What is Taylor polynomial for f(x,y)=(x^2+y^2)^0.5 at (-1,1)?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember what Taylor's polynomial is:

`T_2(-1,1) = f(-1,1) + (1/(1!))*((del f(-1,1))/(del x)*(x+1) + (del f(-1,1))/(del y)*(y-1)) + (1/(2!))*((del^2 f(-1,1))/(del x^2)*(x+1)^2 + 2(del^2 f(-1,1))/(del x*del y)*(x+1)*(y-1) + del^2 f(-1,1))/(del y^2)*(y-1)^2)^2`

You need to evaluate f(-1,1) plugging x=-1 and y=1 in the equation of f(x) such that:

`f(x) = sqrt((-1)^2 + 1^1) = sqrt2`

You need to differentiate the function with respect to x, considering y constant such that:

`(del f(-1,1))/(del x) = (2x)/(2sqrt(x^2 + y^2)) = x/(sqrt(x^2 + y^2)) |_(-1)^1 = -1/sqrt2`

You need to differentiate the function with respect to y, considering x constant such that:

`(del f(-1,1))/(del y) = (2y)/(2sqrt(x^2 + y^2)) = y/(sqrt(x^2 + y^2)) |_(-1)^1 = 1/sqrt2`

`(del^2 f(-1,1))/(del x^2) = (sqrt(x^2 + y^2) - (2x^2)/(2sqrt(x^2 + y^2)))/(x^2+y^2) = (y^2)/(sqrt(x^2 + y^2)^3) = sqrt2/4`

`(del^2 f(-1,1))/(del y^2) = sqrt2/4`

`` `(del^2 f(-1,1))/(del x*del y) = sqrt2/4`

Hence, evaluating the Taylor's polynomial yields:

`T_2(-1,1) =sqrt2 +(sqrt2/2)*((y-1)-(x+1)) + (sqrt2/8)*((x+1)^2 +`

` 2(x+1)(y-1) + (y-1)^2)^2 .`

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