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But now there is a problem: we cannot solve this equation uniquely for f 1 (t) 1 (t) = + t or f 1 (t) = t Thus f 1 is not We do not know whether f a well de ned function Therefore f is not invertible and f 1 does not exist Math Note: There is a simple device which often enables us to obtain an inverse even in situations like Example 142 We change the domain of the function This idea is illustrated in the next example EXAMPLE 143 De ne f : {s : s 0} {t : t 0} by the formula f (s) = s 2 Find f 1 SOLUTION We attempt to solve (f f 1 )(t) = t Writing this out, we have f (f 1 (t)) = t or [f 1 (t)]2 = t

Motion Constraints In de ning a DRD output motion, the kinematic constraints must rst be decided These constraints would be determined both by the speci c requirements of the application and by general requirements for good dynamic behavior For example, the application itself may require only that several displacement constraints be satis ed

22:

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Sometimes you will want to remove leading and trailing spaces from a string This type of operation, called trimming, is often needed by command processors For example, a database might recognize the word print However, a user might enter this command with one or more leading or trailing spaces Any such spaces must be removed before the string can be recognized by the database Conversely, sometimes you will want to pad a string with spaces so that it meets some minimal length For example, if you are preparing formatted output, you might need to ensure that each line is of a certain length in order to maintain alignment Fortunately, C# includes methods that make these types of operations easy To trim a string, use one of these Trim( ) methods:

public string Trim( ) public string Trim(params char[ ] chrs) The first form removes leading and trailing whitespace from the invoking string The second form removes leading and trailing occurrences of the characters specified by chrs In both cases, the resulting string is returned You can pad a string by adding characters to either the left or the right side of the string To pad a string on the left, use one of the methods shown here: public string PadLeft(int len) public string PadLeft(int len, char ch) The first form adds spaces on the left as needed to the invoking string so that its total length equals len The second form adds the character specified by ch as needed to the invoking string so that its total length equals len In both cases, the resulting string is returned If len is less than the length of the invoking string, a copy of the invoking string is returned unaltered To pad a string to the right, use one of these methods: public string PadRight(int len) public string PadRight(int len, char ch) The first form adds spaces on the right as needed to the invoking string so that its total length equals len The second form adds the characters specified by ch as needed to the invoking string so that its total length equals len In both cases, the resulting string is returned If len is less than the length of the invoking string, a copy of the invoking string is returned unaltered The following program demonstrates trimming and padding:

132 10

This looks like the same situation we had in Example 142 But in fact things have improved Now we know that f 1 (t) must be + t, because f 1 must have range S = {s : s 0} Thus f 1 : {t : t 0} {s : s 0} is given by f 1 (t) = + t You Try It: The equation y = x 2 +3x does not describe the graph of an invertible function Find a way to restrict the domain so that it is invertible Now we consider the graph of the inverse function Suppose that f : S T is invertible and that (s, t) is a point on the graph of f Then t = f (s) hence s = f 1 (t) so that (t, s) is on the graph of f 1 The geometrical connection between the points (s, t) and (t, s) is exhibited in Fig 147: they are re ections of each other in the line y = x We have discovered the following important principle: The graph of f 1 is the re ection in the line y = x of the graph of f Refer to Fig 148

// Trimming and padding using System; class TrimPadDemo { static void Main() { string str = "test"; ConsoleWriteLine("Original string: " + str); // Pad on left with spaces str = strPadLeft(10); ConsoleWriteLine("|" + str + "|");

Part II:

[w] [4,1,1,1,1,4] [w] [1,1,2,2,1,1] [w] [09,1,22,22,1,09]

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