Examples MAT-INF1100. Øyvind Ryan

Transcription

1 Examples MAT-INF00 Øyvind Ryan February 9, 20

2 Example 0.. Instead of converting 76 to base 8 let us convert it to base 6. We find that 76//6 = 2 with remainder. In the next step we find 2//6 = 4 with remainder. Finally we have 4//6 = 0 with remainder 4. Displayed in a table this becomes Recall that in the hexadecimal system the letters a f usually denote the values 0. We have therefore found that the number 76 is written eb 6 in the hexadecimal numeral system. Example 0.2. Let us continue to use the decimal number 76 as an example, but now we want to convert it to binary form. If we perform the divisions and record the results as before we find In other words we have 76 = This example illustrates an important property of the binary numeral system: Computations are simple, but long and tedious. This means that this numeral system is not so good for humans as we tend to get bored and make sloppy mistakes. For computers, however, this is perfect as computers do not make mistakes and work extremely fast. Example 0.. Let us convert the hexadecimal number c 6 to binary. We have 6 = 00 2, c 6 = 00 2, 6 = 00 2, 2

3 which means that c 6 = where we have omitted the two leading zeros. Example 0.4. Let us convert the number x = 0.a8 6 to binary. From table?? we find 6 = 00 2, a 6 = 00 2, 8 6 = 000 2, which means that 0.a8 6 = = Example 0.. To convert the binary number to hexadecimal form we note from table?? that 00 2 = c 6, 00 2 = 9 6, 00 2 = 6 6, = 8 6. Note that the last group of binary digits was not complete so we added three zeros. From this we conclude that = 0.c Example 0.6. Let us find the representation of x = 6 in two s complement from (??) when n = 4. In this case x = 6 so x = = 0. The binary representation of the decimal number 0 is 00 and this is the representation of 6 in two s complement. Example 0.7. Let us consider a concrete example of how the UTF-8 code of a code point is determined. The ASCII characters are not so interesting since for these characters the UTF-8 code agrees with the code point. The Norwegian character i smorechalleng ing.i f wechecktheunicodechar t s, we f indthat thi schar acterhasthecodep 97. This is in the range which is covered by rule 2 in fact??. To determine the UTF-8 encoding we must find the binary representation of the code point. This is easy to deduce from the hexadecimal representation. The least significant numeral ( in our case) determines the four least significant bits and the most significant numeral (c) determines the four most significant bits. Since = 00 2 and c 6 = 00 2, the code point in binary is c {}}{{}}{ , The Latin supplement can be found at

4 where we have added three 0s to the left to get the eleven bits referred to by rule 2. We then distribute the eleven bits as in (??) and obtain the two bytes 0000, In hexadecimal this corresponds to the two values c and 8 so the UTF-8 encoding of i sthet wo bytenumber c8 6. Example 0.8. On a typical calculator we compute x = 2, then y = x 2, and finally z = y 2, i.e., the result should be z = ( 2 ) 2 2, which of course is 0. The result reported by the calculator is z = This is a simple example of round-off error. Example 0.9 (Examples of truncation). The number 0. truncated to 4 digits is 0., while 28.4 truncated to 2 digits is 20, and truncated to 4 digits is Example 0.0 (Examples of rounding). The number 0. rounded to 4 digits is 0.. The result of rounding 28.4 to 2 digits is 0, while rounded to 4 digits is Example 0. (Standard case). Suppose that a =.64 and b = We convert the numbers to normal form and obtain a = , b = We add the two significands =.466 so the correct answer is The last step is to convert this to normal form. In exact arithmetic this would yield the result However, this is not in normal form since the significand has five digits. We therefore perform rounding, , and get the final result Example 0.2 (One large and one small number). If a = 42.4 and b = 0.00 we convert the largest number to normal form 42.4 =

