := δ2V5 |
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δ2M |
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:= δ2V1 |
δ2M |
3 |
:= δ2V3 |
δ2M |
4 |
:= δ2V4 |
1 |
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1 |
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(δ2M 0 ) |
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(δ2M |
1 )i0.833 |
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i |
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(δ2M |
2 )0.667 |
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i |
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(δ2M |
3 ) |
0.5 |
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i |
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(δ2M |
4 ) |
i0.333 |
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(δ2M |
5 ) |
i0.167 |
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i 0.1+0.01 |
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δ3V0i |
:= δ1(i 0.1 + 0.01,2,5) |
δ3V2i |
:= δ1(i 0.1 + 0.01,4,5) |
δ3V4i := δ1(i 0.1 + 0.01,6,5) |
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δ3V1i |
:= δ1(i 0.1 + 0.01,3,5) |
δ3V3i |
:= δ1(i 0.1 + 0.01,5,5) |
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δ3M 0 |
:= δ3V0 |
δ3M 2 |
:= δ3V2 |
δ3M 4 := δ3V4 |
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δ3M 1 |
:= δ3V1 |
δ3M 3 |
:= δ3V3 |
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1 |
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(δ3M |
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i0.867 |
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(δ3M |
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i0.733 |
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(δ3M |
2 ) |
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(δ3M 3 ) |
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(δ3M |
4 )i0.467 |
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0.333 |
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0.2 0 |
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i 0.1+0.01 |
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0 ) |
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= 0.52 |
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min δ3M |
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3.41 |
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match |
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δ3M |
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0.1 |
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min |
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+ 0.01 = 3.51 |
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3.61 |
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χminM |
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:= match |
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δ3M |
0 ) |
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δ3M |
0 ) |
1 0.1 |
+ 0.01 |
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( |
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1 ) |
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= 0.439 |
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min δ3M |
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match |
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δ3M |
1 ) |
,δ3M |
1 ) |
0.1 + 0.01 = ( 1.81) |
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χminM |
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:= match |
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δ3M |
1 ) |
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δ3M |
1 ) |
0 0.1 |
+ 0.01 |
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1 |
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( |
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= 0.372 |
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min δ3M |
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match |
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δ3M |
2 ) |
,δ3M |
2 ) |
0.1 + 0.01 = ( 2.21) |
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χminM |
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:= match |
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δ3M |
2 ) |
, |
δ3M |
2 ) |
0 0.1 |
+ 0.01 |
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236
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0.6 |
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(δ5M 0 ) |
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(δ5M |
1 )m20.5 |
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m2 |
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(δ5M |
2 ) 0.4 |
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m2 |
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(δ5M |
3 ) 0.3 |
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m2 |
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(δ5M |
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m20.2 |
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(δ5M |
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m20.1 |
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2.8 |
3.6 |
4.4 |
5.2 |
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m2 |
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4. Построить графические зависимости δ от параметра μ, m; μ = 0 - 5, m = 2 - 6, χ1, χ2. |
t := |
1 |
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8 0.02 |
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λ |
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ω 2 |
− 1 |
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λ |
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χ1m2 := |
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0.4 |
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χ1m2 = |
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0.4 |
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(m2 + 1) |
t |
χ1c(mx) := |
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(mx + 1) t |
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1.633 |
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1.225 |
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0.98 |
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0.816 |
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0.7 |
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δ6V0ω1 |
:= δ1(χ12,2,ω1) |
δ6V2ω1 |
:= δ1(χ14,4,ω1) |
δ6V4ω1 := δ1(χ16,6,ω1) |
δ6V1ω1 |
:= δ1(χ13,3,ω1) |
δ6V3ω1 |
:= δ1(χ15,5,ω1) |
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δ6M 0 := δ6V0 |
δ6M 2 := δ6V2 |
δ6M 4 := δ6V4 |
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δ6M 1 |
:= δ6V1 |
δ6M 3 := δ6V3 |
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0.8 |
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(δ6M |
0 ) |
ω10.667 |
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(δ6M |
1 ) |
ω10.533 |
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(δ6M |
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ω1 0.4 |
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ω10.267 |
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(δ6M |
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ω1 |
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0.133 |
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0 |
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ω1 |
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δ7V0m2 := δ1(χ1m2,m2,0) |
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δ7V2m2 := δ1(χ1m2,m2,2) |
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δ7V4m2 := δ1(χ1m2,m2,4) |
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δ7V1m2 := δ1(χ1m2,m2,1) |
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δ7V3m2 := δ1(χ1m2,m2,3) |
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δ7V5m2 := δ1(χ1m2,m2,5) |
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δ7M 0 := δ7V0 |
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δ7M 2 := δ7V2 |
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δ7M 4 := δ7V4 |
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δ7M 1 |
:= δ7V1 |
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δ7M 3 |
:= δ7V3 |
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δ7M 5 := δ7V5 |
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0.8 |
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(δ7M 0 ) |
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(δ7M |
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m2 |
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(δ7M |
2 ) 0.533 |
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m2 |
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(δ7M |
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0.4 |
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m20.267 |
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(δ7M |
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m20.133 |
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2.8 |
3.6 |
4.4 |
5.2 |
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m2 |
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λ2 + ω2 |
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χ2m2 := (m2 + 1) |
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χ2 |
m2 |
= |
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χ2c(mx) := |
λ2 + ω2 |
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1.7 |
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(mx + 1) |
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1.275 |
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1.02 |
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0.85 |
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0.728 |
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δ8V0ω1 := δ1(χ22,2,ω1) |
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δ8V2ω1 := δ1(χ24,4,ω1) |
δ8V4ω1 := δ1(χ26,6,ω1) |
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δ8V1ω1 := δ1(χ23,3,ω1) |
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δ8V3ω1 := δ1(χ25,5,ω1) |
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δ8M 0 := δ8V0 |
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δ8M 2 := δ8V2 |
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δ8M 4 |
:= δ8V4 |
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δ8M 1 := δ8V1 |
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δ8M 3 := δ8V3 |
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241 |
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0.8 |
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(δ8M |
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ω10.667 |
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(δ8M |
1 ) |
ω10.533 |
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(δ8M |
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ω1 0.4 |
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(δ8M 3 ) |
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ω10.267 |
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(δ8M |
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ω1 |
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0.133 |
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0 |
1 |
2 |
3 |
4 |
5 |
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ω1 |
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δ9V0m2 := δ1(χ2m2,m2,0) |
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δ9V2m2 := δ1(χ2m2,m2,2) |
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δ9V4m2 := δ1(χ2m2,m2,4) |
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δ9V1m2 := δ1(χ2m2,m2,1) |
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δ9V3m2 := δ1(χ2m2,m2,3) |
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δ9V5m2 := δ1(χ2m2,m2,5) |
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δ9M 0 := δ9V0 |
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δ9M 2 := δ9V2 |
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δ9M 4 := δ9V4 |
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δ9M 1 |
:= δ9V1 |
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δ9M 3 |
:= δ9V3 |
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δ9M 5 := δ9V5 |
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0.8 |
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(δ9M 0 ) |
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|||
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1 )m20.667 |
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||
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(δ9M |
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||||
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m2 |
||
(δ9M |
2 ) 0.533 |
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|||||
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m2 |
||
(δ9M |
3 ) |
0.4 |
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||
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|||||
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m2 |
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(δ9M |
4 ) |
|
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||
m20.267 |
|||||
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(δ9M |
5 ) |
m20.133 |
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||
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|
|
2 |
2.8 |
3.6 |
4.4 |
5.2 |
6 |
m2
242