Prova de Métodos Computacionais da Física B - números/caminhantes aleatórios (com programas)

Métodos Computacionais da Física B

•
UFRGS

Luan Bottin De Toni
01/07/2017
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Prova 3
Luan Bottin De Toni
IF-UFRGS
30 de junho de 2017
1 Questão 1
1.1 (a)
A partir de números aleatórios X gerados pela função "drand48"fez-se a
média < X > desses números por:
< X >=
1
N
N∑
i=1
Xi , (1)
onde N é a quantidade de números aleatórios gerados. X varia de 0 a 100.
Vale notar que a função drand48 retorna um número entre 0 e 1 e, aﬁm
de considerar um intervalo de 0 a 100, multiplica-se o valor retornado pela
função por 100, guardando em uma variável inteira; isto acarreta que os
números analisados irão variar de 0 a 99, de modo que sua média analítica é
49,5.
Para analisar o comportamento da média < X > em função de N fez-se
um algoritmo para calcular a média dos números aleatórios gerados variando
N de 1 a 1000 (programa ./Q1/ex1.c). O gráﬁco é apresentado a seguir:
1
 0
 10
 20
 30
 40
 50
 60
 70
 80
 90
 100
 0 100 200 300 400 500 600 700 800 900 1000
<
X>
N
dados
<X>=49,5
Figura 1: Média < X > dos números aleatórios em função da quantidade
N .
Note que a média calculada se aproxima do valor esperado analiticamente
à medida que N cresce. Isso é um sinal que os números aleatórios distribuem-
se de forma uniforme no intervalo considerado.
1.2 (b)
O mesmo programa do item anterior também calculou o desvio padrão
σ por:
σ =
√
< X2 > − < X >2 (2)
Analiticamente obtém-se o valor de σ = 28, 87. O gráﬁco 2 mostra o
comportamento de tal grandeza em função de N .
Novamente nota-se que o desvio padrão aproxima-se do valor analítico
conforme N cresce.
2
 0
 5
 10
 15
 20
 25
 30
 35
 40
 0 100 200 300 400 500 600 700 800 900 1000
σ
(X
)
N
dados
σ(X)=29,87
Figura 2: Desvio padrão de < X > em função da quantidade N .
1.3 (c)
Os itens anteriores dão a entender que a distribuição dos números aleató-
rios gerados é uniforme. Para ver isso mais claramente fez-se um histograma
com a frequência de cada número, no mesmo programa utilizado anterior-
mente.
 0
 1000
 2000
 3000
 4000
 5000
 6000
 0 20 40 60 80 100
n
X
Figura 3: Histograma de frequência dos números aleatórios X.
3
1.4 (d)
Deﬁnindo-se uma nova variável:
< Y >=
1
20
20∑
i=1
Xi , (3)
e considerando tal variável como sendo o número aleatório utilizado nos itens
anteriores (programa ./Q1/ex2.c), obtém-se os gráﬁcos:
 0
 10
 20
 30
 40
 50
 60
 70
 80
 90
 100
 0 100 200 300 400 500 600 700 800 900 1000
<
Y>
N
dados
Figura 4: Média < Y > dos números aleatórios em função da quantidade N .
 0
 2
 4
 6
 8
 10
 0 100 200 300 400 500 600 700 800 900 1000
σ
(Y
)
N
dados
Figura 5: Desvio padrão de < Y > em função da quantidade N .
4
 0
 5000
 10000
 15000
 20000
 25000
 30000
 35000
 0 20 40 60 80 100
fre
qu
ên
cia
Y
Figura 6: Histograma de frequência dos números aleatórios Y .
1.5 (e)
Os gráﬁcos no item "d"mostram comportamentos diferentes, isso era es-
perado, pois a nova variável se trata de uma média de 20 números aleatórios,
ou seja, seu valor estará mais próximo da média calculada nos primeiros
itens da questão. O histograma 6 mostra muito bem isso, a maioria dos
números < Y > gerados estão no valor da média, e a probabilidade de en-
contrar um número aleatório mais afastado de tal valor diminui conforme o
valor se afasta da média, de forma que o histograma assume a forma de uma
gaussiana.
2 Questão 2
Sendo a função densidade de probabilidade fdp(x) deﬁnida como e−x,
sua função cumulativa fdc(x) = R no intervalo x = 1 até x = 7 nos permite
gerar números x de acordo com esta distribuição, com:
x = −ln(e−1 −R.N) , (4)
onde N é a normalização, é é igual a (e6 − 1)/e7; e R é o número aleatório
gerado.
