# Finite arithmetic and error analysis¶

## Decimal and binary systems¶

In the decimal system the number $105.8125$ means:

$$105.8125=1 \cdot 10^2+5\cdot 10^0+8\cdot10^{-1}+1\cdot10^{-2}+2\cdot10^{-3}+5\cdot10^{-4}$$

i.e., it is a linear combination of powers of $10$ multiplied by one of the $10$ digits in $0,1,\ldots,9$.

Computers use the binary system because it is the natural way an electronic device works: switched on or off. Therefore, only $0's$ and $1's$ need to be stored.

In the binary system numbers are represented as linear combinations of powers of $2$, multiplied by $0's$ and $1's$:

$$(105.8125)_{10}=2^6+2^5+2^3+2^0+2^{-1}+2^{-2}+2^{-4}=(1101001.1101)_2 \qquad (1)$$

### Decimal to binary conversion¶

Conversion of the integer part is achieved by sequentially dividing by $2$. The remainings of these divisions are the digits in base $2$, from lesser to larger significancy.

$$\begin{array}{lrrrrrrrr} Quotients & 105 & 52 & 26 & 13 & 6 & 3 & 1\\ Remainders & 1 & 0 & 0 & 1 & 0 & 1 & \swarrow & \end{array}$$

Then, $(105)_{10}$ in the binary system is $(1101001)_2$.

Remark. Python's function bin performs this conversion.

In [1]:
bin(105)

Out[1]:
'0b1101001'

Exercise 1

Write a function, de2bi_a.py, to convert the integer part of a base 10 number into a binary number. Use it for x = 105.8125.

Hint: the following functions may be useful

• np.fix(x): rounds x towards 0.
• a//b : gives the quotient of the division.
• a%b : gives the remainder of the division.
In [2]:
%run Exercise1.py
x = 105.8125
binary1 = de2bi_a(x)
print ''.join(binary1)  # if binary1 is stored as a strings' list

1101001


To convert the decimal part, we multiply by $2$, drop the integer part and repit till reaching $1$.

$$\begin{array}{lrrrr} Decimal & 0.8125& 0.625 & 0.25 & 0.5 \\ Integer & 1 & 1 & 0 & 1 \end{array}$$

Ejercicio 2

Write a function, de2bi_b.py, to convert the decimal part of a base 10 number into a binary number. Use it for x = 105.8125. Limit the maximum number of fracional digits in binary to 23. Try also with x = 14.1

In [3]:
%run Exercise2.py
x = 105.8125
binary2 = de2bi_b(x)
print ''.join(binary2)

1101


And number 105.8125 (in decimal) is written, in binary

In [4]:
print ''.join(binary1 + ['.'] + binary2)

1101001.1101

In [5]:
x = 14.1

binary3 = de2bi_a(x)
binary4 = de2bi_b(x)

print ''.join(binary3 + ['.'] + binary4)

1110.00011001100110011001100


### Binary to decimal conversion¶

Using $(1)$ for converting $(1101001.1101)_2$ to decimal base we get

In [6]:
1*(2**6)+1*(2**5)+0*(2**4)+1*(2**3)+0*(2**2)+0*(2**1)+1*(2**0)+1*(2**-1)+1*(2**-2)+0*(2**-3)+1*(2**-4)

Out[6]:
105.8125

## Integer representation¶

If we have $m$ digits or memory bits then we may store $2^m$ different binary numbers.

### Positive integers¶

If we only consider the positive integers then we may represent numbers between

$$(00\ldots 00)_2=(0)_{10} \quad \mathrm{and} \quad (11\ldots 11)_2=(2^m-1)_{10}.$$

For instance, for $m=3$ bits we may represent positive integers from $0$ to $7$:

$$\begin{array}{|c|c|} \hline Base \; 10 & Base \; 2\\ \hline \mathtt{0} & \mathtt{000}\\ \mathtt{1} & \mathtt{001}\\ \mathtt{2} & \mathtt{010}\\ \mathtt{3} & \mathtt{011}\\ \mathtt{4} & \mathtt{100}\\ \mathtt{5} & \mathtt{101}\\ \mathtt{6} & \mathtt{110}\\ \mathtt{7} & \mathtt{111}\\ \hline \end{array}$$