5 The smaller number b is then written in the same form (same exponent) 0.00 = The significand in this second number must be rounded to four digits, and the result of this is The addition therefore becomes = Example 0. (Subtraction of two similar numbers I). Consider next a case where a = 0.4 and b = 0.27 have opposite signs. We first rewrite the numbers in normal form a = , b = We then add the significands, which really amounts to a subtraction, = Finally, we write the number in normal form and obtain a + b = = () Example 0.4 (Subtraction of two similar numbers II). Suppose that a = 0/7 and b =.42. Conversion to normal form yields Adding the significands yield 0 7 a = , b = = When this is converted to normal form, the result is while the true result rounded to four correct digits is (2)

6 Example 0.. Consider the two numbers a = 2.7 and b = 6.79 which in normalised form are a = , b = To multiply the numbers, we multiply the significands and add the exponents, before we normalise the result at the end, if necessary. In our example we obtain a b = The significand in the result must be rounded to four digits and yields the floatingpoint number i.e., the number , Example 0.6. We use the same numbers as in example 0., but now we perform the division a/b. We have a b = = , i.e., we divide the significands and subtract the exponents. The division yields 0.27/ We round this to four digits and obtain the result a b =.487. Example 0.7. Suppose we are going to evaluate the expression x 2 + x () for a large number like x = 0 8. The problem here is the fact that x and x 2 + are almost equal when x is large, x = 0 8 = , x Even with 64-bit floating-point numbers the square root will therefore be computed as 0 8, so the denominator in () will be computed as 0, and we get division by 0. This is a consequence of floating-point arithmetic since the two terms 6

7 in the denominator are not really equal. A simple trick helps us out of this problem. Note that x 2 + x = ( x x) x 2 ( x 2 + x)( x x) = + + x x 2 + x 2 = x x. This alternative expression can be evaluated for large values of x without any problems with cancellation. The result for x = 0 8 is x x where all the digits are correct. Example 0.8. For most values of x, the expression cos 2 x sin 2 x can be computed without problems. However, for x = π/4 the denominator is 0, so we get division by 0 since cos x and sin x are equal for this value of x. This means that when x is close to π/4 we will get cancellation. This can be avoided by noting that cos 2 x sin 2 x = cos2x, so the expression (4) is equivalent to cos2x. This can be evaluated without problems for all values of x for which the denominator is nonzero, as long as we do not get overflow. Example 0.9. Consider the equation x n+2 2 x n+ x n = 0, x 0 =, x = 0. () Since the two roots of the characteristic equation r 2 2r / / = 0 are r = and r 2 = /, the general solution of the difference equation is ( x n = C + D ) n. The initial conditions yield the equations C + D =, C D/ = 0, which has the solution C = /4 and D = /4. The solution of () is therefore x n = 4( + ( ) n n). (4) 7

8 We observe that x n tends to /4 as n tends to infinity. If we simulate equation () on a computer, the next term is computed by the formula x n+2 = (2x n+ + x n )/. The division by means that floating-point numbers are required to evaluate this expression. If we simulate the difference equation, we obtain the four approximate values x 0 = , x = , x 20 = , x 0 = , (throughout this section we will use x n to denote a computed version of x n ), which agree with the exact solution to 2 digits. In other words, numerical simulation in this case works very well and produces essentially the same result as the exact formula, even if floating-point numbers are used in the calculations. Example We consider the difference equation x n+2 9 x n+ + 2x n = 0, x 0 = 2, x = 8/. (6) The two roots of the characteristic equation are r = / and r 2 = 6, so the general solution of the homogenous equation is x h n = C n + D6 n. To find a particular solution we try a solution x p n = A which has the same form as the right-hand side. We insert this in the difference equation and find A =, so the general solution is x n = x h n + xp n = +C n + D6 n. (7) If we enforce the initial conditions, we end up with the system of equations 2 = x 0 = +C + D, 8/ = x = +C / + 6D. This may be rewritten as C + D =, C + 8D =. which has the solution C = and D = 0. The final solution is therefore (8) (9) x n = n, (0) 8

