Kevin Mitchell Assignment #2 Solutions MACM 316

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1 Machine arithmatic. BF, p29 #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... ( point Suppose decimal machine numbers are of the form ± 0.d d 2 d 3 d 4 0 n d 9 0 d i 9 i = 2, 3, 4 n 5 ( m What is the largest value of m for which the binomial coefficient can be computed for all k directly from the k above definition without overflowing? Since the factorial (! is undefined on negative integers, we have the following restriction on k: It is therefore also true that and and therefore 0 k m k! m! (m k! m! k! (m k! k!(m k! so that ( m m! = k k!(m k! m! ( m and therefore m! is the largest number stored during the computation of regardless of the value of k. k From the specification above, the largest representable number is we therefore must find an integer m such that m! (m +! > to find this value we compute, (2, (2(3,... and check for overflow at each step. This is carried out by the very simple script fact max.m:

2 m=; mfact=m; o f l = ˆ5; while mfact<o f l m=m+; mfact=mfact m; When the while loop is completed m = 8, which is the first number to overflow. Indeed it can be checked that 7! = ! = and therefore m = 7 is the largest value that can be compute without overflowing...2 ( point It can be shown that there is a safer method of computing therefore ( m k. Note that m! (m(m... (m k + (m k(m k... ( = (m k! (m k(m k... ( = (m(m... (m k + m! (m(m... (m k + = k!(m k! (k(k... ( and furthermore the above expression has k factors in both the numerator and the denominator. They can therefore be paired in a one-to-one manner. It is clear that in order to minimise the size of the maximum number stored, it is best to pair the largest numerator with the largest denominator (i.e. m with k, m with k, etc. We are therefore dealing with a sequence of factors like..3 ( point m! k!(m k! = ( ( m k We now use the improved formula to compute ( m ( m = 3 3 ( m... k ( m 2 ( m k + ( m 2 Note that since we have already established k m (and therefore (k (m, etc. each of the factors described above is larger than one. Therefore each additional multiplication increases the result so that the largest number stored in the computation is the final result itself. We therefore seek m such that We know that therefore we bisect the function ( m ( m + 3 f(m = ( ( m on the initial interval m [, 0 6 ] with binom bisect.m > = 6 m ( m 2 3m

3 a=;b=e6 ;%guess f o r i n t e r v a l % m must always be an i n t e g e r. We round down to be on the s a f e s i d e m=floor ( ( a+b / 2 ; fun=i n l i n e ( m (mˆ2 3 m+2/ ˆ5 ; i t s =0; while b a> switch sign ( fun (m sign ( fun ( b case b=m; case a=m; case 0 break ; m=floor ( ( a+b / 2 ; i t s=i t s +; which s m = 8707 as the maximum allowable m after 20 iterations...4 (2 points While we have hitherto been investigating problems with overflow, we now investigate the effects of finite precision. In particular ( we use 4 digit chopping arithmetic to compute the number of possible 5-card hands in a 52-card deck 52 (i.e.. To implement chopping arithmatic in the least tedious manner possible, the following xchop=chop(x,t 5 function is provided function x=chop ( x, t % xchop=chop ( x, t % Loosely based on chop ( pg. 30 o f % RL Burden, JD Faires. Numerical Analysis 7 th ed. Thompson % chops the argument x ( which may be a v e c t o r o f v a l u e s to t % s i g n i f i c a n t decimal p l a c e s. idx=find ( x ; %only operate on nonzero v a l u e s o f x e=floor ( log0 ( abs ( x ( idx + ;% compute the exponent % put a l l the d i g i t s we w i l l keep to the l e f t o f the decimal p l a c e x ( idx=x ( idx. 0. ˆ ( t e ; %t r u n c a t e e v e r y t h i n g to the r i g h t o f the decimal p l a c e x ( idx= f i x ( x ( idx ; % put the exponent back how we found i t x ( idx=x ( idx. 0. ˆ ( e t ; To properly simulate 4 digit ( chopping arithmetic, we must apply this chop function to every intermediate result 52 stored in the computation of. This is implemented in the function bchop=binomchop(m,k,t 5 function [ bchop]=binomchop (m, k, t % [ r e t ]=binomchop (m, k, t % compute the binomial c o e f f i c i e n t % / m \ % \ k / % using a method c a r e f u l to avoid o v e r f l o w % and 4 d i g i t chopping a r i t h m e t i c c o u r t e s y o f the chop ( f u n c t i o n bchop =.0; for i i =0:k

