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---
title: "Análise Numérica - Trabalho Prático 1"
author:
- Diogo Cordeiro
- Hugo Sales
- Pedro Costa
geometry: margin=2cm
output: pdf_document
---
## Motivação
Perceber e analisar técnicas de aproximação de séries numéricas e estratégias
para controlo do erro de cálculos em computador, assim como implementar
algoritmos com recurso a métodos numéricos adequados e interpretar os resultados.
## Questão 1
O _eps_ está definido como a diferença entre 1.0 e o menor valor
representável superior a um, por exemplo, 2^-23^, em precisão simples, e 2^-52^
em precisão dupla (em ambos os casos $Base^{-(precisão-1)}$). O ISO 9899 C define:
"a diferença entre 1 e o menor valor superior a 1 que é
representável num dado tipo de ponto flutuante"
Verificamos a existência de uma definição alternativa: "o menor número que
somado com 1 resulta num número maior do que 1". Não usamos esta definição, uma
vez que devido ao arredondamento para o valor mais próximo, bastaria usar um
valor ligeiramente maior do que metade do nosso _eps_ para satisfazer
a condição. E também porque os standards ISO
optam por "passo entre valores adjacentes".
O código seguinte implementa o cálculo de
$$ (2 - \sum _{n=1}^{\infty}\:\frac{1}{2^n}) - 1 = \textit{eps} $$
onde
$$ \lim _{n\to \infty }\sum _{n=1}^{\infty}\:\frac{1}{2^n} = 1^{-} $$
Logo este cálculo converge para 1 por números superiores.
\pagebreak
double machine_eps() {
// Pela definição, este valor tem que ser superior a 1, e
// como a máquina usa um sistema de vírgula flutuante de base 2,
// este tem que ser uma potência de 2
double epsilon_candidate = 2.0,
epsilon = epsilon_candidate,
// Primeiro termo da série geométrica com proporção 1/2,
// que converge para 1
power = 2.0;
// Diferente de 1 pela definição
while (epsilon_candidate != 1.0) {
epsilon = epsilon_candidate;
// Aproximar o epsilon candidate do epsilon,
// reduzindo este com a acumulação do termo da sucessão
epsilon_candidate -= 1/power;
// Razão da sucessão
power *= 2;
}
return epsilon-1.0;
}
Com a execução deste código obtivemos 2.2204460492503131e-16, o mesmo valor
definido, para aritmética IEEE, na C library `float.h`.
#define DBL_EPSILON 2.2204460492503131e-16
## Questão 2
Utilizamos o seguinte código na linguagem Java para computar a série dada e
obtivemos os resultados apresentados na tabela que segue.
Para evitar o cálculo de fatoriais e de grande magnitude, o que diminuiria a
performace e aumentaria o erro e a possibilidade de overflow,
optamos usar a definição por recorrência e por simplificar algebricamente o
cálculo dos termos da série, obtendo uma expressão mais simples.
$$a_{k+1} = a_k \cdot \frac{k + 1}{4 \cdot k + 6}$$
/**
* Computar a série 2 com um erro absoluto inferior a um dado
* epsilon
*/
public static double compute_series_2(double epsilon) {
// Fator constante
double factor = 9.0/(double)(2.0*Math.sqrt(3));
// Tirado do critério D'Alembert para L = 0.5 < 1
double super_L = 1.0/(double)(1-0.25);
// Index do sumatório
int k = 0;
// Acumulação do sumatório
double acc = 0;
// O valor do termo actual da série
double a = 1.0f; // a with k = 0
// Enquanto o nosso erro absoluto é superior ao
// epsilon dado
while(epsilon < factor*a*super_L) {
// Acumula com o termo anterior
acc += a;
// Computa o termo seguinte e posteriormente
// incrementa k
a = compute_serie_2_term(k++) * a;
}
System.out.println(factor*acc + " " + (k - 1));
}
/**
* Computa o termo k da série 2 dado um anterior
*/
public static double compute_serie_2_term(int k) {
return (double)(k+1.0f)/(double)(4.0f*k+6.0f);
}
### Tabela de resultados
$-log(\epsilon)$ | $S_n$ | Iterações | Tempo (s)
-----------------|---------------------|-------------|----------
8 | 3.141592651 | 13 | 0.094
9 | 3.1415926529 | 14 | 0.099
10 | 3.14159265355 | 16 | 0.100
11 | 3.141592653587 | 18 | 0.100
12 | 3.1415926535892 | 19 | 0.095
13 | 3.14159265358976 | 21 | 0.101
14 | 3.141592653589785 | 22 | 0.097
15 | 3.1415926535897936 | 24 | 0.097
A série aparenta aproximar o valor de $\pi$.
Reparamos que geralmente a ordem de grandeza do erro indica o número de casas
decimais exatas obtidas, contudo o valor obtido para $-log(\epsilon)$ igual a
9 e 14 produz um resultado com menos uma casa decimal exata. Ainda assim
o valor apresentado representa $\pi$ com erro absoluto inferior a
$5 \cdot 10^{-10}$ e $5 \cdot 10^{-15}$ respetivamente.
## Questão 3
Inicialmente implementamos o cálculo do valor aproximado desta série em Java,
mas deparamo-nos com um longo tempo de execução devido ao elevado número de
iterações necessárias para aproximar a série com o $\epsilon$ pretendido, pelo
que decidimos testar uma implementação na linguagem C++, na qual obtivemos maior
performace, o que permitiu o cãlculo para um valor menor de $\epsilon$ em tempo
útil.
double compute_serie_3_term(unsigned long n) {
return -((double)(2*n+1))/(double)(2*n+3);
}
void compute_serie_3(double err) {
// Termo atual
unsigned long k = 0;
// Valor do termo atual
double ak = 1;
// Acumulador do valor da série
double acc = 1;
// Parar quando o erro obtido for inferior ao pretendido
while (err < 4*std::abs(ak)) {
// Calcular o termo seguinte e incrementar k
ak = compute_serie_3_term(k++) * ak;
// Somar o termo ao total
acc += ak;
}
std::cout << k << " " << 4*acc << '\n';
}
### Tabela de Resultados em Java
$-log(\epsilon)$ | $S_n$ | Iterações | Tempo (s)
-----------------|---------------|-------------|-----------
8 | 3.141592659 | 200000000 | 2.237
9 | 3.1415926541 | 2000000000 | 21.522
10 | 3.14159265364 | 20000000012 | 215.442
### Tabela de Resultados em C++
$-log(\epsilon)$ | $S_n$ | Iterações | Tempo (s)
------------------|--------------------|----------------|----------
8 | 3.141592659 | 200000000 | 0.787
9 | 3.1415926541 | 2000000000 | 7.862
10 | 3.14159265364 | 20000000012 | 79.130
11 | 3.141592653593 | 200000002870 | 791.625
12 | 3.1415926535878 | 2000000614027 | 7871.842
Observando as tabelas notamos um padrão, e, motivados por establecer uma relação
entre $\epsilon$ e tanto o números de iterações e o tempo de execução, decidimos
traçar um gráfico com estes valores.
![Tempo em função de $\epsilon$](graficos/questao3_tempos.png){ width=12cm }
![Iterações em função de $\epsilon$](graficos/questao3_iteracoes.png){ width=12cm }
\pagebreak
Destes dois gráficos concluímos que o tempo cresce exponencialmente em função do
valor de $-log(\epsilon)$, sendo que ambos crescem com um fator de 10,
correspondendo aproximadamente a um número da forma $5*10^{-(\epsilon + 1)}$, mas
não exatamente a este valor, sendo que quando $\epsilon$ diminuiu, o valor
de $n$ afasta-se do valor $5*10^{-(\epsilon + 1)}$.
Descobrimos que se utilizarmos mais precisão nos cálculos (nomeadamente através
do uso de `long double` em C++, ou seja, 128 bits), o padrão verifica-se para
todos os casos testados.
Concluímos disto que este desvio se deve ao erro de
arredondamento ($|S_n - \hat{S}_n|$), que neste caso se deve à elevada magnitude
de $n$.
## Questão 4
Usamos como referência
$$ \pi = 3.14159265358979324$$
sendo este valor arredondado a 18 algarismos significativos, para que o erro de
arredondamento aqui utilizado não altere os resultados do cálculo do erro
absoluto efetivo.
Erro efetivo obtido na questão 2:
$-log(\epsilon)$ | $\Delta x$
-----------------|---------------------
8 | $2.6 \cdot 10^{-09}$
9 | $6.9 \cdot 10^{-10}$
10 | $4.0 \cdot 10^{-11}$
11 | $2.8 \cdot 10^{-12}$
12 | $6.0 \cdot 10^{-13}$
13 | $3.4 \cdot 10^{-14}$
14 | $8.0 \cdot 10^{-15}$
15 | $4.5 \cdot 10^{-16}$
Erro efetivo obtido na questão 3:
$-log(\epsilon)$ | $\delta x$
-----------------|---------------------
8 | $5.5 \cdot 10^{-09}$
9 | $5.2 \cdot 10^{-10}$
10 | $5.1 \cdot 10^{-11}$
11 | $3.3 \cdot 10^{-12}$
12 | $2.0 \cdot 10^{-12}$
Notamos que para $\epsilon = 10^{-12}$ o erro efetivamente cometido é maior que
o erro pretendido. Justificamos este facto com a elevada magnitude de $n$, o que
leva a termos de pequena magnitude, o que induz erros de cancelamento.
Concluímos que a série utilizada na questão dois possibilita o cálculo da
aproximação do valor de $\pi$ usando um número de termos muito menor, para uma
mesma precisão. Sendo que a série da questão dois dá resultados
com erro inferior a $10^{-15}$ com um número de iterações na ordem das dezenas,
enquanto que a série da questão três requer iterações na ordem das centenas de
milhões para obter um erro inferior a $10^{-8}$.

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#include <iostream>
#include <cmath>
int main() {
double vals[] = { 3.1415926029605323,
3.1415926506432696,
3.1415926528764806,
3.1415926535477534,
3.1415926535795631,
3.1415926535891852,
3.1415926535897571,
3.1415926535897851};
std::cout.precision(17);
for(double v : vals) {
std::cout << std::abs(3.14159265358979324 - v);
}
}

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#include <iostream>
#include <cmath>
int main() {
// double vals[] = { 3.1415926029605323,
// 3.1415926506432696,
// 3.1415926528764806,
// 3.1415926535477534,
// 3.1415926535795631,
// 3.1415926535891852,
// 3.1415926535897571,
// 3.1415926535897851};
double vals[] = {
// 3.1415925645296694,
// 3.1415926484100587,
// 3.1415926523361879,
// 3.1415926535159437,
// 3.1415926535718257,
// 3.1415926535887251,
// 3.1415926535897296,
// 3.1415926535897780
// 1.2091995566689335,
// 1.2091995750220277
// ,
// 1.209199575881591
// ,
// 1.2091995761399641
// ,
// 1.2091995761522076
// ,
// 1.2091995761559111
// ,
// 1.2091995761561314,
// 1.209199576156142
// 3.1415926413913953,
// 3.1415926506432696,
// 3.1415926534167733,
// 3.1415926535795631,
// 3.1415926535873004,
// 3.1415926535896448,
// 3.1415926535897571,
// 3.1415926535897918
// 3.1415926436246062,
// 3.1415926511835623,
// 3.141592653448583,
// 3.1415926535814478,
// 3.1415926535877601,
// 3.1415926535896723,
// 3.1415926535897643,
// 3.1415926535897927
// 3.1415926441636568
// ,
// 3.1415926513142782
// ,
// 3.1415926534563079
// ,
// 3.141592653581907
// ,
// 3.141592653587872
// ,
// 3.1415926535896794
// ,
// 3.141592653589766
// ,
// 3.1415926535897927
// 3.1415926413913944
// ,
// 3.1415926506432688
// ,
// 3.1415926534167729
// ,
// 3.1415926535795626
// ,
// 3.1415926535873
// ,
// 3.1415926535896443
// ,
// 3.1415926535897567
// ,
// 3.1415926535897909
// 3.1415926413913948
// ,
// 3.1415926506432692
// ,
// 3.1415926534167732
// ,
// 3.1415926535795628
// ,
// 3.1415926535873003
// ,
// 3.1415926535896447
// ,
// 3.141592653589757
// ,
// 3.1415926535897911
// 3.1415926413913948
// ,
// 3.1415926506432692
// ,
// 3.1415926534167733
// ,
// 3.1415926535795628
// ,
// 3.1415926535873003
// ,
// 3.1415926535896447
// ,
// 3.141592653589757
// ,
// 3.1415926535897911
// 3.1415926413913948
// ,
// 3.1415926506432692
// ,
// 3.1415926534167733
// ,
// 3.1415926535795628
// ,
// 3.1415926535873003
// ,
// 3.1415926535896447
// ,
// 3.141592653589757
// ,
// 3.1415926535897911
// 3.1415926585894502,
// 3.1415926540880963,
// 3.1415926536384982,
// 3.1415926535932850,
// 3.1415926535877814
// 3.1415926506432696,
// 3.1415926528764806,
// 3.1415926535477534,
// 3.1415926535873004,
// 3.1415926535891850,
// 3.1415926535897570,
// 3.1415926535897850,
// 3.1415926535897936
3.141592651 ,
3.1415926529 ,
3.14159265355 ,
3.141592653587 ,
3.1415926535892 ,
3.14159265358976 ,
3.141592653589785,
3.1415926535897936
};
std::cout.precision(17);
for(double v : vals) {
std::cout << std::abs(3.14159265358979324 - v) << '\n';
}
}

