MA102: Multivariable Calculus Rupam Barman and Shreemayee Bora - - PowerPoint PPT Presentation

MA102: Multivariable Calculus

Rupam Barman and Shreemayee Bora Department of Mathematics IIT Guwahati

• R. Barman & S. Bora

MA-102 (2017)

Differentiability of f : U ⊂ Rn → Rm

Definition: Let U ⊂ Rn be open. Then f : U ⊂ Rn → Rm is differentiable at X0 ∈ U if there exists a linear map L : Rn → Rm such that lim

H→0

f(X0 + H) − f(X0) − L(H) H = 0.

• R. Barman & S. Bora

MA-102 (2017)

Differentiability of f : U ⊂ Rn → Rm

Definition: Let U ⊂ Rn be open. Then f : U ⊂ Rn → Rm is differentiable at X0 ∈ U if there exists a linear map L : Rn → Rm such that lim

H→0

f(X0 + H) − f(X0) − L(H) H = 0. The linear map L is called the derivative of f at X0 and is denoted by Df(X0), that is, L = Df(X0). Other notations: f ′(X0), df

dX (X0).

• R. Barman & S. Bora

MA-102 (2017)

Characterization of differentiability

Theorem: Consider f : Rn → Rm with f(X) = (f1(X), . . . , fm(X)), where fi : Rn → R. Then f is differentiable at X0 ∈ Rn ⇐ ⇒ fi is differentiable at X0 for i = 1, 2, . . . , m. Further Df(X0)(H) = (∇f1(X0) • H, . . . , ∇fm(X0) • H).

• R. Barman & S. Bora

MA-102 (2017)

Characterization of differentiability

Theorem: Consider f : Rn → Rm with f(X) = (f1(X), . . . , fm(X)), where fi : Rn → R. Then f is differentiable at X0 ∈ Rn ⇐ ⇒ fi is differentiable at X0 for i = 1, 2, . . . , m. Further Df(X0)(H) = (∇f1(X0) • H, . . . , ∇fm(X0) • H). The matrix of Df(X0) is called the Jacobian matrix of f at X0 and is denoted by Jf(X0).

• R. Barman & S. Bora

MA-102 (2017)

Jacobian matrix of f : Rn → Rm

Jf(X0) is an m × n matrix with (i, j)-th entry aij := ∂jfi(X0).

• R. Barman & S. Bora

MA-102 (2017)

Jacobian matrix of f : Rn → Rm

Jf(X0) is an m × n matrix with (i, j)-th entry aij := ∂jfi(X0). Jf(X0) =   ∇f1(X0) . . . ∇fm(X0)  

m×n

= ∂fi ∂xj (X0)

• m×n

.

• R. Barman & S. Bora

MA-102 (2017)

Jacobian matrix of f : Rn → Rm

Jf(X0) is an m × n matrix with (i, j)-th entry aij := ∂jfi(X0). Jf(X0) =   ∇f1(X0) . . . ∇fm(X0)  

m×n

= ∂fi ∂xj (X0)

• m×n

.

• f(x, y) = (f1(x, y), f2(x, y), f3(x, y)

Jf(a, b) =   ∂xf1(a, b) ∂yf1(a, b) ∂xf2(a, b) ∂yf2(a, b) ∂xf3(a, b) ∂yf3(a, b)  

• R. Barman & S. Bora

MA-102 (2017)

Jacobian matrix of f : Rn → Rm

Jf(X0) is an m × n matrix with (i, j)-th entry aij := ∂jfi(X0). Jf(X0) =   ∇f1(X0) . . . ∇fm(X0)  

m×n

= ∂fi ∂xj (X0)

• m×n

.