### Signed integers¶

In signed integers, the first bit is used to store the sign: $0$ for positive and $1$ for negative. The other $m-1$ digits are used for the unsigned number and, therefore, we may write $2^{m-1}-1$ positive numbers, the same amount of negative numbers and two zeros, one with positive sign and another with negative sign. Therefore, the range of representation is $[-2^{m-1}+1, 2^{m-1}-1]$. For instance, if $m=3$ we may represent numbers from $-3$ to $3$:

$$\begin{array}{|c|c|} \hline Base \; 10 & Base \; 2\\ \hline \mathtt{-3} & \mathtt{111}\\ \mathtt{-2} & \mathtt{110}\\ \mathtt{-1} & \mathtt{101}\\ \mathtt{-0} & \mathtt{100}\\ \mathtt{+0} & \mathtt{000}\\ \mathtt{+1} & \mathtt{001}\\ \mathtt{+2} & \mathtt{010}\\ \mathtt{+3} & \mathtt{011}\\ \hline \end{array}$$

Example

Compute how $(80)_{10}$ is stored when using a signed 8-bit binary representation.

Solution: The representation is similar than that of Exercise 4, but with the first digit equal to $1$, due to the negative sign. So we have $(-80)_{10}=(11010000)_2$.

#### Signed integers: Two's complement representation¶

To avoid the double representation of zero we define negative integers by taking the digits of the corresponding positive integers and changing $0's$ to $1's$, and $1's$ to $0's$, and adding $1$ to the result. In this way, the sum of a number and its oposite is always $0$.

$$\begin{array}{|c|c|} \hline Base \; 10 & Base \; 2\\ \hline \mathtt{-4} & \mathtt{100}\\ \mathtt{-3} & \mathtt{101}\\ \mathtt{-2} & \mathtt{110}\\ \mathtt{-1} & \mathtt{111}\\ \mathtt{0} & \mathtt{000}\\ \mathtt{1} & \mathtt{001}\\ \mathtt{2} & \mathtt{010}\\ \mathtt{3} & \mathtt{011}\\ \hline \end{array}$$

Example. To represent $(-2)_{10}$ we start writing $(2)_{10}$ in binary form, $(010)_2$. Then, we

• invert its digits $\rightarrow (101)_2$
• add $001$ (take into account that $0+0=0, \quad 0+1=1, \quad 1+1=10$) $\rightarrow (110)_2$

So the property $(010)_2+(110)_2=(000)_2$ holds.

In this case, the first bit of the negative is $0$ and that of the positive is $1$. We may represent integers in the range $[-2^{m-1},2^{m-1}-1]$.

Example. Compute how $(80)_{10}$ is stored when using a two's complement 8-bit binary representation. Solution: since $(80)_{10}=(01010000)_2$

• we interchange $0's$ and $1's$ $\rightarrow 10101111,$
• and adding $1$ we obtain the representation $(-80)_{10}=(1011000)_2$

#### Signed integers: biased representation¶

Negative numbers are represented as consecutive positive values, starting from the lowest negative integer. Positive numbers are the rest. The representation is obtained by adding the bias $2^{m-1}$ to the number $x$, i.e., setting $x_r=x+2^{m-1}\in[0,2^m-1]$.

$$\begin{array}{|c|c|} \hline Base \; 10 & Base \; 2\\ \hline \mathtt{-4} & \mathtt{000}\\ \mathtt{-3} & \mathtt{001}\\ \mathtt{-2} & \mathtt{010}\\ \mathtt{-1} & \mathtt{011}\\ \mathtt{0} & \mathtt{100}\\ \mathtt{1} & \mathtt{101}\\ \mathtt{2} & \mathtt{110}\\ \mathtt{3} & \mathtt{111}\\ \hline \end{array}$$

We see that the representable rage is again $[-2^{m-1},2^{m-1}-1]$.

Which form is actually used for integer storage? Most machines use two's complement for integer numbers and biased representation for the exponents of floating point numbers (which are integers).

Why? The main reason to use biased representation is its efficiency for comparing numbers. Codes compare floating point numbers very often, and this is done by first comparing their exponents. Only if they are equal, comparison of their mantissas is performed.

### Integer representation in Python¶

In Python, numerical variables may be int, float, long and complex. Boolean variables are a subtype of integers.

Single integers or int have at least 32 bits precission, normally 32 or 64 bits, depending on the computer and operative system we are using.

The maximum single integer value in our computer is given by

In [7]:
import sys
sys.maxint

Out[7]:
9223372036854775807

That is, if $m=64$ is the number of bits, the largest integer is given by $2^{m-1}-1.$

In [8]:
2**(64-1)-1

Out[8]:
9223372036854775807L

And for negative values, the minimum integer is $-2^{m-1}$

In [9]:
-sys.maxint-1

Out[9]:
-9223372036854775808
In [10]:
-2**(64-1)

Out[10]:
-9223372036854775808L

Thus, in our case the single integer has a 64 bits precission.

