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Jim Lambers

MAT 460/560

Fall Semester 2009-10Homework Assignment 2 Solution

Section 1.2

3. Suppose p* must approximate p with relative error at most 10-3. Find the largest interval in

which v: must lie for each value of p.

(a) 150

Solution We must have Ip* - 1501/11501 ~ 10-3, or Ip* - 1501 ~ 0.15, which yields the

interval [149.85,150.15].

(b) 900Solution We must have Ip* - 9001 ~ 0.9, which yields [899.1,900.9].

(c ) 1500

Solution We must have Ip* - 15001 ~ 1.5, which yields [1498.5,1501.5].

(d) 90

Solution We must have Ip* - 901 ~ 0.09, which yields [89.91,90.09].

5. Use three-digit rounding arithmetic to perform the following calculations. Compute the ab-

solute error and relative error with the exact value determined to at least five digits.

(a) 133 + 0.921

Solution p =133.921 and p* =134, so the absolute error is 0.079 and the relative error

is 5.90 x 10-4.

(b) 133 - 0.499

Solution p =132.501 and p* =133, so the absolute error is 0.499 and the relative error

is 3.77 x 10-3.

(c ) (121-0.327)-119

Solution p =1.673 and v:relative error is 0.195.

121 - 119 =2, so the absolute error is 0.327 and the

(d) (121 - 119) - 0.327

Solution p =1.673 and p* =1.67, so the absolute error is 0.003 and the relative error

is 1.79 x 10-3.13 6

()14-'7

e 2e-5.4

Solutionp =1.95354 andp* =(0.929-0.857)/(5.44-5.4) =1.80, so the absolute error

is 0.154 and the relative error is 0.0786.

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(f ) -WIT + 6e - 632

Solution p =-15.1546 and v: =-31.4 + 16.3 - 0.048 =-15.1, so the absolute error is

0.0546 and the relative error is 3.6 x 10-3.

(g) (~). (~)

Solution p=2/7 =0.285714, and p* =(0.222)(1.29) =0.286, so the absolute error is

2.86 x 10-4, and the relative error is 10-3.

7r-~

(h) -1-7

17

Solution p =-0.0214963 and v: =(3.14 - 3.13)/(1/17) =0, so the absolute error is

0.0215 and the relative error is 1.

9. The first three nonzero terms of the Maclaurin series for the arctangent function are x -

(1/3)x3 + (l/5)x5. Compute the absolute error and relative error in the following approxi-

mations of 7 r using the polynomial in place of the arctangent:

(a) 4 [arctan (~) + arctan G)]

Solution We have

Since the exact value of it,to 15 significant digits, is 3.14159265358979, it follows that the

absolute error is 3.983 x 10-3 and the relative error is (3.983 x 1O-3) / 7 r : : : : : 0 1.268 X 10-3.

(b) 16 arctan (~) - 4 arctan ( 2 ~ 9 )

Solution We have

tt " 16 [~ - (I/:l) G ) \ (1/5) G ) ' ] -

4 [2:~9- (1/3) C : ~ S+ (1/5) C : ~ S ]

: : : : : 0 16 [~ - 3~5 + 15~25] - 4 [2~9 - 4095

1

5757 + 389905~325995]: : : : : 0 3.14162102932503.

Since the exact value of it,to 15 significant digits, is 3.14159265358979, it follows that the

absolute error is 2.838 x 10-5 and the relative error is (2.838 x 1O-5) / 7 r : : : : : 0 9.032 X 10-6.

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11. Let

f(x) =xcosx ~ sinx.x - slnx

(a) Find limx---+o(x).

Solution Ifwe substitute x =0, we obtain 0/0, which is an indeterminate form. Using

l'Hospital's Rule three times, we obtain

1. x cos x - sin x

lim f(x) = 1m .x - - - + o x - - - + o X - Slnx

1. cos x - x sin x - cosxIm----------------

x - - - + o 1 - cos x

1. -xsinx. l. F fl . - -- - -- -

x - - - + o 1- cosx

1. - sinx - x cos x.l.Ifl. •

x - - - + o Slnx

1. - cos x - cos x + x sin xIm------------------

x - - - + o cosx

-2.

(b) Use four-digit rounding arithmetic to evaluate f(O.l).

Solution We have

f(O.l)(0.1) cosO.1- sinO.1

0.1 - sin 0.1

(0.1)(0.995) - 0.09983

0.1 - 0.09983

0.0995 - 0.09983

0.00017

-0.00033

0.00017

~ -1.941.

(c) Replace each trigonometric function with its third Maclaurin polynomial, and repeat

part (b).