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#include <iostream>
#include <cmath>
int main() {
double vals[] = {
// 3.1415926335902893,
// 3.1415926515893244,
// 3.1415926533883067,
// 3.1415926535683196,
// 3.1415926535862631
// 3.1415926585894502,
// 3.1415926540880963,
// 3.1415926536384982,
// 3.1415926535932850,
// 3.1415926535877814
3.141592659 ,
3.1415926541 ,
3.14159265364 ,
3.141592653593,
3.1415926535878
};
std::cout.precision(17);
for(double v : vals) {
std::cout << std::abs(3.14159265358979324 - v) << '\n';
}
}

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#!/usr/bin/gnuplot
set output "questao3_tempos.png";
set terminal png size 1200,900 enhanced font "Helvetica,20"
set autoscale;
set xr [7:13];
set logscale y;
set xlabel "-log(epsilon)"
set ylabel "Tempo (s)"
set linetype 1 lc rgb 'black'
set linetype 2 lc rgb 'black
plot "../questao3_tempo_java.txt" with linespoints dashtype 0 lw 4 t "Java", "../questao3_tempo_Cpp.txt" with linespoints dashtype 2 lw 4 t "C++";
set output "questao3_iteracoes.png";
set terminal png size 1200,900 enhanced font "Helvetica,20"
set autoscale;
set xr [7:13];
set logscale y;
set xlabel "-log(epsilon)"
set ylabel "Iterações"
set linetype 1 lc rgb 'black'
set linetype 2 lc rgb 'black
plot "../questao3_iteracoes_java.txt" with linespoints dashtype 0 lw 4 t "Java", "../questao3_iteracoes_Cpp.txt" with linespoints dashtype 2 lw 4 t "C++";

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8 200000000
9 2000000000
10 20000000012
11 200000002870
12 2000000614027

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8 200000000
9 2000000000
10 20000000012

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8 0.787
9 7.862
10 79.130
11 791.625
12 7871.842

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8 2.237
9 21.522
10 215.442

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8 200000000 3.1415926585894502 0.787
9 2000000000 3.1415926540880963 7.862
10 20000000012 3.1415926536384982 79.130
11 200000002870 3.141592653593285 791.625
12 2000000614027 3.1415926535877814 7908.842
14
[0.787]

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8 | 3.1415926029605323 | 12 | 0m0.002s
9 | 3.1415926506432696 | 14 | 0m0.002s
10 | 3.1415926528764806 | 15 | 0m0.002s
11 | 3.1415926535477534 | 17 | 0m0.002s
12 | 3.1415926535795631 | 18 | 0m0.002s
13 | 3.1415926535891852 | 20 | 0m0.002s
14 | 3.1415926535897571 | 22 | 0m0.002s
15 | 3.1415926535897851 | 23 | 0m0.001s
8 | 3.1415926335902893 | 50000000 |0m0.200s
9 | 3.1415926515893244 | 500000000 |0m1.979s
10 | 3.1415926533883067 | 5000000000 |0m19.783s
11 | 3.1415926535683196 | 50000000086 |3m18.003s
12 | ^C | 1m10.621s | 0m0.000s
8 | 50000000 | 3.1415926335902893 | 0.553s
9 | 500000000 | 3.1415926515893244 | 4.608s
10 | 5000000000 | 3.1415926533883067 | 45.069s
11 | 50000000086 | | 450.325s
3.1415926535683196
3.14159265358979324 (arredondado a 18 algorismos significativos, para que o erro de arredondamento aqui utilizado não altere os resultados do cálculo do erro absoluto efetivo)
8 | 3.1415926029605323 | 12 | 0m0.002s
9 | 3.1415926506432696 | 14 | 0m0.002s
10 | 3.1415926528764806 | 15 | 0m0.002s
11 | 3.1415926535477534 | 17 | 0m0.002s
12 | 3.1415926535795631 | 18 | 0m0.002s
13 | 3.1415926535891852 | 20 | 0m0.002s
14 | 3.1415926535897571 | 22 | 0m0.002s
15 | 3.1415926535897851 | 23 | 0m0.002s
8 | 3.1415926122124067 | 12 | 0m0.002s
9 | 3.1415926511835623 | 14 | 0m0.002s
10 | 3.1415926530074607 | 15 | 0m0.001s
11 | 3.1415926535554908 | 17 | 0m0.002s
12 | 3.1415926535814478 | 18 | 0m0.001s
13 | 3.1415926535892975 | 20 | 0m0.002s
14 | 3.1415926535897643 | 22 | 0m0.002s
15 | 3.1415926535897869 | 23 | 0m0.002s

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#include <iostream>
double machine_eps() {
double epsilon_candidate = 2.0,
epsilon = epsilon_candidate,
power = 2.0;
while (epsilon_candidate != 1.0)
{
epsilon = epsilon_candidate;
epsilon_candidate -= 1/power;
power *= 2;
}
return epsilon-1.0;
}
int main() {
std::cout.precision(17);
std::cout << machine_eps() << " - macheps" << '\n';
/*for(double step = 0.0f; step < 100'000'000; step += 1.0)
{
std::cout << step << " " << machine_eps(step) << '\n';
}*/
}

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#include <iostream>
double
machine_eps(double start)
{
double epsilon = 1.0f;
while ((start + 0.5f * epsilon) != start)
{
epsilon = 0.5f * epsilon;
}
return epsilon;
}
int
main()
{
std::cout.precision(17);
std::cout << machine_eps(1) << " - macheps" << '\n';
}

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#include <iostream>
#include <cmath>
#include <cstdlib>
/**
* Compute k term of Serie 2 given the previous one
*/
double _compute_serie_2_term(unsigned long k, double ak) {
return double(k+1)/(double(4*k+6)) * ak;
}
// double c(int k) {
// return std::pow(std::tgamma(k + 1), 2.0)/std::tgamma(2*k + 2);
// }
/**
* Compute serie 2 with absole error below a given epsilon
*/
void compute_serie_2(double epsilon) {
// Constant factor
double factor = 9.0/(2.0*std::sqrt(3));
// Taken from D'Alembert criterion for given that L = 0.5 < 1
double super_L = 1.0/(1.0-0.25);
// Summation index
unsigned long k = 0;
// Acumulation for the Summation
double acc = 0;
// Previous and current value on the series
// double a_prev = 1.0f, // a with k = 0
double a = 1.0; // _compute_serie_2_term(k, a_prev); // compute next
// While our absolute error is higher than the given epsilon
while(epsilon < factor*a*super_L) {
// Accumulate with previous term
acc += a;
// Update previous term to current term
// a_prev = a;
// Compute current term
// a = c(k);
a = _compute_serie_2_term(k++, a);
}
std::cout << k << " " << std::abs(M_PI - factor*acc) << '\n';
std::cout << factor * acc << " | " << k;
}
int main(int, char **argv) {
std::cout.precision(17);
double power_of_ten = std::pow(10.0, -std::atoi(argv[1]));
compute_serie_2(power_of_ten);
}

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#include <iostream>
#include <cmath>
#include <numeric>
#include <ios>
#include <cstdlib>
double compute_serie_3_term(unsigned long n) {
return -((double)(2*n+1))/(double)(2*n+3);
}
double compute_serie_3(double err) {
unsigned long k = 0;
double ak = 1;
double acc = 1;
while (err < 4*std::abs(ak)) {
ak *= compute_serie_3_term(k++);
acc += ak;
}
std::cout << k << " " << 4*acc << " " << std::abs(3.14159265358979324$$ - 4*acc) << '\n';
return 4*acc;
}
int main(int, char **argv) {
const double err = std::pow(10.0, -std::atof(argv[1]));
std::cout.precision(17);
compute_serie_3(err);
}

16
1 - Numerical Series Approximation/source/java for speed comparison/Question1.java Normal file
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@ -0,0 +1,16 @@
public class Question1 {
public static void main(String args[]) {
System.out.println(machine_eps());
}
public static float machine_eps() {
float epsilon = 1.0f;
while ((1.0f + 0.5f * epsilon) != 1.0f){
epsilon = 0.5f * epsilon;
}
return epsilon;
}
}

49
1 - Numerical Series Approximation/source/java for speed comparison/Question2.java Normal file
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@ -0,0 +1,49 @@
public class Question2 {
public static void main(String[] args) {
double power_of_ten = Math.pow(10,-1 * Integer.parseInt(args[0]));
compute_serie_2(power_of_ten);
}
/**
* Compute serie 2 with absole error below a given epsilon
*/
public static void
compute_serie_2(double epsilon)
{
// Constant factor
double factor = 9.0f/(double)(2.0f*Math.sqrt(3));
// Taken from D'Alembert criterion for given that L = 0.5 < 1
double super_L = 1.0f/(double)(1-0.25);
// Summation index
int k = 0;
// Acumulation for the Summation
double acc = 0;
// Previous and current value on the series
double a = 1.0f; // a with k = 0
// While our absolute error is higher than the given epsilon
while(epsilon < factor*a*super_L)
{
// Accumulate with previous term
acc += a;
// Compute current term
a = _compute_serie_2_term(k++) * a;
}
// System.out.println(epsilon);
// System.out.println("Value: " + factor*acc);
// System.out.println("Error: " + (3.1415926535897932 - factor*acc));
// System.out.println("K: " +(k-1) + "\n--------------");
System.out.println(factor*acc + " " + (k - 1));
}
/**
* Compute k term of Serie 2 given the previous one
*/
public static double
_compute_serie_2_term(int k)
{
return (double)(k+1.0f)/(double)(4.0f*k+6.0f);
}
}

28
1 - Numerical Series Approximation/source/java for speed comparison/Question3.java Normal file
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@ -0,0 +1,28 @@
public class Question3 {
public static void main(String[] args) {
double err = Math.pow(10, -Integer.parseInt(args[0].trim()));
compute_serie_2(err);
}
public static double compute_serie_3_term(long n)
{
return -((double)(2*n+1))/(double)(2*n+3);
}
public static void compute_serie_2(double err)
{
long k = 0;
double ak = 1;
double acc = 1;
while (err < 4 * Math.abs(ak)) {
ak = compute_serie_3_term(k++) * ak;
acc += ak;
}
System.out.println(k + " " + 4 * acc);
}
}