• f(x, y) = (f1(x, y), f2(x, y), f3(x, y)

Jf(a, b) =   ∂xf1(a, b) ∂yf1(a, b) ∂xf2(a, b) ∂yf2(a, b) ∂xf3(a, b) ∂yf3(a, b)  

• f(x, y, z) = (f1(x, y, z), f2(x, y, z))

Jf(a, b, c) =

• ∂xf1(a, b, c)

∂yf1(a, b, c) ∂zf1(a, b, c) ∂xf2(a, b, c) ∂yf2(a, b, c) ∂zf2(a, b, c)

• R. Barman & S. Bora

MA-102 (2017)

Examples

• If f(x, y) = (xy, exy, sin y) then

Jf(x, y) =   y x exy ex cos y  

• R. Barman & S. Bora

MA-102 (2017)

Examples

• If f(x, y) = (xy, exy, sin y) then

Jf(x, y) =   y x exy ex cos y  

• If f(x, y, z) = (x + y + z, xyz) then

Jf(x, y, z) =

• 1

1 1 yz xz xy

• R. Barman & S. Bora

MA-102 (2017)

Chain rule

Theorem-A: Let X : R → Rn be differentiable at t0 and f : Rn → R be differentiable at X0 := X(t0). Then f ◦ X is differentiable at t0 and d dt f(X(t))|t=t0 = ∇f(X0) • X ′(t0) =

n

• i=1

∂if(X0)dxi(t0) dt .

• R. Barman & S. Bora

MA-102 (2017)

Chain rule

Theorem-A: Let X : R → Rn be differentiable at t0 and f : Rn → R be differentiable at X0 := X(t0). Then f ◦ X is differentiable at t0 and d dt f(X(t))|t=t0 = ∇f(X0) • X ′(t0) =

n

• i=1

∂if(X0)dxi(t0) dt . Proof: Since f is differentiable at X0, therefore f(X0 + H) = f(X0) + ∇f(X0) • H + E(H)H · · · · · · (∗) and E(H) → 0 as H → 0. Put H := X(t) − X(t0) in (∗) to complete the proof.

• R. Barman & S. Bora

MA-102 (2017)

Chain rule

Theorem-A: Let X : R → Rn be differentiable at t0 and f : Rn → R be differentiable at X0 := X(t0). Then f ◦ X is differentiable at t0 and d dt f(X(t))|t=t0 = ∇f(X0) • X ′(t0) =

n

• i=1

∂if(X0)dxi(t0) dt . Proof: Since f is differentiable at X0, therefore f(X0 + H) = f(X0) + ∇f(X0) • H + E(H)H · · · · · · (∗) and E(H) → 0 as H → 0. Put H := X(t) − X(t0) in (∗) to complete the proof. Remark: d dt f(X(t0)) = ∇f(X0) • X ′(t0) =

n

• i=1

∂if(X0)dxi(t0) dt sometimes referred to as total derivative.

• R. Barman & S. Bora

MA-102 (2017)

Chain rule for partial derivatives

Theorem-B: If X : R2 → Rn, (u, v) → (x1(u, v), . . . , xn(u, v)) has partial derivatives at (a, b) and f : Rn → R is differentiable at Y := X(a, b) then F(u, v) := f(X(u, v)) has partial derivatives at (a, b) and ∂uF(a, b) = ∇f(Y) • ∂uX(a, b) =

n

• j=1

∂f(Y) ∂xj ∂xj(a, b) ∂u , ∂vF(a, b) = ∇f(Y) • ∂vX(a, b) =

n

• j=1

∂f(Y) ∂xj ∂xj(a, b) ∂v .

• R. Barman & S. Bora

MA-102 (2017)

Chain rule for partial derivatives

Case n=2: If x = x(u, v) and y = y(u, v) have first order partial derivatives at the point (u, v), and if z = f(x, y) is differentiable at the point (x(u, v), y(u, v)), then z = f(x(u, v), y(u, v)) has first order partial derivatives at (u, v) given by ∂z ∂u = ∂z ∂x ∂x ∂u + ∂z ∂y ∂y ∂u and ∂z ∂v = ∂z ∂x ∂x ∂v + ∂z ∂y ∂y ∂v . Example: Find ∂w/∂u and ∂w/∂v when w = x2 + xy and x = u2v, y = uv 2.