9 which tends to when n tends to infinity. Let us simulate the equation (6) on the computer. As in the previous example we have to divide by so we have to use floating-point numbers. Some early terms in the computed sequence are x = , x 0 = , x = These values appear to approach as they should. However, some later values are x 20 = , x 0 = , x 40 = , and at least the last two of these are obviously completely wrong! Example 0.2. We consider the third-order difference equation x n+ 6 x n x n+ 4 x n = 0 2 n, x 0 = 2, x = 7, x 2 = The coefficients have been chosen so that the roots of the characteristic equation are r = /, r 2 = and r = 4. To find a particular solution we try with x p n = A2 n. If this is inserted in the equation we find A =, so the general solution is () x n = 2 n + B n +C + D4 n. (2) The initial conditions force B = 0, C = and D = 0, so the exact solution is x n = 2 n. () The discussion above shows that this is bound to lead to problems. Because of round-off errors, the coefficients B and D will not be exactly 0 when the equation is simulated. Instead we will have x n = 2 n + ɛ n + ( + ɛ 2 ) + ɛ 4 n Even if ɛ is small, the term ɛ 4 n will dominate when n becomes large. This is confirmed if we do the simulations. The computed value x 00 is approximately , while the exact value is.8 0 0, rounded to two digits. 9

10 Example Suppose we have the text x = DBACDBD. We note that the frequencies of the four symbols are f (A) =, f (B) = 2, f (C ) = and f (D) =. We assign the shortest codes to the most frequent symbols, c(d) = 0, c(b) =, c(c ) = 0, c(a) = 0. If we replace the symbols in x by their codes we obtain the compressed text z = 00000, altogether 9 bits, instead of the 6 bits (7 bytes) required by a standard text representation. However, we now have a major problem, how can we decode and find the original text from this compressed version? Do the first two bits represent the two symbols D and B or is it the symbol C? One way to get round this problem is to have a special code that we can use to separate the symbols. But this is not a good solution as this would take up additional storage space. Example 0.2. Consider the same four-symbol text x = DBACDBD as in example 0.2. We now use the codes We can then store the text as c(d) =, c(b) = 0, c(c ) = 00, c(a) = 000. (4) z = , () altogether bits, while a standard encoding with one byte per character would require 6 bits. Note also that we can easily decipher the code since the codes have the prefix property. The first bit is which must correspond to a D since this is the only character with a code that starts with a. The next bit is 0 and since this is the start of several codes we read one more bit. The only character with a code that start with 0 is B so this must be the next character. The next bit is 0 which does not uniquely identify a character so we read one more bit. The code 00 does not identify a character either, but with one more bit we obtain the code 000 which corresponds to the character A. We can obviously continue in this way and decipher the complete compressed text. Example In figure?? the tree in figure?? has been turned into a Huffman tree. The tree has been constructed from the text CCDACBDC with the alphabet {A,B,C,D} and frequencies f (A) =, f (B) =, f (C ) = 4 and f (D) = 2. It is easy to see that the weights have the properties required for a Huffman tree, and by following the edges we see that the Huffman codes are given by c(c ) = 0, c(d) = 0, c(a) = 0 and c(b) =. Note in particular that the root of the tree has weight equal to the length of the text. 0

11 Example 0.2. Let us try out algorithm?? on the text then the hen began to eat. This text consists of 2 characters, including the five spaces. We first determine the frequencies of the different characters by counting. We find the collection of one-node trees 4 t h e n b g 2 a o where the last character denotes the space character. Since b and g are two characters with the lowest frequency, we combine them into a tree, 4 t h e n 2 b g 2 a o The two trees with the lowest weights are now the character o and the tree we formed in the last step. If we combine these we obtain 4 t h e n 2 b g o 2 a Now we have several choices. We choose to combine a and h, 4 t 2 h a e n 2 b g o

12 At the next step we combine the two trees with weight, 4 t 2 h a e 6 2 b g o n Next we combine the t and the e, 9 4 t e 2 h a 6 2 b g o n We now have two trees with weight that must be combined 9 4 t e 0 2 h a 6 2 b g o n 2

13 h a 4 n t e 2 o b g Figure : The Huffman tree for the text then the hen began to eat. Again we combine the two trees with the smallest weights, h a 4 n t e 2 o b g By combining these two trees we obtain the final Huffman tree in figure. From this we can read off the Huffman codes as c(h) = 000, c(a) = 00, c( ) = 0, c(b) = 0000, c(g ) = 000, c(o) = 00, c(n) = 0, c(t) = 0, c(e) =.

