1 Solutions to selected problems

Transcription

1 Solutions to selected problems Section., #a,c,d. a. p x = n for i = n : 0 p x = xp x + i end b. z = x, y = x for i = : n y = y + x i z = zy end c. y = (t x ), p t = a for i = : n y = y(t x i ) p t = p t + a i+ y end Section., #a,c,d a. v = a 0 + x n i= a i c. v = n i=0 a ix n i d. v = n i= a ix i, z = x n+ Section., # 3,4 3. Recall ln(.) = ln( + 0 ) = ( ) k= k 0 k k P n (0.) := n ( ) k= k 0 k k. By the alternating series theorem, the error in this approximation is bounded by ln(.) P n (0.) 0 (n+) n +. The RHS is less than 0 8 when n 7. Thus, 7 terms are needed.

2 4. We compute f() = 7, f () = 8, f () = 8 and f () = 6. Thus, f( + h) = 7h 0 /0! + 8h /! + 8h /! + 6h 3 /3! = 7 + 8h + 4h + h 3 or equivalently, f(x) = 7 + 8(x ) + 4(x ) + (x ) 3. Section., #a,b a. Since 30 = ( ) (.00...) and 97 = = (00000), is the 3-bit machine representation b. Since = 6 ( + ) = ( ) ( ) and 33 = = (00000), is the 3-bit machine representation. Section., #3a-e 3-bit machine numbers must be in the range of the machine (roughly 6 to 8 ), and they must have at most 4 signicant digits. 3a. Not a machine number: too large to be in machine range 3b. Not a machine number: binary expansion has more than 4 signicant digits 3c-d. Not machine numbers: their binary expansions are nonterminating e. = 56 8 is a machine number: it is in the range of the machine and its binary expansion has signicant digit. Section., #5a-c, e-h Conversions can be done as in Section. #a,b above. a. +0 b. 0 c. d. e. 6 f. 5.5

3 g. h. 0 Section., #4 WLOG assume x is positive and write x = 0 m.a a.... After chopping to keep only 5 signicant digits we get the machine representation ˆx = 0 m.a a... a 4. Observe that x ˆx = 0 m a 5 a m 0 4. Thus, x ˆx x 0m m.a a... = 0 4.a a Section., #9 Write x = n (.a a... a 3 ) and y = m (.b b... b 3 ). Write x = m ( n m (.a a... a 3 ) ) = n ( a a...). Since x < 5 y, we have n m 5, so in the last display a appears in a spot which is at least 5 places to the rights of the decimal point. Thus, x + y = m (.b b... b a a... a 3 ) where a is 5 or more spots to the right of the decimal point. After chopping (or rounding), we get (x + y) = m (.b b... b 3 ) = y. Section., # Using e x = + x + x /! + x 3 /3! +..., write f(x) = e x x as f(x) = x + x3 6 + x It's easy to check that the rst two terms above are enough for 5 signicant digits: f(0.0) Using e and direct calculation gives f(0.0) = =

4 Section., #3 One option is to write f(x) = ( e) + x + x /x + x 3 / Section., #7 Note that if x 0, x (sinh x tanh x) = sinh x tanh x x(sinh x + tanh x) = tanh x sinh x x(sinh x + tanh x), () where the last equality comes the hyperbolic trig identity cosh x = sinh x and sinh x tanh x = sinh x(cosh x ) cosh x The last expression in () avoids loss of signicance near x = 0. = tanh x sinh x. Section., #4 By rewriting f(x) = x + x as f(x) = x + + x. Section. # Using sin x = x x3 3! + x5 5!... and the alternating series theorem, we get For x < /0 this becomes sin x x x 3 3!. sin x x = Section. Computer Problems, #8 See the script section. #8. 4

5 Section 3. Computer Problems, # See the script section 3. #. Section 3. # Newton's method for nding roots of f(x) = x R is x n+ = x n f(x n) f (x n ) = x n x n R x n = x n x n + R x n = (x n + Rxn ). Section 3. # Using # above, x n+ R = ( x n + R ) R = ) (x n + R + R R 4 x n 4 x n = ) (x n R + R = ( x n R ) ( ) x = n R. 4 4 x n x n Letting e n = x n R be the error in the nth approximation, If e n < R this means e n+ x n e n+ = e 4x n. n 4(R e n ) e n. In particular if e 0 R/, an induction argument shows that e n+ R e n. Section 3. #6 When m = 8, and x 0 =., we have x =.0875 and x = When m = and x 0 =., we have x = and x = Note the slower convergence to r = when m = compared to m = 8. Section 3. #0 For f(x) = tan x R we have f(x) f (x) = tan x R sec x = cos x sin x R cos x. 5