• R. Barman & S. Bora

MA-102 (2017)

Graph and level set

Let f : Rn → R. Then G(f) := {(X, f(X)) : X ∈ Rn} ⊂ Rn+1 is the graph of f. G(f) represents a hyper-surface in Rn+1.

• R. Barman & S. Bora

MA-102 (2017)

Graph and level set

Let f : Rn → R. Then G(f) := {(X, f(X)) : X ∈ Rn} ⊂ Rn+1 is the graph of f. G(f) represents a hyper-surface in Rn+1. The set S(f, α) := {X ∈ Rn : f(X) = α} is called a level set

• f f and represents a hyper-surface in Rn.
• R. Barman & S. Bora

MA-102 (2017)

Graph and level set

Let f : Rn → R. Then G(f) := {(X, f(X)) : X ∈ Rn} ⊂ Rn+1 is the graph of f. G(f) represents a hyper-surface in Rn+1. The set S(f, α) := {X ∈ Rn : f(X) = α} is called a level set

• f f and represents a hyper-surface in Rn.

(e) Graph of

f(x, y) :=

• x2 + y2

(f) Level curve

• x2 + y2 = k
• R. Barman & S. Bora

MA-102 (2017)

Level sets and gradients

Let f : Rn → R be differentiable at X0 ∈ Rn. Suppose that X0 is point on the hyper-surface f(X) = α.

• R. Barman & S. Bora

MA-102 (2017)

Level sets and gradients

Let f : Rn → R be differentiable at X0 ∈ Rn. Suppose that X0 is point on the hyper-surface f(X) = α. Let X : (−ε, ε) → Rn be a curve on the hyper-surface f(X) = α passing through X0, i.e, X(0) = X0 and f(X(t)) = α for t ∈ (−ε, ε).

• R. Barman & S. Bora

MA-102 (2017)

Level sets and gradients

Let f : Rn → R be differentiable at X0 ∈ Rn. Suppose that X0 is point on the hyper-surface f(X) = α. Let X : (−ε, ε) → Rn be a curve on the hyper-surface f(X) = α passing through X0, i.e, X(0) = X0 and f(X(t)) = α for t ∈ (−ε, ε). Suppose that X(t) is differentiable at 0. Then 0 = df(X(t)) dt |t=0 = ∇f(X0) • X ′(0) ⇒ ∇f(X0) ⊥ X ′(0)

• R. Barman & S. Bora

MA-102 (2017)

Level sets and gradients

Let f : Rn → R be differentiable at X0 ∈ Rn. Suppose that X0 is point on the hyper-surface f(X) = α. Let X : (−ε, ε) → Rn be a curve on the hyper-surface f(X) = α passing through X0, i.e, X(0) = X0 and f(X(t)) = α for t ∈ (−ε, ε). Suppose that X(t) is differentiable at 0. Then 0 = df(X(t)) dt |t=0 = ∇f(X0) • X ′(0) ⇒ ∇f(X0) ⊥ X ′(0) Since the line X0 + t X ′(0) is tangent to the curve X(t) at X0, ∇f(X0) is normal to the hyper-surface f(X) = α at X0.

• R. Barman & S. Bora

MA-102 (2017)

Tangent Plane and Normal Line to a Level Surface

f : E ⊆ R3 → R. X0 = (x0, y0, z0) is a point on the level surface S = {(x, y, z) ∈ R3 : f(x, y, z) = c} where c is a fixed real number. Let f be differentiable at X0. The tangent plane to S at X0 is the plane passing through X0 and normal to the gradient vector ∇f at X0. Its equation is fx(X0)(x − x0) + fy(X0)(y − y0) + fz(X0)(z − z0) = 0 . The normal line to S at X0 is the line perpendicular to the tangent plane and parallel to ∇f(X0), given by equations x = x0 + fx(X0)t, y = y0 + fy(X0)t, z = z0 + fz(X0)t for t ∈ R.

