How do you find the quadratic taylor polynomial q(x,y) approximating #f(x,y)= e^(x) cos (5y)# about (0,0)?

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Answer 1 · 2015-09-25T05:02:44

Refer to explanation

The quadratic Taylor approximation is

#q(x,y)=f(0,0) + (df)/dx(0,0) x + (df)/dy(0,0) y +(1/(2!)) *[(d^2f)/(d^2x)(0,0) x^2 + 2 (d^2f)/(dxdy)(0,0) *xy + (d^2f)/(d^2y)(0,0) y^2]#

Hence we have that

#f(0,0)=e^0*cos(0)=1#

# (df)/dx(0,0)=e^0*cos0=1#

#(df)/dy(0,0) = -5*e^0*sin0=0#

#(d^2f)/(d^2x)(0,0)=e^0*cos0=1#

#(d^2f)/(dxdy)(0,0)=-5e^0sin0=0#

#(d^2f)/(d^2y)(0,0)=-25e^0cos0=-25#

Hence we have that

#q(x,y)=1+x+(1/(2!))*[x^2-25y^2]#