Solution The third Maclaurin polynomial for cos x is 1 - ~x2, and the third Maclaurin

polynomial for sin x is x - ix3. Substituting these polynomials for cos x and sin x in

f(x), we obtain the function

h(x)x[1-~x2] - [x_ix3]

X - [x - ix3]

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(d) The actual value is f(O.l) =-1.99899998. Find the relative error for the values obtainedin parts (b) and (c).

Solution The relative error for the value obtained in part (b) is

while the relative error for the value obtained in part (c) is

1- 2 - (-1.99899998) 1

1- 1.998999981 =0.0005.15. Use the 64-bit long real format to find the decimal equivalent of the following floating-point

machine numbers.

(a) 0 10000001010 1001001100000000000000000000000000000000000000000000

Solution The sign bit s is 0, the exponent c is represented by 10000001010 in binary,

which is 210+ 23+ 21=1024+ 8+ 2=1034 in decimal, and the mantissa f is

1 1 1 1 147f =T1 + T4 + T7 + T8 =2 + 16 + 128 + 256 =256'

Therefore, the value of the floating point number, denoted by x, is

x (_1)S2C-1023(1 + f)

(_1)021034-1023 (1 + 147)256

211403

256

211403

288·403

3224.

(b) 1 10000001010 1001001100000000000000000000000000000000000000000000

Solution This number is identical to the number in part (a), except that the sign bit s

is 1 instead of 0, so the value is -3224.

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(c ) 0 0 1 1 1 11 1 1 11 1 0 1 0 10 0 1 1 00 0 0 00 0 0 00 0 0 00 0 0 00 0 0 0 00 0 0 00 0 0 00 0 0 00 0 0 00 0 0 00 0

Solution The sign bit s is 0, the exponent c is given by

9 . 210 - 1" "' 2 '= =1023D 2-1 'i=O

and the mantissa f is

Therefore, the value of the floating point number, denoted by x, is

x (_1)S2C-1023(1 + f)

(_1)021023-1023 (1 + 83)256

339

2561.32421875.

( d ) 0 0 1 1 11 1 1 11 1 1 0 1 0 10 0 1 1 00 0 0 00 0 0 00 0 0 00 0 0 00 0 0 0 00 0 0 00 0 0 00 0 0 00 0 0 00 0 0 00 1

Solution This number is identical to the one in part (c), except that there is an ad-

ditional digit in the mantissa corresponding to 2-52. It follows that the value of this

number, denoted by x, is

x=

1.32421875 +T2

: : : : : 0 1.3242187500000002220446049250313.

17. Suppose two points (xo , YO ) and (Xl, Yl) are on a straight line with Y l # Yo . Two formulas

are available to find the x-intercept of the line:

Y l - Y o

and X =Xo _ (X l - X o )yo .

Y l - Y o

XOYI - X lYOX=

(a) Show that both formulas are algebraically correct.

Solution The equation of the line is

Y l - YoY = (X - X o ) + Yo ·

X l - X o

Setting Y =0 and solving for x, we obtainX l - X o

-y o =X - XoY l - Y o

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or(X l - XO )YO

X =Xo - ,YI - Yo

which is precisely the second formula. Ifwe use a common denominator, then we obtain

YI - Yo (X l - XO )YOX 0 "------"-

YI - Yo YI - Yo

XO (YI - Yo ) - (X l - XO )YO

YI - Yo

(XOYI - XOYO ) - (X IYO - XOYO )

YI - Yo

XOYI - X IYO

YI - Yo

which is precisely the first formula.

(b) Use the data (xo ,yo) =(1.31,3.24) and (XI,YI) =(1.93,4.76) and three-digit roundingarithmetic to compute the x-intercept both ways. Which method is better and why?

X =

Solution Using the first formula, we obtain

X =1.31 .4.76 - 1.93 . 3.24

4.76 - 3.24

6.24 - 6.25

1.52

0.00658.

Using the second formula, we obtain

X = 1.31 _ (1.93 - 1.31)3.24

4.76 - 3.24

1.31 _ 0.62 . 3.24

1.52

1.31 _ 2.01

1.52

1.31 - 1.32

-0.01.

The exact value, to three significant digits, is -0.0116, so clearly the second formula is

better. The first formula suffers from catastrophic cancellation in the numerator.

19. The two-by-two linear system

ax + bye,

ex + dy j,

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where a, b, c, d , e, f are given, can be solved for x and y a follows:

y

c, provided a

#0;

ad-mb;

f-me;

h

d1'

e - by

set m

x =a

Implement this algorithm using MATLAB and solve the following linear systems.