• R. Barman & S. Bora

MA-102 (2017)

Normal Line and Tangent Plane to a Surface

• R. Barman & S. Bora

MA-102 (2017)

Example

Find the tangent plane and normal line to the surface x2 + y2 + z2 = 3 at the point (−1, 1, 1). The given surface can be written as a level surface f(x, y, z) = 3 where f(x, y, z) = x2 + y2 + z2. fx(x, y, z) = 2x, fy(x, y, z) = 2y and fz(x, y, z) = 2z fx(−1, 1, 1) = −2, fy(−1, 1, 1) = 2, fz(−1, 1, 1) = 2. Equation of tangent plane: −2(x − (−1)) + 2(y − 1) + 2(z − 1) = 0

• r equivalently, −x + y + z = 3.

The normal line to the surface at (−1, 1, 1) is given by x = −1 − 2t, y = 1 + 2t, z = 1 + 2t for t ∈ R .

• R. Barman & S. Bora

MA-102 (2017)

Tangent Plane and Normal line for a Surface z = f(x, y)

The equation for a surface S : z = f(x, y) can be written in the form f(x, y) − z = 0. Hence, the surface z = f(x, y) is also the level surface F(x, y, z) = 0 of the function w = F(x, y, z) = f(x, y) − z. If X0(x0, y0, z0) is a point on the surface z = f(x, y) and fx and fy are continuous at (x0, y0) then the tangent plane to the surface S at X0 is the plane fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0) − (z − z0) = 0

• r equivalently

z = z0 + fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0).

• R. Barman & S. Bora

MA-102 (2017)

Tangent Plane and Normal line for a Surface z = f(x, y)

The normal line to the surface S at P0 is the line x = x0+fx(x0, y0)t, y = y0+fy(x0, y0)t, z = z0−t for t ∈ R. Example: Find equations for the tangent plane and normal line to the surface z = 9 − x2 − y2 at the point P0(1, 2, 4). Observe that fx(1, 2) = −2 and fy(1, 2) = −4. Therefore, the equation for the tangent plane is (−2)(x−1)+(−4)(y−2)−(z−4) = 0 = ⇒ 2x+4y+z = 14 . The equation for the normal line is x = 1 − 2t, y = 2 − 4t, z = 4 − t for t ∈ R.

• R. Barman & S. Bora

MA-102 (2017)

Partial Derivatives of Higher Order

Let f : R2 → R be defined by f(x, y) = x4y + y3x for (x, y) ∈ R2. Then ∂f ∂x = 4x3y + y3 and ∂f ∂y = x4 + 3y2x . Now, these first order partial derivatives are again functions from R2 to R. Again, we can do partial differentiation with respect to the variables x and y. For example, ∂ ∂x ∂f ∂x

• = ∂

∂x

• 4x3y + y3

= 12x2y = ∂2f ∂x2 ∂ ∂y ∂f ∂x

• = ∂

∂y

• 4x3y + y3

= 4x3 + 3y2 = ∂2f ∂y ∂x ∂ ∂x ∂f ∂y

• = ∂

∂x

• x4 + 3y2x
• = 4x3 + 3y2 =

∂2f ∂x ∂y ∂ ∂y ∂f ∂y

• = ∂

∂y

• x4 + 3y2x
• = 6xy = ∂2f

∂y2

• R. Barman & S. Bora

MA-102 (2017)

Notations for Partial Derivatives of Higher Order

Let f : S ⊆ Rn → R where S is an open set in Rn. ∂2f ∂xj∂xi = ∂ ∂xj ∂f ∂xi

• = fxixj = ∂ijf = ∂xixjf

∂3f ∂xk∂xj∂xi = ∂ ∂xk ∂ ∂xj ∂f ∂xi

• = ∂ijkf = fxixjxk

and so on. Note: In the notation fxixjxk, the variable xi which is close to f is first and then next variable xj and so on. Mixed Partial Derivatives: The partial derivatives ∂ij and ∂ji with i = j are called the mixed (partial) derivatives.

