1. Taylor Expansion Let U be an open subset of R 2 and f

1. Taylor Expansion

Let U be an open subset of R² and f :U → Rbe a function. Suppose that f ∈C^j(U).

Letp(x0, y0) be a point ofU.We chooseR >0 such thatB(p, R)⊆U.Letq(x, y) be a point inB(p, R).Then (x−x₀)²+ (y−y₀)² < R².Denoteh=x−x₀ andk=y−y₀ and define a function F : [−1,1]→Rby

F(t) =f(x0+th, y0+th).

Then F isj-times continuously differentiable. By mean value theorem, we have F(1) =

j−1

X

i=0

F⁽ⁱ⁾(0)

i! +F^(j)(c) j! . Let us compute F⁽ⁱ⁾(0) for 1≤i≤j−1 :

F⁰(0) =f_x(p)h+f_y(p)k

F⁰⁰(0) =f_xx(p)h²+ 2f_xy(p)hk+f_yy(p)k² ...

F^(l)(0) =

l

X

i=0

l i

f_xiy^l−j(p)hⁱk^l−i ...

We set H0(f)(p)(h, k) =f(p) and for each 1≤l≤j set H_l(f)(p)(h, k) =

l

X

i=0

l i

f_xiy^l−i(p)hⁱk^l−i.

Definition 1.1. A functionf :Rⁿ→Ris called a homogeneous function of degreed∈Z≥0

iff(λx) =λ^df(x) for λ >0.

ThenH_l(f)(p) is a homogeneous polynomial of degreelin (h, k).Hence forh²+k² < R², f(x₀+h, y₀+k) =

j−1

X

i=0

1

i!H_i(f)(p)(h, k) + 1

j!H_j(x₀+ch, y₀+ck)(h, k).

Definition 1.2. The polynomial

l

X

i=0

1

i!H_i(f)(p)(h, k) in (h, k) is called the l-th Taylor polynomial off atp.

This leads to the definition of analyticity of function in two variables.

Definition 1.3. Let f : U → R be a smooth function. We say that f is analytic at p if there existsR >0 such thatB(p, R)⊂U and that

f(x0+h, y0+k) =

∞

X

i=0

1

i!Hi(f)(x0, y0)(h, k)

for anyh²+k²< R².In this case, the above series representation of f is called the Taylor expansion forf on B(p, R).

1

2

Theorem 1.1. The Taylor expansion of a functionf :U →R at a pointp of U is unique if it exists.

Example 1.1. Find the Taylor expansion of f(x, y) =x²y at (1,1).

Solution:

By direct computation, f_x(1,1) = 2 and f_y(1,1) = 1 and f_xx(1,1) = 2 and f_xy(1,1) = 2 andf_yy(1,1) = 0 andf_x3(1,1) =f_y3(1,1) =f_xy3(1,1) = 0 andf_x2y(1,1) = 2.Leth=x−1 and k=y−1.We have

H₀(f)(1,1)(h, k) =f(1,1) = 1

H₁(f)(1,1)(h, k) =f_x(1,1)h+f_y(1,1)k= 2h+k

H2(f)(1,1)(h, k) =fxx(1,1)h²+ 2fxy(1,1)hk+fyy(1,1)k² = 2h²+ 4hk

H₃(f)(1,1)(h, k) =f_x³(1,1)h³+ 3f_x²_y(1,1)h²k+ 3f_xy²(1,1)hk²+f_y³(1,1)k³ = 6h²k H_n(f)(1,1)(h, k) = 0, n≥4.

Therefore the Taylor expansion off at (1,1) is given by f(x, y) =

∞

X

n=0

1

n!H_n(f)(1,1)(h, k)

= 1 + 1

1!(2h+k) + 1

2!(2h²+ 4hk) + 1 3!(6h²k)

= 1 + 1

1!(2(x−1) + (y−1)) + 1

2!(2(x−1)²+ 4(x−1)(y−1)) + 1

6!(6(x−1)²(y−1)).

Example 1.2. Find the 2nd Taylor polynomial of f(x, y) =p

1 + 4x²+y² atp(1,2).

Example 1.3. Compute the 3rd Taylor polynomial of the function f(x, y) =e^2xsin(3y) at (0,0) directly from all of the partial derivatives off at (0,0) up to the 3rd order and from the multiplication of the Taylor expansion ofe^2x and of the Taylor expansion of sin(3y).