Assignment 1: Solutions

Applied Math 205

October 17, 2007

1. (30 points) The length of the hypotenuse (h) of a right-angle triangle is given by

h =√

a2 + b2, (1)

where a and b are the sides of the triangle constituting the right angle. Write a MATLAB functionmyhypot(a,b), which computes h given a and b using only double precision arithmetic. In yoursolutions, draw a flow chart of the algorithm used. Notice that a straight-forward implementation of(1) is susceptible to numerical overflow, i.e., if either |a| or |b| are close to the largest (or smallest)representable machine number, a2 will be too big to be represented and the algorithm will fail even ifthe expected result can in fact be represented. As a test in double precision, use the values a = 1×10300

and b = 2 × 10300 for which your function should yield accurate results. The problem of computingthe absolute value of a complex number is exactly the same.

A generalized version of this algorithm, computes the 2-norm of an n-dimensional vector, defined as

‖x‖2

=

√

√

√

√

n∑

i=0

x2i . (2)

In the same spirit as myhypot, write a MATLAB function mynorm(v) to compute the 2-norm of vectorrepresented as an array in the variable v.

Useful MATLAB built-in functions are max, min and sqrt.

Solution:

myhypot.m

function h = myhypot(a, b)

m = min(a,b);

n = max(a,b);

h = n.*sqrt(1 + (m/n).^2);

mynorm.m

function h = mynorm(v)

vmax = max(abs(v));

u = v./vmax;

h = vmax.*sqrt(sum(u.^2));

2. (40 points) Write a double precision MATLAB function myexp(x) to evaluate ex using the seriesexpansion

ex = 1 + x +x2

2!+

x3

3!+ · · · =

∞∑

n=0

xn

n!. (3)

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Provide a flow chart for the algorithm you implemented in the function.

Plot the absolute and relative error (compared with the built-in function exp(x) for a range of x from-100 to 100 on a log-log scale. Briefly explain the result.

The way established numerical libraries compute the exponential is not through a series expansion butby using the identities:

ex+y = exey and eax = (ex)a. (4)

For example, a way of calculating e123 is

e123 = e1×102+2×10

1+3×10

0

= e100 × (e10)2 × (e1)3, (5)

provided e100, e10 and e are known. Develop and write a function myexp2(x) that evaluates theexponential using decomposition of decimal numbers in the exponent similar to shown in (5). Yourfunction should also work for negative values of x.

The values of the exponentials of powers of 10 are provided on the course website through the fileexpm1.dat..Since ex ≈ 1 for x ≪ 1, ex−1 is tabulated in the file to provide better accuracy if required.This table can be loaded in MATLAB by the command load ’expm1.dat’, or you can hardwire thetable in your function by copy-pasting. You may find the MATLAB built-in functions like rem, mod,fix, round, ceil and floor useful for decomposing x in multiples of powers of 10. Also include theresults of this function in the previous plot.

Standard libraries use a binary version of this algorithm for exponential, trigonometric and hyperbolicfunctions., which you may verify, does not require the use of the power-of (ˆ-operator in MATLAB)function. An efficient implementation of the power-of function is instructive. How the table itself isgenerated for the first time, is also an interesting question. However, we will not bother ourselves toomuch with these details, even though you have enough background to answer these issues.

Solution:

myexp.m

function mysum = myexp(x)

% Does not work for vectors x

mysum = 1;

myterm = x;

n = 2;

while(mysum+myterm~=mysum)

mysum = mysum + myterm;

myterm = myterm.*x./n;

n = n+1;

end

myexp2.m

function y = myexp2(x)

if(x<0) % For negative values of x, compute 1/exp(-x)

y = 1./myexp2(-x);

elseif(x==0)

y = 1;

else

dat = [ ... % expm1.dat copied verbatim

9.999999999999999999805e-10 1.000000000500000136887e-09; % 1

1.000000000000000000001e-08 1.000000005000000105234e-08; % 2

9.999999999999999999846e-08 1.000000050000001591961e-07; % 3

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Figure 1: Flowchart for series summation

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−100 −80 −60 −40 −20 0 20 40 60 80 10010

−60

10−40

10−20

100

1020

1040

1060

1080

x

Err

or

Absolute in myexpAbsolute in myexp2Relative in myexpRelative in myexp2

Figure 2: Absolute and relative errors in the various functions.

1.000000000000000000036e-06 1.000000500000166516403e-06; % 4

1.000000000000000000078e-05 1.000005000016666821257e-05; % 5

1.000000000000000000078e-04 1.000050001666708378648e-04; % 6

1.000000000000000000064e-03 1.000500166708341662908e-03; % 7

1.000000000000000000064e-02 1.005016708416805841508e-02; % 8

1.000000000000000000081e-01 1.051709180756476291752e-01; % 9

1.000000000000000000108e+00 1.718281828459045312840e+00; % 10

1.000000000000000000087e+01 2.202546579480671789497e+04; % 11

1.000000000000000000069e+02 2.688117141816135609425e+43; % 12

];

dat = double(dat);

n = floor(log10(x)); % x = quotient x 10^n + remain

divisor = dat(10 + n, 1); % divisor = 10^n

quotient = floor(x/divisor); %

remain = x - (quotient*divisor);