• R. Barman & S. Bora

MA-102 (2017)

∂ijf(X0) = ∂jif(X0) is possible

Let f(x, y) =

• xy(x2−y2)

x2+y2

if (x, y) = (0, 0) , if (x, y) = (0, 0) . Prove that

• f, fx, fy are continuous in R2.
• ∂12f and ∂21f exist at every point of R2, and are

continuous except at (0, 0).

• ∂12f(0, 0) = 1 and ∂21f(0, 0) = −1 (Second Order

Mixed Partial Derivatives are not equal at origin).

• If X0 = (x0, y0) = (0, 0) then ∂12f(X0) = ∂21f(X0).
• R. Barman & S. Bora

MA-102 (2017)

When mixed partial derivatives are equal?

Theorem Let f : S ⊆ Rn → R where S is an open set in Rn. Let X0 ∈ S. Assume that ∂ijf and ∂jif exist in a neighborhood

• f the point X0.

If the mixed partial derivatives ∂ijf and ∂jif are continuous at X0 then ∂ijf(X0) = ∂jif(X0) .

• R. Barman & S. Bora

MA-102 (2017)

Continuous partial derivatives

Let S be an open subset of Rn and f : S ⊂ Rn → R. Suppose that ∂if(X) exists for all X ∈ S and i = 1, . . . , n. Then each ∂if defines a function on S. If ∂if : S → R, X → ∂if(X) is continuous for i = 1, . . . , n then f is said to be continuously differentiable on S (in short, C1).

• R. Barman & S. Bora

MA-102 (2017)

Continuous partial derivatives

Let S be an open subset of Rn and f : S ⊂ Rn → R. Suppose that ∂if(X) exists for all X ∈ S and i = 1, . . . , n. Then each ∂if defines a function on S. If ∂if : S → R, X → ∂if(X) is continuous for i = 1, . . . , n then f is said to be continuously differentiable on S (in short, C1). Fact: f is C1 ⇐ ⇒ ∇f : S ⊂ Rn → Rn, X → ∇f(X) is continuous. Recall: f is C1 ⇒ f is differentiable ⇒ f is C1.

• R. Barman & S. Bora

MA-102 (2017)

Examples:

• Consider f : R2 → R given by f(x, y) = x2 + exy + y2.

Then f is C1.

• Consider f : R2 → R given by f(0, 0) = 0 and

f(x, y) := (x2 + y2) sin(1/(x2 + y2)) if (x, y) = (0, 0). Then f is differentiable but NOT C1.

• R. Barman & S. Bora

MA-102 (2017)

Continuous partial derivatives

Let S be an open subset of Rn. Let f : S ⊆ Rn → R be differentiable so that ∂if : S → R for i = 1, . . . , n. If the partial derivatives of ∂jf exist at X0 ∈ S for j = 1, . . . , n, that is, ∂i∂jf(X0) exists for i, j = 1, 2, . . . , n, then f is said to have second order partial derivatives at X0.

• R. Barman & S. Bora

MA-102 (2017)

Continuous partial derivatives

Let S be an open subset of Rn. Let f : S ⊆ Rn → R be differentiable so that ∂if : S → R for i = 1, . . . , n. If the partial derivatives of ∂jf exist at X0 ∈ S for j = 1, . . . , n, that is, ∂i∂jf(X0) exists for i, j = 1, 2, . . . , n, then f is said to have second order partial derivatives at X0. f is said to be C2 (twice continuously differentiable) if ∂i∂jf(X) exists for X ∈ S and ∂i∂jf : S → R is continuous for i, j = 1, 2, . . . , n.