14 so we see that the Huffman coding of the text then the hen began to eat is The spaces and the new line have been added to make the code easier to read; on a computer these will not be present. The original text consists of 2 characters including the spaces. Encoding this with standard eight-bit encodings like ISO Latin or UTF-8 would require 400 bits. Since there are only nine symbols we could use a shorter fixed width encoding for this particular text. This would require five bits per symbol and would reduce the total length to 2 bits. In contrast the Huffman encoding only requires 7 bits. Example Let us return to example 0.26 and compute the entropy in this particular case. From the frequencies we obtain the probabilities c(t) = 4/2, c(h) = /2, c(e) = /, c(n) = /2, c(b) = /2, c(g ) = /2, c(a) = 2/2, c(o) = /2, c( ) = /. We can then compute the entropy to be H 2.9. If we had a compression algorithm that could compress the text down to this number of bits per symbol, we could represent our 2-symbol text with 74 bits. This is only one bit less than what we obtained in example 0.26, so Huffman coding is very close to the best we can do for this particular text. Example Suppose that we have a two-symbol alphabet A = {0,} with the probabilities p(0) = 0.9 and p() = 0.. Huffman coding will then just use the obvious codes c(0) = 0 and c() =, so the average number of bits per symbol is, i.e., there will be no compression at all. If we compute the entropy we obtain H = 0.9log log So while Huffman coding gives no compression, there may be coding methods that will reduce the file size to less than half the original size. Example 0.28 (Determining an arithmetic code). We consider the two-symbol text As for Huffman coding we first need to determine the probabilities of the two symbols which we find to be p(0) = 0.8 and p() = 0.2. The idea is to allocate different parts of the interval [0,) to the different symbols, and let the length of the subinterval be proportional to the probability of the symbol. In our case we allocate the interval [0,0.8) to 0 and the interval [0.8,) to. Since 4

15 our text starts with 0, we know that the floating-point number which is going to represent our text must lie in the interval [0,0.8), see the first line in figure??. We then split the two subintervals according to the two probabilities again. If the final floating point number ends up in the interval [0,0.64), the text starts with 00, if it lies in [0.64,0.8), the text starts with 0, if it lies in [0.8,0.96), the text starts with 0, and if the number ends up in [0.96,) the text starts with. This is illustrated in the second line of figure??. Our text starts with 00, so the arithmetic code we are seeking must lie in the interval [0,0.64). At the next level we split each of the four sub-intervals in two again, as shown in the third line in figure??. Since the third symbol in our text is, the arithmetic code must lie in the interval [0.2,0.64). We next split this interval in the two subintervals [0.2,0.644) and [0.644,0.64). Since the fourth symbol is 0, we select the first interval. This interval is then split into [0.2, 0.992) and [0.992, 0.644). The final symbol of our text is 0, so the arithmetic code must lie in the interval [0.2,0.992). We know that the arithmetic code of our text must lie in the half-open interval [0.2,0.992), but it does not matter which of the numbers in the interval we use. The code is going to be handled by a computer so it must be represented in the binary numeral system, with a finite number of bits. We know that any number of this kind must be on the form i /2 k where k is a positive integer and i is an integer in the range 0 i < 2 k. Such numbers are called dyadic numbers. We obviously want the code to be as short as possible, so we are looking for the dyadic number with the smallest denominator that lies in the interval [0.2,0.992). In our simple example it is easy to see that this number is 9/6 = In binary this number is , so the arithmetic code for the text 0000 is 00. Example Suppose we have the text x = {ACBBC AAB AA} and we want to encode it with arithmetic coding. We first note that the probabilities are given by p(a) = 0., p(b) = 0., p(c ) = 0.2, so the cumulative probabilities are F (A) = 0., F (B) = 0.8 and F (C ) =.0. This means that the interval [0,) is split into the three subintervals [0,0.), [0.,0.8), [0.8,). The first symbol is A, so the first subinterval is [a,b ) = [0,0.). The second symbol is C so we must find the part of [a,b ) that corresponds to C. The mapping from [0,) to [0,0.) is given by g 2 (z) = 0.z so [0.8,] is mapped to [a 2,b 2 ) = [ g 2 (0.8), g 2 () ) = [0.4,0.).

