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1. Find the gradient of the function: \[ f\left(x_{1}, x_{2}\right)=\frac{c_{1}...
Jun 14, 2024
Solution by Steps
step 1
To find the gradient of the function f(x1,x2)=c11+e−x1+c21+e−x2 f(x_1, x_2) = \frac{c_1}{1 + e^{-x_1}} + \frac{c_2}{1 + e^{-x_2}} , we need to compute the partial derivatives with respect to x1 x_1 and x2 x_2
step 2
The partial derivative of f f with respect to x1 x_1 is given by: ∂f∂x1=∂∂x1(c11+e−x1)=c1e−x1(1+e−x1)2 \frac{\partial f}{\partial x_1} = \frac{\partial}{\partial x_1} \left( \frac{c_1}{1 + e^{-x_1}} \right) = \frac{c_1 e^{-x_1}}{(1 + e^{-x_1})^2} Similarly, the partial derivative of f f with respect to x2 x_2 is: ∂f∂x2=∂∂x2(c21+e−x2)=c2e−x2(1+e−x2)2 \frac{\partial f}{\partial x_2} = \frac{\partial}{\partial x_2} \left( \frac{c_2}{1 + e^{-x_2}} \right) = \frac{c_2 e^{-x_2}}{(1 + e^{-x_2})^2}
step 3
Therefore, the gradient of the function f f is: ∇f=(∂f∂x1,∂f∂x2)=(c1e−x1(1+e−x1)2,c2e−x2(1+e−x2)2) \nabla f = \left( \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2} \right) = \left( \frac{c_1 e^{-x_1}}{(1 + e^{-x_1})^2}, \frac{c_2 e^{-x_2}}{(1 + e^{-x_2})^2} \right)
Answer
The gradient of the function f(x1,x2)=c11+e−x1+c21+e−x2 f(x_1, x_2) = \frac{c_1}{1 + e^{-x_1}} + \frac{c_2}{1 + e^{-x_2}} is (c1e−x1(1+e−x1)2,c2e−x2(1+e−x2)2) \left( \frac{c_1 e^{-x_1}}{(1 + e^{-x_1})^2}, \frac{c_2 e^{-x_2}}{(1 + e^{-x_2})^2} \right) .
Key Concept
Gradient of a function
Explanation
The gradient of a function is a vector that contains all of its partial derivatives. It points in the direction of the greatest rate of increase of the function.
Generated Graph
Solution by Steps
step 1
To find the gradient of the function f(x1,x2)=x12+x22 f(x_1, x_2) = x_1^2 + x_2^2 at xˉ=(1,1) \bar{x} = (1, 1) , we need to compute the partial derivatives with respect to x1 x_1 and x2 x_2
step 2
The partial derivative of f f with respect to x1 x_1 is given by: ∂f∂x1=∂∂x1(x12+x22)=2x1 \frac{\partial f}{\partial x_1} = \frac{\partial}{\partial x_1} (x_1^2 + x_2^2) = 2x_1 At x1=1 x_1 = 1 , this becomes: ∂f∂x1∣x1=1=2⋅1=2 \frac{\partial f}{\partial x_1} \bigg|_{x_1=1} = 2 \cdot 1 = 2
step 3
The partial derivative of f f with respect to x2 x_2 is given by: ∂f∂x2=∂∂x2(x12+x22)=2x2 \frac{\partial f}{\partial x_2} = \frac{\partial}{\partial x_2} (x_1^2 + x_2^2) = 2x_2 At x2=1 x_2 = 1 , this becomes: ∂f∂x2∣x2=1=2⋅1=2 \frac{\partial f}{\partial x_2} \bigg|_{x_2=1} = 2 \cdot 1 = 2
step 4
Therefore, the gradient of f f at xˉ=(1,1) \bar{x} = (1, 1) is: ∇f(1,1)=(∂f∂x1,∂f∂x2)∣(1,1)=(2,2) \nabla f(1, 1) = \left( \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2} \right) \bigg|_{(1, 1)} = (2, 2)
Answer
The gradient of the function f(x1,x2)=x12+x22 f(x_1, x_2) = x_1^2 + x_2^2 at xˉ=(1,1) \bar{x} = (1, 1) is (2,2) (2, 2) .
Key Concept
Gradient Calculation
Explanation
The gradient of a function at a given point is found by computing the partial derivatives with respect to each variable and evaluating them at that point.
