作者gfs777 (成人展三萬嫖妓)
看板MATLAB
標題[討論] 牛頓法的求值與繪圖問題>"<
時間Tue Nov 15 11:03:36 2011
是這樣的>"< 這個問題是老師後天考試前發下來 給大家作為範例參考的題目
如下:
================================================================
Newton's method is an efficient algorithm for finding approximations to the
zeros (or roots) of a real-valued function. The algorithm can be written
as follows
Xn+1 = Xn - [f(Xn) / f'(Xn)]
Newton’s method iteratively approximates the root(s) for function f(x)
until Xn+1 and Xn difference is smaller than real number e
for example |Xn+1 - Xn| <e
(e沒給 我自設 1e-4)
write a script that that performs the Newton method to find a root for
y = xn^3 - 5*xn^2 - xn + 5
Plot the Xn on the y-axis with number of iterations on the x-axis.
===========================================================
目前我寫成這樣:
xn = input('first x?\n');
y = xn^3 - 5*xn^2 - xn + 5;
yplum = 3*xn^2 - 10*xn - 1;
cnt = 1;
e = 1e-4;
while cnt<=10000;
xn1 = xn - y / yplum
if abs(xn1 - xn)<= e
fprintf('converged')
break;
end
xn = xn1
cnt = cnt + 1
end
===========================================================
問一下這樣是對的嗎?
xn = -1.6062e+005 這答案是對的嗎? 我起始x = 50
xn = -2.9412e+005 起始x = 90
xn = -4.9415e+005 起始x= 150
跑幾次都不一樣><
另外畫圖要怎麼畫? 跪求~感謝><
那時候作業有個神人教我用clf hold figure等
現在我又不會用了....當時畫出來的都跟同學不一樣><
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◆ From: 140.112.76.121
1F：→ DKer:迴圈很有問題 自己想像模擬一下馬上就會看出來了 11/15 12:59
2F：→ DKer:你在迴圈裡沒有重新計算xn,y,yplum 跑再多次都一樣 11/15 13:00
3F：→ DKer:然後記得每一行結尾要加 ; 11/15 13:00
4F：→ gfs777:感謝大大。目前解答改成 11/15 15:03
xn = input('first x?\n');
cnt = 1;
e = 1e-4;
while cnt<=10000;
y = xn^3 - 5*xn^2 - xn + 5;
yplum = 3*xn^2 - 10*xn - 1;
xn1 = xn - y / yplum
if abs(xn1 - xn)<= e
fprintf('converged')
break;
end
xn = xn1
cnt = cnt + 1
end
預祝大大事事順利
※ 編輯: gfs777 來自: 140.112.183.66 (11/15 15:04)