4 bchop=chop ( bchop chop ( chop (m i i, t / chop ( k i i, t, t, t ; In order to assess the accuracy of the computation performed using chopping arithmetic, we compute a much more accurate result by providing binomchop( with a t argument greater than 6. In this case, our result will be dominated by limitations on machine precision, rather than the artificial limitation provided by chop(. octave :> format long octave :2> binomchop ( 5 2, 5, 4 ans = octave :3> binomchop ( 5 2, 5, 2 0 ans = octave :4> a b s e r r=abs ( binomchop (52,5,20 binomchop ( 5 2, 5, 4 a b s e r r = octave :4> r e l e r r=abs ( binomchop (52,5,20 binomchop ( 5 2, 5, 4 / binomchop ( 5 2, 5, 2 0 r e l e r r = e 04 We should not be too surprised by the fact that the relative error is the same order of magnitude as the smallest significant digit stored. 2 Rootfinding Write a function that solves f(x = 0 x (a, b in a safe way using Newton s method when stable, and otherwise resorting to bisection; a non-zero error code or print an error message if the method fails. This scheme is implemented in [soln,rc]=newtonsafe(fun,a,b,tol

5 function [ p, rc ] = newtonsafe ( f dx, a, b, t o l % [ soln, rc ] = newtonsafe ( fun, a, b, t o l % Use Newton s method to f i n d a root f ( x=0 on the i n t e r v a l xe(a, b % f a l l i n g back to b i s e c t i o n i f the s o l u t i o n g e t s taken o u t s i d e the % s p e c i f i e d i n t e r v a l. % % Loosely based on r t s a f e ( on pg.366 o f % WH Press, SA Teukolsky, WT V e t t e r l i n g, BP Flannery. Numerical Recipes % in C. 2nd Ed. Cambridge U n i v e r s i t y Press 988. % % My most n o t a b l e omission from above i s t h a t t h i s code only f a l l s back % to b i s e c t i o n i f Newton f a i l s, but not i f i t i s merely converging % s l o w l y. I found t h i s c h o i c e l e d to fewer t o t a l i t e r a t i o n s f o r the % quantum problem c o n s i d e r e d. % % INPUTS % f d x : f u n c t i o n p o i n t e r o f the form [ f, dx]= f d x ( x. % f i s the v a l u e o f the f u n c t i o n whose root we are l o o k i n g f o r % e v a l u a t e d at x. dx=f ( x / f ( x i s the newton c o r r e c t i o n at x. % Returning dx i n s t e a d o f f ( x d i r e c t l y a v o i d s the problem where % f ( x i s zero at the r o ot ( though you should p r o b a b l y r e t h i n k % your problem i f t h i s i s r e a l l y the case. % a, b : l e f t and r i g h t e x c l u s i v e p o i n t s f o r the i n t e r v a l in which to search % t o l : t o l e r a n c e d e s i r e d f o r the s o l u t i o n % % OUTPUTS % s o l n : computed v a l u e f o r root o f f ( s o l n =0 solne (a, b % rc : code i n d i c a t i n g s u c c e s s or mode o f f a i l u r e % 0 : s u c c e s s % : got NaN from t he f u n c t i o n e v a l u a t i o n ( can t b i s e c t t h a t 2 : % f u n c t i o n e v a l u a t i o n has non n e g l i g i b l e imaginary part ( can t % b i s e c t t h a t % 3 : the f u n c t i o n has the same s i g n on e i t h e r s i d e % o f the i n t e r v a l ( can t b i s e c t t h a t % 4 : v a l u e o f r o ot i s no % l o n g e r changing, but t o l e r a n c e was not met % SUCCESS =0; NOTFINITE=; COMPLEX =2; NOBRACKET=3; STALLED =4; %the i n t e r v a l i s e x c l u s i v e, but b i s e c t i o n i s i n c l u s i v e, so add or s u b t r a c t %the s m a l l e s t number p o s s i b l e from the points a=a+eps ( a ; b=b eps ( b ; r c=success;% optimism! p=(a+b / 2 ; % i n i t i a l guess i s the midpoint o f the provided i n t e r v a l % t h i s i s the only p l a c e we e x p l i c i t l y e v a l u a t e the f u n c t i o n at the % points % WARNING: t h i s i s a l s o the only p l a c e t h a t f a=f ( a. % As the a l g o r i t h m proceeds to r e f i n e the i n t e r v a l, and in p a r t i c u l a r, % change the p o s i t i o n o f a, then f a =f ( a s i n c e f a w i l l s i l l be f i x e d % the e v a l u a t i o n immediately below. Keep t h i s in mind i f f o r some