16 The third symbol is B which corresponds to the interval [0.,0.8). We map [0,) to the interval [a 2,b 2 ) with the function g (z) = a 2 + z(b 2 a 2 ) = z so [0.,0.8) is mapped to [a,b ) = [ g (0.), g (0.8) ) = [0.4,0.48). Let us now write down the rest of the computations more schematically in a table, g 4 (z) = z, x 4 = B, [a 4,b 4 ) = [ g 4 (0.), g 4 (0.8) ) = [0.46,0.474), g (z) = z, x = C, [a,b ) = [ g (0.8), g () ) = [0.4722,0.474), g 6 (z) = z, x 6 = A, [a 6,b 6 ) = [ g 6 (0), g 6 (0.) ) = [0.4722,0.47), g 7 (z) = z, x 7 = A, [a 7,b 7 ) = [ g 7 (0), g 7 (0.) ) = [0.4722,0.4726), g 8 (z) = z, x 8 = B, [a 8,b 8 ) = [ g 8 (0.), g 8 (0.8) ) = [ ,0.4726), g 9 (z) = z, x 9 = A, g 0 (z) = z, x 0 = A, The midpoint M of this final interval is [a 9,b 9 ) = [ g 9 (0), g 9 (0.) ) = [ , ), [a 0,b 0 ) = [ g 0 (0), g 0 (0.) ) = [ , ). M = = , and the arithmetic code is M rounded to log 2 ( p(a) p(b) p(c ) 2) + = 6 bits. The arithmetic code is therefore the number C (x) = = , but we just store the 6 bits In this example the arithmetic code therefore uses.6 bits per symbol. In comparison the entropy is.49 bits per symbol. Example 0.0 (Decoding of an arithmetic code). Suppose we are given the arithmetic code 00 from example 0.29 together with the probabilities p(0) = 0.8 and p() = 0.2. We also assume that the length of the code is known, the probabilities, and how the probabilities were mapped into the interval [0, ]; this is the typical output of a program for arithmetic coding. Since we are going to do this manually, we start by converting the number to decimal; if we were to program arithmetic coding we would do everything in binary arithmetic. 6

17 The arithmetic code 00 corresponds to the binary number which is the decimal number z = Since this number lies in the interval [0,0.8) we know that the first symbol is x = 0. We now map the interval [0,0.8) and the code back to the interval [0,) with the function We find that the code becomes h (y) = y/0.8. z 2 = h (z ) = z /0.8 = relative to the new interval. This number lies in the interval [0,0.8) so the second symbol is x 2 = 0. Once again we map the current interval and arithmetic code back to [0,) with the function h 2 and obtain z = h 2 (z 2 ) = z 2 /0.8 = This number lies in the interval [0.8,), so our third symbol must be a x =. At the next step we must map the interval [0.8,) to [0,). From observation?? we see that this is done by the function h (y) = (y 0.8)/0.2. This means that the code is mapped to z 4 = h (z ) = (z 0.8)/0.2 = This brings us back to the interval [0,0.8), so the fourth symbol is x 4 = 0. This time we map back to [0,) with the function h 4 (y) = y/0.8 and obtain z = h 4 (z 4 ) = 0.942/0.8 = Since we remain in the interval [0,0.8) the fifth and last symbol is x = 0, so the original text was Example 0.. Let us test a naive compression strategy based on the above idea. The plots in figure?? illustrate the principle. A signal is shown in (a) and its DCT in (b). In (d) all values of the DCT with absolute value smaller than 0.02 have been set to zero. The signal can then be reconstructed with the inverse DCT of theorem??; the result of this is shown in (c). The two signals in (a) and (b) visually look almost the same even though the signal in (c) can be represented with less than 2 % of the information present in (a). We test this compression strategy on a data set that consists of points. We compute the DCT and set all values smaller than a suitable tolerance to 0. With a tolerance of 0.04, a total of 42 4 values are set to zero. When we then 7