Faz-se um histograma (ﬁgura 7) da frequência dos números gerados em
um intervalo de modo que o histograma possua 100 bins entre o valor mínimo
e máximo de x, e colocou-se juntamente a função e−x multiplicada por uma
constante a ﬁm de ilustrar o comportamente similar do histograma com
a função. Mostrando que os números gerados obedecem tal distribuição.
(programa ./Q2/exp.c)
5
 0
 100
 200
 300
 400
 500
 600
 1 2 3 4 5 6 7
fre
qu
ên
cia
x
hist
c.e−x
Figura 7: Histograma dos números gerados segundo a distribuição da
fdp(x) = e−x no intervalo 1 < x < 7.
3 Questão 3
3.1 (a)
Para fdp(x) = x4 no intervalo de x = −3 até x = 3, os números gerados
de acordo com tal distribuição são:
x = (5.R.N − 243)(1/5) , N = 486
5
. (5)
O programa que gera tais números está em "./Q3/dist.c".
O histograma de frequências gerado de maneira análoga à questão ante-
rior possui a cara da função fdp(x) = x4, como mostrado na ﬁgura 8. Onde
a função densidade de probabilidade foi multiplicada por uma constante a
ﬁm de ilustrar suas semelhanças.
6
 0
 5000
 10000
 15000
 20000
 25000
−3 −2 −1 0 1 2 3
fre
qu
ên
cia
x
hist
c.x4
Figura 8: Histograma dos números gerados segundo a distribuição da
fdp(x) = x4 no intervalo −3 < x < 3.
3.2 (b)
O valor esperado de xk para esta função é dado por:
< xk >=
∫ 3
−3
xkx4dx. (6)
Assim, tem-se que < x >= 0. Fex-se um gráﬁco dos valores esperados
calculados numericamente em função da quantidade de números aleatórios
gerados e compara-se com o valor analítico (ﬁgura 9).
−1.5
−1
−0.5
 0
 0.5
 1
 1.5
 2
 2.5
 0 100 200 300 400 500 600 700 800 900 1000
<
x>
N
dados
<X>=0
Figura 9: Valores esperados de x para a fdp(x) = x4 em função da quanti-
dade N de números aleatórios gerados.
7
3.3 (c)
O valor esperado de x2 calculada pela expressão do item anterior é, após
normalizada, < x2 >= 6, 4286. O programa também calculou tal valor, e o
gráﬁco 10 mostra seu comportamento em função de N .
 5
 6
 7
 8
 9
 10
 11
 12
 13
 0 100 200 300 400 500 600 700 800 900 1000
<
x2
>
N
dados
<X>=6,4286
Figura 10: Valores esperados de x2 para a fdp(x) = x4 em função da quan-
tidade N de números aleatórios gerados.
3.4 (d)
O desvio padrão é σ = sqrt(< x2 > − < x >2) = sqrt(6, 4286) = 2, 5355.
Compara-se com o valor calculado pelo programa, na ﬁgura 11.
 2
 2.2
 2.4
 2.6
 2.8
 3
 3.2
 3.4
 3.6
 0 100 200 300 400 500 600 700 800 900 1000
σ
(x)
N
dados
σ(x)=2,5355
Figura 11: Desvio padrão em função da quantidade N de números aleatórios
gerados.
8
3.5 (e)
Vimos que o cálculo numérico de < x >, < x2 e σ(x) a partir de valores
aleatórios segunda a distribuição fdp(x) = x4 no intervalo de −3 < x < 3
converge para o valor analítico conforme aumentamos a quantidade N de
números gerados. E, antes, pôde-se comprovar o comportamento semelhante
da frequência de números aleatórios com a função distribuição de probabili-
dade, permitindo que se calcula-se tais valores e esperados e desvio padrão
para a função analisada.
4 Questão 4
Utilizando o método de resolução de integrais por monte carlo por amos-
tragem uniforme (programa ./Q4/mont.c), tentou-se estimar um valor
para
pi. Este sendo:
pi = 4
∫ 1
0
√
(1− x2)dx . (7)
4.1 (a)
 1
 1.5
 2
 2.5
 3
 3.5
 4
 4.5
 5
 0 100 200 300 400 500 600 700 800 900 1000
pi
e
st
N
dados
pi
Figura 12: Gráﬁco do valor de pi estimado pela integral em função de N .