• p-th order partial derivatives of f are defined similarly.
• R. Barman & S. Bora

MA-102 (2017)

Continuous partial derivatives

Fact: f : Rn → R is C2 ⇒ ∇f : Rn → Rn is differentiable. Hessian: Suppose that f(x, y) has second order partial derivatives at X0 = (a, b). Then the matrix Hf(X0) :=

• ∂x∂xf(X0)

∂y∂xf(X0) ∂x∂yf(X0) ∂y∂yf(X0)

• =
• fxx(X0)

fxy(X0) fyx(X0) fyy(X0)

• is called the Hessian of f at X0.
• R. Barman & S. Bora

MA-102 (2017)

Hessian

Fact: Suppose that f : Rn → R is C2 and X0 ∈ Rn. Then the Hessian Hf(X0) :=   ∂1∂1f(X0) · · · ∂n∂1f(X0) . . . · · · . . . ∂1∂nf(X0) · · · ∂n∂nf(X0)   is symmetric. Also Hf(X0) = J∇f(X0) = Jacobian of ∇f at X0.

• R. Barman & S. Bora

MA-102 (2017)

Hessian

Fact: Suppose that f : Rn → R is C2 and X0 ∈ Rn. Then the Hessian Hf(X0) :=   ∂1∂1f(X0) · · · ∂n∂1f(X0) . . . · · · . . . ∂1∂nf(X0) · · · ∂n∂nf(X0)   is symmetric. Also Hf(X0) = J∇f(X0) = Jacobian of ∇f at X0. Example: Consider f(x, y) = x2 − 2xy + 2y2. Then Hf(x, y) =

• fxx(x, y)

fxy(x, y) fyx(x, y) fyy(x, y)

• =
• 2

−2 −2 4

• .
• R. Barman & S. Bora

MA-102 (2017)

Mean Value Theorem

Let f : S ⊆ Rn → R where S is an open and convex set in

• Rn. If f is differentiable in S then for any two points X1 and

X2 in S, there exists a point X0 on the line segment L joining X1 and X2 such that f(X2) − f(X1) = ∇f(X0) • (X2 − X1). Proof: Let Φ : R → Rn be defined by Φ(t) = (1 − t)X1 + tX2 for t ∈ R . Consider the function g(t) = f(Φ(t)) for t ∈ [0, 1] and invoke chain rule and apply the mean value theorem of single variable calculus to g.

• R. Barman & S. Bora

MA-102 (2017)

Taylor’s Theorem

We can also write the MVT in the following way: Let f : S → R be differentiable. Let X0 ∈ S. Then there exists 0 < θ < 1 such that f(X0 + H) = f(X0) +

n

• i=1

∂if(X0 + θH)hi. Let f : S ⊆ Rn → R where S is an open set in Rn. Let X0 = (x1, · · · , xn) ∈ S and let H = (h1, · · · , hn) ∈ Rn be such that the line segment L joining X0 and X0 + H lies inside S. Suppose that f and its partial derivatives through order (n + 1) are continuous in S.

• R. Barman & S. Bora

MA-102 (2017)

Taylor’s Theorem

Then, for some 0 < θ < 1, we have f(X0 + H) = f(X0) +

n

• k=1

1 k! (h1∂1 + · · · + hn∂n)k f

• X0

+ 1 (n + 1)! (h1∂1 + · · · + hn∂n)n+1 f

• X0+θH

. Case n=2: Let f : R2 → R. Set the differential operators as: F ′ =

• h ∂

∂x + k ∂ ∂y

• ,

F ′′ =

• h ∂

∂x + k ∂ ∂y 2 = h2 ∂2 ∂x2 + 2hk ∂2 ∂x∂y + k2 ∂2 ∂y2 and F (n) =

• h ∂

∂x + k ∂ ∂y n (Expand it by binomial theorem).