18 reconstruct the sound with the inverse DCT, we obtain a signal that differs at most 0.09 from the original signal. We can store the signal by storing a gzip ed version of the DCT-values (as 2-bit floating-point numbers) of the perturbed signal. This gives a file with 622 bytes, which is 88 % of the gzip ed version of the original data. Example 0.2. Suppose we want to find the zero 2 with error less than 0 0 by solving the equation f (x) = x 2 2. We have f () = and f (2) = 2, so we can use the Bisection method, starting with the interval [,2]. To get the error to be smaller than 0 0, we know that N should be larger than ln(b a) lnɛ ln2 = 0ln0 ln Since N needs to be an integer this shows that N = is guaranteed to make the error smaller than 0 0. If we run algorithm?? we find m 0 = , m = , m 2 = ,. m = We have with eleven correct digits, and the actual error in m is approximately Example 0.. Let us see if the predictions above happen in practice. We test the Secant method on the function f (x) = x 2 2 and attempt to compute the zero c = We start with x 0 = 2 and x =. and obtain x , e , x , e , x , e , x , e.2 0 0, x , e This confirms the claim in observation??. Example 0.4. The equation is f (x) = x 2 2 which has the solution c = If we run Newton s method with the initial value x 0 =.7, 8

19 we find x , e , x , e , x , e , x , e.4 0 8, x , e We see that although we only use one starting value, which is further from the root than the best of the two starting values used with the Secant method, we still end up with a smaller error than with the Secant method after five iterations. Example 0.. As mentioned in the beginning of this chapter, it may be that the position of an object is known only at isolated instances in time. Assume that we have a file with GPS data. In the file we are looking at, the positions are stored in terms of elevation, latitude, and longitude. Essentially these are what we call spherical coordinates. From the spherical coordinates one can easily compute cartesian coordinates, and also coordinates in a system where the three axes point towards east, north and upwards, respectively, as in a 2D and D map. Accompanying time instances are also stored in the file. Since the derivative of the position with respect to time is the speed, with the time instances one can approximate the speed by taking the absolute value of the Newton difference quotient. The position is, however, given in terms of the three coordinates. If we call the corresponding samples x n, y n, z n, and we apply Newton s difference quotient to each of these at all time instances, we get vectors v x,n, v y,n, v z,n, representing approximations to the speed in different directions. The speed vector at time instace n is the vector (v x,n, v y,n, v z,n ), and we define the speed at time instance n as (v x,n, v y,n, v z,n ) = vx,n 2 + v 2 y,n + vz,n. 2 Let us test this on some actual GPS data. In Figure 2(a)) we have plotted the GPS data in a coordinate system where the axes represent the east and north directions. In this system we can t see the elevation information in the data. In (b) we have plotted the data in a system where the axis represent the east and upward directions instead. Finally, in (c) we have also plotted the speed using the approximation we obtain from Newton s difference quotient. When visualized together with geographical data, such as colour indicating sea, forest, or habitated areas, this gives very useful information. Example 0.6. Assume that we read samples from a digital sound file, and compute the Newton difference quotient for all samples in the file. The x-axis now represents time, and h is the sampling period (the difference in time between two time samples). We can consider the set of all Newton difference quotients 9

20 North 00 Up East (a) The x axis points east, the y-axis points north East (b) The x axis points east, the y-axis points upwards x 0 4 (c) Time plotted againts speed Figure 2: Experiments with GPS data in a file as samples in another sound, and we can listen to it. When we do this we hear a sound where the bass has been reduced. To see why, in chapter?? we argued that we could reduce the bass in sound by using a row in Pascals triangle with alternating sign, and (,) is the first row in Pascals triangle. But f (a+h) f (a) = h(f (a +h) f (a))/h, so that the Newton difference quotient is equivalent to the procedure for reducing bass, up to multiplication with a constant. In summary, when we differentiate a sound we reduce the bass in the sound. Example 0.7. Let us test the approximation (??) for the function f (x) = sin x at a = 0. (using 64-bit floating-point numbers). In this case we know that the exact derivative is f (x) = cos x so f (a) with 0 correct digits. This makes it is easy to check the accuracy of the numerical method. We try 20