6 Thus, the iteration formula x n+ = x n cos x n sin x n + R cos x n can be used for nding solutions to tan x = R. Section 3. Computer Problems, #7,9 See the scripts section 3. #7 and section 3. #9. Section 4. #6 The polynomial, in Newton form, is p(x) = + x 3x(x ) + 4x(x )(x 3). Section 4. #7b The columns can be lled top to bottom with third column: 3, 5, fourth column: 7, 9.5, fth column: 0.5. Thus, the interpolating polynomial is p(x) = 3(x + ) + 7(x + )(x ) + (x + )(x )(x 3). Section 4. #0 Newton's interpolation polynomial is p(x) = 7 + x + 5x(x ) + x(x )(x 3). A nested form for ecient computation is p(x) = 7 + x( + (x )(5 + (x 3))). Section 4. #0 6

7 With x = (,, 0,, ), y = (, 4, 4,, ) we have P 0 (x) = a 0 P (x) = a 0 + a (x + ) P (x) = a 0 + a (x + ) + a (x + )(x + ) P 3 (x) = a 0 + a (x + ) + a (x + )(x + ) + a 3 (x + )(x + )x P 4 (x) = a 0 + a (x + ) + a (x + )(x + ) + a 3 (x + )(x + )x + a 4 (x + )(x + )x(x ), for our Newton polynomials. Using P i (x i ) = y i, i = 0,,, 3, 4, we can solve to get a 0 =, a =, a =, a 3 = 5, a 4 =.5. Alternatively, we may write Q(x) = b 0 + b x + b x + b 3 x 3 + b 4 x 4 and solve b 0 b b b 3 = b 4 to get Of course, P 4 (x) = Q(x). b 0 = 4, b = 8, b = 5.5, b 3 =, b 4 =.5. Section 4. # To interpolate the additional point (3, 0), we look at R(x) = Q(x) + c(x + )(x + )x(x )(x ), and plug in 0 = R(3) to get c = /5. Now R(x) interpolates the points from #0, along with the extra point (3, 0). Section 4. Computer Problems #,,9 See scripts online. Section 4. #43 The Newton form of the interpolating polynomial at x 0, x is P (x) = f[x 0 ] + f[x 0, x ](x x 0 ). 7

8 Plugging in x = x and solving for f[x 0, x ], we nd f[x 0, x ] = P (x ) f[x 0 ] x x 0 = f(x ) f(x 0 ) x x 0. By the mean value theorem, there is ξ (x 0, x ) such that f (ξ) = f(x ) f(x 0 ) x x 0 = f[x 0, x ]. Section 4. #45 Let Q(x) = g(x) + x x i x n x 0 [g(x) h(x)]. Since both g and h interpolate f at i =,..., n, Q(x i ) = g(x i ) + x 0 x i x n x 0 [g(x i ) h(x i )] = f(x i ) + x 0 x i x n x 0 [f(x i ) f(x i )] = f(x i ) for i =,..., n. Since g interpolates x 0, Q(x 0 ) = g(x 0 ) + x 0 x 0 x n x 0 [g(x 0 ) h(x 0 )] = f(x 0 ) + 0 = f(x 0 ), and since h interpolates x n, Q(x n ) = g(x n ) + x 0 x n x n x 0 [g(x n ) h(x n )] = g(x n ) [g(x n ) f(x n )] = f(x n ). Note that this is the main step in proving Theorem in Section 4.. Section 4. # Without loss of generality we can assume x j = 0 and x j+ = h. We must show that (x 0)(x h) h 4 for all x [0, h]. To do this, consider Q(x) := (x 0)(x h) = x hx. Since Q (x) = x h = 0 when x = h/, the function Q has its smallest and largest values on [0, h] at h/ or at the endpoints 0, h. Since Q(0) = Q(h) = 0 we conclude max 0 x h Q(x) = Q(h/) = h /4. Section 4.3 #3 8

