Throughout this module we use x to denote the positive square root of x; for example, 4 = 2.

Transcription

1 Throughout this module we use x to denote the positive square root of x; for example, 4 = 2.

2 You may often see (although not in FLAP) the notation sin 1 used in place of arcsin.

3 sinh and cosh are pronounced as shine, cosh ; while tanh is pronounced than (with a th as in the word thin ).

4 A function f1(x) is odd if f1(x) = f1( x) even if f1(x) = f1( x) for all values of x.

5 The same notation is used for positive powers of hyperbolic functions as for trigonometric functions. For example, sinh 2 0(x0) means sinh1(x) sinh1(x). We will refer back to the result given in Question T1 part (c) shortly.

6 The right-hand side of Equation 9 is actually a Taylor expansion about x = 0 (as are the other series considered in this subsection). An introduction to the Taylor expansion is given elsewhere in FLAP.

7 The complete series for tanh1(x) and tan1(x) involve Bernoulli numbers which are not covered in FLAP. Consequently only the first few terms of each series are given here. The Bernoulli numbers can be defined by the series x (e x 1) = 1 x 2 + B 1 x 2 2! B 2 x 4 4!+ B 3 x 6 6!+

8 The usual (and somewhat suspect) approach adopted by physicists is to stop calculating when terms in the series no longer contribute to the required accuracy. Accurate approximations for the error do exist, but they are beyond the scope of FLAP.

9 Your calculator may not have buttons for hyperbolic functions. If it doesn t, then use the basic definition of tanh1(x) in Equation 5 together with the e x or EXP button.

10 Some computer algebra packages use csch in place of cosech.

11 This is precisely the same problem that we have when defining the inverse of the function F(x) = x 2. We solve it by assigning the unique positive value to the square root in order to make x a function. Remember that a function must give rise to a unique value f1(x) for each value of x.

12 If y = arccosh1(x) then cosh1(y) = x.

13 See Figure 3. cosh (x) x Figure 34Graph of cosh1(x).

14 Alternatively, if we suppose that e y = x x 2 1, then we would have x x 2 1 1, which would imply (x 1) 2 x 2 1, which would give 2x + 1 1, or x 1. But we know that x 1, so this is a contradiction.

15 Don t forget that in general there are two solutions to equations such as cosh1(x) = 2.

16 It is not suggested that you should be able to quote all of these identities. Most people just learn the few that they use most frequently and look the rest up in tables of such identities. The really important point is to know what kind of identities do exist.

17 The symmetry relations show the oddness and evenness of the function.

18 The corresponding formulae for trigonometric functions are usually known as the addition formulae.

19 The corresponding formulae for trigonometric functions are usually known as the double-argument formulae.

20 The corresponding formulae for trigonometric functions are usually known as the half-argument formulae.

21 The identities can be derived quite simply from their trigonometric counterparts using Osborne s rules: change sin to sinh, cos to cosh, etc. in the trigonometric identities and then whenever sin sin occurs change the sign (i.e. in sin 2, tan 2, cosec 2 and cot 2 ).

22 To justify the first step, let y = 3x and use the chain rule to write d dy d sinh (3x) = sinh (y) dx dx dy = 3 cosh (y) = 3 cosh (3x)

23 This result is of great importance in integration, as discussed elsewhere in FLAP.

24 Here we use the standard notation that f (a) is df evaluated at x = a, dx df (x) i.e. f (a) = dx x = a

25 This is the example mentioned in the Introduction (Figure 1). The shape of the curve taken up by a cable suspended between two points is known as a catenary (from the Latin catena, a chain).

26 Since e 0 = 1 we have sinh1(0) = 0 and cosh1(0) = 1.4

27 tanh1(0) = 0 since (as we saw before) sinh1(0) = 0 and cosh1(0) = 1.4

28 sinh(1) = e e cosh(1) = e + e tanh(1) = sinh(1) cosh(1) and cosh 2 (1) sinh 2 (1) (1.543) 2 (1.175)

29 (a) Using Equation 11 x 2n+1 sinh (x) = n=0 (2n +1)! = x 1! + x 3 3! + x 5 5! + x 7 7! + x 9 9! + and Equation 12 ( 1) n x 2n+1 sin (x) = n=0 (2n +1)! = x 1! x 3 3! + x 5 5! x 7 7! + x 9 9!

30 we have sinh(x) + sin(x) = x 1! + x 3 3! + x 5 5! + x 7 7! + x 9 9! + + x 1! x 3 3! + x 5 5! x 7 7! + x 9 9! =2 x 1! + x 5 5! + x 9 9! + x 4n+1 =2 n=0 (4n+1)!

31 (b) Similarly from Equation 13 x 2n cosh (x) = n=0 (2n)! = 1+ x 2 and Equation 14 cos(x) = n=0 = 1 x 2 2! + x4 4! + x6 6! + x8 8! + ( 1) n x 2n (2n)! 2! + x4 4! x6 6! + x8 8!

32 we find cosh(x) cos(x) = 1+ x 2 2! + x4 4! + x6 6! + x8 8! + 1 x2 2! + x4 4! x6 6! + x8 8! 2! + x6 6! + x10 10! + =2 x2 x 4n+2 =2 n=0 (4n+2)! 4

33 x is approximately 2.0.4

34 The values of x are approximately 0, ±1.3 and ±1.8.4

35 (a) In this module we always use the positive square root, so that x 2 = x and there are no restrictions on the value of x. On the other hand, x only makes sense if x 0, in which case ( x ) 2 = x. (b) arccosh always gives rise to a positive value, and cosh1(x) is defined for all values of x, so that arccosh1[cosh1(x)] = 1x1 for all x. On the other hand, arccosh1(x) only makes sense if x 1, in which case cosh1[arccosh1(x)] = x.4

36 We find arccosh1(2) 1.3 and arccosh1(3) 1.8. These values are consistent with Figure 11.4 arccosh (x) x 1 2

37 Our first thought might be that we have just shown that arccosh1(2) 1.3 and therefore the solution is approximately 1.3. However, from Figure 3 we know that there are two solutions of the equation cosh1(x) = 2. These solutions are given by x = ± arccosh1(2) ±1.34

38 cosh (x) x Figure 34Graph of cosh1(x).

39 The left-hand side can be written as 1 2 (ex+y e x y ) The right-hand side seems complicated but reduces to the left-hand side after some algebraic manipulation: e x e x e y +e y ex +e x e y e y 2 2 = 1 4 [(ex+y + e x y e x+y e x y ) + (e x+y e x y +e x+y e x y )] = (ex+y e x y )