Part a) Approximations and Maclaurin Series of f(x) = cos(x)
i) Linear and Quadratic Approximations
We are given: f(x) = cos(x)
To find linear and quadratic approximations, we use the Taylor series around x = 0 (Maclaurin series).
Step 1: Derivatives of cos(x)
f(x) = cos(x)f'(x) = -sin(x)f''(x) = -cos(x)f'''(x) = sin(x)
At x = 0:
f(0) = 1f'(0) = 0f''(0) = -1
Linear Approximation:
L(x) = f(0) + f'(0)x = 1 + 0 = 1
Quadratic Approximation:
Q(x) = f(0) + f'(0)x + (f''(0)/2!)x²
= 1 + 0 − (1/2)x² = 1 − x²/2
Therefore:
- Linear Approximation:
f(x) ≈ 1 - Quadratic Approximation:
f(x) ≈ 1 − x²/2
ii) Maclaurin Series Expansion of cos(x)
Maclaurin Series:
f(x) = cos(x) = Σn=0∞ [ (−1)n x2n / (2n)! ]
First few terms:
cos(x) ≈ 1 − x²/2! + x⁴/4! − x⁶/6! + ...≈ 1 − x²/2 + x⁴/24 − x⁶/720 + ...
Graphical Insight:
For small values of x (near 0), the quadratic approximation 1 − x²/2 gives a good estimate of cos(x). Higher-order terms improve the accuracy over a larger interval.
Part b) Jacobian of f(x, y) = (e2xy, 2x² + 3y²)
Step 1: Define the vector function
f(x, y) = (f₁(x, y), f₂(x, y)) = (e2xy, 2x² + 3y²)
Jacobian Matrix Jf:
The Jacobian matrix is:
Jf = [[∂f₁/∂x, ∂f₁/∂y], [∂f₂/∂x, ∂f₂/∂y]]
Compute Partial Derivatives:
∂f₁/∂x = 2y * e2xy∂f₁/∂y = 2x * e2xy∂f₂/∂x = 4x∂f₂/∂y = 6y
Jacobian Matrix:
Jf(x, y) = [ [2y * e2xy, 2x * e2xy], [4x, 6y] ]
Evaluate at (x, y) = (2, 1):
e2*2*1 = e⁴ ≈ 54.598∂f₁/∂x = 2*1*54.598 = 109.196∂f₁/∂y = 2*2*54.598 = 218.392∂f₂/∂x = 4*2 = 8∂f₂/∂y = 6*1 = 6
Final Jacobian Matrix at (2, 1):
Jf(2,1) = [ [109.196, 218.392], [8, 6] ]
Conclusion
We derived the linear and quadratic approximations and the Maclaurin series expansion for cos(x), which are useful for estimating function values. We also computed the Jacobian matrix of a multivariable function and evaluated it at a specific point, which is essential in multivariable calculus and optimization. These tools are vital in economics, engineering, and applied mathematics for approximating and analyzing functions and models.