(a)

1.130x - 6.990y 14.20

8.110x + 12.20y -0.1370

(b)

1.013x - 6.099y 14.22

-18.11x + 112.2y -0.1376

Solution The following function solves the general two-by-two linear system, given the values

of a, b, c, d , e and f.

function [x,yJ=hwlprob 1 2 1 9 ( a,b,c,d,e,f )

% display error message if a is zero

if a==O,

error( ' a must be nonzero' )

end

m=c/a;

dl=d-m*b;

fl=f-m*e;

y=fl/dl;

x=(e-b*y)/a;

In the following MATLAB session, this function is used to solve the systems in parts (a) and

(b).

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» [x,yJ=hw 1 pr o b1 2 1 9 ( 1 . 1 3 0 , - 6 . 9 9 0 , 8 . 1 1 0 , 1 2 . 2 0 , 1 4 . 2 0 , - 0 . 1 3 7 0 )

x=

2 . 4 4 4 5 9 1 9 0 4 3 5 1 7 6

y

- 1 . 6 3 6 2 8 1 9 9 5 4 3 3 8 4

» [x,yJ=hw 1 pr o b1 2 1 9 ( 1 . 0 1 3 , - 6 . 0 9 9 , - 1 8 . 1 1 , 1 1 2 . 2 , 1 4 . 2 2 , - 0 . 1 3 7 6 )

x=

4 . 9 7 4 3 8 8 7 5 5 0 6 5 1 9 1 e + 0 0 2

y

8 0 . 2 8 9 4 8 6 9 4 6 7 2 9 5 8

25. The binomial coefficient

(m) m!

k - k !(m - k)!

describes the number of ways of choosing a subset of k objects from a set of m elements.

(a) Suppose decimal machine numbers are of the form

What is the largest value ofmfor which the binomial coefficient ( 7 ) can be computedfor all k by the definition without causing overflow?

Solution The largest number that can be represented in this floating-point system is

0.9999 x 1015=999,900,000,000,000. Using the definition of the binomial coefficient,

overflow will occur if m! is larger than this number. This is the case if m! 2 ': 18, since

18! =6,402,373, 705, 728, 000 and 17! =355,687,428,096,000. Therefore the largest

value ofm for which the binomial coefficient can be computed without causing overflow

is 17.

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(b) Show that ( 7 ) can also be computed by

( 7 ) =(7)(7~:). . .(m-:+1).

Solution We have

( 7 )m!

k!(m - k)!

1·2·3· .... (m - 1) . m

(1·2·3· (k - 1) . k)( l ·2·3···· . (m - k - 1) . (m - k))

(1·2·3· (m - k - 1) . (m - k))((m - k + 1) ..... (m - 1) .m)(1·2·3· (k - 1) . k)( l ·2·3···· . (m - k - 1) . (m - k))

(m - k + 1) (m - 1) .m

1·2·3· (k - 1) . k

(7)(7~:). . .(m-:+1).

(c) What is the largest value ofm for which the binomial coefficient ( 7 ) can be computedby the formula in part (b) without causing overflow?

Solution We have

(m ) =mm-1m-2 =m(m-1)(m-2).

3 3 2 1 6

To avoid overflow, this coefficient must not exceed 0.9999 x 1015, which implies that m

must satisfy

m(m - l)(m - 2) ~ 6(0.9999) X 1015 ~ 5.9994 X 1015.

Since m(m - l)(m - 2) ~ m3 for any nonnegative integer m, it follows that the largest

value of m for which the binomial coefficient can be computed is not less than the

largest value of m for which m3 ~ 5.9994 x 1015. This value is (5.9994 x 1015)1/3 : : : : : 0

(1.81706 x 105) : : : : : 0 181, 706.

Ifwe let m=181,706, we obtain m(m-1)(m-2) : : : : : 0 5.9993x 1015. To see if m can be any

larger, we try m=181,707 and obtain m(m-1)(m-2) : : : : : 0 5.9993998 x 1015, so this value

is acceptable as well. However, if we try m =181,708, we obtain m(m - l)(m - 2) : : : : : 0

5.9994988 which is too large, so we conclude that the largest value of m is 181,707.

(d) Use the equation in part (b) and four-digit chopping arithmetic to compute the number

of possible 5-card hands in a 52-card deck. Compute the actual and relative errors.

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Solution The number of possible 5-card hands in a 52-card deck is

(52 ) =52 51 504948.

5 543 2 1

Using four-digit chopping arithmetic, we obtain

( 552) : : : : : 0 (1004) (12.75) (16.66) (24.5) (48)

: : : : : 0 (132.6) (16.66) (24.5) (48)

: : : : : 0 (2209) (24.5) (48)

: : : : : 0 (54,120)(48)

: : : : : 0 2,597,000.

The actual value is 2,598,960, so the absolute error is 1,960 and the relative error is

7.541 x 10-4.

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