If we need to represent a larger integer, Python uses long integers. Their precission is only limited by the memory available. They are recognized by its last character: L.

In [11]:
sys.maxint+1

Out[11]:
9223372036854775808L
In [12]:
-sys.maxint-2

Out[12]:
-9223372036854775809L

## Real number representation¶

For real numbers, floating point representation of base $\beta=2$ is used:

$$x_r=(-1)^s \cdot m \cdot \beta^e=(-1)^s \cdot (a_1.a_2\ldots a_t) \cdot \beta^e$$

where:

• $s$ is the sign: $1$ for negative numbers and $0$ for positive.
• $m$ is the mantissa which when normalized has values in $1\leq m < \beta$ and a nonzero first digit. In binary base this digit only can be $1$, i.e. in a normalized mantissa we always have $a_1=1$, so it is unnecesary to store it (hidden bit method).
• $e$ is the exponent, a signed integer (biased binary representation).

Numbers are stored either in words of 32 bits (single precission), 64 bits (double precission), or 128 bits (extended precission). In most computers, for Pyhthon, the default float precission is double precission. In this case bits are used as follows:

• 1 bit for the sign.
• 11 bits for the exponent,
• 52 bits for the mantissa.
In [13]:
%matplotlib inline
%run Double_precision.py


With $11$ bits for the signed exponent, we have room for $2^{11}=2048$ binary numbers, $0 < E < 2047$. The first number, $00000000000$ is reserved for zeros and non-normalized numbers; and the last, $11111111111$, for Inf and NaN.

Thus, the exponent take values $1 < E < 2046$, and since the bias is $1023$, these values correspond to the exponents $-1022 < E - 1023 < 1023$. Therefore, the maximum exponent is $E_{max}=1023$, and the minimum is $E_{min}=-1022$. Dividing by the minimum value, we get

$$\frac{1}{x_{min}}=\frac{1}{m\beta^{E_{min}}}=\frac{1}{m\beta^{-1022}}=\frac{1}{m}\beta^{1022}<\frac{1}{m}\beta^{1023}$$

so the maximum value is not reached, i.e. there is not overflow.

Exercise 3

Write a script Exercise3.py that calculates the single binary representation of 105.8125, 120.875, 7.1 and -1.41, following the standard IEEE 754. Use the scrips written in exercises 1 and 2. Consider only the case where the integer part is greater than zero (the number, in absolute value, is greater than $1$). Round the numbers using truncation.

Note: Remember that single precision uses 32 bits: 1 for the sign , 8 for the exponent and 23 bits for the mantissa. The exponent bias is 127.

In [14]:
%run Exercise3.py

105.8125 --->  1101001.1101
[Sign]:  0  [Exponent]:  10000101  [Mantissa]:  10100111010000000000000

120.875  --->  1111000.111
[Sign]:  0  [Exponent]:  10000101  [Mantissa]:  11100011100000000000000

7.1      --->  111.00011001100110011001100
[Sign]:  0  [Exponent]:  10000001  [Mantissa]:  11000110011001100110011

-1.41    --->  1.01101000111101011100001
[Sign]:  1  [Exponent]:  01111111  [Mantissa]:  01101000111101011100001


The features of ours machine float type are obtained using the command sys.float_info

In [15]:
sys.float_info

Out[15]:
sys.float_info(max=1.7976931348623157e+308, max_exp=1024, max_10_exp=308, min=2.2250738585072014e-308, min_exp=-1021, min_10_exp=-307, dig=15, mant_dig=53, epsilon=2.220446049250313e-16, radix=2, rounds=1)

The number of digits in the mantissa, mant_dig, is 53 (52 bits but 53 digits because the hidden bit). That is, Python's float uses 64 bits and is double precision.

But

max_exp=1024 and min_exp=-1021

does not seems to agree with what it was written above.