674
2 - Newton and Iterative methods/LICENSE Normal file
View File

@ -0,0 +1,674 @@
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WILL ANY COPYRIGHT HOLDER, OR ANY OTHER PARTY WHO MODIFIES AND/OR CONVEYS
THE PROGRAM AS PERMITTED ABOVE, BE LIABLE TO YOU FOR DAMAGES, INCLUDING ANY
GENERAL, SPECIAL, INCIDENTAL OR CONSEQUENTIAL DAMAGES ARISING OUT OF THE
USE OR INABILITY TO USE THE PROGRAM (INCLUDING BUT NOT LIMITED TO LOSS OF
DATA OR DATA BEING RENDERED INACCURATE OR LOSSES SUSTAINED BY YOU OR THIRD
PARTIES OR A FAILURE OF THE PROGRAM TO OPERATE WITH ANY OTHER PROGRAMS),
EVEN IF SUCH HOLDER OR OTHER PARTY HAS BEEN ADVISED OF THE POSSIBILITY OF
SUCH DAMAGES.
17. Interpretation of Sections 15 and 16.
If the disclaimer of warranty and limitation of liability provided
above cannot be given local legal effect according to their terms,
reviewing courts shall apply local law that most closely approximates
an absolute waiver of all civil liability in connection with the
Program, unless a warranty or assumption of liability accompanies a
copy of the Program in return for a fee.
END OF TERMS AND CONDITIONS
How to Apply These Terms to Your New Programs
If you develop a new program, and you want it to be of the greatest
possible use to the public, the best way to achieve this is to make it
free software which everyone can redistribute and change under these terms.
To do so, attach the following notices to the program. It is safest
to attach them to the start of each source file to most effectively
state the exclusion of warranty; and each file should have at least
the "copyright" line and a pointer to where the full notice is found.
<one line to give the program's name and a brief idea of what it does.>
Copyright (C) <year> <name of author>
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation, either version 3 of the License, or
(at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with this program. If not, see <https://www.gnu.org/licenses/>.
Also add information on how to contact you by electronic and paper mail.
If the program does terminal interaction, make it output a short
notice like this when it starts in an interactive mode:
<program> Copyright (C) <year> <name of author>
This program comes with ABSOLUTELY NO WARRANTY; for details type `show w'.
This is free software, and you are welcome to redistribute it
under certain conditions; type `show c' for details.
The hypothetical commands `show w' and `show c' should show the appropriate
parts of the General Public License. Of course, your program's commands
might be different; for a GUI interface, you would use an "about box".
You should also get your employer (if you work as a programmer) or school,
if any, to sign a "copyright disclaimer" for the program, if necessary.
For more information on this, and how to apply and follow the GNU GPL, see
<https://www.gnu.org/licenses/>.
The GNU General Public License does not permit incorporating your program
into proprietary programs. If your program is a subroutine library, you
may consider it more useful to permit linking proprietary applications with
the library. If this is what you want to do, use the GNU Lesser General
Public License instead of this License. But first, please read
<https://www.gnu.org/licenses/why-not-lgpl.html>.

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---
title: "Análise Numérica - Trabalho Prático 2"
author:
- Diogo Cordeiro
- Hugo Sales
- Pedro Costa
geometry: margin=2cm
output: pdf_document
---
### Motivação
Pretende-se usar os métodos de Newton e iterativo simples para determinar um valor aproximado de um zero de
$$x^2 - cos(x)^2$$
### 1.a)
Optamos por implementar o algoritmo pedido em C++ devido à possiblidade da aplicação de
templates e lambdas. Deste modo, foi-nos possível implementar os dois métodos pedidos partindo
de um algoritmo genérico pois a diferença entre o método de Newton
e o método iterativo consiste apenas na fórmula de recorrência. Assim, o método de Newton
pode ser visto como uma forma do método iterativo simples.
template<typename step_func>
double find_root(double x0, double epsilon, step_func step, long iter_limit) {
double x1 = x0, err;
long iter = 1;
do {
x0 = x1;
x1 = step(x0);
err = std::abs(x1 - x0);
} while(err > epsilon && iter++ < iter_limit);
return x1;
}
template<typename F_t, typename dF_t>
double newton(double x0, double epsilon, F_t F, dF_t dF, long iter_limit) {
return find_root(x0, epsilon,
[&F, &dF](double x0){ return x0 - F(x0)/dF(x0); },
iter_limit);
}
int main() {
auto F = [](double x){ return std::pow(x, 2.0) - std::pow(std::cos(x), 2.0); };
auto dF = [](double x){ return 2.0 * x + std::sin(2.0 * x); };
double epsilon = 5.0 * std::pow(10.0, -12.0);
std::cout << newton(0.8, epsilon, F, dF, 100000) << '\n';
}
\pagebreak
### 1.b)
![Gráfico para $x \in [-1;1]$](./graph.png){ width=8cm }
Como $\forall{x}: cos^2(x) \in [0:1]$, temos que $x^2 \geq cos^2(x)$ para $|x| > 1$, sabemos que o comportamento de $f(x)$ é dominado pelo comportamento de $x^2$, a qual só tem duas raízes.
A menor das raízes encontra-se no intervalo $]-\infty;0]$ e a maior destas em $[0;\infty[$.
Através da análise do gráfico, verificamos que o intervalo $[0.7;0.8]$ contém uma raíz. Definimos então $a = 0.7$ e $b = 0.8$.
![Gráfico para $x \in [0.7;0.8]$](./graph_small.png){ width=6cm }
### 1.c)
Queremos mostrar que as condições de aplicabilidade do método de Newton são satisfeitas no intervalo. Assim,
$$F(x) = x^2 - cos(x)^2$$
$$F'(x) = 2 \cdot x + sin(2 \cdot x)$$
$$F''(x) = 2 + 2 \cdot cos(2 \cdot x)$$
Como todas estas funções são compostas partindo de somas de funções contínuas em $\mathbb{R}$, são também continuas no intervalo considerado, verificando-se assim o primeiro critério.
Dado que,
$$F(a) < 0, F(b) < 0 \Rightarrow F(a) \cdot F(b) < 0$$
Verifica-se também o segundo critério.
Temos que $F''(x) = 2 + 2 \cdot cos(2 \cdot x)$, logo $cos(2 \cdot x) \in [-1;1]$, por isso $2 \cdot cos(2 \cdot x) \in [-2; 2]$, e $2 \cdot cos(2 \cdot x) + 2 \in [0;4]$ o que implica que $F''(x) \geq 0$, para $x \in \mathbb{R}$ e por isso $F'(x)$ é não decrescente em $\mathbb{R}$, ou seja, $F'(b) \geq F'(a)$ e $F'(a) > 3$ e por isso $F'(x) \neq 0 \forall{x} : \in [a;b]$, o que verifica o terceiro critério.
Como foi dito anteriormente, $F''(x) \geq 0$ em $\mathbb{R}$ e, por isso, também para $\forall{x} : \in [a;b]$, verificando-se a quarta condição.
Para $x_0 = b$, temos que $F(x_0) > 0$ e $F''(x_0) > 0$, logo $F(x_0) \cdot F''(x_0) > 0$, verificando-se a quinta condição.
Então a sucessão gerada converge para a unica raíz no intervalo $[a;b]$.
### 1.d)
Aplicamos o programa apresentado em $1.a$ e obtivemos os seguintes resultados.
Iterações | Erro estimado | Valor
----------|---------------|---------------
1 | 5.9e-02 | 0.740528800196
2 | 1.4e-03 | 0.739086050826
3 | 9.2e-07 | 0.739085133216
4 | 3.7e-13 | 0.739085133215
Verficamos que o erro estimado é aproximadamente metade do erro estimado da iteração amterior,
o que justificamos com o facto de que o resultado teórico
nos dizer que o erro converge segundo uma secessão de segunda ordem.
### 1.e)
Aplicando a fórmula
$$M = \frac{1}{2} \cdot \frac{\max\limits_{a \leq x \leq b} | F''(x) |}{\min\limits_{a \leq x \leq b} | F'(x) |}$$
Como $F''(x)$ é decrescente em $[0;\pi/2]$, então $\max\limits_{a \leq x \leq b} | F''(x) | = |F''(a)|$ e como $F'(x)$ é não decrescente em $\mathbb{R}$, $\min\limits_{a \leq x \leq b} | F'(x) | = F'(a)$.
Assim, obtemos $M = \frac{F''(0.7)}{2F'(0.7)} \leq \frac{2.4}{2 \cdot 2.3} \leq 0.53$ e
$n \geq \frac{ln(\alpha)}{ln(2)}, \alpha = \frac{ln(5 \cdot 10^{-14}) + ln(0.53)}{ln(0.53) + ln(10^{-1})}$, logo $\alpha \leq \frac{-32}{2.9} \leq 12$ e $n \geq 4$, ou seja que com 4 iterações conseguimos um erro absoluto inferior a $5 \cdot 10^{-14}$.
### 2.a)
Dada a implementação genérica do algoritmo, o método iterativo simples pode ser
implementado sucintamente como:
template<typename F_t>
double fixed_point(double x0, double epsilon, F_t f, long iter_limit) {
return find_root(x0, epsilon, f, iter_limit);
}
e usado como
auto f = [](double x){ return std::cos(x); };
std::cout << fixed_point(0.8, epsilon, f, 100000) << '\n';
A expressão de $\texttt{f}$ foi obtida por manipulação algébrica do seguinte modo:
$$F(x) = 0 \iff x^2 - cos^2(x) = 0 \iff x^2 = cos^2(x) \iff x = \pm cos(x)$$
Logo no intervalo $[a;b]$, temos que $x = cos(x)$ ou seja $f(x) = cos(x)$.
### 2.b)
Aplicando o programa supra apresentado, obtivemos os seguintes erros:
![Gŕafico da comparação do erro estimado dos dois métodos](err_fixed_point.png){ width=12cm }
Partindo destes resultados, verificamos empiricameente a diferença na ordem de convergência dos dois métodos.
Obtivemos como valor final para este método o valor $0.739085133217$, que difere do valor calculado
como o método de Newton em $-2 \cdot 10^{-12}$, por isso concluimos numericamente que, para os valores iniciais usados, este método pode ser aplicado, sendo que aproxima o valor real.
Verificando as condições de aplicabilidade deste método, temos que $f(x)$ é continua em $[a;b]$, verificando-se a primeira condição,
que $f(0.7) \approx 0.76484219$ e $f(0.7) \approx 0.69670671$ o que implica que $f([a;b]) \notin [a;b]$, o implica que não se verifica a segunda condição
e que $\forall x \in [0.7;0.8] : |-sin(x)| < 1$, o que significa que se verifica a terceira condição.
Apesar de não se verificar a segunda condição, verificamos na mesma a convergência da sucessão, o que não contradiz os resultados teóricos, uma vez que estas condições são suficientes, mas não necessárias para esta convergência.

