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métodos numéricos matlab, Esquemas y mapas conceptuales de Métodos Numéricos
métodos numéricos que sirve para resolver ejercicios basicos en matlab
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CHAPTER 2
IF x < 10 THEN
IF x < 5 THEN
x = 5
ELSE
PRINT x
END IF
ELSE
DO
IF x < 50 EXIT
x = x - 5
END DO
END IF
Step 1: Start
Step 2: Initialize sum and count to zero
Step 3: Examine top card.
Step 4: If it says “end of data” proceed to step 9; otherwise, proceed to next step.
Step 5: Add value from top card to sum.
Step 6: Increase count by 1.
Step 7: Discard top card
Step 8: Return to Step 3.
Step 9: Is the count greater than zero?
If yes, proceed to step 10.
If no, proceed to step 11.
Step 10: Calculate average = sum / count
Step 11: End
start
sum = 0
count = 0
INPUT
value
value =
“end of data”
value =
“end of data”
sum = sum + value
count = count + 1
T
F
count > 0
average = sum/count
end
T
F
2.5 The development of the algorithm hinges on recognizing that the series approximation of the
sine can be represented concisely by the summation,
x
i
i
i
n 2 1
1
−
=
∑
where i = the order of the approximation. The following algorithm implements this
summation:
Step 1: Start
Step 2: Input value to be evaluated x and maximum order n
Step 3: Set order (i) equal to one
Step 4: Set accumulator for approximation (approx) to zero
Step 5: Set accumulator for factorial product (fact) equal to one
Step 6: Calculate true value of sin(x)
Step 7: If order is greater than n then proceed to step 13
Otherwise, proceed to next step
Step 8: Calculate the approximation with the formula
approx approx ( 1)
x
factor
i- 1
2i- 1
Step 9: Determine the error
true
true approx
%error
Step 10: Increment the order by one
Step 11: Determine the factorial for the next iteration
factor = factor • (2 • i − 2) • (2 • i −1)
Step 12: Return to step 7
Step 13: End
start
INPUT
x, n
i > n
end
i = 1
true = sin(x)
approx = 0
factor = 1
approx approx
x
factor
i
i -
- 1
2 1
error
true approx
true
OUTPUT
i,approx,error
i = i + 1
F
T
factor=factor(2i-2)(2i-1)
Pseudocode:
SUBROUTINE Sincomp(n,x)
i = 1
true = SIN(x)
approx = 0
factor = 1
DO
IF i > n EXIT
approx = approx + (-1)
i-
- x
2
i-
/ factor
error = Abs(true - approx) / true) * 100
PRINT i, true, approx, error
i = i + 1
factor = factor • (2 • i-2) • (2 • i-1)
END DO
END
Interpretation: The absolute percent relative error drops until at n = 6, it actually yields a
perfect result (pure luck!). Beyond, n = 8, the errors starts to grow. This occurs because of
round-off error, which will be discussed in Chap. 3.
2.8 AQ = 442/5 = 88.
AH = 548/6 = 91.
without final
AG =
with final
AG =
The following pseudocode provides an algorithm to program this problem. Notice that the
input of the quizzes and homeworks is done with logical loops that terminate when the user
enters a negative grade:
INPUT number, name
INPUT WQ, WH, WF
nq = 0
sumq = 0
DO
INPUT quiz (enter negative to signal end of quizzes)
IF quiz < 0 EXIT
nq = nq + 1
sumq = sumq + quiz
END DO
AQ = sumq / nq
PRINT AQ
nh = 0
sumh = 0
PRINT "homeworks"
DO
INPUT homework (enter negative to signal end of homeworks)
IF homework < 0 EXIT
nh = nh + 1
sumh = sumh + homework
END DO
AH = sumh / nh
PRINT "Is there a final grade (y or n)"
INPUT answer
IF answer = "y" THEN
INPUT FE
AG = (WQ * AQ + WH * AH + WF * FE) / (WQ + WH + WF)
ELSE
AG = (WQ * AQ + WH * AH) / (WQ + WH)
END IF
PRINT number, name$, AG
END
n F
2.10 Programs vary, but results are
Bismarck = −10.842 t = 0 to 59
Yuma = 33.040 t = 180 to 242
n A
Step v (12)
t
Error is halved when step is halved
Fortran 90 VBA
Dim V1 As Single, v2 As Single, pi As Single
pi = 4 * Atn(1)
If d < R Then
Vol = pi * d ^ 3 / 3
ElseIf d <= 3 * R Then
V1 = pi * R ^ 3 / 3
v2 = pi * R ^ 2 * (d - R)
Vol = V1 + v
Else
Vol = "overtop"
End If
End Function
The results are
R d Volume
1 3.1 overtop
2.15 Here is a flowchart for the algorithm:
Function Polar(x, y)
2 2
r = x + y
x < 0
y > 0 y > 0
θ + π
−
x
y
1
tan
θ − π
−
x
y 1
tan
θ −π
−
x
y 1
tan
θ = 0
π
θ =−
y < 0
θ = π
y < 0
Polar
Polar
End Polar
T
T
T
T
T
F
F
F
F
And here is a VBA function procedure to implement it:
Option Explicit
Function Polar(x, y)
Dim th As Single, r As Single
Const pi As Single = 3.
r = Sqr(x ^ 2 + y ^ 2)
If x < 0 Then
If y > 0 Then
th = Atn(y / x) + pi
ElseIf y < 0 Then
th = Atn(y / x) - pi
Else
th = pi
End If
Else
If y > 0 Then
th = pi / 2
ElseIf y < 0 Then
th = -pi / 2
Else
th = 0
End If
End If
Polar = th * 180 / pi
End Function
The results are:
x y θ
4.18 f(x) = x-1-1/2*sin(x)
f '(x) = 1-1/2*cos(x)
f ''(x) = 1/2*sin(x)
f '''(x) = 1/2*cos(x)
f
IV
(x) = -1/2*sin(x)
Using the Taylor Series Expansion (Equation 4.5 in the book), we obtain the following
st
nd
rd
, and 4
th
Order Taylor Series functions shown below in the Matlab program-
f1, f2, f4. Note the 2
nd
and 3
rd
Order Taylor Series functions are the same.
From the plots below, we see that the answer is the 4
th
Order Taylor Series expansion.
x=0:0.001:3.2;
f=x-1-0.5*sin(x) ;
subplot(2,2,1);
plot(x,f);grid;title('f(x)=x-1-0.5*sin(x)');hold on
f1=x-1.5 ;
e1=abs(f-f1); %Calculates the absolute value of the
difference/error
subplot(2,2,2);
plot(x,e1);grid;title('1st Order Taylor Series Error');
f2=x-1.5+0.25.((x-0.5pi).^2);
e2=abs(f-f2);
subplot(2,2,3);
plot(x,e2);grid;title('2nd/3rd Order Taylor Series Error');
f4=x-1.5+0.25.((x-0.5pi).^2)-(1/48)((x-0.5pi).^4);
e4=abs(f4-f);
subplot(2,2,4);
plot(x,e4);grid;title('4th Order Taylor Series Error');hold off