6 % reason you d e c i d e you want to l o o k at f a during i t e r a t i o n. f a=f d x ( a ; fb=f d x ( b ; i f f a==0 disp ( l e f t point i s the exact s o l u t i o n! p=a ; e l s e i f fb==0 disp ( r i g h t point i s the exact s o l u t i o n! p=b ; % Since t h e r e are d i s t i n c t f l o a t i n g p o i n t numbers I n f/ I n f % with o p p o s i t e signs, t he r oot i s s t i l l b i s e c t a b l e i f f ( a or f ( b i s i n f i n i t e. % Therefore, we only check f o r e v i l NaN which does not have a s i g n. i f any( isnan ( [ fa, fb ] disp ( got NaN e v a l u a t i n g f u n c t i o n at points! r c=notfinite; % Make sure t h a t f ( a and f ( b have zero imaginary p a r t s to w i t h i n % machine e p s i l o n o f the r e a l p a rt. e l s e i f or ( i s c o m p l e x l o o s e ( f a, i s c o m p l e x l o o s e ( f a disp ( f u n c t i o n i s complex at points! r c=complex; % In case t h e r e are imaginary p a r t s t h a t we have j u s t decided are n e g l i g i b l e % by a l l o w i n g e x e c u t i o n to proceed, we s e t them to zero now. f a=real ( f a ; fb=real ( fb ; % check t h a t the r o ot i s indeed b r a c k e t e d i f sign ( f a sign ( fb== disp ( root i s not bracketed by points! r c=nobracket; done=f a l s e ; newt=0; b i s c =0; % loop u n t i l we have marked done=t r u e ; while done [ pn, fp, absdx ]=newton ( f dx, p ; % same checks as f o r the i n i t i a l point e v a l u a t i o n s i f isnan ( fp disp ( got NaN f o r f ( p r c=notfinite; break ; e l s e i f i s c o m p l e x l o o s e ( fp disp ( got complex value f o r f ( p r c=complex; break ; % d i s c a r d n e g l i g i b l e imaginary components as b e f o r e f o r the points fp=real ( fp ; [ pb, a, b, fb ]= b i s e c t (p, a, b, fp, fb ; % i f we meet the t o l e r a n c e, t h i s i s the l a s t i t e r a t i o n