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1.0e-01 1
7.0e-02 2
4.7e-02 3
3.2e-02 4
2.1e-02 5
1.4e-02 6
9.7e-03 7
6.5e-03 8
4.4e-03 9
3.0e-03 10
2.0e-03 11
1.3e-03 12
9.0e-04 13
6.1e-04 14
4.1e-04 15
2.8e-04 16
1.9e-04 17
1.3e-04 18
8.5e-05 19
5.7e-05 20
3.8e-05 21
2.6e-05 22
1.7e-05 23
1.2e-05 24
7.9e-06 25
5.3e-06 26
3.6e-06 27
2.4e-06 28
1.6e-06 29
1.1e-06 30
7.4e-07 31
5.0e-07 32
3.3e-07 33
2.3e-07 34
1.5e-07 35
1.0e-07 36
6.9e-08 37
4.6e-08 38
3.1e-08 39
2.1e-08 40
1.4e-08 41
9.6e-09 42
6.4e-09 43
4.3e-09 44
2.9e-09 45
2.0e-09 46
1.3e-09 47
8.9e-10 48
6.0e-10 49
4.1e-10 50
2.7e-10 51
1.8e-10 52
1.2e-10 53
8.3e-11 54
5.6e-11 55
3.8e-11 56
2.6e-11 57
1.7e-11 58
1.2e-11 59
7.8e-12 60
5.3e-12 61
3.5e-12 62

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1.4e-03 1
9.2e-07 2
3.7e-13 3

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#!/usr/bin/gnuplot
set output "err_fixed_point.png";
set terminal png size 1200,900 enhanced font "Helvetica,20"
set autoscale;
set logscale x;
set xlabel "Erro"
set ylabel "Iterações"
set xrange [] reverse;
set linetype 1 lc rgb 'black'
set linetype 2 lc rgb 'black
plot "err_fixed_point.txt" with points pointtype 1 pointsize 2 lw 4 t "Fixed Point", "err_newton.txt" with points pointtype 2 pointsize 2 lw 4 t "Newton";
set terminal png size 1200,900 enhanced font "Helvetica,35"
set output "graph.png";
unset logscale x;
set xr [-1:1];
set zeroaxis ls 1 lw 3;
set xlabel "x"
set ylabel "y"
f(x)= x**2 - cos(x)**2;
plot f(x) lw 3 t "x^2 - cos(x)^2";
set output "graph_small.png";
set xr [0.7:0.8];
plot f(x) lw 3 t "x^2 - cos(x)^2";

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src: local('Quattrocento Bold'), local('Quattrocento-Bold'), url(https://fonts.gstatic.com/s/quattrocento/v10/OZpbg_xvsDZQL_LKIF7q4jP_eE3vcKnY.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* latin-ext */
@font-face {
font-family: 'Quattrocento Sans';
font-style: italic;
font-weight: 400;
src: local('Quattrocento Sans Italic'), local('QuattrocentoSans-Italic'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9a4lja2NVIDdIAAoMR5MfuElaRB0zMh0P2GEHJ.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Quattrocento Sans';
font-style: italic;
font-weight: 400;
src: local('Quattrocento Sans Italic'), local('QuattrocentoSans-Italic'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9a4lja2NVIDdIAAoMR5MfuElaRB0zMh032GA.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* latin-ext */
@font-face {
font-family: 'Quattrocento Sans';
font-style: italic;
font-weight: 700;
src: local('Quattrocento Sans Bold Italic'), local('QuattrocentoSans-BoldItalic'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9X4lja2NVIDdIAAoMR5MfuElaRB0zMj_bTDXDojYsJ.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Quattrocento Sans';
font-style: italic;
font-weight: 700;
src: local('Quattrocento Sans Bold Italic'), local('QuattrocentoSans-BoldItalic'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9X4lja2NVIDdIAAoMR5MfuElaRB0zMj_bTDX7ojQ.woff2) format('woff2');
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}
/* latin-ext */
@font-face {
font-family: 'Quattrocento Sans';
font-style: normal;
font-weight: 400;
src: local('Quattrocento Sans'), local('QuattrocentoSans'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9c4lja2NVIDdIAAoMR5MfuElaRB0zHt0_uHA.woff2) format('woff2');
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}
/* latin */
@font-face {
font-family: 'Quattrocento Sans';
font-style: normal;
font-weight: 400;
src: local('Quattrocento Sans'), local('QuattrocentoSans'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9c4lja2NVIDdIAAoMR5MfuElaRB0zJt08.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* latin-ext */
@font-face {
font-family: 'Quattrocento Sans';
font-style: normal;
font-weight: 700;
src: local('Quattrocento Sans Bold'), local('QuattrocentoSans-Bold'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9Z4lja2NVIDdIAAoMR5MfuElaRB0RyklrfPXzwiQ.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Quattrocento Sans';
font-style: normal;
font-weight: 700;
src: local('Quattrocento Sans Bold'), local('QuattrocentoSans-Bold'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9Z4lja2NVIDdIAAoMR5MfuElaRB0RyklrRPXw.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* cyrillic */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 400;
src: local('Spectral Italic'), local('Spectral-Italic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCt-xNNww_2s0amA9M8on7mTMuk.woff2) format('woff2');
unicode-range: U+0400-045F, U+0490-0491, U+04B0-04B1, U+2116;
}
/* vietnamese */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 400;
src: local('Spectral Italic'), local('Spectral-Italic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCt-xNNww_2s0amA9M8onXmTMuk.woff2) format('woff2');
unicode-range: U+0102-0103, U+0110-0111, U+1EA0-1EF9, U+20AB;
}
/* latin-ext */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 400;
src: local('Spectral Italic'), local('Spectral-Italic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCt-xNNww_2s0amA9M8onTmTMuk.woff2) format('woff2');
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}
/* latin */
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font-family: 'Spectral';
font-style: italic;
font-weight: 400;
src: local('Spectral Italic'), local('Spectral-Italic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCt-xNNww_2s0amA9M8onrmTA.woff2) format('woff2');
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}
/* cyrillic */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 700;
src: local('Spectral Bold Italic'), local('Spectral-BoldItalic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCu-xNNww_2s0amA9M8qsHDWfCFXUIJ.woff2) format('woff2');
unicode-range: U+0400-045F, U+0490-0491, U+04B0-04B1, U+2116;
}
/* vietnamese */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 700;
src: local('Spectral Bold Italic'), local('Spectral-BoldItalic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCu-xNNww_2s0amA9M8qsHDWfuFXUIJ.woff2) format('woff2');
unicode-range: U+0102-0103, U+0110-0111, U+1EA0-1EF9, U+20AB;
}
/* latin-ext */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 700;
src: local('Spectral Bold Italic'), local('Spectral-BoldItalic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCu-xNNww_2s0amA9M8qsHDWfqFXUIJ.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 700;
src: local('Spectral Bold Italic'), local('Spectral-BoldItalic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCu-xNNww_2s0amA9M8qsHDWfSFXQ.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* cyrillic */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 400;
src: local('Spectral Regular'), local('Spectral-Regular'), url(https://fonts.gstatic.com/s/spectral/v5/rnCr-xNNww_2s0amA9M9knj-SA.woff2) format('woff2');
unicode-range: U+0400-045F, U+0490-0491, U+04B0-04B1, U+2116;
}
/* vietnamese */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 400;
src: local('Spectral Regular'), local('Spectral-Regular'), url(https://fonts.gstatic.com/s/spectral/v5/rnCr-xNNww_2s0amA9M2knj-SA.woff2) format('woff2');
unicode-range: U+0102-0103, U+0110-0111, U+1EA0-1EF9, U+20AB;
}
/* latin-ext */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 400;
src: local('Spectral Regular'), local('Spectral-Regular'), url(https://fonts.gstatic.com/s/spectral/v5/rnCr-xNNww_2s0amA9M3knj-SA.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 400;
src: local('Spectral Regular'), local('Spectral-Regular'), url(https://fonts.gstatic.com/s/spectral/v5/rnCr-xNNww_2s0amA9M5kng.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* cyrillic */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 700;
src: local('Spectral Bold'), local('Spectral-Bold'), url(https://fonts.gstatic.com/s/spectral/v5/rnCs-xNNww_2s0amA9uCt23FafadWQ.woff2) format('woff2');
unicode-range: U+0400-045F, U+0490-0491, U+04B0-04B1, U+2116;
}
/* vietnamese */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 700;
src: local('Spectral Bold'), local('Spectral-Bold'), url(https://fonts.gstatic.com/s/spectral/v5/rnCs-xNNww_2s0amA9uCt23OafadWQ.woff2) format('woff2');
unicode-range: U+0102-0103, U+0110-0111, U+1EA0-1EF9, U+20AB;
}
/* latin-ext */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 700;
src: local('Spectral Bold'), local('Spectral-Bold'), url(https://fonts.gstatic.com/s/spectral/v5/rnCs-xNNww_2s0amA9uCt23PafadWQ.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 700;
src: local('Spectral Bold'), local('Spectral-Bold'), url(https://fonts.gstatic.com/s/spectral/v5/rnCs-xNNww_2s0amA9uCt23BafY.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}

