Taylor's Theorem and proof

Taylor's Theorem and proof:

Theorem :

                    If f(x,y) is a function which possesses continuous partial derivatives of order n in any domain of a point (a,b) , and the domain is large enough to contain a point (a+h,b+k) with it, then there exists a positive number 0<θ<1, such that 

f(a+h,b+k) = f(a,b) + (h∂/∂x + k ∂/∂y)f(a,b)

            + (1/2!)(h∂/∂x + k∂/∂y)² f(a,b)

+ ....+{1/(n-1)!}(h∂/∂x + k∂/∂y)ⁿ⁻¹f(a,b)+Rₙ,

 where
 Rₙ = (1/n!)(h∂/∂x + k∂/∂y)ⁿf(a+θh,b+θk)

                                                       0<θ<1.

proof:

Let x = a+th , y = b+tk , where 0≤t≤1 is a parameter, and f(x,y) = f(a+th,b+tk) = φ(t)

Since the partial derivatives of f(x,y) of order n are continuous in the domain under consideration , φⁿ(t) is continuous in [0,1], and also 

φ'(t) = df/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt)

                     = h∂f/∂x + k∂f/∂y = (h∂/∂x+ k∂/∂y)f

φ"(t) = (h∂/∂x + k∂/∂y)² f

.

.

φⁿ(t) = (h∂/∂x + k∂/∂y)ⁿ f

therefore by Maclaurin's theorem

 φ(t) = φ(0) + tφ'(0) + (t²/ 2!) φ"(0)

             + ...+tⁿ⁻¹φⁿ⁻¹(0) / (n-1)! + tⁿ φⁿ (θt)/n!,

where 0<θ<1.

Now on putting t=1, we get

  φ(1) = φ(0) + φ'(0) + (1/2!)φ"(0) + ...

                        + {1 /(n-1)!}φⁿ⁻¹(0)+ (1/n!)φⁿ(0)

But φ(1) = f(a+h,b+k) , and φ(0) = f(a,b)

                  φ'(0) = (h∂/∂x + k∂/∂y) f(a,b)

                  φ"(0) = (h∂/∂x + k∂/∂y)²f(a,b)
                   :
                   :

                φⁿ(θ) = (h∂/∂x + k∂/∂y)ⁿ f(a+θh,b+θk)

Therefore f(a+h,b+k) = f(a,b)

                               +  (h∂/∂x + k∂/∂y)f(a,b)

                         + (1/2!)(h∂/∂x + k∂/∂y)²f(a,b) +...

            + {1/(n-1)!}(h∂/∂x + k∂/∂y)ⁿ⁻¹f(a,b)/+ Rₙ,

where

         Rₙ  = (1/n!)(h∂/∂x +k∂/∂y)ⁿf(a+θh,b+θk),                                                                                                                                                     0<θ<1
Rₙ is called the remainder after n terms , and the theorem ,Taylor's theorem with remainder or Taylor's expansion about the point (a,b) .

If we put a=b = 0 ; h= x , k= y , we get

f(x,y) = f(0,0) + (x∂/∂x + y∂/∂y)f(0,0)

           +( 1/2!) ( x∂/∂x + y ∂/∂y)² f(0,0) + ...

           +{ 1/(n-1)!} (x∂/∂x + y∂/∂y)ⁿ⁻¹ f(0,0) + Rₙ

where Rₙ =( 1/n!) (x∂/∂x +y∂/∂y)ⁿ f(θx,θy) ,                                                                  0<θ<1,

is the Maclaurin's theorem or Maclaurin's expansion .

It is easy to see that Taylor's theorem can also be put in the form :

f(a+h,b+k) = f(a,b) + df(a,b)

                                  + (1/2!)d² f(a,b) + ...

+ {1/(n-1)!} dⁿ⁻¹f(a,b)+( 1/n!)dⁿf(a+θh,+ b+θk)

The reasoning in the general case of several variables is precisely the same and so the theorem can be easily extend any number of variables.
Graphically it can be repesented by

See Also :

Maclaurin's Series Expansion

About Taylor : 

Brook Taylor FRS was an English mathematician who is best known for Taylor's theorem and the Taylor series continue reading