Question

Expand x² y²+2x²y+3xy² in powers of (x+2)
and (y-1) using Taylor's series upto third degree forms​

Answer 1

Answer:

Functions of Two Variables

Last updatedDec 21, 2020

Classifying Critical Points

Book: Method of Lagrange Multipliers (Trench)

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Contributed by Paul Seeburger

Professor (Mathematics) at Monroe Community College

Earlier this semester, we saw how to approximate a function f(x,y) by a linear function, that is, by its tangent plane. The tangent plane equation just happens to be the 1st-degree Taylor Polynomial of f at (x,y), as the tangent line equation was the 1st-degree Taylor Polynomial of a function f(x).

Now we will see how to improve this approximation of f(x,y) using a quadratic function: the 2nd-degree Taylor polynomial for f at (x,y).

Review of Taylor Polynomials for a Function of One Variable

Do you remember Taylor Polynomials from Calculus II?

Definition: Taylor polynomials for a function of one variable, y=f(x)

If f has n derivatives at x=c, then the polynomial,

Pn(x)=f(c)+f′(c)(x−c)+f′′(c)2!(x−c)2+⋯+f(n)(c)n!(x−c)n(1)

is called the nth-degree Taylor Polynomial for f at c.

Now a function of one variable f(x) can be approximated for x near c using its 1st-degree Taylor Polynomial (i.e., using the equation of its tangent line at the point (c,f(c)). This 1st-degree Taylor Polynomial is also called the linear approximation of f(x) for x near c.

That is:

Answer 2

Answer:

The expansion is

$f(x,y) = 6 + [(x+2)(2xy^2 + 4xy + 3y^2) + (y-1)(2x^2y +2x^2+6xy )] \\\\+ [(x+2)^2(y^2+2y) + (y-1)^2(x^2 + 3x) + 2(x+2)(y-1)(2xy + 2x+3y)] \\\\+ [(x+2)^2(y-1)(y+1) + (x+2)(y-1)^2(x+1)]$

Step-by-step explanation:

The given equation is x² y² + 2x²y + 3xy².

We have to expand the equation in powers of (x+2) and (y-1) using Taylor's series up to third-degree forms.

​$f(x,y) = f(a,b) + \frac{1}{1!}[(x-a)\frac{\partial f}{\partial x} + (y-b)\frac{\partial f}{\partial y}] \\\\+ \frac{1}{2!}[(x-a)^2\frac{\partial ^2f}{\partial x^2} + (y-b)^2\frac{\partial ^2f}{\partial y^2} + 2(x-a)(y-b)\frac{\partial ^2f}{\partial x\partial y}] \\\\+ \frac{1}{3!}[(x-a)^3\frac{\partial ^3f}{\partial x^3} + (y-b)^3\frac{\partial ^3f}{\partial y^3} + 3(x-a)^2(y-b)\frac{\partial ^3f}{\partial x^2\partial y} + 3(x-a)(y-b)^2\frac{\partial ^3f}{\partial x\partial y^2} ]$The value of a & b are

The values can be found by equating to 0

$\implies x + 2 = 0\ \ \ and, \ \ \ y - 1 = 0\\\implies x = -2 \ \ \ and, \ \ \ y = 1\\\therefore a = -2 \ \ \ and, \ \ \ b = 1$

Substituting the value of a & b in Taylor's series expansion

$f(x,y) = f(-2,1) + \frac{1}{1!}[(x+2))\frac{\partial f}{\partial x} + (y-1)\frac{\partial f}{\partial y}] \\\\+ \frac{1}{2!}[(x+2)^2\frac{\partial ^2f}{\partial x^2} + (y-1)^2\frac{\partial ^2f}{\partial y^2} + 2(x+2)(y-1)\frac{\partial ^2f}{\partial x\partial y}] \\\\+ \frac{1}{3!}[(x+2)^3\frac{\partial ^3f}{\partial x^3} + (y-1)^3\frac{\partial ^3f}{\partial y^3} + 3(x+2)^2(y-1)\frac{\partial ^3f}{\partial x^2\partial y} + 3(x+2)(y-1)^2\frac{\partial ^3f}{\partial x\partial y^2} ]$

Now, we will find f(-2,1)

f(x,y) = x² y² + 2x²y + 3xy²    ----(i)

f(-2,1) = (-2)²(1)² + 2(-2)²1 + 3(-2)(1)²

f(-2,1) = 4 + 8 - 6

f(-2,1) = 6

Now, partially differentiating (i) w.r.t 'x', we get

$\frac{\partial f}{\partial x} = 2xy^2 + 4xy + 3y^2 \\\\\frac{\partial ^2f}{\partial x^2} = 2y^2 + 4y \\\\\frac{\partial ^3f}{\partial x^3} = 0\\\\\frac{\partial f}{\partial y} = 2x^2y + 2x^2 + 6xy \\\\\frac{\partial ^2f}{\partial y^2} = 2x^2 + 6x\\\\\frac{\partial ^3f}{\partial x^3} = 0\\\\\frac{\partial ^2f}{\partial x\partial y} = 4xy + 4x + 6y\\\\\frac{\partial ^3f}{\partial x^2\partial y} = 4y + 4\\\\\frac{\partial ^3f}{\partial x\partial y^2} = 4x+4$

Putting all the value in the above Taylor's series, we get

$f(x,y) = 6 + \frac{1}{1!}[(x+2)(2xy^2 + 4xy + 3y^2) + (y-1)(2x^2y +2x^2+6xy )] \\\\+ \frac{1}{2!}[(x+2)^2(2y^2+4y) + (y-1)^2(2x^2 + 6x) + 2(x+2)(y-1)(4xy + 4x+6y)] \\\\+ \frac{1}{3!}[(x+2)^3(0) + (y-1)^3(0) + 3(x+2)^2(y-1)(4y+4) + 3(x+2)(y-1)^2(4x+4)]\\\\f(x,y) = 6 + \frac{1}{1!}[(x+2)(2xy^2 + 4xy + 3y^2) + (y-1)(2x^2y +2x^2+6xy )] \\\\+ \frac{1}{2!}\times 2[(x+2)^2(y^2+2y) + (y-1)^2(x^2 + 3x) + 2(x+2)(y-1)(2xy + 2x+3y)] \\\\+ \frac{1}{3!}\times 12[(x+2)^2(y-1)(y+1) + (x+2)(y-1)^2(x+1)]$

Therefore, the expansion upto third degree is

$f(x,y) = 6 + [(x+2)(2xy^2 + 4xy + 3y^2) + (y-1)(2x^2y +2x^2+6xy )] \\\\+ [(x+2)^2(y^2+2y) + (y-1)^2(x^2 + 3x) + 2(x+2)(y-1)(2xy + 2x+3y)] \\\\+ [(x+2)^2(y-1)(y+1) + (x+2)(y-1)^2(x+1)]$

To learn more about Taylor's series, click on the link below:

https://brainly.in/question/32510207

To learn more about third-degree forms, click on the link below:

https://brainly.in/question/32080842

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