7 % but we might as w e l l go ahead and update our computed % s o l u t i o n with t he work we ve a l r e a d y done so we don t break % the loop here i f errorbound ( absdx, a, b< t o l done=true ; i f and ( i s s e n s i b l e r o o t ( pn, i s i n i n t e r v a l ( a, pn, b % Newton s o l u t i o n i s good i f and ( p==pn, done disp ( s o l u t i o n no l o n g e r changing, but t o l e r a n c e was not met r c=stalled; break ; p=real ( pn ;% d i s c a r d n e g l i g i b l e imaginary components newt=newt+;%increment number o f Newton i t e r a t i o n s else % Newton s o l u t i o n isn t good, use b i s e c t i o n s o l u t i o n i f and ( p==pb, done disp ( s o l u t i o n no l o n g e r changing, but t o l e r a n c e was not met r c=stalled; break ; p=pb ; b i s c=b i s c +;%increment number o f b i s e c t i o n i t e r a t i o n s % show s t a t i s t i c s disp ( [ newt b i s c t o t ] disp ( [ newt, bisc, newt+b i s c ] function e r r=errorbound ( absdx, a, b % Both our e s t i m a t e and t he e x a c t s o l u t i o n must be in the i n t e r v a l % [ a, b ]. This g i v e s ( b a as an upper bound on the e r r o r. abs ( dx w i l l % almost always be smaller, but j u s t to be c a r e f u l, we l l check both. e r r=min( absdx, ( b a ; function bool=i s c o m p l e x l o o s e ( x % check i f v a l u e has a s i g n i f i c a n t imaginary component bool=eps ( real ( x<abs (imag( x ; function bool=i s i n i n t e r v a l ( a, pn, b % check t h a t a<=pn<=b bool=0<=(pn a ( b pn ; function bool=i s s e n s i b l e r o o t ( pn % i s t h i s a s e n s i b l e number to c a l l a root? % I t should not be I n f or NaN and i t should have n e g l i g i b l e imaginary part. bool=and ( i s f i n i t e ( pn, i s c o m p l e x l o o s e ( pn ; function [ p, fp, dx]=newton ( f dx, p % Do a newton e v a l u a t i o n. This i s the only p l a c e where % we need to a c t u a l l y e v a l u a t e the provided f u n c t i o n

8 % i n s i d e the loop [ fp, dx]= f d x ( p ; p=p dx ; dx=abs ( dx ; %we only care about the s i z e, not the s i g n o f dx from here on function [ p, a, b, fb ] = b i s e c t (p, a, b, fp, fb % perform b i s e c t i o n with provided f u n c t i o n e v a l u a t i o n s at the % e s t i m a t e f o r the r o o t and the l e f t point only % We assume t h a t i t has a l r e a d y been checked t h a t % s i g n ( f ( a s i g n ( f ( b<0 % so we only need to check the s i g n o f % s i g n ( f ( p s i g n ( f ( b switch sign ( fp sign ( fb case a=p ; % we don t e v a l u a t e f ( a because we never a c t u a l l y use i t case b=p ; % don t r e e v a l u a t e f ( b, s i n c e we a l r e a d y have f ( p and now b=p fb=fp ; case 0 % We ve found the e x a c t r o ot. Make a zero s i z e d i n t e r v a l around the s o l u t i o n i f fp==0 a=p ; b=p ; fb=fp ; e l s e i f fb==0 % t h i s should p r o b a b l y never happen s i n c e we would have n o t i c e d a l r e a d y a=b ; else disp ( t h i s should never happen! p=(a+b / 2 ; 2. testing: f(x = x We confirm that newtonsafe( behaves as expected with the test function f(x = x. In this case f (x = 2 x /2 and since newtonsafe( requires a function that evaluate dx dx = f(x f (x x = 2 x /2 = 2 ( x x This demonstrates the advantage of having the function provided to newtonsafe dx instead of simply f (x. At x = 0 f f(x (x is undefined, however, the above expression instead automatically evaluates dx = lim x 0 f (x. The function we supply to newtonsafe is [f,dx]=sqrttest(x