But, if we execute

help(sys.float_info)

we obtain

| max_exp | DBL_MAX_EXP -- maximum int e such that radix**(e-1) is representable

| min_exp | DBL_MIN_EXP -- minimum int e such that radix**(e-1) is a normalized float

Ejercicio 4

The largest double precission normalized number that Matlab may store in binary representation is

$$(+1)\cdot (1.11\ldots 11) \cdot 2^{1023}$$

Since we do not have to keep the first $1$, there are 52 bits left to store the $1's$ of $0.11\ldots 11$. Therefore, in base $10$ this number is

$$(1+1\cdot 2^{-1}+1\cdot 2^{-2}+1\cdot 2^{-3}+\cdots+1\cdot 2^{-52})\cdot 2^{1023}$$

Write a code, Exerxise4.py, to compute this sum. Perform the sum from lowest to largest terms (we shall see why later on). Its value should coincide with that obtained with sys.float_info.max. Define the variable output with the output value.

In [16]:
import sys

%run Exercise4a.py
print '%.16e' % output      # exponential format with 16 decimal digits

1.7976931348623157e+308

In [17]:
sys.float_info.max

Out[17]:
1.7976931348623157e+308

We check that they are equal

In [18]:
print output == sys.float_info.max

True


(b) Using a similar procedure than in Exercise4a, write a code Exercise4b.py to compute the lowest representable normalized floating point number using the expression

$$(+1)\cdot (1.00\ldots 00) \cdot 2^{-1022}$$

Its value should coincide with that obtained from sys.float_info.min.

In [19]:
%run Exercise4b.py
print '%.16e' % output

2.2250738585072014e-308

In [20]:
sys.float_info.min

Out[20]:
2.2250738585072014e-308

The largest consecutive integer which can be stored exactly in binary form with floating point representation is

$$(+1)\cdot (1.11\ldots 11) \cdot 2^{52}$$

Therefore, we have $53$ precission digits in binary form. The next integer,

$$(+1)\cdot (1.00\ldots 00) \cdot 2^{53}$$

can also be stored exactly. But not the next one, since we would need an extra bit in the mantissa.

In [21]:
print(2**53)

9007199254740992


Hence, all 15-digits integers and most of 16-digits integers may be stored exactly.

Observe that, actually, the number

$$(+1)\cdot (1.11\ldots 11) \cdot 2^{62}$$

is integer, storable and larger than the largest integer which can be stored!! However, the ten last digits of this number are necessarily zeros. Therefore, from the point of view of precission they are negligible, since they can not be changed.

Denormalized numbers

What happens to numbers with exponents out of the range $[-1022,1023]$?

If the exponent is lower than $-1022$ then the number is non-normalized or zero, and we are using for the bits corresponding to the exponent the special value $00000000000$. Then, the hidden bit is now $0$, instead of $1$.

In [22]:
x = 0.5*(2**-1023)
print(x)               # denormalized number
print(1/x)

5.56268464627e-309
inf

In [23]:
print(2**-1080)        # Underflow

0.0


The lowest non-normalized number is

$$(+1)(0.000\ldots 01)\times 2^{-1022}$$

that is, $2^{-1022-52}$

In [24]:
num_min = 2.**(-1022-52)
print(num_min)

4.94065645841e-324


For any other smaller value, we get zero (underflow):

In [25]:
2.**(-1022-53)    # Underflow

Out[25]:
0.0

If the exponent is larger than $1023$ we get OverflowError.

In [26]:
2.**1024          # Overflow

---------------------------------------------------------------------------
OverflowError                             Traceback (most recent call last)
<ipython-input-26-38590efc1dcd> in <module>()
----> 1 2.**1024          # Overflow

OverflowError: (34, 'Result too large')

### Machine precission¶

Let us compute the lowest number we may add to $1$ using double precission floating point binary representation.

The number $1$, in normalized double precission floating point representation, is

$$(+1)\cdot (1.00\ldots 00) \cdot 2^{0}$$

with $52$ zeros in the mantissa. The lowest number we may add in floating point non-normalized representation is

$$\epsilon = (+1)\cdot (0.00\ldots 01) \cdot 2^{0}$$

which in base $10$ is

In [27]:
1*2.**(-52)

Out[27]:
2.220446049250313e-16

which is obtained from Python by

In [28]:
sys.float_info.epsilon

Out[28]:
2.220446049250313e-16

This value, the lowest number $\epsilon$ such that $1+\epsilon>1$ is called the machine precission, which gives the floating point representation precission.

Since, in double precission, $\epsilon \approx 2.22\cdot 10^{-16}$ we have that it corresponds to, approximately, $16$ decimal digits.

Between

$$1=(+1)\cdot (1.00\ldots 00) \cdot 2^{0}$$

and

$$1+\epsilon=(+1)\cdot (1.00\ldots 01) \cdot 2^{0}$$

we can not represent exactly (and in floating point) any other real number.

