Question

Given `\ \ f(x)=1/2x^4-x^3+x-3\ \`... Show the equation?

Answer 1

a) The function `f(x) = 1/2x^4-x^3+x-3` is continuous.

Now `f(2) = -1` and `f(2.5) =+3.40625`.

Since `f(2)` and `f(2.5)` have opposite signs, according to the intermediate value theorem, there is at least one root of `f(x)=0` between 2 and 2.5.

An interval like `[2,2.5]`, where a continuous function takes different signs at the two endpoints, is called a bracketing interval. In both the bisection and the false-position methods, the aim is to reduce this bracketing interval so that the root (At least one is guaranteed to be in such an interval) can be determined with higher and higher accuracy.

b) (i) In the bisection method, we successively halve the interval, until it becomes sufficiently small to meet our accuracy goal. The steps for this particular function will go as follows :

• Halve the interval at `(2+2.5)/2 = 2.25`
  `qquad f(2.25)~~0.6738 > 0`,
  and thus the root is between 2 and 2.25

• Bisect `[2,2.25]` at `2.125`
  `qquad f(2.125) ~~ -0.2753 < 0`,
  and thus the root is between 2.125 and 2.25

• Bisect `[2.125,2.25]` at `2.1875`
  `qquad f(2.1875) ~~ 0.1688 > 0`,
  and thus the root is between 2.125 and 2.1875

• Bisect `[2.125,2.1875]` at 2.15625
  `qquad f(2.15625) ~~ -0.0605 < 0`,
  and thus the root is between 2.15625 and 2.1875

At this stage, we have narrowed down the root to within the interval `[2.15625, 2.1875]`. Carrying on for a few more steps will give us the root to the accuracy we want.

Note that though theoretically we may make the bracketing interval as small as we want, practical considerations (like the precision of the machine you are using tom calculate) may limit this to within reasonable values.

(ii) The problem with the bisection method is that it pays no attention to the values `f(a)` and `f(b)` at the two endpoints of a bracketing interval `[a,b]` - just their sign! If, for example, `|f(a)|` is much smaller than `|f(b)|`, we would expect the root to be closer to `a` than `b` - but bisection simply ignores this fact and goes right to the middle of the interval. Sometimes a better choice is to approximate the curve of `f(x)` versus `x` by the straight line joining the points `(a,f(a))` and `(b,f(b))` and get the next endpoint by finding where this straight line intersects the `X` axis :

`c = (af(b)-bf(a))/(f(b)-f(a))`

In the false location method this replaces `(a+b)/2`. We once again start with `[a,b] = [2,2.5]`

• `c = (2f(2.5)-2.5f(2))/(f(2.5)-f(2)) ~~ 2.1135`
  `qquad f(c) ~~ -0.3509<0`
  The new bracketing interval is [2.1135,2.5]

• For this interval, `c ~~ 2.1496`
  `qquad f(c) ~~ -0.1076<0`
  The new bracketing interval is [2.1496,2.5]

• For this interval, `c ~~ 2.1603`
  `qquad f(c) ~~ -0.0316<0`
  The new bracketing interval is [2.1603,2.5]

Note that unlike in the bisection method, where the bracketing interval is bound to become smaller by a factor of 2 on each step, for the false location method it is possible, as in this case, `c` will approach the actual root from one side - so that the bracketing interval will never become very small. In this case we should stop our process either when the magnitude of `f(c)` becomes sufficiently small, or when successive values of `c` are very close to each other.