9 function [ f, dx]= s q r t t e s t ( x % f u n c t i o n e v a l u a t i o n f ( x f=sqrt ( x ; i f nargout> % e v a l u a t i o n o f dx=f ( x / f ( x dx=2 (x sqrt ( x ; We can then find the root using octave :> [ p, rc ]= newtonsafe 0,. 3, eps newt b i s c t o t p = rc = 0 octave :2> p ans = 0 In this case we find the exact root after 6 Newton iterations. Note that even though Octave is warning about a division by zero, the 2.2 testing: f(x = log(x We next make sure that newtonsafe( finds the root of log(x by providing it with the function [f,dx]=logtest(x function [ f, dx]= l o g t e s t ( x % f u n c t i o n e v a l u a t i o n f ( x f=log ( x ; i f nargout> % e v a l u a t i o n o f dx=f ( x / f ( x dx=f. / x ; octave :3> [ p, rc ]= newtonsafe 0,. 3, eps newt b i s c t o t 5 6 p = rc = 0 octave :4> p ans = 0 This time we also perform six iterations to find the exact root, however one of them is a bisection rather than a Newtons step. 2.3 Quantum finite square well Now we seek the solution to the more challenging problem (z0 2 tan z = ( z

10 0 8 y=tan(z y=sqrt((z0/z 2 - roots 6 y z / π Figure : Finite square well roots at the intersections of the left and right-hand-side of Eq. (. where z 0 = a 2mV0 m = kg a = 30nm = Js V 0 = J 2.3. (2 points We approach the problem by plotting the left and right hand sides of Eq. ( as shown in. The roots are the points at which the two functions intersect. As seen in the figure, there are 5 real roots. There are infinitely many complex roots which are not shown graphically. We observe (and probably already knew that tan(z is singular at odd multiples of π/2 approaching + from ( the left and form the right. We also note that z0 2 z approaches + from both sides of z = 0. It is also purely imaginary when z > z 0. With the calculation of z 0 above, we find that 5 2 π < z 0 < 3π and that all real ( roots occur in the interval 5 2 π < z < 5 2π with exactly one in each interval between the singularities of tan(z: 5 2 π, 3 2 π, ( 3 2 π, 2 π, ( 2 π, 2 π, ( 2 π, 3 2 π, ( 3 2 π, 5 2 π (2 points We wish to recast Eq. ( in a form for which the roots solve f(z = 0. This is easily done by subtracting the right hand side (z0 2 f(z = tan z z Of course, in order to use Newton s method, we will need to know the derivative ( (z0 f (z = sec 2 z + z2 2 /2 0 z 3 z

11 z± 8.9e-6 λ(m x 0 (m e e e e e e e e e e 09 Table : Roots of the finite square well problem Eq. ( and the associated decay constants. We define the physical constants in the script quantum constants.m a=30e 9; m= e 3; h= e 34; V0=.7 e 20; z0=a/h sqrt (2 m V0 ; which is simply sourced into the function [f,dx]=quantum(x to be passed into newtonsafe function [ f, dx]=quantum ( z % Function to pass to newtonsafe f o r root f i n d i n g % %source in the quantum c o n s t a n t s quantum constan ts % the f u n c t i o n v a l u e i t s e l f f=tan ( z sqrt ( ( z0. / z.ˆ2 ; i f nargout> % i f we are c a l l e d with two output arguments, % go ahead and compute dx=f ( x / f ( x dx=f. / ( sec ( z.ˆ2+ z0 ˆ 2. / ( z. ˆ 3. sqrt ( ( z0. / z. ˆ 2 ; We could be more careful about computing dx, but in the present case we are content with letting Newton fail spectacularly at singularities so that we will fall back to bisection. All the roots shown in Fig. and tabulated in Table. are computed by the script doquantum.m. % S c r i p t to f i n d the r o o t s in t he quantum mechanics problem on assignment 2 % we know where tan has s i n g u l a r i t i e s in the r egion where the s q r t f u n c t i o n i s r e a l s i n g u l a r i t i e s =[ 5:2:5] pi / 2 ; a=s i n g u l a r i t i e s ( : ; b=s i n g u l a r i t i e s ( 2 : ; % We can t e x p e c t to g e t s o l u t i o n s more accurate than machine e p s i l o n o f % the l a r g e s t p o s s i b l e s o l u t i o n t o l=eps (max( abs ( s i n g u l a r i t i e s ; for i i =:length ( a % f i n d the r oot in t he c u r r e n t i n t e r v a l [ p ( i i, rc ( i i ]= newtonsafe (@quantum, a ( i i, b ( i i, t o l ; %save the r e s u l t s save quantum roots We find each root by restricting newtonsafe to the appropriate interval between the singularities of tan z.