21 with a few values of h and find h ( f (a + h) f (a) )/ h E(f ; a,h) where E(f ; a,h) = f (a) ( f (a+h) f (a) )/ h. We observe that the approximation improves with decreasing h, as expected. More precisely, when h is reduced by a factor of 0, the error is reduced by the same factor. Example 0.8. Let us check that the error formula (??) agrees with the numerical values in example 0.8. We have f (x) = sin x, so the right-hand side in (??) becomes E(sin;0.,h) = h 2 sinξ h, where ξ h (0.,0.+h). We do not know the exact value of ξ h, but for the values of h in question, we know that sin x is monotone on this interval. For h = 0. we therefore have that the error must lie in the interval [0.0sin0., 0.0sin0.6] = [ , ], and we see that the right end point of the interval is the maximum value of the right-hand side in (??). When h is reduced by a factor of 0, the number h/2 is reduced by the same factor, while ξ h is restricted to an interval whose width is also reduced by a factor of 0. As h becomes even smaller, the number ξ h will approach 0. so sinξ h will approach the lower value sin For h = 0 n, the error will therefore tend to 0 n 2 sin n, which is in close agreement with the numbers computed in example 0.8. Example 0.9. Recall that we estimated the derivative of f (x) = sin x at a = 0. and that the correct value with ten digits is f (0.) If we check 2

22 values of h for 0 7 and smaller we find ( )/ h f (a + h) f (a) h E(f ; a,h) This shows very clearly that something quite dramatic happens. when we come to h = 0 7, the derivative is computed as zero. Ultimately, Example We test the approximation (??) with the same values of h as in examples 0.8 and Recall that f (0.) with 0 correct digits. The results are ( )/ h f (a + h) f (a h) (2h) E(f ; a,h) If we compare with examples 0.8 and 0.40, the errors are generally smaller for the same value of h. In particular we note that when h is reduced by a factor of 0, the error is reduced by a factor of 00, at least as long as h is not too small. However, when h becomes smaller than about 0 6, the error starts to increase. It therefore seems like the truncation error is smaller than for the original method based on Newton s quotient, but as before, the round-off error makes it impossible to get accurate results for small values of h. The optimal value of h seems to be h 0 6, which is larger than for the first method, but the error is then about 0 2, which is smaller than the best we could do with the asymmetric Newton s quotient. 22

23 Example 0.4. Let us try the midpoint rule on an example. As usual, it is wise to test on an example where we know the answer, so we can easily check the quality of the method. We choose the integral 0 cos x dx = sin where the exact answer is easy to compute by traditional, symbolic methods. To test the method, we split the interval into 2 k subintervals, for k =, 2,..., 0, i.e., we halve the step length each time. The result is h I mid (h) Error By error, we here mean 0 f (x)dx I mid (h). Note that each time the step length is halved, the error seems to be reduced by a factor of 4. Example We test the trapezoidal rule on the same example as the midpoint rule, 0 cos x dx = sin As in example 0.42 we split the interval into 2 k subintervals, for k =, 2,..., 0. 2

24 The resulting approximations are where the error is defined by h I trap (h) Error f (x)dx I trap (h). We note that each time the step length is halved, the error is reduced by a factor of 4, just as for the midpoint rule. But we also note that even though we now use two function values in each subinterval to estimate the integral, the error is actually twice as big as it was for the midpoint rule. Example 0.4. Let us test Simpson s rule on the same example as the midpoint rule and the trapezoidal rule, 0 cos x dx = sin As in example 0.42, we split the interval into 2 k subintervals, for k =, 2,..., 0. The result is h I Simp (h) Error

