Q. 의학연구에의하면 4초후에낙하속도가 36 m/s를초과하면척추손상을입을가능성이높다. 항력계수가 kg/m 로주어질때이러한판정기준이초과하는질량을구하시오.
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1 수치해석 Ch5. Roots: Bracketing Methods 1
2 앞의낙하속도문제에서 v( t) gm gcd = tanh t c d m Q. 의학연구에의하면 4초후에낙하속도가 36 m/s를초과하면척추손상을입을가능성이높다. 항력계수가 0.25 kg/m 로주어질때이러한판정기준이초과하는질량을구하시오. gm gc d f( m) = tanh t v( t) c d m t=4, v=36 m/s f(m) = 0 을만족하는 m 을구하기
3 5.1 공학과과학에서의근 외재적 (Eplicit) 표현과내재적 (Implicit) 표현 gm gcd v( t) = tanh t 는파라미터가주어지고독립변수가 t 주어 c d m 지면종속변수 v(t) 를구할수있다. 외재적표현 y=f() gm gc d f( m) = tanh t vt () = 0 c d m 는와 t v(t) 가주어져도, 독립변수 m을바로구할수없다. 내재적표현 f(,y)=0
4 5.2 그래프를사용하는방법 (1/2) 방정식 f()=0 에대한근의근사값을구하는방법 f() 를그려 축과만나는점을찾는다. (a) 같은부호, 근없음 (b) 다른부호, 근 1개 (c) 같은부호, 근 2개 (d) 다른부호, 근3개 같은부호 : 짝수근 다른부호 : 홀수근 그림 5.1 하한값 l 과상한값 u 사이의구간에서근이존재할수있는몇가지경우
5 5.2 그래프를사용하는방법 (2/2) 일반적이지않은경우 (a) 중근을갖는경우 (b) 불연속인경우
6 예제 5.1 (1/2) Q. 자유낙하 4 초후의속도를 36 m/s 로되게하는번지점프하는사람의질량을그래프적인접근법으로구하라. ( 항력계수는 0.25 kg/m 이고, 중력가속도는 9.81 m/s2 이다.)
7 예제 5.1 (2/2) >> cd = 0.25; g = 9.81; v = 36; t = 4; >> mp = linspace (50, 200); >> fp = sqrt(g*mp/cd).*tanh(sqrt(g*cd./mp)*t) - v; >> plot(mp,fp), grid 근 m=145 m=145 를확인하면, >> sqrt(g*145/cd)*tanh(sqrt(g*cd/145)*t) ans =
8 < 수치해석의근찾는방법 > 초기가정값의유형에따른분류 구간법 - 근을포함하고있는구간의양끝을나타내는초기가정값에서부터시작. - 항상근을찾지만수렴이느리다. 개방법 - 한개또는그이상의초기가정값에서출발. - 근을포함한구간의양끝값은아님, - 발산하는경우도있지만, 수렴이빠르다
9 5.3 구간법과초기가정 (1/2) 증분탐색법 (incremental search method) f( l) f( u ) < 0 이면적어도 과 l u 사이에실근이하나이상존재 증분구간이너무작으면 : 계산시간이많이소요너무크면 : 근을놓치게됨 증분구간의크기에관계없이중근은놓칠위험이많음
10 5.3 구간법과초기가정 (2/2) function b = incsearch(func, min, ma, ns) % finds brackets of that contain sign changes of a function on an interval % input: % func= name of function % min, ma = endpoints of interval % ns = (optional) number of subintervals along % output: % b(k,1) is the lower bound of the kth sign changes % b(k,2) is the upper bound of the kth sign changes % If no brackets found, kb =[]. if nargin <4, ns =50; end % if ns blank set to 50 % Incremental search = linspace(min, ma, ns); f = feval(func,); nb = 0, b =[]; % b is null unless sign change detected for k = 1:length()-1 if sign(f(k)) ~= sign(f(k+1)) % check for sign change nb = nb + 1; b(nb,1) = (k); b(nb,2) = (k+1); end end if isempty(b) % display that no brackets were found disp('no brackets found') disp('check interval or increase ns') else disp('number of brackets:') %display number of brackets disp(nb) end
11 예제 5.2 (1/2) Q. incsearch 를사용하여구간 [3,6] 사이에서다음함수의부호가바뀌는구간을찾아라. Sol) f ( ) = sin(10) + cos(3) >> incsearch(@() sin(10*)+cos(3*), 3, 6) number of brackets: 5 ans = 소구간이너무넓어서 =4.25 와 5.2 사이의근을놓쳤다.
12 예제 5.2 (2/2) >> incsearch sin(10*)+cos(3*), 3,6, 100) number of brackets: 9 ans =
13 5.4 이분법 (bisection method) (1/4) 증분탐색법의변형으로구간폭을항상반으로나누는방법이다. 함수의부호가구간내에서바뀌면구간의중간점에서함수값을계산한다. 나뉜소구간중에서부호가바뀌는소구간에위치한근을구한다. 절대오차는매반복마다반으로감소 r = 2 l + u
14 5.4 이분법 (2/4) 이분법을마치기위한객관적인판단기준은? 근의참값을모르므로를이용할수없다. ε t 근사상대오차, a new old r r new r ε = 100% <ε s
15 예제 5.4 (1/3) Q. 이분법을이용하여자유낙하 4 초후의속도를 36 m/s 로되게하는번지점프하는사람의질량을구하라. 근사오차가 s = 0.5% 의종료판정기준이하가될때까지계산을반복하라. ( 단, 항력계수는 0.25 kg/m이고, 중력가속도는 9.81 m/s 2 이다. )
16 예제 5.4 (2/3) 반 복 구간추정근오차 (%) l u r εa εt f( 50) f(125) = 4.579( 0.409) = = 125 = 12.43% = = f( 125) f(162.5) = 0.409(0.359) = 0.147
17 예제 5.4 (3/3) 왜참오차는들쭉날쭉한형태를갖는가? 구간내의어느점이나참근이될수있기때문. 참근이구간의중앙에위치할때는 εt 와 εa 의차이가크다 참근이구간의끝쪽에위치할때는 ε 와 ε 의차이가작다. t a
18 5.4 이분법 (3/4) ε a ε t 는와같이줄어드는경향을보인다. ε a 는참오차의상한이므로 ε t 보다항상크다. 절대오차 : 반복을시작하기전 E 0 a = 0 = 0 u 0 l 1 번반복후 1 E a = 2 0 n 차례반복후 E n a = 2 n 0 만약 E a,d 가원하는오차라면, n = log( 0 log2 / E a, d ) = log 2 Ea, 0 d
19 5.4 이분법 (4/4) sin(10*)+cos(3*) >> bisection('func1', 50, 200, 0.001, 50) ans =
20 5.5 가위치법 (false position method) (1/3) 선형보간법이라고도하는구간법이다. 이분법과매우유사하다. f( l ) 과 f( u ) 를연결하는직선과 축의교점을찾아개선된추정값으로이용. 가위치법공식 r = u f( u f( l )( ) l f( u u ) )
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