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f(x) = e– x – x |
By Secant (i =
1) |
التعويض
(i)x |
f(x) |
xn+1 – xn |
│ϵα│% |
by Excel |
Solver |
Opation ↓ |
Use:
Goal Seek |
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1 |
Value ®xi - 1 = x1 |
0 |
1.000000 |
X2 - X1 |
↓ |
x |
0.56714329 |
Preision |
0.5671 |
8E-06 |
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2 |
Value ® xi = x2 |
1 |
-0.632121 |
1.0000 |
1.00000 |
f(x)=e– x – x |
0.00000000 |
0.000000007 |
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3 |
Secant law®x3
= |
0.6127 |
-0.070814 |
-0.3873 |
-0.63212 |
Secant law |
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4 |
®x4 = |
0.56383839 |
0.005182 |
-0.0489 |
-0.08666 |
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5 |
®x5 = |
0.56717036 |
-0.000042 |
0.0033 |
0.00587 |
See Example8 |
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6 |
®x6 = |
0.56714331 |
0.000000 |
0.0000 |
-0.00005 |
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7 |
®x7 = |
0.56714329 |
0.000000 |
0.0000 |
0.00000 |
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8 |
®x8 = |
0.56714329 |
0.000000 |
0.0000 |
0.00000 |
ßß |
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↓ use excel ↑ |
↑ stop because the last two answers has been Equaled ↑ |
Newton law |
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نظير المصفوفة |
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f(x)=e– x – x |
By Newton |
f /(x) = – e– x – 1 |
القسمة |
قانون حساب (i+1)x |
Newton law |
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x(i) |
f x(i) |
f /x(i) |
fx(i)/f/x(i) |
x(i) – fx(i)/f/x(i) |
↓ |
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0.2 |
0.618730753 |
-1.818731 |
-0.340199 |
0.54019920 |
↓ |
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0.5401992032 |
0.042432975 |
-1.582632 |
-0.026812 |
0.56701085 |
↓ |
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0.5670108503 |
0.000207558 |
-1.567218 |
-0.000132 |
0.56714329 |
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0.5671432872 |
4.97416E-09 |
-1.567143 |
0.000000 |
0.56714329 |
Secant law |
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0.5671432904 |
0 |
-1.567143 |
0.000000 |
0.56714329 |
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Program in
Mthlab (Secant) |
Secant for ® |
f(x) = e– x – x |
قيم نقاط رسم المنحنى |
f(x) = e–
x – x |
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clc |
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ans = |
x |
f(x) |
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disp('Secant for e^(–x)–x') |
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0.00000000 |
-2 |
9.4 |
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x=[0 1]; |
لاحظ الرسم جهة اليسار |
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1.00000000 |
-1 |
3.7 |
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format long |
ثماية ارقام عشرية |
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Resault ® |
0.61269984 |
0 |
1.0 |
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for i=2:8; |
بدء من 2 حتى 8 |
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0.56383839 |
1 |
-0.6 |
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fx=exp(-x(i-1))-x(i-1); |
قيمة الدالة عند i-1 |
0.56717036 |
2 |
-1.9 |
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fxx=exp(-x(i))-x(i); |
قيمة الدالة عند i |
0.56714331 |
3 |
-3.0 |
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x(i+1)=x(i)-((x(i)-x(i-1))*fxx)/(fxx-fx); |
القيمة الجديدة |
0.56714329 |
4 |
-4.0 |
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end |
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نهاية اللوب |
0.56714329 |
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الرسم في Excel |
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x' |
حتى تتساوي آخر قيمتين |
القيم التي حُسبت |
0.56714329 |
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¾® |
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disp(' stop because the last two answers has
been Equaled') |
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stop because the last two answers has
been Equaled |
Inteval [0 , 1] |
0 |
0 |
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----------------------------------------------------------------------- |
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1 |
® |
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Program in
Mthlab (Newton-Raphson ) |
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Newton for e– x–x |
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| لرسم الفترة ¬ اللون الأحمر في الشكل من صفر للواحد |
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%e^(–x)–x') |
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التوضيح أعلاه |
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ans = |
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disp('Newton for e^(-x)-x') |
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0.00000000 |
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x=[0];tol=0.0000000007;format
long |
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0.50000000 |
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for i=1:6; |
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Resault ® |
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0.56631100 |
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fx=exp(-x(i))- x(i); |
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0.56714317 |
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fxx=-exp(-x(i))-1; |
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0.56714329 |
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fxxx=exp(-x(i)); |
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0.56714329 |
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x(i+1)=x(i)-(fx/fxx); |
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0.56714329 |
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end |
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if abs(x(i+1)-x(i))<tol |
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x' |
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الرسم من Derive |
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disp(' stop
because the last two answers has been
Equaled') |
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stop because the last two answers has
been Equaled |
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end |
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By: Mathematica 9 |
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By: Derive 6 |
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By: MathCad 15 |
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By: Maple 18 |
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MthLab 11 |
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NSolve[E^-x - x, {x}] |
NSOLVE(e-x-x, x,
Real) |
Ctrl+Shift+. Before solve, Ctrl + = To have = |
f(x) := exp(-x)-x = 0: |
See up B20:B46 |
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{{x ->
0.567143}} |
-> Mean ® |
0.5671432904 |
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e-x -x =
0 solve ® LambertW(1) |
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fsolve(f(x)) |
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Excel 2007 |
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{{x ®
0.567143}} |
أو نحصل مباشرة على |
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LambertW(1) float, 9®0.5671432904 |
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0.5671432904 |
See up B1:F18 |
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By: Mathematica 10 |
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clc |
MthLab 11, 15
(Also) |
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MthLab 11
(Also) |
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NSolve[E^(-x) - x == 0,
x, Reals] |
syms x real |
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syms x real |
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{{x -> 0.567143}} |
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f1=input('please type the equation ','S'); |
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vpa(solve(exp(-x)-x)) |
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By: Maple 18 |
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[x]=solve(f1); |
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ans |
0.56714329 |
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Examples |
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x = vpa(x(1),8) % for real number(first answer) |
حل معادلة تحوي exp ¬ |
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with(Student[NumericalAnalysis]); |
%exp(-x)-x=0
® 0.5671432904 |
وتوجد هنا ثلاث منها ¬ |
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f := exp(-x)-x; |
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%exp(-2*x)-exp(-x)-1=0
® -0.4812118246 |
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exp(-x) - x |
%exp(-x)-exp(2*x)+1=0
® 0.2811995743 |
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Newton(f, x =
2.0, tolerance = 10^(-2)); |
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0.5671432904 |
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