Taller3
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18 DE OCTUBRE
SOLUCION ANALITICA primera forma
enf :=
for x ∈ 0 .. 200
(
y ← −50⋅ e x
) + 100
En esta forma x esun subindice
− 0.045 ⋅ x
y i := 0 .. 200
0 0 1 2 3 4 5 6 enf = 7 i 8 9 10 11 12 13 14 15
50 52.2 54.303 56.314 58.236 60.074 61.831 63.511 65.116 66.651 68.119 69.521 70.863 72.145 73.37 74.542
100
80 enfi 60
40
0
50
100
150
200
i
Segunda forma i := 0 , 2 .. 200
se incrementa de 2 en 2 y puede tomar varios valores
enf1 := h ← 2 for x ∈ 0 , 2 .. 200 y ← −50⋅ e x
x← x+ h y
− 0.045 ⋅ x
+ 100
enf1 = i
100
50 54.303 58.236 61.831 65.116 68.119 70.863 73.37 75.662 77.757 79.672 81.421 83.02 84.482 85.817 87.038
80 enf1i 60
40
0
50
100
150
200
i
Tercera Forma
enf2( n ) :=
for x ∈ 1 .. n enf2 ← −50⋅ e
− 0.045 ⋅ x
la variable x ya no esta como subindice y puede ser valor decimal
+ 100
enf2 n := 0 , 2.5.. 200 0 0 1 2 3 4 5 6 enf2( n ) = 7 8 9 10 11 12 13 14 15
SOLUCION NUMERICA
50 54.303 60.074 63.511 68.119 70.863 74.542 76.733 79.672 81.421 83.767 85.164 87.038 88.154 89.65 90.54
100
80 enf2( n) 60
40
0
50
100 n
150
200
Error aproximado
Verdadero
enf3 := x ← 0 enf3 − enf30 i+ 1 i EA := ⋅ 100 T ← 100 i 0 enf3 i+ 1 k ← 0.045
EV :=
enf − enf3 i
i
enf
i
⋅ 100
i
ta ← 100 h ← 0.5
EA = i
2.2 2.132 2.067 2.005 1.946 1.889 1.835 1.783 1.733 1.685 1.64 1.595 1.553 1.512 1.473 1.435
EV = i
for i ∈ 0 .. 100 T
i+ 1
x
i+ 1
← ( −k⋅ i + ta⋅ k) ⋅ h + T
100 95.881 92.396 89.442 86.936 84.813 83.019 81.509 80.247 79.202 78.347 77.661 77.124 76.721 76.437 76.259
i
←x +h i
T i := 0 .. 100 x := 0 0
h := 0.5 x
i+ 1
:= x + h i
x =
enf3 =
i
i
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5
100 102.25 104.478 106.683 108.865 111.025 113.163 115.278 117.37 119.44 121.487 123.513 125.515 127.495 129.453 131.388
100
80 enfi 60
40
0
10
20
30 xi
40
50
SOLUCION ANALITICA primera forma
enf :=
for x ∈ 0 .. 200
(
y ← −50⋅ e x
) + 100
En esta forma x esun subindice
− 0.045 ⋅ x
y i := 0 .. 200
0 0 1 2 3 4 5 6 enf = 7 i 8 9 10 11 12 13 14 15
50 52.2 54.303 56.314 58.236 60.074 61.831 63.511 65.116 66.651 68.119 69.521 70.863 72.145 73.37 74.542
100
80 enfi 60
40
0
50
100
150
200
i
Segunda forma i := 0 , 2 .. 200
se incrementa de 2 en 2 y puede tomar varios valores
enf1 := h ← 2 for x ∈ 0 , 2 .. 200 y ← −50⋅ e x
x← x+ h y
− 0.045 ⋅ x
+ 100
enf1 = i
100
50 54.303 58.236 61.831 65.116 68.119 70.863 73.37 75.662 77.757 79.672 81.421 83.02 84.482 85.817 87.038
80 enf1i 60
40
0
50
100
150
200
i
Tercera Forma
enf2( n ) :=
for x ∈ 1 .. n enf2 ← −50⋅ e
− 0.045 ⋅ x
la variable x ya no esta como subindice y puede ser valor decimal
+ 100
enf2 n := 0 , 2.5.. 200 0 0 1 2 3 4 5 6 enf2( n ) = 7 8 9 10 11 12 13 14 15
SOLUCION NUMERICA
50 54.303 60.074 63.511 68.119 70.863 74.542 76.733 79.672 81.421 83.767 85.164 87.038 88.154 89.65 90.54
100
80 enf2( n) 60
40
0
50
100 n
150
200
Error aproximado
Verdadero
enf3 := x ← 0 enf3 − enf30 i+ 1 i EA := ⋅ 100 T ← 100 i 0 enf3 i+ 1 k ← 0.045
EV :=
enf − enf3 i
i
enf
i
⋅ 100
i
ta ← 100 h ← 0.5
EA = i
2.2 2.132 2.067 2.005 1.946 1.889 1.835 1.783 1.733 1.685 1.64 1.595 1.553 1.512 1.473 1.435
EV = i
for i ∈ 0 .. 100 T
i+ 1
x
i+ 1
← ( −k⋅ i + ta⋅ k) ⋅ h + T
100 95.881 92.396 89.442 86.936 84.813 83.019 81.509 80.247 79.202 78.347 77.661 77.124 76.721 76.437 76.259
i
←x +h i
T i := 0 .. 100 x := 0 0
h := 0.5 x
i+ 1
:= x + h i
x =
enf3 =
i
i
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5
100 102.25 104.478 106.683 108.865 111.025 113.163 115.278 117.37 119.44 121.487 123.513 125.515 127.495 129.453 131.388
100
80 enfi 60
40
0
10
20
30 xi
40
50
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