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<html>
<head>
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class: title, smokescreen, shelf, no-footer
# Análise Numérica - Trabalho Prático 1
### Diogo Cordeiro | Hugo Sales | Pedro Costa | Ricardo Pimenta
---
name: motivacao
class: roomy
# Motivação
Pretende-se usar os métodos de Newton e iterativo simples para determinar um valor aproximado de um zero de
$$x^2 - cos(x)^2$$
---
name: 1a
class: compact
# 1.a)
Optamos por implementar o algoritmo pedido em C++ devido à possiblidade da aplicação de
templates e lambdas. Deste modo, foi-nos possível implementar os dois métodos pedidos partindo
de um algoritmo genérico pois a diferença entre o método de Newton
e o método iterativo consiste apenas na fórmula de recorrência. Assim, o método de Newton
pode ser visto como uma forma do método iterativo simples.
int main() {
auto F = [](double x){ return std::pow(x, 2.0) - std::pow(std::cos(x), 2.0); };
auto dF = [](double x){ return 2.0 * x + std::sin(2.0 * x); };
double epsilon = 5.0 * std::pow(10.0, -12.0);
std::cout << newton(0.8, epsilon, F, dF, 100000) << `\n`;
}
---
name: 1a
class: compact
# 1.a)
template<typename step_func>
double find_root(double x0, double epsilon, step_func step, long iter_limit) {
double x1 = x0, err;
long iter = 1;
do {
x0 = x1;
x1 = step(x0);
err = std::abs(x1 - x0);
} while(err > epsilon && iter++ < iter_limit);
return x1;
}
template<typename F_t, typename dF_t>
double newton(double x0, double epsilon, F_t F, dF_t dF, long iter_limit) {
return find_root(x0, epsilon,
[&F, &dF](double x0){ return x0 - F(x0)/dF(x0); },
iter_limit);
}
---
name: 1b
class: compact, img-right
# 1.b)
![Gráfico para `\(x \in [-1;1]\)`](./graph.png)
Como `\(\forall{x}: cos^2(x) \in [0:1]\)`, temos que `\(x^2 \geq cos^2(x)\)` para `\(|x| > 1\)`, sabemos que o comportamento de `\(f(x)\)` é dominado pelo comportamento de `\(x^2\)`, a qual só tem duas raízes.
A menor das raízes encontra-se no intervalo `\(]-\infty;0]\)` e a maior destas em `\([0;\infty[\)`.
---
name: 1b-cont
class: compact, img-right
# 1.b)
![Gráfico para `\(x \in [0.7;0.8]\)`](./graph_small.png)
Através da análise do gráfico, verificamos que o intervalo `\([0.7;0.8]\)` contém uma raíz. Definimos então `\(a = 0.7\)` e `\(b = 0.8\)`.
---
name: 1c
class: compact
# 1.c)
Queremos mostrar que as condições de aplicabilidade do método de Newton são satisfeitas no intervalo. Assim,
$$F(x) = x^2 - cos(x)^2$$
$$F'(x) = 2 \cdot x + sin(2 \cdot x)$$
$$F''(x) = 2 + 2 \cdot cos(2 \cdot x)$$
Como todas estas funções são compostas partindo de somas de funções contínuas em `\(\mathbb{R}\)`, são também continuas no intervalo considerado, verificando-se assim o primeiro critério.
Dado que,
$$F(a) < 0, F(b) < 0 \Rightarrow F(a) \cdot F(b) < 0$$
---
name: 1c-cont
class: compact
# 1.c)
Verifica-se também o segundo critério.
Temos que `\(F''(x) = 2 + 2 \cdot cos(2 \cdot x)\)`, logo `\(cos(2 \cdot x) \in [-1;1]\)`, por isso `\(2 \cdot cos(2 \cdot x) \in [-2; 2]\)`, e `\(2 \cdot cos(2 \cdot x) + 2 \in [0;4]\)` o que implica que `\(F''(x) \geq 0\)`, para `\(x \in \mathbb{R}\)` e por isso `\(F'(x)\)` é não decrescente em `\(\mathbb{R}\)`, ou seja, `\(F'(b) \geq F'(a)\)` e `\(F'(a) > 3\)` e por isso `\(F'(x) \neq 0 \forall{x} : \in [a;b]\)`, o que verifica o terceiro critério.
Como foi dito anteriormente, `\(F''(x) \geq 0\)` em `\(\mathbb{R}\)` e, por isso, também para `\(\forall{x} : \in [a;b]\)`, verificando-se a quarta condição.
Para `\(x_0 = b\)`, temos que `\(F(x_0) > 0\)` e `\(F''(x_0) > 0\)`, logo `\(F(x_0) \cdot F''(x_0) > 0\)`, verificando-se a quinta condição.
Então a sucessão gerada converge para a unica raíz no intervalo `\([a;b]\)`.
---
name: 1d
class: compact
# 1.d)
Aplicamos o programa apresentado em `\(1.a\)` e obtivemos os seguintes resultados.
Iterações | Erro estimado | Valor
----------|---------------|---------------
1 | 5.9e-02 | 0.740528800196
2 | 1.4e-03 | 0.739086050826
3 | 9.2e-07 | 0.739085133216
4 | 3.7e-13 | 0.739085133215
Verficamos que o erro estimado é aproximadamente metade do erro estimado da iteração amterior,
o que justificamos com o facto de que o resultado teórico
nos dizer que o erro converge segundo uma secessão de segunda ordem.
---
name: 1e
class: compact
# 1.e)
Aplicando a fórmula
<center>\( M = \frac{1}{2} \cdot \frac{\max \limits_{a \leq x \leq b} | F''(x) | }{\min \limits_{a \leq x \leq b} | F'(x) | } \)</center>
Como `\(F''(x)\)` é decrescente em `\([0;\pi/2]\)`, então `\(\max\limits_{a \leq x \leq b} | F''(x) | = |F''(a)|\)` e como `\(F'(x)\)` é não decrescente em `\(\mathbb{R}\)`, `\(\min\limits_{a \leq x \leq b} | F'(x) | = F'(a)\)`.
Assim, obtemos `\(M = \frac{F''(0.7)}{2F'(0.7)} \leq \frac{2.4}{2 \cdot 2.3} \leq 0.53\)` e
`\(n \geq \frac{ln(\alpha)}{ln(2)}, \alpha = \frac{ln(5 \cdot 10^{-14}) + ln(0.53)}{ln(0.53) + ln(10^{-1})}\)`, logo `\(\alpha \leq \frac{-32}{2.9} \leq 12\)` e `\(n \geq 4\)`, ou seja que com 4 iterações conseguimos um erro absoluto inferior a `\(5 \cdot 10^{-14}\)`.
---
name: 2a
class: compact
# 2.a)
Dada a implementação genérica do algoritmo, o método iterativo simples pode ser
implementado sucintamente como:
template<typename F_t>
double fixed_point(double x0, double epsilon, F_t f, long iter_limit) {
return find_root(x0, epsilon, f, iter_limit);
}
---
name: 2a-cont
class: compact
# 2.a)
e usado como
auto f = [](double x){ return std::cos(x); };
std::cout << fixed_point(0.8, epsilon, f, 100000) << `\n`;
A expressão de `\(\texttt{f}\)` foi obtida por manipulação algébrica do seguinte modo:
$$F(x) = 0 \iff x^2 - cos^2(x) = 0 \iff x^2 = cos^2(x) \iff x = \pm cos(x)$$
Logo no intervalo `\([a;b]\)`, temos que `\(x = cos(x)\)` ou seja `\(f(x) = cos(x)\)`.
---
name: 2b
class: img-right
![Gŕafico da comparação do erro estimado dos dois métodos](err_fixed_point.png)
# 2.b)
Partindo destes resultados, verificamos empiricameente a diferença na ordem de convergência dos dois métodos.
Obtivemos como valor final para este método o valor `\(0.739085133217\)`, que difere do valor calculado
como o método de Newton em `\(-2 \cdot 10^{-12}\)`, por isso concluimos numericamente que, para os valores iniciais usados, este método pode ser aplicado, sendo que aproxima o valor real.
---
name: 2b
class: img-right
![Gŕafico da comparação do erro estimado dos dois métodos](err_fixed_point.png)
# 2.b)
Verificando as condições de aplicabilidade deste método, temos que `\(f(x)\)` é continua em `\([a;b]\)`, verificando-se a primeira condição,
que `\(f(0.7) \approx 0.76484219\)` e `\(f(0.7) \approx 0.69670671\)` o que implica que `\(f([a;b]) \notin [a;b]\)`, o implica que não se verifica a segunda condição
e que `\(\forall x \in [0.7;0.8] : |-sin(x)| < 1\)`, o que significa que se verifica a terceira condição.
---
name: 2b
class: img-right
![Gŕafico da comparação do erro estimado dos dois métodos](err_fixed_point.png)
# 2.b)
Apesar de não se verificar a segunda condição, verificamos na mesma a convergência da sucessão, o que não contradiz os resultados teóricos, uma vez que estas condições são suficientes, mas não necessárias para esta convergência.

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#include <iostream>
#include <cmath>
#include <iomanip>
template<typename step_func>
double find_root(double x0, double epsilon, step_func step, long iter_limit) {
double x1 = x0, err;
long iter = 1;
std::cout << "Erros: ";
do {
x0 = x1;
x1 = step(x0);
err = std::abs(x1 - x0);
std::cout << iter << " "
<< std::setprecision(1) << std::scientific << err << " "
<< std::defaultfloat << std::setprecision(12) << x1 << '\n';
} while(err > epsilon && iter++ < iter_limit);
std::cout << '\n';
// std::cout << std::defaultfloat << std::setprecision(12);
return x1;
}
template<typename F_t, typename dF_t>
double newton(double x0, double epsilon, F_t F, dF_t dF, long iter_limit) {
return find_root(x0, epsilon, [&F, &dF](double x0){ return x0 - F(x0)/dF(x0); }, iter_limit);
}
template<typename F_t>
double fixed_point(double x0, double epsilon, F_t F, long iter_limit) {
return find_root(x0, epsilon, F, iter_limit);
}
int main() {
auto F = [](double x){ return std::pow(x, 2.0) - std::pow(std::cos(x), 2.0); };
auto dF = [](double x){ return 2.0 * x + std::sin(2.0 * x); };
// std::cout.precision(17);
double epsilon = 5.0 * std::pow(10.0, -12.0);
std::cout << newton(0.8, epsilon, F, dF, 100000) << '\n';
auto F2 = [](double x){ return std::cos(x); };
std::cout << fixed_point(0.8, epsilon, F2, 100000) << '\n';
}

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3 - Splines/LICENSE Normal file
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---
title: "Análise Numérica - Trabalho Prático 3"
author:
- Diogo Cordeiro
- Hugo Sales
- Pedro Costa
geometry: margin=2cm
output: pdf_document
---
### Motivação
Pretende-se interpolar uma função através do método de Newton em diferenças divididas,
construir o spline cúbico natural e proceder a uma comparação e interpretação dos resultados obtidos.
### 1.a)
using points_t = std::pair<std::vector<double>, std::vector<double>>;
using matrix_t = std::vector<std::vector<double>>;
std::vector<double> newton_differences(points_t points) {
std::vector<double> factors{};
auto &[x, fx] = points;
int n = points.first.size();
for (int i = 1; i <= n; ++i) {
factors.push_back(fx[0]);
for (int j = 0; j < (n - i); ++j) {
fx[j] = (fx[j + 1] - fx[j]) / (x[j + i] - x[j]);
}
}
return factors;
}
double newton_polynomial(points_t points, std::vector<double> factors, double x) {
auto xs = points.first;
int n = points.second.size() - 1;
double val = 0;
for (int k = 0; k <= n; ++k) {
double acc = 1;
for(int i = 0; i < k; ++i) {
acc *= (x - xs[i]);
}
val += acc * factors[k];
}
return val;
}
\pagebreak
void exercise_a(points_t points) {
auto &[xs, fx] = points;
auto factors = newton_differences(points);
std::ofstream poly{"a_polynomial.txt"};
for(double x = 0; x < 4; x += 0.001)
poly << x << " " << newton_polynomial(points, factors, x) << '\n';
std::ofstream spline{"a_spline.txt"};
unsigned long n = xs.size();
matrix_t mat(n, std::vector<double>(n + 1));
calculate_natural_cubic_spline_matrix(points, mat);
for(double x = 0; x < 4; x += 0.001)
spline << x << " " << natural_cubic_spline(points, mat, x) << '\n';
}
### 1.b)
void calculate_natural_cubic_spline_matrix(points_t points, matrix_t &mat) {
auto &[xs, fx] = points;
int n = xs.size();
// Construção da matriz
for (int i = 1; i < n - 1; ++i) {
mat[i][i - 1] = (xs[i] - xs[i - 1])/6;
mat[i][i] = (xs[i + 1] - xs[i - 1])/3;
mat[i][i + 1] = (xs[i + 1] - xs[i])/6;
mat[i][n] = (fx[i + 1] - fx[i])/(xs[i + 1] - xs[i]) -
(fx[i] - fx[i - 1])/(xs[i] - xs[i - 1]);
}
mat[0][0] = 1;
mat[0][n - 1] = 0;
mat[n - 1][n - 1] = 1;
mat[n - 1][n] = 0;
// Passar para a forma triangular
for (int k = 0; k < n; ++k) {
for (int i = k + 1; i < n; ++i) {
if (mat[k][k] != 0) {
double mul = mat[i][k]/mat[k][k];
for (int j = k; j < n; ++j) {
mat[i][j] -= mul * mat[k][j];
}
mat[i][n] -= mul * mat[k][n];
}
}
}
\pagebreak
// Resolução da matriz
for (int i = n - 1; i > 0; --i) {
if (mat[i][i] != 0) {
double mul = mat[i - 1][i]/mat[i][i];
for (int j = 0; j < n + 1; ++j) {
mat[i - 1][j] -= mul * mat[i][j];
}
mat[i][n] /= mat[i][i];
mat[i][i] = 1;
}
}
}
double natural_cubic_spline(points_t points, matrix_t &mat, double x) {
auto &[xs, fx] = points;
int n = xs.size();
int i = 0;
for (int i_ = 0; i_ < n; ++i_) {
if (xs[i_] > x){
i = i_;
break;
}
}
double hi = xs[i] - xs[i - 1];
return mat[i - 1][n] * std::pow((xs[i] - x), 3)/(6 * hi) +
mat[i][n] * std::pow((x - xs[i - 1]), 3)/(6 * hi) +
(fx[i - 1] - mat[i - 1][n] * (hi * hi)/6)*(xs[i] - x)/hi +
(fx[i] - mat[i][n] * (hi * hi)/6)*(x - xs[i - 1])/hi;
}
void exercise_b() {
points_t points;
auto &[xs, fx] = points;
auto f = [](double x) { return 4 * std::pow(x, 2) + std::sin(9 * x); };
for (double x = -1; x <= 1; x += (1 - -1)/8.0) {
xs.push_back(x);
fx.push_back(f(x));
}
std::ofstream poly{"b_polynomial.txt"};
auto factors = newton_differences(points);
for(double x = -1; x < 1; x += 0.001)
poly << x << " " << newton_polynomial(points, factors, x) << '\n';
std::ofstream spline{"b_spline.txt"};
unsigned long n = xs.size();
matrix_t mat(n, std::vector<double>(n + 1));
calculate_natural_cubic_spline_matrix(points, mat);
for(double x = -1; x < 1; x += 0.001)
spline << x << " " << natural_cubic_spline(points, mat, x) << '\n';
}
### 2.a)
![Comparação Polinómio e Spline](a.png){ width=10cm }
Através dos graficos é possivel verificar que os valores das funções coincidem
no pontos de interpolação, como era esperado. Não é possivel determinal qual a
melhor aproximação sem conhecer a função original. O programa foi modificado para
imprimir a matriz inicial assim como o resultado final, que foi utilizado para calcular
o vetor resíduo e a norma deste. Este script encontra-se em baixo.
#!/usr/bin/env python3
import numpy as np
from scipy.linalg import solve
from numpy.linalg import norm
# Matriz construída pelo programa em C++
A = np.matrix([[1, 0, 0, 0, 0, 0],
[0.166667, 0.666667, 0.166667, 0, 0, 0],
[0, 0.166667, 0.5, 0.0833333, 0, 0],
[0, 0, 0.0833333, 0.333333, 0.0833333, 0],
[0, 0, 0, 0.0833333, 0.5, 0.166667],
[0, 0, 0, 0, 0, 1]])
b = np.array([0, 1.2, -1.2, 0.8, 0.4, 0])
x2 = np.array([0, 2.76846, -3.87386, 3.30622, 0.248963, 0])
x = solve(A, b)
print(norm(x - x2))
Obteu-se para a norma do vetor resíduo o valor $1.0e-05$ que
é considerado aceitável visto que os dados são apresentados
com dois algarismos significativos.
\pagebreak
### 2.b)
#### i)
$$
\begin{array}{c|ccccccccc}{x_{i}} & {-1} & {-0.75} & {-0.5} & {-0.25} & {0} &
{0.25} & {0.5} & {0.75} & {1} \\ \hline {f_{i}} & {3.58788} & {1.79996} &
{1.97753} & {-0.528073} & {0} & {1.02807} & {0.0224699} & {2.70004} & {4.41212}
\end{array}
$$
#### ii)
![Comparação de polinomio, spline e a função](b.png){ width=10cm }
De forma análoga ao exercício 2.a) foi calculada a norma do vetor resíduo
usando o seguinte script:
A = np.matrix(
[[1, 0, 0, 0, 0, 0, 0, 0, 0],
[0.0416667, 0.166667, 0.0416667, 0, 0, 0, 0, 0, 0],
[0, 0.0416667, 0.166667, 0.0416667, 0, 0, 0, 0, 0],
[0, 0, 0.0416667, 0.166667, 0.0416667, 0, 0, 0, 0],
[0, 0, 0, 0.0416667, 0.166667, 0.0416667, 0, 0, 0],
[0, 0, 0, 0, 0.0416667, 0.166667, 0.0416667, 0, 0],
[0, 0, 0, 0, 0, 0.0416667, 0.166667, 0.0416667, 0],
[0, 0, 0, 0, 0, 0, 0.0416667, 0.166667, 0.0416667],
[0, 0, 0, 0, 0, 0, 0, 0, 1]])
b = np.array([0, 7.862, -10.7327, 12.1347, 2, -8.13471, 14.7327, -3.862, 0])
x2 = np.array([0, 73.9996, -107.31, 97.6564, 7.91753, -81.3265, 122.156, -53.7109, 0])
x = solve(A, b)
print(norm(x - x2))
Obteu-se para a norma do vetor resíduo o valor $9.0e-04$ que
é considerado aceitável visto que este valor tem uma magnitude pequena.
\pagebreak
#### iii)
x | f(x) | p(x) | abs(f(x)-p(x)) | s(x) | f(x)-s(x)
-----|------------|----------|--------------------|-------------------|-------------------
0.30 | 0.78737988 | 0.923318 | $$ 1.4*10^{-1} $$ | 0.826621 | $$ 4.0*10^{-2} $$
0.83 | 3.68278040 | 4.834190 | 1.2 | $$ 2.4*10^{-1} $$ | $$ 2.4*10^{-1} $$
Verificamos que o erro do spline cubico é inferior em ambos os casos. Além disso
em ambos os casos o erro é maior quando a abcissa é mais distante do centro do
intervalo.
#### iv)
É possivel observar que a interpolação pelo spline aproxima, em geral, melhor a
função dada do que o polinomio. Em particular verifica-se que o erro do polinomio
acentua-se à medida que as abcissas se afastam do centro do intervalo de
interpolação.