9
4.2 (b)
 0
 0.5
 1
 1.5
 2
 2.5
 3
 0 100 200 300 400 500 600 700 800 900 1000
pi
e
st
−
pi
N
dados
Figura 13: Gráﬁco do valor de piest − pi em função de N .
4.3 (c)
O valor piest calculado numericamente converge para o valor analítico
à medida que N aumenta. Ainda assim, há certa dispersão, mesmo para
valores próximo de N = 1000. Isso pode ser visto também no gráﬁco 13
que consiste na diferença entre o valor analítico e numérico. O método,
ainda que simples, retorna um valor razoavelmente próximo, mas dependendo
da aplicação pode ser necessário um método mais preciso para cálculo de
integrais.
5 Questão 5
Para encontrar o valor de a que minimiza a variância, deve-se encontrar
a curva mais parecida com a função a qual deseja-se calcular a integral. Para
isso, fez-se um gráﬁco com alguns valores diferentes de a.
Nota-se que a melhor opção é quando a = 1, ou e( − x). Normalizando
a função obtem-se o valor de A = 0, 95679 que será utilizada no programa
"./Q5/montecarlo.c"para o cálculo da integral pelo método de monte carlo
por amostragem seletiva. Para gerar os números aleatórios segundo a distri-
buição de e−x utilizou-se a mesma função da questão 2.
O gráﬁco 15 mostra o valor da integral ao variar-se N . O valor analítico
é de aproximadamente 1,58119.
10
 0
 0.1
 0.2
 0.3
 0.4
 0.5
 0.6
 0.7
 0.8
 0.9
 1
 0 0.5 1 1.5 2 2.5 3
f(x
)
x
1/(x2+cos(x)2)
exp(−x/2)
exp(−x)
exp(−2x)
Figura 14: Funções a serem analisadas.
 1.5
 1.52
 1.54
 1.56
 1.58
 1.6
 1.62
 1.64
 1.66
 1.68
 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
I
N
dados
I=1,58119
Figura 15: Valor da integral I em função de N .
Percebe-se que, assim como o método utilizado na questão anterior, o va-
lor da integral ﬂutua um pouco em torno do valor analítico. Por se tratarem
de funções diferentes, não é razoável admitir que um método seja melhor que
o outro.
11
6 Questão 6
6.1 (a)
O programa "./Q6/caminhante.c"calcula a posição ﬁnal de 10000 cami-
nhantes após 1000 passos. Não foi possível calcular mais que isso devido ao
pouco espaço de memória. Os histogramas abaixo mostram a distribuição
espacial dos caminhantes quando estes se deslocam 1 posição (ﬁgura 16), 2
posições (ﬁgura 17) e 3 posições (ﬁgura 18) a cada passo. Nota-se que ao au-
mentar o deslocamento dos caminhantes, a curva alarga sua base, indicando
uma probabilidade menor destes pararem no ponto inicial após 1000 passos.
 0
 5
 10
 15
 20
 25
 30
 35
−400 −200 0 200 400
fre
qu
ên
cia
posição
Andando 1 posição
histograma
Figura 16: Histograma da posição ﬁnal de 1000 caminhantes, após 1000
passo, deslocando-se 1 posição por passo.
12
 0
 5
 10
 15
 20
 25
 30
 35
−400 −200 0 200 400
fre
qu
ên
cia
posição
Andando 2 posições
histograma
Figura 17: Histograma da posição ﬁnal de 1000 caminhantes, após 1000
passo, deslocando-se 2 posições por passo.
 0
 5
 10
 15
 20
 25
 30
 35
−400 −200 0 200 400
fre
qu
ên
cia
posição
Andando 3 posições
histograma
Figura 18: Histograma da posição ﬁnal de 1000 caminhantes, após 1000
passo, deslocando-se 3 posições por passo.
6.2 (b)
A seguir estão os gráﬁcos das posições dos caminhantes em função do
tempo, para 100 caminhantes (mais que isso gera arquivos muito grandes).
Nota-se um espalhamento maior quando os caminhantes deslocam-se 3 po-
sições por passo.
13
−300
−200
−100
 0
 100
 200
 300
 0 200 400 600 800 1000
po
siç
ão
tempo
Andando 1 posição
−300
−200
−100
 0
 100
 200
 300
 0 200 400 600 800 1000
po
siç
ão
tempo
Andando 2 posições
−300
−200
−100
 0
 100
 200
 300
 0 200 400 600 800 1000
po
siç
ão
tempo
Andando 3 posições
Figura 19: Gráﬁco da posição de 100 caminhantes em função do tempo.