25 where the error is defined by 0 f (x)dx I Simp (h). When we compare this table with examples 0.42 and 0.4, we note that the error is now much smaller. We also note that each time the step length is halved, the error is reduced by a factor of 6. In other words, by introducing one more function evaluation in each subinterval, we have obtained a method with much better accuracy. This will be quite evident when we analyse the error below. Example We consider the differential equation x = t 2x, x(0) = 0.2. (6) Suppose we want to compute an approximation to the solution at the points t = 0., t 2 = 0.2,..., t 0 =, i.e., the points t k = kh for k =, 2,..., 0, with h = 0.. We start with the initial point (t 0, x 0 ) = (0,0.2) and note that x 0 = x (0) = 0 2x(0) = 0.. The tangent T 0 (t) to the solution at t = 0 is therefore given by T 0 (t) = x(0) + t x (0) = t. To advance the approximate solution to t = 0., we just follow this tangent, x(0.) x = T 0 (0.) = = 0.2. At (t, x ) = (0.,0.2) the derivative is x = f (t, x ) = t 2x = = 0.99, so the tangent at t is T (t) = x + (t t )x = x + (t t )f (t, x ) = 0.2 (t 0.)0.99. The approximation at t 2 is therefore x(0.2) x 2 = T (0.2) = x + h f (t, x ) = = If we continue in the same way, we find (we only print the first 4 decimals) x = 0.289, x 4 = 0.08, x = 0.090, x 6 = 0.08, x 7 = , x 8 = 0.062, x 9 = 0.62, x 0 = This is illustrated in figure?? where the computed points are connected by straight line segments. 2

26 Example 0.4. Let us consider the differential equation x = f (t, x) = F (t, x) = t, x(0) =, (7) + x which we want to solve on the interval [0,]. To illustrate the method, we choose a large step length h = 0. and attempt to find an approximate numerical solution at x = 0. and x = using a quadratic Taylor method. From (20) we obtain x (t) = F 2 (t, x) = + x (t) ( + x(t) ) 2. (8) To compute an approximation to x(h) we use the quadratic Taylor polynomial x(h) x = x(0) + hx (0) + h2 2 x (0). The differential equation (20) and (2) give us the values which leads to the approximation x(0) = x 0 =, x (0) = x 0 = 0 /2 = /2, x (0) = x 0 = /8 = 7/8, x(h) x = x 0 + hx 0 + h2 2 x 0 = h 2 + 7h2 6 = To prepare for the next step we need to determine approximations to x (h) and x (h) as well. From the differential equation (20) and (2) we find x (h) x = F (t, x ) = t /( + x ) = , x (h) x = F 2(t, x ) = + x /( + x ) 2 = , rounded to eight digits. From this we can compute the approximation x() = x(2h) x 2 = x + hx + h2 2 x = The result is shown in figure??a. 26

27 Example 0.46 (Euler s method for a system). We consider the equations in example??, x = f (t, x), x(a) = x 0, where f (t, x) = ( f (t, x, x 2, x ), f 2 (t, x, x 2, x ), f (t, x, x 2, x ) ) = (x x 2 + cos x,2 t 2 + x 2 x 2,sin t x + x 2 ). Euler s method is easily generalised to vector equations as x k+ = x k + h f (t k, x k ), k = 0,,..., n. (9) If we write out the three components explicitly, this becomes x k+ = x k + h f (t k, x k, xk 2, xk ) = xk + h( x k xk 2 + cos ) xk, x2 k+ = x2 k + h f 2(t k, x k, xk 2, xk ) = xk 2 + h( 2 t 2 k + (xk )2 x2 k ), x k+ = x k + h f (t k, x k, xk 2, xk ) = xk + h( sin t k x k + ) xk 2, (20) for k = 0,,..., n, with the starting values (a, x 0, x0 2, x0 ) given by the initial condition. Although they look rather complicated, these formulas can be programmed quite easily. The trick is to make use of the vector notation in (22), since it nicely hides the details in (2). Example 0.47 (System of higher order equations). Consider the system of differential equations given by x = t + x + y, x(0) =, x (0) = 2, y = x y + x, y(0) =, y (0) =, y (0) = 2. We introduce the new functions x = x, x 2 = x, y = y, y 2 = y, and y = y. Then the above system can be written as x = x 2, x (0) =, x 2 = t + x 2 + y 2, x 2 (0) = 2, y = y 2, y (0) =, y 2 = y, y 2 (0) =, y = x 2y + x, y (0) = 2. 27