Math, asked by sasikala986532, 22 days ago

Expand f(x, y) =x²+3y²-9x-9y+26 in Taylor's series upto second order about the point (2,2)​

Answers

Answered by kumarikajal111296
0

lets put (2,2) in the given cubic function :

2^2 + 2x2^2 - 9x2 - 9x2 + 26

or, (4 + 8) + 26 - 18

or, 12 + 8

or, 20. Answer.

Answered by Afreenakbar
0

The expansion of function in Taylor's series up to second order about the point (2,2) is the function it self that is x²+3y² -9x-9y+26.

Given that,

The function is f(x,y) = x²+3y²-9x-9y+26

We have to find the expansion of function in Taylor's series up to second order about the point (2,2).

We know that,

The Taylor's series expansion for a function up to second order is

F(x,y) = f(x₀,y₀)+ (\frac{df}{dx})of(x₀,y₀) (x-x₀) + (\frac{df}{dy})of(x₀,y₀) (y-y₀) + \frac{1}{2!}(\frac{d^{2}f }{dx^{2} })of(x₀,y₀) (x-x₀)² + \frac{1}{2!}(\frac{d^{2}f }{dy^{2} })of(x₀,y₀) (y-y₀)² + \frac{1}{2!}(\frac{d^{2}f }{dxdy })of(x₀,y₀) (x-x₀)(y-y₀)+......

So,

(x₀,y₀) = (2,2)

f(x₀,y₀) = 4+3(4)-9(2)-9(2)+26=6

\frac{df}{dx} = 2x-9

(\frac{df}{dx})of(x₀,y₀) = -5

\frac{df}{dy} = 6y-9

(\frac{df}{dy})of(x₀,y₀) = 3

\frac{d^{2}f }{dx^{2} } = 2

(\frac{d^{2}f }{dx^{2} })of(x₀,y₀) =2

\frac{d^{2}f }{dy^{2} } = 6

(\frac{d^{2}f }{dy^{2} })of(x₀,y₀) =6

(\frac{d^{2}f }{dxdy })of(x₀,y₀) =0

So,

F(x,y) = 6 - 5(x-2) + 3(y-2) + \frac{1}{2!}×2(x-2)² + \frac{1}{2!}×6(y-2)²+0

F(x,y) = 6 - 5x + 10 + 3y - 6 + x² - 4x + 4 + 3(y²-4y+4)

F(x,y) = -5x + 3y + x² - 4x + 3y² - 12y + (6+10-6+4+12)

F(x,y) = x²+3y² -9x-9y+26

Therefore, the expansion of function in Taylor's series up to second order about the point (2,2) is the function it self that is x²+3y² -9x-9y+26.

To learn more about function visit:

https://brainly.in/question/37271546

https://brainly.in/question/3089321

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