Consider the curve defined by y[sup]2[/sup]−x[sup]5[/sup]=0.[br]Write down and depict the first four Taylor polynomials approximating y at the point (1,1).[br]Explain why you cannot write down Taylor polynomials for the curve at (0,0).[br][br][math]f_0\left(x\right)=x^{\frac{5}{2}}\longrightarrow f_0\left(1\right)=1[/math][br][math]f_1\left(x\right)=\frac{5}{2}x^{\frac{3}{2}}\longrightarrow f_1\left(1\right)=\frac{5}{2}[/math][br][math]f_2\left(x\right)=\frac{15}{4}x^{\frac{1}{2}}\longrightarrow f_2\left(1\right)=\frac{15}{4}[/math][br][math]f_3\left(x\right)=\frac{15}{8}x^{-\frac{1}{2}}\longrightarrow f_3\left(1\right)=\frac{15}{8}[/math][br][math]f_4\left(x\right)=-\frac{15}{16}x^{-\frac{3}{2}}\longrightarrow f_4\left(1\right)=-\frac{15}{16}[/math]

Taylor polynomials cannot be written down for the curve at (0,0) because the Taylor polynomial will be undefined at n=3 and beyond.[br][math]f_1\left(x\right)=\frac{5}{2}x^{\frac{3}{2}}\longrightarrow f_1\left(0\right)=0[/math][br][math]f_2\left(x\right)=\frac{15}{4}x^{\frac{1}{2}}\longrightarrow f_2\left(0\right)=0[/math][br][math]f_3\left(x\right)=\frac{15}{8}x^{-\frac{1}{2}}\longrightarrow f_3\left(0\right)=\frac{15}{8\left(0\right)^{\frac{1}{2}}}=undefined[/math]