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#!/usr/bin/gnuplot
set output "a.png";
set terminal png size 1200,900 enhanced font "Helvetica, 20";
plot "a_polynomial.txt" with l lw 4 t "Polinomial", "a_spline.txt" with l lw 4 t "Spline Cúbico";
set output "b.png";
f(x) = 4*x**2 + sin(9*x)
plot f(x) with l lw 4 t "f(x)", "b_polynomial.txt" with l lw 4 t "Polinomial", "b_spline.txt" with l lw 4 t "Spline Cúbico";

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0 1.4
1 0.6
2 1.0
2.5 0.6
3 0.6
4 1.0

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mat:
1,0,0,0,0,0,
0.166667,0.666667,0.166667,0,0,0,
0,0.166667,0.5,0.0833333,0,0,
0,0,0.0833333,0.333333,0.0833333,0,
0,0,0,0.0833333,0.5,0.166667,
0,0,0,0,0,1,
b:
0
1.2
-1.2
0.8
0.4
0
x:
0
2.76846
-3.87386
3.30622
0.248963
0
mat:
1,0,0,0,0,0,0,0,0,
0.0416667,0.166667,0.0416667,0,0,0,0,0,0,
0,0.0416667,0.166667,0.0416667,0,0,0,0,0,
0,0,0.0416667,0.166667,0.0416667,0,0,0,0,
0,0,0,0.0416667,0.166667,0.0416667,0,0,0,
0,0,0,0,0.0416667,0.166667,0.0416667,0,0,
0,0,0,0,0,0.0416667,0.166667,0.0416667,0,
0,0,0,0,0,0,0.0416667,0.166667,0.0416667,
0,0,0,0,0,0,0,0,1,
b:
0
7.862
-10.7327
12.1347
2
-8.13471
14.7327
-3.862
0
x:
0
73.9996
-107.31
97.6564
7.91753
-81.3265
122.156
-53.7109
0

0
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#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <iomanip>
#include <ios>
#include <iostream>
#include <numeric>
#include <string>
#include <utility>
#include <vector>
using namespace std::literals;
using points_t = std::pair<std::vector<double>, std::vector<double>>;
using matrix_t = std::vector<std::vector<double>>;
std::vector<double> newton_differences(points_t points) {
std::vector<double> factors{};
auto &[x, fx] = points;
int n = points.first.size();
for (int i = 1; i <= n; ++i) {
factors.push_back(fx[0]);
for (int j = 0; j < (n - i); ++j) {
fx[j] = (fx[j + 1] - fx[j]) / (x[j + i] - x[j]);
}
}
return factors;
}
double newton_polynomial(points_t points, std::vector<double> factors, double x) {
auto xs = points.first;
int n = points.second.size() - 1;
double val = 0;
for (int k = 0; k <= n; ++k) {
double acc = 1;
for(int i = 0; i < k; ++i) {
acc *= (x - xs[i]);
}
val += acc * factors[k];
}
return val;
}
void calculate_natural_cubic_spline_matrix(points_t points, matrix_t &mat) {
auto &[xs, fx] = points;
int n = xs.size();
// Construção da matriz
for (int i = 1; i < n - 1; ++i) {
mat[i][i - 1] = (xs[i] - xs[i - 1])/6;
mat[i][i] = (xs[i + 1] - xs[i - 1])/3;
mat[i][i + 1] = (xs[i + 1] - xs[i])/6;
mat[i][n] = (fx[i + 1] - fx[i])/(xs[i + 1] - xs[i]) - (fx[i] - fx[i - 1])/(xs[i] - xs[i - 1]);
}
mat[0][0] = 1;
mat[0][n - 1] = 0;
mat[n - 1][n - 1] = 1;
mat[n - 1][n] = 0;
std::ofstream matriz("matriz.txt", std::ios_base::app);
matriz << "mat:\n";
for (auto i : mat){
for (auto j = i.begin(); j != i.end() - 1; ++ j)
matriz << *j << ",";
matriz << "\n";
}
matriz << "b:\n";
for (auto i : mat){
matriz << i[n] << '\n';
}
// Passar para a forma triangular
for (int k = 0; k < n; ++k) {
for (int i = k + 1; i < n; ++i) {
if (mat[k][k] != 0) {
double mul = mat[i][k]/mat[k][k];
for (int j = k; j < n; ++j) {
mat[i][j] -= mul * mat[k][j];
}
mat[i][n] -= mul * mat[k][n];
}
}
}
// Resolução da matriz
for (int i = n - 1; i > 0; --i) {
if (mat[i][i] != 0) {
double mul = mat[i - 1][i]/mat[i][i];
for (int j = 1; j < n + 1; ++j) {
mat[i - 1][j] -= mul * mat[i][j];
}
mat[i - 1][i] = 0;
mat[i][n] /= mat[i][i];
mat[i][i] = 1;
}
}
matriz << "x:\n";
for (auto i : mat){
matriz << i[n] << '\n';
}
}
double natural_cubic_spline(points_t points, matrix_t &mat, double x) {
auto &[xs, fx] = points;
int n = xs.size();
int i = 0;
for (int i_ = 0; i_ < n; ++i_) {
if (xs[i_] > x){
i = i_;
break;
}
}
double hi = xs[i] - xs[i - 1];
return mat[i - 1][n] * std::pow((xs[i] - x), 3)/(6 * hi) +
mat[i][n] * std::pow((x - xs[i - 1]), 3)/(6 * hi) +
(fx[i - 1] - mat[i - 1][n] * (hi * hi)/6)*(xs[i] - x)/hi +
(fx[i] - mat[i][n] * (hi * hi)/6)*(x - xs[i - 1])/hi;
}
void exercise_a(points_t points) {
auto &[xs, fx] = points;
auto factors = newton_differences(points);
std::ofstream poly{"a_polynomial.txt"};
for(double x = 0; x < 4; x += 0.001)
poly << x << " " << newton_polynomial(points, factors, x) << '\n';
std::ofstream spline{"a_spline.txt"};
unsigned long n = xs.size();
matrix_t mat(n, std::vector<double>(n + 1));
calculate_natural_cubic_spline_matrix(points, mat);
for(double x = 0; x < 4; x += 0.001)
spline << x << " " << natural_cubic_spline(points, mat, x) << '\n';
}
void exercise_b() {
points_t points;
auto &[xs, fx] = points;
auto f = [](double x) { return 4 * std::pow(x, 2) + std::sin(9 * x); };
for (double x = -1; x <= 1; x += (1 - -1)/8.0) {
xs.push_back(x);
fx.push_back(f(x));
}
std::ofstream poly{"b_polynomial.txt"};
auto factors = newton_differences(points);
for(double x = -1; x < 1; x += 0.001)
poly << x << " " << newton_polynomial(points, factors, x) << '\n';
std::ofstream spline{"b_spline.txt"};
unsigned long n = xs.size();
matrix_t mat(n, std::vector<double>(n + 1));
calculate_natural_cubic_spline_matrix(points, mat);
for(double x = -1; x < 1; x += 0.001)
spline << x << " " << natural_cubic_spline(points, mat, x) << '\n';
std::cout << "Erros poly: "
<< std::abs(std::round((f(0.3) - newton_polynomial(points, factors, 0.3)) * 100.0) / 100.0) << " "
<< std::abs(std::round((f(0.83) - newton_polynomial(points, factors, 0.83)) * 100.0) / 100.0) << '\n';
std::cout << "Erros spline: "
<< std::abs(std::round((f(0.3) - natural_cubic_spline(points, mat, 0.3))*100.0)/100.0) << " "
<< std::abs(std::round((f(0.83) - natural_cubic_spline(points, mat,0.83))*100.0)/100.0) << '\n';
std::cout << "p(0.3) = " << newton_polynomial(points, factors, 0.3) << ' '
<< "p(0.83) = " << newton_polynomial(points, factors, 0.83) << '\n';
std::cout << "p(0.3) = " << natural_cubic_spline(points, mat, 0.3) << ' '
<< "p(0.83) = " << natural_cubic_spline(points, mat, 0.83) << '\n';
}
int main(int argc, char **argv) {
std::istream *s;
std::ifstream file;
if (argc == 2) {
if (argv[1] == "-"s) {
s = &std::cin;
} else {
file.open(argv[1]);
s = &file;
}
} else{
std::cerr << "An argument required\n";
return 1;
}
points_t points;
std::pair<double, double> p;
while(true) {
*s >> p.first >> p.second;
if(s->eof())
break;
points.first.push_back(p.first);
points.second.push_back(p.second);
}
exercise_a(points);
exercise_b();
}

