1 Nonlinear Equations

A linear equation is one related to a straight line, for example f ( x ) = m x + c  describes a straight line with slope m and the linear equation f ( x ) = 0 , involving such an f , is easily solved to give x = c m (as long as m 0 ). If a function f is not represented by a straight line in this way we say it is nonlinear .

The nonlinear equation f ( x ) = 0 may have just one solution, like in the linear case, or it may have no solutions at all, or it may have many solutions. For example if f ( x ) = x 2 9 then it is easy to see that there are two solutions x = 3 and x = 3 . The nonlinear equation f ( x ) = x 2 + 1 has no solutions at all (unless the application under consideration makes it appropriate to consider complex numbers).

Our aim in this Section is to approximate (real-valued) solutions of nonlinear equations of the form f ( x ) = 0 . The definitions of a root of an equation and a zero of a function have been gathered together in Key Point 13.

Key Point 13

If the value x is such that f ( x ) = 0 we say that

1.  x is a root of the equation f ( x ) = 0

2.  x is a zero of the function f .

Example 22

Find any (real valued) zeros of the following functions. (Give 3 decimal places if you are unable to give an exact numerical value.)

  1. f ( x ) = x 2 + x 20
  2. f ( x ) = x 2 7 x + 5
  3. f ( x ) = 2 x 3
  4. f ( x ) = e x + 1
  5. f ( x ) = sin ( x )
Solution
  1. This quadratic factorises easily into f ( x ) = ( x 4 ) ( x + 5 ) and so the two zeros of this f are x = 4 , x = 5 .
  2. The nonlinear equation x 2 7 x + 5 = 0 requires the quadratic formula and we find that the two zeros of this f are x = 7 ± 7 2 4 × 1 × 5 2 = 7 ± 29 2   which are equal to x = 0.807 and x = 6.193 , to 3 decimal places.
  3. Using the natural logarithm function we see that

    x ln ( 2 ) = ln ( 3 )

     from which it follows that x = ln ( 3 ) ln ( 2 ) = 1.585 , to 3 decimal places.

  4. This f has no zeros because e x + 1 is always positive.
  5. sin ( x ) has an infinite number of zeros at x = 0 , ± π , ± 2 π , ± 3 π , . To 3 decimal places these

     are x = 0.000 , ± 3.142 , ± 6.283 , ± 9.425 , .

Task!

Find any (real valued) zeros of the following functions.

  1. f ( x ) = x 2 + 2 x 15 ,
  2. f ( x ) = x 2 3 x + 3 ,
  3. f ( x ) = ln ( x ) 2 ,
  4. f ( x ) = cos ( x ) .

    For parts 1. to 3. give your answers to 3 decimal places if you cannot give an exact answer; your answers to part 4. may be left in terms of π .

  1. This quadratic factorises easily into f ( x ) = ( x 3 ) ( x + 5 ) and so the two zeros of this f are x = 3 , x = 5 .

  2. The equation x 2 3 x + 3 = 0 requires the quadratic formula and the two zeros of this f are

    x = 3 ± 3 2 4 × 1 × 3 2 = 3 ± 3 2

    which are complex values. This f has no real zeros.

  3. Solving ln ( x ) = 2 gives x = e 2 = 7.389 , to 3 decimal places.
  4. cos ( x ) has an infinite number of zeros at x = π 2 , π 2 ± π , π 2 ± 2 π , .

Many functions that crop up in engineering applications do not lend themselves to finding zeros directly as was achieved in the examples above. Instead we approximate zeros of functions, and this Section now goes on to describe some ways of doing this. Some of what follows will involve revision of material you have seen in HELM booklet  12 concerning Applications of Differentiation.