9 Solution. We must show that n! (j + )!(n j)! = ( ) n, j = 0,..., n. j + j When j =,..., n, we have and ( n j+ n j) n so that ) ) j + ( n j And when j = 0, we get j+( n j) =. = ( ) n j n ( n j n n =. Section 4. #4 Let x [a, b] and a = x 0 < x <... < x n = b be any choice of nodes. We must show n i=0 x x i h n+ n!, h := max i n (x i x i ). () Notice h is the largest spacing between adjacent nodes. For x j x x j+, x x j x x j+ 4 x j+ x j 4 h. where the rst inequality comes from Section 4., #. Since x x j+, j j j j x x i = (x x i ) (x j+ x i ) = (j i + )h = (j + )! h j. i=0 i=0 Similarly, since x x j, n i=j+ x x i = n (x i x) i=j+ i=0 n (x i x j ) = i=j+ Combining the last three inequalities, we get i=0 n (i j)h = (n j)! h n j. i=j+ n x x i 4 h (j + )!h j (n j)!h n j 4 hn+ n! i=0 Section 4. #8 Let P (x) be the degree interpolant of f at x 0, x, x : P (x) = f[x 0 ] + f[x 0, x ](x x 0 ) + f[x 0, x, x ](x x 0 )(x x ), 9

10 and dene Q(x) = f(x) P (x). Then Q has at least 3 roots namely x 0, x, x so, due to Rolle's theorem, Q has at least roots and Q has at least root, call it ξ. Thus, which shows 0 = Q (ξ) = f (ξ) P (ξ) = f (ξ) f[x 0, x, x ], f[x 0, x, x ] = f (ξ). Section 4. Computer Problem #9 See script online. Section 4.3 #3 Using Taylor's theorem, f(x + h) = f(x) + f (x)h + f (x)h + 6 f (x)h f(x + h) = f(x) + f (x)h + f (x)h f (x)h Straightforward calculations give h [4f(x + h) 3f(x) f(x + h)] = f (x) 3 f (x)h +... Terminating this expansion at second order, we nd the error term h [4f(x + h) 3f(x) f(x + h)] f (x) = 3 f (ξ) h. Section 4.3 #5 We have f(x + h) = f(x) + f (x)h + f (x)h + 6 f (α)h 3, f(x h) = f(x) f (x)h + f (x)h 6 f (β)h 3. Thus, f(x + h) f(x) f (x) = h f (x)h + 6 f (α)h, f(x) f(x h) f (x) = h f (x)h + 6 f (β)h, and using Taylor's theorem 0

11 so ( f(x + h) f(x) + h ) f(x) f(x h) f (x) = h h 6 [f (α) + f (β)]. Section 4.3 #7 The problem with the analysis is that the expansions do not go to high enough order. If we include one more term in the expansions for f(x + h) f(x) and f(x h) f(x), we nd the error is in fact O(h ). Section 4.3 #0 a) We write the interpolating polynomial in Newton form: and compute P (x) = f(x 0 ) + f[x 0, x ](x x 0 ) + f[x 0, x, x ](x x 0 )(x x ) P (x) = f[x 0, x, x ]. By the divided dierence recursion theorem, f[x 0, x, x ] = f[x, x ] f[x 0, x ] x x 0 = f(x ) f(x ) x x f(x ) f(x 0 ) x x 0 x x 0 f(x ) f(x ) f(x ) f(x 0 ) αh h = h + αh = [ f(x0 ) h + α f(x ) α + f(x ) α(α + ) b) Consider P 0 (x) =, P (x) = x x and P (x) = (x x ). We will solve the equations for A, B, C. The equations are P i (x) = AP (x 0 ) + BP (x ) + CP (x ), i = 0,,, 0 = A + B + C 0 = ha + αhc = h A + α h C. It is straightforward to verify that A = h (+α), B = h α, and C = h α(+α). ]. Section 4.3 #4

12 Write f(x + h) = f(x) + f (x)h + f (x)h + 6 f (x)h f (4) (x)h f (5) (x)h f(x h) = f(x) f (x)h + f (x)h 6 f (x)h f (4) (x)h 4 0 f (5) (x)h and similarly f(x + h) = f(x) + f (x)h + f (x)h f (x)h f (4) (x)h f (5) (x)h f(x h) = f(x) f (x)h + f (x)h 4 3 f (x)h f (4) (x)h f (5) (x)h Thus, h [f(x + h) f(x h)] = f (x) + 6 f (x)h + 0 f (5) (x)h (3) and h [f(x + h) f(x h)] = 3 f (x) + 9 f (x)h + 45 f (5) (x)h (4) Notice that multiplying equation (3) by 4/3 and subtracting (4) gives [f(x + h) f(x h)] 3 h [f(x + h) f(x h)] = f (x) 30 f (5) (x)h Terminating this expansion at fourth order, we nd [f(x + h) f(x h)] 3 h [f(x + h) f(x h)] f (x) = 30 f (5) (ξ) h 4. Section 4.3 Computer problems #4 See script online.