Question 1- Find the secondβorder Taylor formula for π(π₯,π¦)=(9π₯+8π¦)^2 at x0=(0,0). Note that β2(0,π‘)=0 in this case. (Use symbolic notation and fractions where needed. Give your answer in the form of π(β1,β2)=f(l,m) where l=h1 and m=h2.) Answer: π(π,π)= 2-Find the secondβorder Taylor formula for π(π₯,π¦)=8sin(π₯π¦)+2cos(π₯π¦) at (0,0). (Use symbolic notation and fractions where needed. Give your answer in the form of π(β1,β2)=π(π,π) where l=h1 and m=h2.) Answer: π(π,π)=f(l,m)= ....... + β2(0,π‘) 3-Let π(π₯,π¦)=7sin(π₯π¦)β9(π₯^2)ln(π¦)+8.. Find the degree 2 polynomial, π, which best approximates π near the point (π2,1). (Use symbolic notation and fractions where needed.) Answer: π(π₯,π¦)=
1- Find the second‑order Taylor formula for 𝑓(𝑥,𝑦)=(9𝑥+8𝑦)^2 at x0=(0,0). Note that ℝ2(0,𝐡)=0 in this case.
(Use symbolic notation and fractions where needed. Give your answer in the form of 𝑓(ℎ1,ℎ2)=f(l,m) where l=h1 and m=h2.)
Answer: 𝑓(𝑙,𝑚)=
2-Find the second‑order Taylor formula for 𝑓(𝑥,𝑦)=8sin(𝑥𝑦)+2cos(𝑥𝑦) at (0,0).
(Use symbolic notation and fractions where needed. Give your answer in the form of 𝑓(ℎ1,ℎ2)=𝑓(𝑙,𝑚) where l=h1 and m=h2.)
Answer: 𝑓(𝑙,𝑚)=f(l,m)= ....... + ℝ2(0,𝐡)
3-Let 𝑔(𝑥,𝑦)=7sin(𝑥𝑦)−9(𝑥^2)ln(𝑦)+8.. Find the degree 2 polynomial, 𝑝, which best approximates 𝑔 near the point (𝜋2,1).
(Use symbolic notation and fractions where needed.)
Answer: 𝑝(𝑥,𝑦)=
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\begin{array}{l}f(x, y)=(9 x+8 y)^{2} \quad, x_{0} \rightarrow(0,0) \\ \text { 7. } f(x, y)=f\left(h_{1}, h_{2}\right)+f_{x}\left(h_{1}, h_{2}\right)\left(x-h_{1}\right)+f_{y}\left(h_{1}, h_{2}\right)\left(y-h_{2}\right)+ \\ \frac{f_{x x}\left(h_{1}, h_{2}\right)}{2}\left(x-h_{1}\right)^{2}+\frac{f_{y y}\left(h_{1}, h_{2}\right)\left(y-h_{2}\right)^{2}}{2}+ \\ f_{x y}\left(h_{1}, h_{1}\right)\left(x-h_{1}\right)\left(y-h_{2}\right) \\\end{array}\begin{array}{l} \rightarrow f^{\prime}(0,0)=0 \\ \rightarrow f_{x}=2(9 x+8 y) \cdot(9) \\ f_{x}(0,0)=0 \\ \rightarrow f_{y}=2(9 x+8 y) 8 \\ \Rightarrow f_{y}(0,0)=0 \\ \rightarrow f_{x x}=(9)(9)(2)=162 \\ \rightarrow f_{y y}=64(2)=168 \\ f_{x y}=8(2)(9)=144\end{array}A2, \begin{aligned} f(\lim )= & 0+0(x-0)+0(y-0)+ \\ & \frac{162}{2}(x-0)^{2}+\frac{128(y-0)^{2}}{2}+144(x-0)(y-0) \\ \Rightarrow f(1, m)= & 81 x^{2}+64 y^{2}+144 x y\end{aligned} PLZZ like if satisfied ...