2. Find Out the type of extreme points in
the following figures.
3. Now what about this?
• This one has no
minima or maxima.
• The minima or
maxima are defined in
a specific region
which, in other words,
we call as constrained
problems.
4. What is minimum or maximum?
▪ Minimum, in mathematical terms, for a function can be defined as
▪ f (X0 + h) > f (X0) V X0 ∈ domain
▪ Similarly, maximum can be defined, for a function, as the following
▪ f (X0 + h) < f (X0) V X0∈ domain
▪ Here “h” is a small value and tends to zero.
▪ These are local maxima and minima, because we are not basing them
for the whole domain.
5. ▪ But if we take the smallest value of all the local minimas, then the
value is called a Global Minima.
▪ If we take the largest of all the local maximas, then it is called as
Global Maxima.
6. Necessary Conditions
▪ We are going to develop necessary and sufficient conditions for an n-
variable function f(X) to have extrema.
▪ It is assumed that the first and second order partial derivatives of f(X)
are continuous for all X.
▪ Theorem 1: A necessary condition for X0 to be an extreme point of f(X)
is that ∇ f(X0) = 0
7. Sufficient conditions
▪ Theorem: A sufficient condition for a stationary point X0 to be an
extreme point is that the Hessian matrix H evaluated at X0 satisfy the
following conditions
i. H is positive definite, if X0 is a minimum point
ii. H is negative definite, if X0 is a maximum point
9. The Newton Raphson Method
▪ The necessary condition, sometimes ∇ f(X) = 0 , can be difficult to
solve numerically.
▪ So we use an iterative method called Newton Raphson method,
which helps solving simultaneous nonlinear equations.
▪ The method is mentioned in the next slides.
10. ▪ Consider the simultaneous equation f i(X) = 0, i = 1,2,3 … m
▪ ByTaylor’s expression at a given point Xk , we can write the whole
expression in the following form
f i(X) ~ f i(Xk) + ∇ fi(Xk)(X - Xk)
Changing the equation will give us the following expression
f i(Xk) + ∇ fi(Xk)(X - Xk) = 0
This can be written as Ak +Bk(X - Xk) = 0
OR X = Xk - B-1
k Ak (Bkis non singular)
11. ▪ The whole idea of this method is to start from an initial point and
then move on by using the above equation to find a point until it
converges.
▪ This process is done until 2 successive points are almost equal.
▪ For a single variable function this can be shown as
xk + 1 = xk -
𝒇(𝒙k)
𝒇′(𝒙k)
12. Example
Demonstrate Newton Raphson Method on the following
g(x) = (3x - 2)2(2x - 3)2
First find out f(x) = g’(x) = 72x3 - 234x2 + 241x – 78
Then follow the newton Raphson equation for a single variable that is
shown below.
xk + 1 = xk -
𝒇(𝒙k)
𝒇′(𝒙k)
14. ▪ It converges at 1.5
▪ Taking some other initial value we can converge at the other points.
Initial values 1 and 0.5 should give the other 2 extreme points.