17
3 - Splines/source/old.foo Normal file
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Mat:
1,0,0,0,0,0,0,
0,1,-7.14916e-18,2.73077e-18,-4.18062e-19,0,2.76846,
0,0,1,-7.50963e-18,1.14967e-18,0,-3.87386,
0,0,-3.57458e-18,1,-4.38965e-18,0,3.30622,
0,0,0,-7.50963e-18,1,0,0.248963,
0,0,0,0,0,1,0,
Mat:
1,0,0,0,0,0,0,0,0,0,
0,1,-1.73472e-18,4.64061e-19,2.85733e-19,5.30686e-20,-1.26382e-20,-4.49207e-21,0,73.9996,
0,0,1,-1.74023e-18,-1.0715e-18,-1.99007e-19,4.73931e-20,1.68453e-20,0,-107.31,
0,0,-1.73472e-18,1,4.00026e-18,7.4296e-19,-1.76934e-19,-6.2889e-20,0,97.6564,
0,0,0,-1.74023e-18,1,-2.77283e-18,6.60344e-19,2.34711e-19,0,7.91753,
0,0,0,0,4.00026e-18,1,-2.46444e-18,-8.75954e-19,0,-81.3265,
0,0,0,0,0,-2.77283e-18,1,3.26911e-18,0,122.156,
0,0,0,0,0,0,-2.46444e-18,1,0,-53.7109,
0,0,0,0,0,0,0,0,1,0,

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#!/usr/bin/env python3
import numpy as np
from scipy.linalg import solve
from numpy.linalg import norm
# Matriz construída pelo programa em C++
A = np.matrix([[1, 0, 0, 0, 0, 0],
[0.166667, 0.666667, 0.166667, 0, 0, 0],
[0, 0.166667, 0.5, 0.0833333, 0, 0],
[0, 0, 0.0833333, 0.333333, 0.0833333, 0],
[0, 0, 0, 0.0833333, 0.5, 0.166667],
[0, 0, 0, 0, 0, 1]])
b = np.array([0, 1.2, -1.2, 0.8, 0.4, 0])
x2 = np.array([0, 2.76846, -3.87386, 3.30622, 0.248963, 0])
x = solve(A, b)
print(norm(x - x2))
A = np.matrix([[1, 0, 0, 0, 0, 0, 0, 0, 0],
[0.0416667, 0.166667, 0.0416667, 0, 0, 0, 0, 0, 0],
[0, 0.0416667, 0.166667, 0.0416667, 0, 0, 0, 0, 0],
[0, 0, 0.0416667, 0.166667, 0.0416667, 0, 0, 0, 0],
[0, 0, 0, 0.0416667, 0.166667, 0.0416667, 0, 0, 0],
[0, 0, 0, 0, 0.0416667, 0.166667, 0.0416667, 0, 0],
[0, 0, 0, 0, 0, 0.0416667, 0.166667, 0.0416667, 0],
[0, 0, 0, 0, 0, 0, 0.0416667, 0.166667, 0.0416667],
[0, 0, 0, 0, 0, 0, 0, 0, 1]])
b = np.array([0, 7.862, -10.7327, 12.1347, 2, -8.13471, 14.7327, -3.862, 0])
x2 = np.array([0, 73.9996, -107.31, 97.6564, 7.91753, -81.3265, 122.156, -53.7109, 0])
x = solve(A, b)
print(norm(x - x2))

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Mat:
1,0,0,0,0,0,0,
0,1,0,0,0,0,2.76846,
0,0,1,0,0,0,-3.87386,
0,0,-3.57458e-18,1,0,0,3.30622,
0,0,0,-7.50963e-18,1,0,0.248963,
0,0,0,0,0,1,0,
Mat:
1,0,0,0,0,0,0,0,0,0,
0,1,0,0,0,0,0,0,0,73.9996,
0,0,1,0,0,0,0,0,0,-107.31,
0,0,-1.73472e-18,1,0,0,0,0,0,97.6564,
0,0,0,-1.74023e-18,1,0,0,0,0,7.91753,
0,0,0,0,4.00026e-18,1,0,0,0,-81.3265,
0,0,0,0,0,-2.77283e-18,1,0,0,122.156,
0,0,0,0,0,0,-2.46444e-18,1,0,-53.7109,
0,0,0,0,0,0,0,0,1,0,

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---
title: "Análise Numérica - Trabalho Prático 3"
author:
- Diogo Cordeiro
- Hugo Sales
- Pedro Costa
geometry: margin=2cm
output: pdf_document
---
### Motivação
Pretende-se compreender o funcionamento conceptual bem como os desafios da implementação de dois métodos de integração
numéricos: `Regra de Simpson` e `Regra dos Trapézios.`
### 1
<!-- $$ \frac{d^4}{dx^4}_{x \in [0;2]}sin(sin(sin(sin(x)))) $$ -->
Através da análise do gráfico abaixo, verificamos que o majorante, em valor absoluto, da 4ª derivada
da função é menor que 12, este valor é usado para majorar a formula do erro para o calculo de n.
![Quarta derivada da função enunciada](majorante_erro_1.png){ width=8cm }
/* Função que implementa o método de Simpson */
double simpson (Function f, double a, double b, int n)
{
// Intervalo de passo
double h = (b - a)/n;
// Valor de f nos pontos de indice par
double evens = summation(f, 2, n - 2, 2, h, a);
// Valor de f nos pontos de indice impar
double odds = summation(f, 1, n - 1, 2, h, a);
// Aplicação da método de Simpson
value = (h/3)*(f(a) + f(b) + 2 * evens + 4 * odds);
}
\pagebreak
/* Função para calcular o sumatório de F entre os pontos de indice init
e stop saltando step pontos, usando h como intervalo de passo */
double summation (Function f, int init, int stop, int step, double h, double a)
{
// Acumulador
double total = 0;
for (int i = init; i <= stop; i += step) {
// Adicionamos o valor de f correspondente ao x de indice i
total += f(a + i * h);
}
return total;
}
/* Função para calcular o número de pontos necessários para o calculo
do integral com erro menor que error */
int calculateN (double A, double B, double error) {
int n = ceil( (B - A) / pow((15.0 * error) / 2.0, 1.0 / 4) );
return n + (n % 2);
}
void main ()
{
Function f(x) = sin(sin(sin(sin(x))));
// com 7 casas decimais correctas
print(simpson(f, 0, 2, calculateN(0, 2, pow(10, -7))));
// com 12 casas decimais correctas
print(simpson(f, 0, 2, calculateN(0, 2, pow(10, -12))));
}
#### Output
Erro | Resultado
---------------|-------------------
$$ 10^{-7} $$ | 1.0548418906594816
$$ 10^{-12} $$ | 1.0548418772492483
### 2)
// Valor exacto do integral calculado com o WolframAlpha, arredondado
// com 15 algarismos significativos, um a mais do que o erro majorado
// máximo para o caso de 2^20 pontos.
double I = 1.05484187724912;
/* Função para calcular o integral recorrendo ao método do Trapézio */
double trapezio (Function f, double a, double b) {
// Acumulador
double summation = 0;
// Intervalo de passo com n inicial = 2
double h = (b - a)/2;
// Valor constante
double fa_fb = f(a) + f(b);
// Para cada expoente de 1 a 20 (com passo 1)
for (int k = 1; k <= 20; ++k) {
// Guarda sumatorio dos pontos anteriores
double partial_sum = 0;
// Número de intervalos
// Com o left shift fazemos a potência de 2^k
int n = 1 << k;
// Evitamos recalcular pontos da função previamente computados
// guardando o sumatório destes em partial_sum e adicionamos a cada
// iteração os pontos novos, sendo estes de indice impar
for (int i = 1; i < n; i += 2) {
// Adicionamos o valor de f correspondente ao x de indice i
partial_sum += f(a + i * h);
}
summation += partial_sum;
// Aplicação da formula do Trapézio
double value = (h/2)*fa_fb + h * summation;
print(k + "\t| " + value + "\t| " + (I - value));
// Dividimos o intervalo por 2
h /= 2;
}
}
void main () {
Function f(x) = sin(sin(sin(sin(x))));
print(trapezio(f, 0, 2);
}
#### Output
k | $$ I_{n_{k}} $$ | $$ \mid I - I_{n_{k}} \mid $$
----|---------------------|--------------------------
1 | 0.9533749638740736 | $$ 1.1 \cdot 10^{-1} $$
2 | 1.0308378382617962 | $$ 2.5 \cdot 10^{-2} $$
3 | 1.0489039934457873 | $$ 6.0 \cdot 10^{-3} $$
4 | 1.053360809734676 | $$ 1.5 \cdot 10^{-3} $$
5 | 1.0544718169560368 | $$ 3.8 \cdot 10^{-4} $$
6 | 1.054749374997165 | $$ 9.3 \cdot 10^{-5} $$
7 | 1.0548187524860693 | $$ 2.4 \cdot 10^{-5} $$
8 | 1.0548360961083287 | $$ 5.8 \cdot 10^{-6} $$
9 | 1.054840431967042 | $$ 1.5 \cdot 10^{-6} $$
10 | 1.0548415159287925 | $$ 3.6 \cdot 10^{-7} $$
11 | 1.0548417869190467 | $$ 9.1 \cdot 10^{-8} $$
12 | 1.0548418546666 | $$ 2.3 \cdot 10^{-8} $$
13 | 1.054841871603487 | $$ 5.7 \cdot 10^{-9} $$
14 | 1.0548418758377078 | $$ 1.5 \cdot 10^{-9} $$
15 | 1.0548418768962615 | $$ 3.5 \cdot 10^{-10} $$
16 | 1.0548418771608998 | $$ 8.9 \cdot 10^{-11} $$
17 | 1.0548418772270582 | $$ 2.3 \cdot 10^{-11} $$
18 | 1.0548418772436017 | $$ 5.6 \cdot 10^{-12} $$
19 | 1.0548418772477444 | $$ 1.4 \cdot 10^{-12} $$
20 | 1.0548418772487873 | $$ 3.3 \cdot 10^{-13} $$

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/* cyrillic-ext */
@font-face {
font-family: 'Merriweather';
font-style: normal;
font-weight: 400;
src: local('Merriweather Regular'), local('Merriweather-Regular'), url(https://fonts.gstatic.com/s/merriweather/v19/u-440qyriQwlOrhSvowK_l5-cSZMZ-Y.woff2) format('woff2');
unicode-range: U+0460-052F, U+1C80-1C88, U+20B4, U+2DE0-2DFF, U+A640-A69F, U+FE2E-FE2F;
}
/* cyrillic */
@font-face {
font-family: 'Merriweather';
font-style: normal;
font-weight: 400;
src: local('Merriweather Regular'), local('Merriweather-Regular'), url(https://fonts.gstatic.com/s/merriweather/v19/u-440qyriQwlOrhSvowK_l5-eCZMZ-Y.woff2) format('woff2');
unicode-range: U+0400-045F, U+0490-0491, U+04B0-04B1, U+2116;
}
/* vietnamese */
@font-face {
font-family: 'Merriweather';
font-style: normal;
font-weight: 400;
src: local('Merriweather Regular'), local('Merriweather-Regular'), url(https://fonts.gstatic.com/s/merriweather/v19/u-440qyriQwlOrhSvowK_l5-cyZMZ-Y.woff2) format('woff2');
unicode-range: U+0102-0103, U+0110-0111, U+1EA0-1EF9, U+20AB;
}
/* latin-ext */
@font-face {
font-family: 'Merriweather';
font-style: normal;
font-weight: 400;
src: local('Merriweather Regular'), local('Merriweather-Regular'), url(https://fonts.gstatic.com/s/merriweather/v19/u-440qyriQwlOrhSvowK_l5-ciZMZ-Y.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Merriweather';
font-style: normal;
font-weight: 400;
src: local('Merriweather Regular'), local('Merriweather-Regular'), url(https://fonts.gstatic.com/s/merriweather/v19/u-440qyriQwlOrhSvowK_l5-fCZM.woff2) format('woff2');
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}

