Math Problem Statement
quadratic taylor approximation of f(x,y)=2+x^2+ycosx at (0,0)
Solution
The quadratic Taylor approximation of a function around a point is given by:
+ \frac{1}{2} \left[ f_{xx}(x_0, y_0)(x - x_0)^2 + 2f_{xy}(x_0, y_0)(x - x_0)(y - y_0) + f_{yy}(x_0, y_0)(y - y_0)^2 \right]$$ ### Step-by-Step Solution: #### 1. **Function:** $$f(x, y) = 2 + x^2 + y \cos x$$ #### 2. **Evaluate $$f(0, 0)$$:** $$f(0, 0) = 2 + (0)^2 + 0 \cdot \cos(0) = 2$$ #### 3. **First Partial Derivatives:** - $$f_x = \frac{\partial}{\partial x} \left( 2 + x^2 + y \cos x \right) = 2x - y \sin x$$ - $$f_y = \frac{\partial}{\partial y} \left( 2 + x^2 + y \cos x \right) = \cos x$$ Evaluate at $$(0, 0)$$: - $$f_x(0, 0) = 2(0) - 0 \sin(0) = 0$$ - $$f_y(0, 0) = \cos(0) = 1$$ #### 4. **Second Partial Derivatives:** - $$f_{xx} = \frac{\partial}{\partial x} (2x - y \sin x) = 2 - y \cos x$$ - $$f_{yy} = \frac{\partial}{\partial y} (\cos x) = 0$$ - $$f_{xy} = \frac{\partial}{\partial y} (2x - y \sin x) = -\sin x$$ Evaluate at $$(0, 0)$$: - $$f_{xx}(0, 0) = 2 - 0 \cdot \cos(0) = 2$$ - $$f_{yy}(0, 0) = 0$$ - $$f_{xy}(0, 0) = -\sin(0) = 0$$ #### 5. **Quadratic Approximation:** Substitute everything into the formula: $$f(x, y) \approx f(0, 0) + f_x(0, 0) x + f_y(0, 0) y + \frac{1}{2} \left[ f_{xx}(0, 0) x^2 + 2f_{xy}(0, 0) xy + f_{yy}(0, 0) y^2 \right]$$ $$f(x, y) \approx 2 + 0 \cdot x + 1 \cdot y + \frac{1}{2} \left[ 2x^2 + 2(0)xy + 0 \cdot y^2 \right]$$ $$f(x, y) \approx 2 + y + x^2$$ ### Final Answer: The quadratic Taylor approximation is: $$f(x, y) \approx 2 + y + x^2$$ --- Do you want further details, or have any questions? ### Related Questions: 1. How do we compute higher-order Taylor approximations? 2. What are the applications of quadratic approximations in optimization? 3. Can we expand a function centered at points other than $$(0, 0)$$? 4. How does the approximation differ for functions with more complex cross-terms? 5. What happens when higher-order derivatives vanish? ### Tip: For multivariable Taylor approximations, always carefully compute cross-derivatives $$f_{xy}$$, as they play a crucial role in the approximation.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Multivariable Calculus
Taylor Series
Partial Derivatives
Formulas
Taylor approximation formula: f(x, y) ≈ f(x₀, y₀) + fₓ(x₀, y₀)(x - x₀) + fᵧ(x₀, y₀)(y - y₀) + (1/2) [fₓₓ(x₀, y₀)(x - x₀)² + 2fₓᵧ(x₀, y₀)(x - x₀)(y - y₀) + fᵧᵧ(x₀, y₀)(y - y₀)²]
Theorems
Taylor Series Expansion Theorem
Suitable Grade Level
College (Undergraduate, Calculus II or III)
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