To compute the lowest number comparable to $x$ we must add the number eps

In [29]:
x = 10.
eps = np.spacing(x)
print(eps)

1.7763568394e-15

In [30]:
x = 100.
eps = np.spacing(x)
print(eps)

1.42108547152e-14

In [31]:
x = 1000.
eps = np.spacing(x)
print(eps)

1.13686837722e-13


We see that eps increases with the absolute value of $x$. This means that the difference between two consecutive numbers exactly representable in floating point representation increases as we depart from zero. Therefore, the density of exact floating point numbers is not uniform. It is larger close to zero and smaller far from it. Remind that floating point numbers are actually rational numbers.

### Special cases¶

Some operations lead to special results

In [32]:
x = 1.e200
y = x*x
print(y)

inf

In [33]:
y/y

Out[33]:
nan

nan means "not a number". In general, it arises when a non valid operation has been performed.

Overflow. It happens when the result of the operation is finite but not representable (larger than the largest representable number).

In [34]:
x = 1.e300
x**2

---------------------------------------------------------------------------
OverflowError                             Traceback (most recent call last)
<ipython-input-34-35043931f28e> in <module>()
1 x = 1.e300
----> 2 x**2

OverflowError: (34, 'Result too large')

## Approximation and rounding error¶

Often, there is no exact floating point representation of a real number $x$, which lies between two consecutive representable numbers $x^-rounding method. IEEE admits five rounding methods: • Round toward 0 ("truncation"). • Round toward$+ \infty$. • Round toward$- \infty$. • Round toward the nearest number which is furthest from zero. • Round toward the nearest even number ("rounding"). The last method is the most usual. ### Loss of digits¶ Suppose that we are using a three digits decimal representation and we want to sum the numbers$a=123$and$b=1.25$. The exact result is$s=124.25$but, since only three digits are available, we shall round to, for instance,$s_r=124$. An error arises. In general, finite arithmetic operations involve errors. Example. The operation$3\times 0.1 \times 100$returns: In [35]: 3*0.1*100  Out[35]: 30.000000000000004 The result is not the right one ($30$) due to rounding error. These errors may propagate and, in some cases, affect substantially the result. In the previous case the error is due to, as we already saw, the periodicity of$x=0.1$in the binary base. Thus, its representation is not exact in the computer. Example. Computing $$\sum_{k=1}^{10000} 0.1$$ gives In [36]: Sum = 0. for i in range(1,10001): Sum = Sum + 0.1 print '%.16f'% Sum # Decimal format with 16 decimals  1000.0000000001588205  with absolute error given by In [37]: abs(Sum-1000.)  Out[37]: 1.588205122970976e-10 Example If we add or substract numbers which are very different in magnitude we always lose accuracy due to rounding error. For instance, In [38]: a = 1.e+9 epsilon = 1.e-8 Sum = a for i in range(1,10001): Sum = Sum + epsilon print '%.16e' % Sum # Exponencial format with 16 decimals  1.0000000000000000e+09  However, the result should have been In [39]: Sum = a + 10000*epsilon print '%.16e'% Sum  1.0000000000001000e+09  Example Let us sum the$N$-first term of the harmonic series $$\sum_{n=1}^N \frac{1}{n}.$$ We use single precission (function float32) to make the effect more evident. The exact result for the first$1000$-terms is In [40]: import numpy as np from scipy import special N = 10000 Sum = special.polygamma(0,N+1)-special.polygamma(0,1) print '%.20f'% Sum  9.78760603604438372827  Starting the sum from the term$n=1$, we get In [41]: Sum1 = 0.; for n in range(1,N+1): Sum1 += 1./n print '%.20f'% Sum1  9.78760603604434820113  In [42]: error1 = abs(Sum1 - Sum) print error1  3.5527136788e-14  Starting from the last term, we get In [43]: Sum2 = 0.; for n in range(N,0,-1): Sum2 += 1./n print '%.20f'% Sum2  9.78760603604438550462  Thus, the absolute difference is In [44]: error2 = abs(Sum2 - Sum) print error2  1.7763568394e-15  The error is largest in the first sum because the addition starts with the largest term and continues adding subsequent lower terms. Thus we lose accuracy because the different magnitudes between the accumulated sum and the new added terms. ### Cancellation error¶ It