14
6.3 (c)
Ajustando a curva f(x) = bx2+log(a) sobre o logaritmo dos histogramas.
 0
 1
 2
 3
 4
 5
 6
−150 −100 −50 0 50 100 150
lo
g 
fre
qu
ên
cia
posição
Andando 1 posição
histograma
f(x)=b*x2 + log(a)
Figura 20: Ajuste de f(x) sobre o histograma do deslocamento de 1 posição.
(a=238,328; b=-0,000483677)
 0
 1
 2
 3
 4
 5
 6
−250 −200 −150 −100 −50 0 50 100 150 200 250 300
lo
g 
fre
qu
ên
cia
posição
Andando 2 posições
histograma
f(x)=b*x2 + log(a)
Figura 21: Ajuste de f(x) sobre o histograma do deslocamento de 2 posições.
(a=205,678; b=-0,000106759)
15
 0
 1
 2
 3
 4
 5
 6
−400 −300 −200 −100 0 100 200 300 400
lo
g 
fre
qu
ên
cia
posição
Andando 3 posições
histograma
f(x)=b*x2 + log(a)
Figura 22: Ajuste de f(x) sobre o histograma do deslocamento de 3 posições.
(a=214,236; b=-0,0000496592)
6.4 (d)
Os desvios quadráticos médios foram calculados pelo gnuplot em "rms
of residuals". Para o caso do deslocamento de 1 posição tem-se rms1 =
0, 377156. Para 2 posições: rms2 = 0, 41226. Para 3 posições: rms3 =
0, 410843.
6.5 (e)
As legendas das ﬁguras 20, 21 e 22 mostram o valor de b para cada
caso. Nota-se que tal valor aumenta conforme se aumenta o deslocamento
por passo, indicando um alargamento da curva. Ou seja, quanto maior o
deslocamento, mais provável é de encontrar um caminhante mais distante do
ponto de partida.
6.6 (f)
Como já dito anteriormente, o aumento no deslocamento por passo dos
caminhantes causa uma maior dispersão destes em torno do ponto de partida.
16
		Questão 1
		(a)
		(b)
		(c)
		(d)
		(e)
		Questão 2
		Questão 3
		(a)
		(b)
		(c)
		(d)
		(e)
		Questão 4
		(a)
		(b)
		(c)
		Questão 5
		Questão 6
		(a)
		(b)
		(c)
		(d)
		(e)
		(f)
prova/prova.out
\BOOKMARK [1][-]{section.1}{Quest\343o 1}{}% 1
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\BOOKMARK [2][-]{subsection.1.3}{\(c\)}{section.1}% 4
\BOOKMARK [2][-]{subsection.1.4}{\(d\)}{section.1}% 5
\BOOKMARK [2][-]{subsection.1.5}{\(e\)}{section.1}% 6
\BOOKMARK [1][-]{section.2}{Quest\343o 2}{}% 7
\BOOKMARK [1][-]{section.3}{Quest\343o 3}{}% 8
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\BOOKMARK [2][-]{subsection.6.5}{\(e\)}{section.6}% 24
\BOOKMARK [2][-]{subsection.6.6}{\(f\)}{section.6}% 25
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prova/prova.tex
prova/prova3_tb_2016-1.pdf
Me´todos Computacionais da F´ısica B
Prova 3 / 2016-1 – Turma U
Aluno: Matr´ıcula:
1. Utilize um gerador uniforme e produza nu´meros (pseudo-)aleato´rios no intervalo de 0 ate´
100.
(a) Fac¸a um gra´fico da me´dia dos nu´meros gerados em func¸a˜o do nu´mero N (de nu´meros
aleato´rios gerados). Escreva a expressa˜o utilizada.
(b) Fac¸a um gra´fico do desvio padra˜o dos nu´meros em func¸a˜o do nu´mero N (de nu´meros
aleato´rios gerados). Escreva a expressa˜o utilizada.
(c) Fac¸a um histograma (100 bins) que mostre que a distribuic¸a˜o e´ uniforme.
(d) Define uma varia´vel
y =
1
20
20∑
i=1
xi,
onde xi sa˜o os nu´meros aleato´rios dos itens anteriores. Repita os itens anteriores para
esta varia´vel.