• R. Barman & S. Bora

MA-102 (2017)

Taylor’s Formula:

Suppose f(x, y) and its partial derivatives through order (n + 1) are continuous throughout an open rectangular region R centered at (a, b) in R2. Then, throughout R, f(a+h, b+k) = f(a, b)+

n

• r=1

1 r!

• h ∂

∂x + k ∂ ∂y r f|(a, b)+Remainder term The Remainder term is 1 (n + 1)!

• h ∂

∂x + k ∂ ∂y n+1 f|(a+ch, b+ck) where ((a + ch), (b + ck)) for some c (0 < c < 1), is a point on the line segment joining (a, b) and (a + h, b + k).

• R. Barman & S. Bora

MA-102 (2017)

Taylor’s formula and Polynomial Approximations

Taylor’s formula provides polynomial approximations of f near the point (a, b). The first n derivative terms yield the n-the degree polynomial approximation to f and the last term (remainder term) gives the approximation error. Therefore, an upper bound of the remainder term is called as an upper bound for error term while approximating f by the n-th degree polynomial. If n = 1, we get linear approximation of f near (a, b). If n = 2, we get quadratic approximation of f near (a, b). If n = 3, we get cubic approximation of f near (a, b).

• R. Barman & S. Bora

MA-102 (2017)

Example

Let f(x, y) =

1 xy if xy = 0. Let (a, b) = (1, −1). Compute

the first two terms in the Taylor’s formula of f near (1, −1). fx = −x−2y−1, fy = −x−1y−2, fxx = 2x−3y−1, fxy = x−2y−2 and fyy = 2x−1y−3. 1 (1 + h)(−1 + k) = −1 + (h − k) +

• h2

(1 + θh)3(−1 + θk) + hk (1 + θh)2(−1 + θk)2 + k2 (1 + θh)(−1 + θk)3

• where 0 < θ < 1.
• R. Barman & S. Bora

MA-102 (2017)

Maxima/Minima

Let f : E ⊆ Rn → R. If f is continuous on E and E is a closed & bounded, then f attains its maximum and minimum value on E. How to find the (extremum) points at which f attains the maximum value or the minimum value on E? Before finding answer to this question, we formally define maximum and minimum of f.

• R. Barman & S. Bora

MA-102 (2017)

Local minimum/maximum and Global minimum/maximum

Let f : E ⊆ Rn → R. A point X ∗ ∈ E is said to be a point of relative/local minimum of f if there exists a r > 0 such that f(X ∗) ≤ f(X) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local minimum of f.

• R. Barman & S. Bora

MA-102 (2017)

Local minimum/maximum and Global minimum/maximum

Let f : E ⊆ Rn → R. A point X ∗ ∈ E is said to be a point of relative/local minimum of f if there exists a r > 0 such that f(X ∗) ≤ f(X) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local minimum of f. A point X ∗ ∈ E is said to be a point of relative/local maximum of f if there exists a r > 0 such that f(X) ≤ f(X ∗) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local maximum of f.

• R. Barman & S. Bora

MA-102 (2017)

Local minimum/maximum and Global minimum/maximum

Let f : E ⊆ Rn → R. A point X ∗ ∈ E is said to be a point of relative/local minimum of f if there exists a r > 0 such that f(X ∗) ≤ f(X) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local minimum of f. A point X ∗ ∈ E is said to be a point of relative/local maximum of f if there exists a r > 0 such that f(X) ≤ f(X ∗) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local maximum of f. A point X ∗ ∈ E is said to be a point of absolute/ global minimum of f if f(X ∗) ≤ f(X) for all X ∈ E.

• R. Barman & S. Bora

MA-102 (2017)

Local minimum/maximum and Global minimum/maximum

Let f : E ⊆ Rn → R. A point X ∗ ∈ E is said to be a point of relative/local minimum of f if there exists a r > 0 such that f(X ∗) ≤ f(X) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local minimum of f. A point X ∗ ∈ E is said to be a point of relative/local maximum of f if there exists a r > 0 such that f(X) ≤ f(X ∗) for all X ∈ E with X − X ∗ < r. In such case, the value f(X ∗) is called the relative/local maximum of f. A point X ∗ ∈ E is said to be a point of absolute/ global minimum of f if f(X ∗) ≤ f(X) for all X ∈ E. A point X ∗ ∈ E is said to be a point of absolute/ global maximum of f if f(X) ≤ f(X ∗) for all X ∈ E.