arises in the substraction of large numbers of similar value. For instance, $$\sqrt{x^2+\epsilon^2}-x$$ In [45]: import numpy as np x = 1000000. ep = 0.0001 a = np.sqrt(x**2 + ep) - x print a  1.16415321827e-10  In this case, cancellation error may be avoided by using an equivalent mathematical expression $$\sqrt{x^2+\epsilon^2}-x=\frac{(\sqrt{x^2+\epsilon^2}-x)(\sqrt{x^2+\epsilon^2}+x)}{\sqrt{x^2+\epsilon^2}+x}=\frac{\epsilon^2}{\sqrt{x^2+\epsilon^2}+x}\approx \frac{\epsilon^2}{\sqrt{x^2}+x}=\frac{\epsilon^2}{2x}$$ In [46]: b = ep / (np.sqrt(x**2 + ep) + x) print b  5e-11  The relative error with the first expression is In [47]: Er = abs((a - b)/b) print Er  1.32830643654  This example shows that equivalent mathematical expressions are not necessarily equivalent computational expressions. Ejercicio 5 (a) When solving the second order equation $$x^2+10^{8}x+1=0$$ using the well known formula $$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ one of the solutions is In [48]: a = 1.; b = 10.**8; c = 1. x1 = (-b + np.sqrt(b**2 - 4*a*c)) / (2*a) print x1  -7.45058059692e-09  But after substitution back into the polynomial we get that the residual In [49]: residual1 = a*x1**2 + b*x1 + c print residual1  0.254941940308  is relatively large. Find an equivalent mathematical expression to improve the result and write a code for it Exercise5a.py In [50]: %run Exercise5a.py print x2 print residual2  -1e-08 1.11022302463e-16  (b) When solving $$x^2-10^{8}x+1=0$$ with the formula $$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$ we obtain In [51]: a = 1.; b = -10.**8; c = 1 x3 = (- b - np.sqrt(b**2 - 4*a*c)) / (2*a) print x3  7.45058059692e-09  but is we substitute this value in the equation In [52]: residual3 = a*x3**2 + b*x3 + c print residual3  0.254941940308  that, again, it is an big relative error. (b) Find an equivalent mathematical expression to improve the result and write a code for it Exercise5b.py. In [53]: %run Exercise5b.py print x4 print residual4  1e-08 1.11022302463e-16  Example Given $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},$$ let's approximate the value of the number$e$, using$x = 1$, with a finite number of addition terms and see how the error evolves In [1]: import math suma = 0. x = 1. for n in range(100): suma += x**n / math.factorial(n) if n in np.array([5,10,15,20,40,60,80,100])-1: error = abs(np.exp(1.) - suma) print "Number of terms %i" % (n+1) print "Error %.16e" % error print ""  Number of terms 5 Error 9.9484951257120535e-03 Number of terms 10 Error 3.0288585284310443e-07 Number of terms 15 Error 8.1490370007486490e-13 Number of terms 20 Error 4.4408920985006262e-16 Number of terms 40 Error 4.4408920985006262e-16 Number of terms 60 Error 4.4408920985006262e-16 Number of terms 80 Error 4.4408920985006262e-16 Number of terms 100 Error 4.4408920985006262e-16  The error first decreases but then, it stagnates and does not decrease anymore. Why? If we would be using symbolic calculus, the error would decrease as iterations increase. But we are using finite arithmetic, and the error is limited by the spacing between representable numbers, which around number$e$is In [54]: Ea = np.spacing(np.exp(1.)) print Ea  4.4408920985e-16  Let's compute the number of terms we need to get$e^1$with the lowest possible error. In [55]: import math Ea = np.spacing(np.exp(1.)) n = 0 Sum = 1. itermax = 100 error = np.abs(np.exp(1.) - Sum) while (error > Ea and n < itermax): n += 1 Sum += 1. / math.factorial(n) error = np.abs(np.exp(1.) - Sum) print "Number of iterations %i" % n print "Error %.16e" % error print "Ea %.16e" % Ea  Number of iterations 17 Error 4.4408920985006262e-16 Ea 4.4408920985006262e-16  So after$17$iterations the truncation error is machine-neglegible. Exercise 6 Taking into account that $$\log (1+x) = \dfrac{x^1}{1}-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}-\cdots \quad \mathrm{para} \quad |x|\lt 1$$ write a code (Exercise6.py) to compute the number of terms we need to get$\log 1.5\$ with the lowest possible error.

In [2]:
%run Exercise6.py
print "Number of iterations     %i" % n
print "Error                    %.16e" % error
print "Spacing around ln(1+x)   %.16e" % Ea

Number of iterations     45
Error                    5.5511151231257827e-17
Spacing around ln(1+x)   5.5511151231257827e-17