Question

expand f(x,y)=x^2y+siny+e^x in powers of (x-1) and(y-pi) through quadratic terms and find the remainder term​

Answer 1

Hey Friend,

I don't know the answer but I know the formula

* This is a long essay*

Some other examples are in the images

Taylor Formula -:

Consider the function f(x, y). Recall that we can approximate f(x, y) with a linear function

in x and y:

f(x, y) ⇡ f(a, b) + fx(a, b) (x  a) + fy(a, b) (y  b)

Notice that again this is just a linear polynomial in two-variables that does a good job of

approximating f near the point (x, y)=(a, b). It’s also exactly the equation of the tangent

plane to the surface f at the point (a, b).

Example 53

Find the linear approximation to f(x, y) = xey at the point (0, 0).

We need evaluate the function and its first partial derivatives at the point (0, 0). We have

f = xey f (0, 0) = 0

fx = ey fx (0, 0) = 1

fy = xey fy (0, 0) = 0

Then the linear approximation is

f(x, y) ⇡ 0 + (1 · (x  0) + 0 · (y  0)) = x = L(x, y)

Example 54

Use L(x, y) to approximate f(x, y) = key at the point (0.05, 0.05) and find the error in

the approximation.

L(0.05, 0.05) = 0.05 f(0.05, 0.05) = 0.05e0.05 = 0.052564

|E(0.05, 0.05)| = |L(0.05.0.05)  f(0.05, 0.05)| = 2.5 ⇥ 103

OK, that’s pretty good. But what if we need to do better? The linearization is the best

approximation by a linear polynomial of f(x, y) near the point (0, 0). It’s natural to ask if

we can get a better approximation if we use a quadratic polynomial.

It turns out that we can. The quadratic approximation of f(x, y) near the general point (a, b)

is given by

f(x, y) ⇡ f(a, b) + fx(a, b) (x  a) + fy(a, b) (y  b) +

1

2

⇥

fxx(a, b) (x  a)

2 + 2fxy(a, b) (x  a) (y  b) + fyy(a, b) (y  b)

2⇤

Notice that the first three terms in the approximation are just the linearization of f(x, y)

about the point (a, b). The additional terms are quadratic in x and y and involve the second

partial derivatives of f evaluated at the point (a, b).

Answer 2

Step-by-step explanation:

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