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/* cyrillic */
@font-face {
font-family: 'Oswald';
font-style: normal;
font-weight: 400;
src: local('Oswald Regular'), local('Oswald-Regular'), url(https://fonts.gstatic.com/s/oswald/v16/TK3iWkUHHAIjg752HT8Ghe4.woff2) format('woff2');
unicode-range: U+0400-045F, U+0490-0491, U+04B0-04B1, U+2116;
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/* vietnamese */
@font-face {
font-family: 'Oswald';
font-style: normal;
font-weight: 400;
src: local('Oswald Regular'), local('Oswald-Regular'), url(https://fonts.gstatic.com/s/oswald/v16/TK3iWkUHHAIjg752Fj8Ghe4.woff2) format('woff2');
unicode-range: U+0102-0103, U+0110-0111, U+1EA0-1EF9, U+20AB;
}
/* latin-ext */
@font-face {
font-family: 'Oswald';
font-style: normal;
font-weight: 400;
src: local('Oswald Regular'), local('Oswald-Regular'), url(https://fonts.gstatic.com/s/oswald/v16/TK3iWkUHHAIjg752Fz8Ghe4.woff2) format('woff2');
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}
/* latin */
@font-face {
font-family: 'Oswald';
font-style: normal;
font-weight: 400;
src: local('Oswald Regular'), local('Oswald-Regular'), url(https://fonts.gstatic.com/s/oswald/v16/TK3iWkUHHAIjg752GT8G.woff2) format('woff2');
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}

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/* cyrillic-ext */
@font-face {
font-family: 'Ubuntu Mono';
font-style: normal;
font-weight: 400;
src: local('Ubuntu Mono'), local('UbuntuMono-Regular'), url(https://fonts.gstatic.com/s/ubuntumono/v8/KFOjCneDtsqEr0keqCMhbCc3CsTKlA.woff2) format('woff2');
unicode-range: U+0460-052F, U+1C80-1C88, U+20B4, U+2DE0-2DFF, U+A640-A69F, U+FE2E-FE2F;
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/* cyrillic */
@font-face {
font-family: 'Ubuntu Mono';
font-style: normal;
font-weight: 400;
src: local('Ubuntu Mono'), local('UbuntuMono-Regular'), url(https://fonts.gstatic.com/s/ubuntumono/v8/KFOjCneDtsqEr0keqCMhbCc-CsTKlA.woff2) format('woff2');
unicode-range: U+0400-045F, U+0490-0491, U+04B0-04B1, U+2116;
}
/* greek-ext */
@font-face {
font-family: 'Ubuntu Mono';
font-style: normal;
font-weight: 400;
src: local('Ubuntu Mono'), local('UbuntuMono-Regular'), url(https://fonts.gstatic.com/s/ubuntumono/v8/KFOjCneDtsqEr0keqCMhbCc2CsTKlA.woff2) format('woff2');
unicode-range: U+1F00-1FFF;
}
/* greek */
@font-face {
font-family: 'Ubuntu Mono';
font-style: normal;
font-weight: 400;
src: local('Ubuntu Mono'), local('UbuntuMono-Regular'), url(https://fonts.gstatic.com/s/ubuntumono/v8/KFOjCneDtsqEr0keqCMhbCc5CsTKlA.woff2) format('woff2');
unicode-range: U+0370-03FF;
}
/* latin-ext */
@font-face {
font-family: 'Ubuntu Mono';
font-style: normal;
font-weight: 400;
src: local('Ubuntu Mono'), local('UbuntuMono-Regular'), url(https://fonts.gstatic.com/s/ubuntumono/v8/KFOjCneDtsqEr0keqCMhbCc0CsTKlA.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Ubuntu Mono';
font-style: normal;
font-weight: 400;
src: local('Ubuntu Mono'), local('UbuntuMono-Regular'), url(https://fonts.gstatic.com/s/ubuntumono/v8/KFOjCneDtsqEr0keqCMhbCc6CsQ.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}

224
4 - Simpson and Trapezoidal rules/slides/css/css4.css Normal file
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@ -0,0 +1,224 @@
/* latin-ext */
@font-face {
font-family: 'Quattrocento';
font-style: normal;
font-weight: 400;
src: local('Quattrocento'), url(https://fonts.gstatic.com/s/quattrocento/v10/OZpEg_xvsDZQL_LKIF7q4jP3zWj6T4g.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Quattrocento';
font-style: normal;
font-weight: 400;
src: local('Quattrocento'), url(https://fonts.gstatic.com/s/quattrocento/v10/OZpEg_xvsDZQL_LKIF7q4jP3w2j6.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* latin-ext */
@font-face {
font-family: 'Quattrocento';
font-style: normal;
font-weight: 700;
src: local('Quattrocento Bold'), local('Quattrocento-Bold'), url(https://fonts.gstatic.com/s/quattrocento/v10/OZpbg_xvsDZQL_LKIF7q4jP_eE3vfqnYgXc.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Quattrocento';
font-style: normal;
font-weight: 700;
src: local('Quattrocento Bold'), local('Quattrocento-Bold'), url(https://fonts.gstatic.com/s/quattrocento/v10/OZpbg_xvsDZQL_LKIF7q4jP_eE3vcKnY.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* latin-ext */
@font-face {
font-family: 'Quattrocento Sans';
font-style: italic;
font-weight: 400;
src: local('Quattrocento Sans Italic'), local('QuattrocentoSans-Italic'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9a4lja2NVIDdIAAoMR5MfuElaRB0zMh0P2GEHJ.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Quattrocento Sans';
font-style: italic;
font-weight: 400;
src: local('Quattrocento Sans Italic'), local('QuattrocentoSans-Italic'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9a4lja2NVIDdIAAoMR5MfuElaRB0zMh032GA.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* latin-ext */
@font-face {
font-family: 'Quattrocento Sans';
font-style: italic;
font-weight: 700;
src: local('Quattrocento Sans Bold Italic'), local('QuattrocentoSans-BoldItalic'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9X4lja2NVIDdIAAoMR5MfuElaRB0zMj_bTDXDojYsJ.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Quattrocento Sans';
font-style: italic;
font-weight: 700;
src: local('Quattrocento Sans Bold Italic'), local('QuattrocentoSans-BoldItalic'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9X4lja2NVIDdIAAoMR5MfuElaRB0zMj_bTDX7ojQ.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* latin-ext */
@font-face {
font-family: 'Quattrocento Sans';
font-style: normal;
font-weight: 400;
src: local('Quattrocento Sans'), local('QuattrocentoSans'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9c4lja2NVIDdIAAoMR5MfuElaRB0zHt0_uHA.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Quattrocento Sans';
font-style: normal;
font-weight: 400;
src: local('Quattrocento Sans'), local('QuattrocentoSans'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9c4lja2NVIDdIAAoMR5MfuElaRB0zJt08.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* latin-ext */
@font-face {
font-family: 'Quattrocento Sans';
font-style: normal;
font-weight: 700;
src: local('Quattrocento Sans Bold'), local('QuattrocentoSans-Bold'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9Z4lja2NVIDdIAAoMR5MfuElaRB0RyklrfPXzwiQ.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Quattrocento Sans';
font-style: normal;
font-weight: 700;
src: local('Quattrocento Sans Bold'), local('QuattrocentoSans-Bold'), url(https://fonts.gstatic.com/s/quattrocentosans/v11/va9Z4lja2NVIDdIAAoMR5MfuElaRB0RyklrRPXw.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* cyrillic */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 400;
src: local('Spectral Italic'), local('Spectral-Italic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCt-xNNww_2s0amA9M8on7mTMuk.woff2) format('woff2');
unicode-range: U+0400-045F, U+0490-0491, U+04B0-04B1, U+2116;
}
/* vietnamese */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 400;
src: local('Spectral Italic'), local('Spectral-Italic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCt-xNNww_2s0amA9M8onXmTMuk.woff2) format('woff2');
unicode-range: U+0102-0103, U+0110-0111, U+1EA0-1EF9, U+20AB;
}
/* latin-ext */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 400;
src: local('Spectral Italic'), local('Spectral-Italic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCt-xNNww_2s0amA9M8onTmTMuk.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 400;
src: local('Spectral Italic'), local('Spectral-Italic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCt-xNNww_2s0amA9M8onrmTA.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* cyrillic */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 700;
src: local('Spectral Bold Italic'), local('Spectral-BoldItalic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCu-xNNww_2s0amA9M8qsHDWfCFXUIJ.woff2) format('woff2');
unicode-range: U+0400-045F, U+0490-0491, U+04B0-04B1, U+2116;
}
/* vietnamese */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 700;
src: local('Spectral Bold Italic'), local('Spectral-BoldItalic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCu-xNNww_2s0amA9M8qsHDWfuFXUIJ.woff2) format('woff2');
unicode-range: U+0102-0103, U+0110-0111, U+1EA0-1EF9, U+20AB;
}
/* latin-ext */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 700;
src: local('Spectral Bold Italic'), local('Spectral-BoldItalic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCu-xNNww_2s0amA9M8qsHDWfqFXUIJ.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Spectral';
font-style: italic;
font-weight: 700;
src: local('Spectral Bold Italic'), local('Spectral-BoldItalic'), url(https://fonts.gstatic.com/s/spectral/v5/rnCu-xNNww_2s0amA9M8qsHDWfSFXQ.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* cyrillic */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 400;
src: local('Spectral Regular'), local('Spectral-Regular'), url(https://fonts.gstatic.com/s/spectral/v5/rnCr-xNNww_2s0amA9M9knj-SA.woff2) format('woff2');
unicode-range: U+0400-045F, U+0490-0491, U+04B0-04B1, U+2116;
}
/* vietnamese */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 400;
src: local('Spectral Regular'), local('Spectral-Regular'), url(https://fonts.gstatic.com/s/spectral/v5/rnCr-xNNww_2s0amA9M2knj-SA.woff2) format('woff2');
unicode-range: U+0102-0103, U+0110-0111, U+1EA0-1EF9, U+20AB;
}
/* latin-ext */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 400;
src: local('Spectral Regular'), local('Spectral-Regular'), url(https://fonts.gstatic.com/s/spectral/v5/rnCr-xNNww_2s0amA9M3knj-SA.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 400;
src: local('Spectral Regular'), local('Spectral-Regular'), url(https://fonts.gstatic.com/s/spectral/v5/rnCr-xNNww_2s0amA9M5kng.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}
/* cyrillic */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 700;
src: local('Spectral Bold'), local('Spectral-Bold'), url(https://fonts.gstatic.com/s/spectral/v5/rnCs-xNNww_2s0amA9uCt23FafadWQ.woff2) format('woff2');
unicode-range: U+0400-045F, U+0490-0491, U+04B0-04B1, U+2116;
}
/* vietnamese */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 700;
src: local('Spectral Bold'), local('Spectral-Bold'), url(https://fonts.gstatic.com/s/spectral/v5/rnCs-xNNww_2s0amA9uCt23OafadWQ.woff2) format('woff2');
unicode-range: U+0102-0103, U+0110-0111, U+1EA0-1EF9, U+20AB;
}
/* latin-ext */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 700;
src: local('Spectral Bold'), local('Spectral-Bold'), url(https://fonts.gstatic.com/s/spectral/v5/rnCs-xNNww_2s0amA9uCt23PafadWQ.woff2) format('woff2');
unicode-range: U+0100-024F, U+0259, U+1E00-1EFF, U+2020, U+20A0-20AB, U+20AD-20CF, U+2113, U+2C60-2C7F, U+A720-A7FF;
}
/* latin */
@font-face {
font-family: 'Spectral';
font-style: normal;
font-weight: 700;
src: local('Spectral Bold'), local('Spectral-Bold'), url(https://fonts.gstatic.com/s/spectral/v5/rnCs-xNNww_2s0amA9uCt23BafY.woff2) format('woff2');
unicode-range: U+0000-00FF, U+0131, U+0152-0153, U+02BB-02BC, U+02C6, U+02DA, U+02DC, U+2000-206F, U+2074, U+20AC, U+2122, U+2191, U+2193, U+2212, U+2215, U+FEFF, U+FFFD;
}

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<!DOCTYPE html>
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