• R. Barman & S. Bora

MA-102 (2017)

Extremum Points and Extremum Values

A point X ∗ ∈ E is said to be a point of extremum of f if it is either (local/global) minimum point or maximum point of f. The function value f(X ∗) at the extremum point X ∗ is called an extremum value of f.

• R. Barman & S. Bora

MA-102 (2017)

Absolute/Global Extremum

f(x, y) = −x2 − y2 has an absolute maximum at (0, 0) in R2. f(x, y) = x2 + y2 has an absolute minimum at (0, 0) in R2.

• R. Barman & S. Bora

MA-102 (2017)

Relative/Local Extremum and Saddle Point

• R. Barman & S. Bora

MA-102 (2017)

Critical Points

Let f : E ⊆ Rn → R. A point X ∗ ∈ E is said to be a critical point of f if

• either

∂f ∂x1 (X ∗) = ∂f ∂x2 (X ∗) = · · · = ∂f ∂xn (X ∗) = 0 ,

• or at least one of the first order partial derivatives of f

does not exist. The function value f(X ∗) at the critical point X ∗ is called a critical value of f.

• R. Barman & S. Bora

MA-102 (2017)

Critical Points

Let f : E ⊆ Rn → R. A point X ∗ ∈ E is said to be a critical point of f if

• either

∂f ∂x1 (X ∗) = ∂f ∂x2 (X ∗) = · · · = ∂f ∂xn (X ∗) = 0 ,

• or at least one of the first order partial derivatives of f

does not exist. The function value f(X ∗) at the critical point X ∗ is called a critical value of f. Examples: The point (0, 0) is the critical point of the function f(x, y) = x2 + y2 and the critical value is 0 corresponding to this critical point. The point (0, 0) is a critical point of the function h(x, y) = x sin(1/x) + y if x = 0 and h(x, y) = y if x = 0.

• R. Barman & S. Bora

MA-102 (2017)

Saddle Points

Let X ∗ be a critical point of f. If every neighborhood N(X ∗) of the point X ∗ contains points at which f is strictly greater than f(X ∗) and also contains points at which f is strictly less than f(X ∗). That is, f attains neither relative maximum nor relative minimum at the critical point X ∗. Example: Let f(x, y) = y2 − x2 for (x, y) ∈ R2. Then (0, 0) is a saddle point of f.

• R. Barman & S. Bora

MA-102 (2017)

To find the extremum points of f, where to look for?

If f : E ⊆ Rn → R, then where to look in E for extremum values of f? The maxima and minima of f can occur only at

• boundary points of E,
• critical points of E
• interior point of E where all the first order partial derivatives of f are

zero,

• interior point of f where at least one of the first order partial

derivatives of f does not exist.

Example: In the closed rectangle R = {(x, y) ∈ R2 : −1 ≤ x ≤ 1 and − 1 ≤ y ≤ 1}, the function f(x, y) = x2 + y2 attains

• its minimum value at (0, 0),
• its maximum value at (±1, ±1).
• R. Barman & S. Bora

MA-102 (2017)

Necessary Condition for Extremum

Let f : E ⊆ Rn → R. If an interior point X ∗ of E is a point

• f relative/absolute extremum of f, and if the first order

partial derivatives of f at X ∗ exists then ∂1f(X ∗) = · · · = ∂nf(X ∗) = 0. That is, the gradient vector at X ∗ is the zero vector. Further, the directional derivative of f at X ∗ in all directions is zero, if f is differentiable at X ∗.

• R. Barman & S. Bora

MA-102 (2017)