====== Analiza numeryczna (M) - Lista 10. ======
{{:analiza_numeryczna:09.listaM10.pdf|Lista 10}}\\
{{:analiza_numeryczna:m10.pdf|Rozwiązania zadań 2, 3, 4, 7}}

===== Zadanie 1. =====
===== Zadanie 2. =====
===== Zadanie 3. =====
[[http://pl.wikipedia.org/wiki/Metody_Newtona-Cotesa]]
===== Zadanie 4. =====
===== Zadanie 5. =====

<code python># -*- coding: utf-8 -*- #
from math import *

eps = pow(10,-4)
a = 1
b = 10
h = 0.01 # tu się można bawić wartością, dla 0.01 jest już ok

n = (b-a)/h
m = int(n/2)

res = 14.02585092994046	# wynik z wolframa

def sum2(list):
	return list[0]*0.5 + sum(list[1:][:len(list)-2]) + list[len(list)-1]*0.5

print "h=",h
S1 = sum2([log(a + h*2*k) for k in xrange(1,m+1)])
print "S1=", S1
S2 = sum([log(a + h*(2*k-1)) for k in xrange(1,m+1)])
print "S2=", S2
Sn = h/3 * (2*S1 + 4*S2)
print "Sn=", Sn
print "Diff= %.6f" % abs(res-Sn)</code>

===== Zadanie 6. =====
$\displaystyle S_n(f)=\frac 13[4T_n(f) - T_{n/2}(f)]\ (n=2, 4, \cdots)$\\
Podprzedziały w $T_{n/2}$ są $2 $ razy większe, więc $ h' =  2h $\\
$\displaystyle =\frac 13[4\cdot h\sum^n_{k=0}" f(t_k) - 2\cdot h\sum^{n/2}_{k=0}" f(t_{2k}) ]$\\
$\displaystyle  =\frac h3[4\cdot \sum^n_{k=0}" f(t_k) - 2\cdot \sum^{n/2}_{k=0}" f(t_{2k}) ]$\\
$\displaystyle  =\frac h3[2\cdot f(t_0) + 4\cdot \sum^{n-1}_{k=1} f(t_k) + 2\cdot f(t_n)+$\\
$\displaystyle  2\cdot(-\frac 12\cdot f(t_0) - \sum^{n/2-1}_{k=1} f(t_{2k}) - \frac 12\cdot f(t_{n}))]$\\
$\displaystyle  =\frac h3[2\cdot f(t_0) + 4\cdot (f(t_1)+f(t_2)+\cdots +f(t_{n-1})) + 2\cdot f(t_n)$\\
$\displaystyle  -f(t_0) - 2(f(t_2) + f(t_4 )+\cdots f(t_{n-2}) ) - f(t_{n}))]$\\
$\displaystyle  =\frac h3[f(t_0) + 2\cdot(f(t_1)+f(t_2)+\cdots +f(t_{n-1})) + f(t_n)$\\
$\displaystyle  +2\cdot(f(t_1)+f(t_3)+\cdots + f(t_{n-1}))]$\\
$\displaystyle =\frac h3[f(t_0) + 2\cdot(f(t_2)+f(t_4)+\cdots +f(t_{n-2})) + f(t_n)$\\
$\displaystyle  +4\cdot(f(t_1)+f(t_3)+\cdots + f(t_{n-1}))]$\\
$\displaystyle =\frac h3[f(t_0) + 2\cdot \sum^{n/2 -1}_{k=1} f(t_{2k}) + f(t_n) + 4\cdot \sum^{n/2}_{k=1} f(t_{2k-1})]$\\
$\displaystyle =\frac h3[2\cdot \sum^{n/2}_{k=0}" f(t_{2k}) + 4\cdot \sum^{n/2}_{k=1} f(t_{2k-1})]$\\


===== Zadanie 7. =====

{{tag>[listy_zadan]}}