y = (1+dat(10 + n, 2))^quotient;

x = remain;

n = floor(log10(x));

while(remain>0 && n>-10) % repeat the same process with the

divisor = dat(10 + n, 1); % remainder and multiply the result

quotient = floor(x/divisor); % to the answer at each step

remain = x - (quotient*divisor);

y = y * (1+ dat(10 + n, 2))^quotient;

x = remain;

n = floor(log10(x));

end %while

end %if

Explanation: See figure 2. The absolute error using myexp.m is seen to be increasing exponentially with|x|. But it is incorrect to conclude in general that the computation is not accurate. For positive valuesof x, the absolute error may seem large, but in comparison to the value of theexp(x), it is accurateto machine precision. However, for negative values of x the value of the exponential itself is small,thus the relative error is also large. Hence we conclude that the series summation loses accuracy for

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negative values of x. This loss of accuracy can be attributed to the alternation of signs in the series,the large magnitudes of terms and the fact that these large terms are supposed to add up to a smallnumber.

The myexp2.m function is always accurate to machine precision. Of course that is so because we havetreated negative values of x separately.

3. (30 points) You are asked to derive a finite difference expression for the fourth derivative of a functionf(x) using uniformly spaced points a distance h apart from each other. The expression must be atleast second order accurate. What is the minimum number of points you must use in your stencil?

Derive such an expression (using the minimum number of points you estimated).

The fourth derivative of exp(x) at x = 1 is e = exp(1). Use your expression to compute the fourthderivative in a MATLAB subroutine for values of h = 2−1, 2−2 . . . 2−45 and calculate the error in yourcomputation. Plot this error on a log-log scale as a function of h. Explain as many features on thisgraph as quantitatively as possible for you.

Solution: We expect there to be 5 point in a second-order accurate finite difference expression forthe fourth derivative. In general, for the fourth derivative on a non-uniform grid, we would require 6points for second order accuracy. But on an uniform grid, the grid points can be chose symmetricallyabout the point in question, thus automatically getting rid of odd powers of h in the error, where h isthe grid spacing. This is demonstrated in the derivation below.

Expand the neighbouring 4 points in a Taylor series about xk, the kth grid point as

fk+1 = fk + hf ′

k +h2

2f ′′

k +h3

6f ′′′

k +h4

24f iv

k +h5

60fv

k +h6

360fvi

k + · · · , (6)

fk−1 = fk − hf ′

k +h2

2f ′′

k −h3

6f ′′′

k +h4

24f iv

k −h5

60fv

k +h6

360fvi

k + · · · , (7)

fk+2 = fk + 2hf ′

k +4h2

2f ′′

k +8h3

6f ′′′

k +16h4

24f iv

k +32h5

60fv

k +64h6

360fvi

k + · · · , (8)

fk−2 = fk − 2hf ′

k +4h2

2f ′′

k −8h3

6f ′′′

k +16h4

24f iv

k −32h5

60fv

k +64h6

360fvi

k + · · · . (9)

The odd derivatives can be easily eliminated by adding fk+1 to fk−1 and fk+2 to fk−2,

fk+1 + fk−1 − 2fk = h2f ′′

k +h4

12f iv

k +h6

180fvi

k + · · · , (10)

fk+2 + fk−2 − 2fk = 4h2f ′′

k +4h4

3f iv

k +16h6

45fvi

k + · · · . (11)

And now, eliminating the second derivative gives

fk+2 − 4fk+1 + 6fk − 4fk−1 + fk−2 = h4f ivk +

h6

3fvi

k + · · · , (12)

which can be solved for the fourth derivative. The final expression is:

f ivk =

fk+2 − 4fk+1 + 6fk − 4fk−1 + fk−2

h4−

h2

3fvi

k + · · · . (13)

The error in computing the derivative also has contribution from round off error in evaluation of thevarious fk’s. The round off error in evaluation of each fk is approximately ǫ|fk|. Using the furtherapproximation that fk+1 ≈ fk+2 ≈ fk, gives an estimate for round off error plus truncation error to be

Error = E(h) = 16ǫ|fk|

h4+ |fvi

k |h2

3+ · · · ..., (14)

5

10−14

10−12

10−10

10−8

10−6

10−4

10−2

100

10−5

100

105

1010

1015

1020

1025

1030

1035

1040

1045

h

Abs

olut

e er

ror

ComputedAnalytical estimate

Figure 3: Error in the finite difference expression for the fourth derivative.

(especially notice the absolute signs since in the worst case, errors add up). We compare this expressionwith the numerical computations in figure 3.

For values of h bigger than 10−2, the error decreases with decreasing h. The rate of this decrease ish2, For h below 10−3, the error to grows with decreasing h. The slope of that part of the curve is -4.These are the two limits of E(h) derived. The minimum occurs at

h = 2

(

3ǫ

2

)1/6

≈ 0.005, (15)

and the error there is about 4 × 10−5. This is about as accurate a finite difference approximation forthe fourth derivative can get. As the graph shows, the prediction matches with the computation quitewell.

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