Use taylors formula to find quadratic and cubic approximations of f(x, y) =exsin y at origin. Estimate the error of the approximations if modx less than equal to 0.1 and mod y less than equal to 0.1
Solution:
Calculate First partial derivatives at "(0,0)" .
"f\nx\n=e^xsin(y)"
"f\nx\n(\n0\n,\n0\n)\n=\n0"
"f\ny\n=\ne^xcos(y)"
"f\ny\n(\n0\n,\n0\n)\n=\n1"
Calculate second derivatives at "(0,0)"
"f\nx\nx\n=\ne^xsin(y)"
"f\nx\n(\n0\n,\n0\n)\n=\n0"
"f\ny\n=\ne^xcos(y)"
"f\ny\n(\n0\n,\n0\n)\n=\n1"
"f\nx\ny\n=\n-e^xsin(y)"
"f\nx\ny\n(\n0\n,\n0\n)\n=\n0"
Therefore quadratic approximation of "f(x,y)" at origin using Taylor formula is:
"T\n(\nx\n,\ny\n)\n\u2248\n9\n+\n\\frac{1}{2}\n[\n\u2212\n9\nx^2\n\u2212\n9\ny^2\n]\n\u2248\n9\n\u2212\n\\frac{9}{2}\n[\nx^2\n+\ny^2\n]"
Next Calculating error at the origin using "|\nx\n|\n\u2264\n0.1" and "|\ny\n|\n\u2264\n0.1" as:
"E\n(\nx\n,\ny\n)\n\u2264\n\\frac{1}{2}\n\u00d7\n9\n[\n(\n0.1\n)^2\n+\n(\n0.1\n)\n^2\n]\n\u2264\n0.08685"
Thus, the Quadratic polynomial using the Taylor formula is "9\n\u2212\n\\frac{9}{2}\n[\nx\n^2\n+\ny\n^2\n]" and error in the approximation is less than 0.08685.
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