Solution by Steps
step 1
To find the eigenvalues of matrix A=[3amp;00amp;0] A = \left[\begin{array}{cc} 3 & 0 \\ 0 & 0 \end{array}\right] , we solve the characteristic equation det⁡(A−λI)=0 \det(A - \lambda I) = 0
step 2
The characteristic equation for matrix A A is det⁡([3amp;00amp;0]−λ[1amp;00amp;1])=0 \det\left(\left[\begin{array}{cc} 3 & 0 \\ 0 & 0 \end{array}\right] - \lambda \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\right) = 0
step 3
This simplifies to det⁡([3−λamp;00amp;−λ])=(3−λ)(−λ)=0 \det\left(\left[\begin{array}{cc} 3 - \lambda & 0 \\ 0 & -\lambda \end{array}\right]\right) = (3 - \lambda)(-\lambda) = 0
step 4
Solving (3−λ)(−λ)=0 (3 - \lambda)(-\lambda) = 0 , we get the eigenvalues λ1=3 \lambda_1 = 3 and λ2=0 \lambda_2 = 0
step 5
To find the eigenvectors corresponding to λ1=3 \lambda_1 = 3 , we solve (A−3I)v=0 (A - 3I)\mathbf{v} = 0 . This gives [0amp;00amp;−3][v1v2]=[00] \left[\begin{array}{cc} 0 & 0 \\ 0 & -3 \end{array}\right]\left[\begin{array}{c} v_1 \\ v_2 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \end{array}\right] , leading to v1=1 v_1 = 1 and v2=0 v_2 = 0 . Thus, v1=[10] \mathbf{v}_1 = \left[\begin{array}{c} 1 \\ 0 \end{array}\right]
step 6
For λ2=0 \lambda_2 = 0 , we solve (A−0I)v=0 (A - 0I)\mathbf{v} = 0 . This gives [3amp;00amp;0][v1v2]=[00] \left[\begin{array}{cc} 3 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{c} v_1 \\ v_2 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \end{array}\right] , leading to v1=0 v_1 = 0 and v2=1 v_2 = 1 . Thus, v2=[01] \mathbf{v}_2 = \left[\begin{array}{c} 0 \\ 1 \end{array}\right]
step 7
To find the eigenvalues of matrix B=[3amp;−11amp;0] B = \left[\begin{array}{cc} 3 & -1 \\ 1 & 0 \end{array}\right] , we solve the characteristic equation det⁡(B−λI)=0 \det(B - \lambda I) = 0
step 8
The characteristic equation for matrix B B is det⁡([3amp;−11amp;0]−λ[1amp;00amp;1])=0 \det\left(\left[\begin{array}{cc} 3 & -1 \\ 1 & 0 \end{array}\right] - \lambda \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\right) = 0
step 9
This simplifies to det⁡([3−λamp;−11amp;−λ])=(3−λ)(−λ)−(−1)(1)=λ2−3λ+1=0 \det\left(\left[\begin{array}{cc} 3 - \lambda & -1 \\ 1 & -\lambda \end{array}\right]\right) = (3 - \lambda)(-\lambda) - (-1)(1) = \lambda^2 - 3\lambda + 1 = 0
step 10
Solving λ2−3λ+1=0 \lambda^2 - 3\lambda + 1 = 0 , we get the eigenvalues λ1=3+52 \lambda_1 = \frac{3 + \sqrt{5}}{2} and λ2=3−52 \lambda_2 = \frac{3 - \sqrt{5}}{2}
step 11
To find the eigenvectors corresponding to λ1=3+52 \lambda_1 = \frac{3 + \sqrt{5}}{2} , we solve (B−λ1I)v=0 (B - \lambda_1 I)\mathbf{v} = 0 . This gives [3−3+52amp;−11amp;−3+52][v1v2]=[00] \left[\begin{array}{cc} 3 - \frac{3 + \sqrt{5}}{2} & -1 \\ 1 & -\frac{3 + \sqrt{5}}{2} \end{array}\right]\left[\begin{array}{c} v_1 \\ v_2 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \end{array}\right] , leading to v1=[3+521] \mathbf{v}_1 = \left[\begin{array}{c} \frac{3 + \sqrt{5}}{2} \\ 1 \end{array}\right]
step 12
For λ2=3−52 \lambda_2 = \frac{3 - \sqrt{5}}{2} , we solve (B−λ2I)v=0 (B - \lambda_2 I)\mathbf{v} = 0 . This gives [3−3−52amp;−11amp;−3−52][v1v2]=[00] \left[\begin{array}{cc} 3 - \frac{3 - \sqrt{5}}{2} & -1 \\ 1 & -\frac{3 - \sqrt{5}}{2} \end{array}\right]\left[\begin{array}{c} v_1 \\ v_2 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \end{array}\right] , leading to v2=[3−521] \mathbf{v}_2 = \left[\begin{array}{c} \frac{3 - \sqrt{5}}{2} \\ 1 \end{array}\right]
Answer
The eigenvalues and eigenvectors of matrix A A are λ1=3 \lambda_1 = 3 , λ2=0 \lambda_2 = 0 , v1=[10] \mathbf{v}_1 = \left[\begin{array}{c} 1 \\ 0 \end{array}\right] , and v2=[01] \mathbf{v}_2 = \left[\begin{array}{c} 0 \\ 1 \end{array}\right] . The eigenvalues and eigenvectors of matrix B B are λ1=3+52 \lambda_1 = \frac{3 + \sqrt{5}}{2} , λ2=3−52 \lambda_2 = \frac{3 - \sqrt{5}}{2} , v1=[3+521] \mathbf{v}_1 = \left[\begin{array}{c} \frac{3 + \sqrt{5}}{2} \\ 1 \end{array}\right] , and v2=[3−521] \mathbf{v}_2 = \left[\begin{array}{c} \frac{3 - \sqrt{5}}{2} \\ 1 \end{array}\right] .
