Question

Expand x^2y+3y-2 in power of (x-1) & (y + 2)
using Taylor's theorem.​

Answer 1

ANSWER:

Solution:

According to the question:

Given Data: Expand $x^{2} y+3 y-2$ in power of (x-1) & (y + 2)

Step-by-step explanation:

Step 1:

Using Taylor's theorem.

Let $f(x, y)=x^{2} y+3 y-2$

So, f (1, -2) = -10.  

Step 2:

Differentiating:  

f(x) = 2xy ==> f(x)(1, -2) = -4  

$f(y)=x^{2}+3=\Rightarrow f(y)(1,-2)=4$

Step 3:

f(xx) = 2y ==> f(xx) (1, -2) = -4  

f(xy) = 2x ==> f(xy) (1, -2) = 2  

f(yy) = 0  

Step 4:

f(xxy) = 2  

(All other partial derivatives equal 0.)  

Step 5:

Result:

Hence,  

$\begin{array}{l}{f(x, y)=-10+[-4(x-1)+4(y+2)]+(1 / 2 !)\left[-4(x-1)^{\wedge} 2+2 * 2(x-1)(y+2)+0(y)\right.} \\ {\left.+2)^{2}\right]} \\ {+(1 / 3 !)\left[3 * 2(x-1)^{2}(y+2)+0\right]+0}\end{array}$

.........$=-10-4(\mathrm{x}-1)+4(\mathrm{y}+2)-2(\mathrm{x}-1)^{2}+2(\mathrm{x}-1)(\mathrm{y}+2)+(\mathrm{x}-1)^{2}(\mathrm{y}+2)$

Answer 2

Step-by-step explanation:

Solution:

According to the question:

Given Data: Expand x^{2} y+3 y-2x

2

y+3y−2 in power of (x-1) & (y + 2)

Step-by-step explanation:

Step 1:

Using Taylor's theorem.

Let f(x, y)=x^{2} y+3 y-2f(x,y)=x

2

y+3y−2

So, f (1, -2) = -10.

Step 2:

Differentiating:

f(x) = 2xy ==> f(x)(1, -2) = -4

f(y)=x^{2}+3=\Rightarrow f(y)(1,-2)=4f(y)=x

2

+3=⇒f(y)(1,−2)=4

Step 3:

f(xx) = 2y ==> f(xx) (1, -2) = -4

f(xy) = 2x ==> f(xy) (1, -2) = 2

f(yy) = 0

Step 4:

f(xxy) = 2

(All other partial derivatives equal 0.)

Step 5:

Result:

Hence,

\begin{gathered}\begin{array}{l}{f(x, y)=-10+[-4(x-1)+4(y+2)]+(1 / 2 !)[-4(x-1)^{\wedge} 2+2 * 2(x-1)(y+2)+0(y).} \\ {.+2)^{2}]} \\ {+(1 / 3 !)[3 * 2(x-1)^{2}(y+2)+0]+0}\end{array}\end{gathered}

f(x,y)=−10+[−4(x−1)+4(y+2)]+(1/2!)[−4(x−1)

∧

2+2∗2(x−1)(y+2)+0(y).

.+2)

2

]

+(1/3!)[3∗2(x−1)

2

(y+2)+0]+0