(e) O resultado do item anterior era esperado? Comente o seu resultado.
2. Produza nu´meros aleato´rios de acordo com a distribuic¸a˜o
p(x) = exp(−x)
no intervalo de x = 1 ate´ x = 7.
Fac¸a um histograma (100 bins) com a distribuic¸a˜o dos nu´meros gerados e a func¸a˜o p(x).
3. Produza nu´meros aleato´rios de acordo com a distribuic¸a˜o
p(x) = x4
no intervalo de x = −3 ate´ x = 3.
(a) Fac¸a um histograma (100 bins) dos nu´meros gerados.
(b) Calcule o valor esperado de x, < x >, e compare com o resultado anal´ıtico. Fac¸a um
gra´fico de < x > ×N .
(c) Calcule o valor esperado de x2, < x2 >, e compare com o resultado anal´ıtico.
(d) Calcule o desvio padra˜o, e compare com o resultado anal´ıtico.
(e) Interprete os resultados.
4. Calcule o valor de pi por meio do ca´lculo da integral de
pi
4
=
∫ 1
0
√
1− x2dx.
Utilize o me´todo da tentativa e erro e obtenha piest sorteando dois nu´meros aleato´rios xi e
yi. Repita, ou utilize os mesmos xi, para estimar a me´dia e o desvio padra˜o da func¸a˜o no
intervalo.
1
(a) Fac¸a um gra´fico de piest ×N .
(b) Fac¸a um gra´fico de (piest − pi)×N .
(c) Interprete os seus resultados.
5. Utilize a func¸a˜o
p(x) = A exp(−a x)
como func¸a˜o de amostragem para calcular a integral∫ pi
0
1
x2 + cos2(x)
dx.
Determine o valor de a que minimiza a variaˆncia da integral. Compare o resultado com os
outros me´todos. Interprete seu resultado.
6. Fac¸a um programa que simule a difusa˜o de M caminhantes aleato´rios em um espac¸o unidi-
mensional.
(a) Fac¸a histogramas da distribuic¸a˜o espacial de 105 caminhantes apo´s 10000 passos quando
esses se deslocarem respectivamente de 1, 2 e 3 unidades de distaˆncia a cada passo de
tempo.
(b) Fac¸a gra´ficos no gnuplot da distribuic¸a˜o espacial dos caminhantes.
(c) Ajuste o logaritmo dos histogramas encontrados a para´bolas do tipo: f(x) = b(x2) + log(a).
(d) Determine o desvio quadra´tico me´dio por passo em cada caso.
(e) Relacione o valor do paraˆmetro b ajustado com a distaˆncia percorrida por passo e com
o tempo.
(f) Interprete os seus resultados. Como estes se comparam aos resultados anal´ıticos conhe-
cidos?
2
prova/Q5/int-eps-converted-to.pdf
 1.5
 1.52
 1.54
 1.56
 1.58
 1.6
 1.62
 1.64
 1.66
 1.68
 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
I
N
dados
I=1,58119
prova/Q5/funcoes-eps-converted-to.pdf
 0
 0.1
 0.2
 0.3
 0.4
 0.5
 0.6
 0.7
 0.8
 0.9
 1
 0 0.5 1 1.5 2 2.5 3
f(x
)
x
1/(x2+cos(x)2)
exp(−x/2)
exp(−x)
exp(−2x)
prova/Q5/int.eps
prova/Q5/funcoes.eps
prova/Q5/script.plot
#Luan Bottin De Toni - 02/07/16
reset
set encoding utf8
set xlabel "x"
set ylabel "f(x)"
f(x)=1/(x**2+cos(x)**2)
g(x)=exp(-x)
set xrange[0:pi]
set term post eps enhan color solid
set output 'funcoes.eps'
plot f(x) t "1/(x^2+cos(x)^2)", g(0.5*x) t "exp(-x/2)", g(x) t "exp(-x)", g(2*x) t "exp(-2x)"
################################
reset
set encoding utf8
set xlabel "N"
set ylabel "I"
set term post eps enhan color solid
set output 'int.eps'
plot './int.dat' u 1:2 t "dados", 1.58119 t "I=1,58119"
prova/Q5/int.dat
100 1.564925
101 1.602177
102 1.548194
103 1.574336
104 1.567577
105 1.580543
106 1.633019
107 1.573536
108 1.633092
109 1.583907
110 1.598378
111 1.595165