12 2.3.3 ( point The roots shown in the first column Table can be used to compute the spatial decay rates λ using the equation λ = z a tan z where λ is the decay constant for the wave function of electrons in the well φ(x = exp( λx x a This suggests a more intuitive measure of the decay rate x 0 λ yielding φ(x = exp ( xx0 x a So that when x = x 0, the wave function will have decayed to φ(x 0 = /e 0.37 and when x = 2x 0 it will have decayed to φ(2x 0 = /e Since the probability density function is equal to the square of the wave function, P (2x 0 = φ(2x 0 2 = /e , so it would seem reasonable to space the wells 2x 0 apart to give 2% overlap between wells. Both λ and x 0 are tabulated in the second and third columns of Table.. Presumably the negative values do not correspond to bound states. The largest positive x 0 is 7nm. We should therefore place the wells 4nm apart. 2.4 BF p82 #0 (5 points Suppose p is a zero of multiplicity m of the function f(x, where the mth derivative of f(x is continuous on an open interval containing p. The following fixed-point method has g (p = 0, g(x = x m f(x f (x Since f(x has a continuous mth derivative, we can rewrite it as where q(p 0 then and so that taking the derivative g (x = f(x = (x p m q(x f (x = m(x p m q(x + (x p m q (x f(x f (x = (x pq(x mq(x + (x pq (x g(x = x = x + (x p q (x m q(x which we can evaluate at x = p since q(x 0 g (p = + 0 q (p m q(p = + 0 = 0 m(x pq(x mq(x + (x pq (x (x p + ( + (x p q (x m q(x + (x p + (x p m 0 ( + 0 m q (x q(x q (p q(p This therefore ensures quadratic convergence of the method of Eq. (2. 2 ( + 2 ( + 0 m (x p m q (p q(p q (x q(x (2

13 2.5 BF p74 #33 (5 points We must find p such that 2 = + p ( p 2 p + p 2 ( = ( + p p p + p We can reformulate this in terms of a zero finding problem ( 2 p f(p = ( + p p + p 2 The derivative of this function is ( 2 ( ( 20 f p p (p = p + p 2 2 ( + p p + p 2 p + p 2 + p ( p + p 2 2 ( 2p We use newtonsafe( again with the function [f,dx]=raquet(x function [ f, dx]= raquet ( x f =(+x. ( x./( x+x. ˆ 2. ˆ 2 ; i f nargout> df =2 (+x. ( x./( x+x. ˆ 2. ˆ 2 0. (. / ( x+x.ˆ2+ x./( x+x.ˆ2. ( 2 x ; df=df+(x./( x+x. ˆ 2. ˆ 2 ; dx=f. / df ; octave :> format long octave :2> newtonsafe 0,, eps newt b i s c t o t ans = BF p72 #9 (5 points The enticing alternative method for implementing the secant method p n = f(p n p n 2 f(p n 2 p n f(p n f(p n 2 is a classic example of how numerical precision is lost by subtraction. When subtracting two finite precision numbers, the number of accurate digits in the result is the number of digits that can be stored minus the number of digits that the two terms have in common. This becomes a big problem when two very nearby numbers are subtracted. as in both the numerator and the denominator of this expression as n. The standard method p n p n 2 p n = p n f(p n f(p n f(p n 2 p n p n 2 also has such a quotient of nearby subtractions ( f(p n f(p n 2, it however has two distinct factors in the numerator that are going to zero (f(p n and p n p n 2, so that the whole problematic term ts to zero as we approach the root. It s inaccuracy therefore matters less and less the closer we get.

Math 473: Practice Problems for Test 1, Fall 2011, SOLUTIONS

Math 473: Practice Problems for Test 1, Fall 011, SOLUTIONS Show your work: 1. (a) Compute the Taylor polynomials P n (x) for f(x) = sin x and x 0 = 0. Solution: Compute f(x) = sin x, f (x) = cos x, f