Key Concept
Eigenvalues and Eigenvectors
Explanation
Eigenvalues are the special set of scalars associated with a linear system of equations (i.e., a matrix equation) that provide insights into the system's properties. Eigenvectors are the corresponding vectors that, when multiplied by the matrix, result in a vector that is a scalar multiple of the original vector.
Solution by Steps
step 1
To find the inverse of the matrix AA, we start with the given matrix A=[1amp;11amp;2]A = \left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]
step 2
The formula for the inverse of a 2x2 matrix A=[aamp;bcamp;d]A = \left[\begin{array}{ll}a & b \\ c & d\end{array}\right] is A−1=1ad−bc[damp;−b−camp;a]A^{-1} = \frac{1}{ad - bc} \left[\begin{array}{ll}d & -b \\ -c & a\end{array}\right]
step 3
For the given matrix A=[1amp;11amp;2]A = \left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right], we have a=1a = 1, b=1b = 1, c=1c = 1, and d=2d = 2
step 4
Calculate the determinant of AA: ad−bc=(1)(2)−(1)(1)=2−1=1ad - bc = (1)(2) - (1)(1) = 2 - 1 = 1
step 5
Substitute the values into the inverse formula: A−1=11[2amp;−1−1amp;1]=[2amp;−1−1amp;1]A^{-1} = \frac{1}{1} \left[\begin{array}{ll}2 & -1 \\ -1 & 1\end{array}\right] = \left[\begin{array}{ll}2 & -1 \\ -1 & 1\end{array}\right]
Answer
A−1=[2amp;−1−1amp;1]A^{-1} = \left[\begin{array}{ll}2 & -1 \\ -1 & 1\end{array}\right]
Key Concept
Matrix Inversion
Explanation
The inverse of a 2x2 matrix is found using the formula A−1=1ad−bc[damp;−b−camp;a]A^{-1} = \frac{1}{ad - bc} \left[\begin{array}{ll}d & -b \\ -c & a\end{array}\right], where ad−bcad - bc is the determinant of the matrix.
Generated Graph
Solution by Steps
step 1
We need to find the global maximum of the function f(x1,x2)=x1−x12−x22f(x_1, x_2) = x_1 - x_1^2 - x_2^2 where x2∈[0,1]x_2 \in [0, 1]
step 2
To find the critical points, we first take the partial derivatives of ff with respect to x1x_1 and x2x_2 and set them to zero
step 3
The partial derivative with respect to x1x_1 is ∂f∂x1=1−2x1\frac{\partial f}{\partial x_1} = 1 - 2x_1. Setting this to zero gives 1−2x1=0⇒x1=121 - 2x_1 = 0 \Rightarrow x_1 = \frac{1}{2}
step 4
The partial derivative with respect to x2x_2 is ∂f∂x2=−2x2\frac{\partial f}{\partial x_2} = -2x_2. Setting this to zero gives −2x2=0⇒x2=0-2x_2 = 0 \Rightarrow x_2 = 0
step 5
We now evaluate the function ff at the critical point (x1,x2)=(12,0)(x_1, x_2) = \left(\frac{1}{2}, 0\right)
step 6
Substituting x1=12x_1 = \frac{1}{2} and x2=0x_2 = 0 into ff, we get f(12,0)=12−(12)2−02=12−14=14f\left(\frac{1}{2}, 0\right) = \frac{1}{2} - \left(\frac{1}{2}\right)^2 - 0^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}
step 7
We also need to check the boundary values of x2x_2. For x2=0x_2 = 0, we already have f(12,0)=14f\left(\frac{1}{2}, 0\right) = \frac{1}{4}. For x2=1x_2 = 1, we evaluate ff at x1=12x_1 = \frac{1}{2} and x2=1x_2 = 1
step 8
Substituting x1=12x_1 = \frac{1}{2} and x2=1x_2 = 1 into ff, we get f(12,1)=12−(12)2−12=12−14−1=−34f\left(\frac{1}{2}, 1\right) = \frac{1}{2} - \left(\frac{1}{2}\right)^2 - 1^2 = \frac{1}{2} - \frac{1}{4} - 1 = -\frac{3}{4}
step 9
Comparing the values, the maximum value of ff is 14\frac{1}{4} at (x1,x2)=(12,0)(x_1, x_2) = \left(\frac{1}{2}, 0\right)
Answer
The global maximum of f(x1,x2)=x1−x12−x22f(x_1, x_2) = x_1 - x_1^2 - x_2^2 for x2∈[0,1]x_2 \in [0, 1] is 14\frac{1}{4} at (x1,x2)=(12,0)(x_1, x_2) = \left(\frac{1}{2}, 0\right).
Key Concept
Global Maximum
Explanation
The global maximum of a function within a specified domain is found by evaluating the function at critical points and boundary values, then comparing these values to determine the highest one.
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