Question
A basic identity for quadratics shows $y=A^{-1} b$ as minimizing:$$P(y)=\frac{1}{2} y^{\mathrm{T}} A y-y^{\mathrm{T}} b=\frac{1}{2}\left(y-A^{-1} b\right)^{\mathrm{T}} A\left(y-A^{-1} b\right)-\frac{1}{2} b^{\mathrm{T}} A^{-1} b$$The minimum over a subspace of trial functions is at the y nearest to $A^{-1} b$. (That makes the first term on the right as small as possible; it is the key to convergence of $U$ to $u$.) If $A=I$ and $b=(1,0,0)$, which multiple of $V=(1,1,1)$ gives the smallest value of $P(y)=\frac{1}{2} y^{\top} y-y_{1} ?$
Step 1
Given that $A=I$ (the identity matrix) and $b=(1,0,0)$, we have $A^{-1}b = I^{-1}(1,0,0) = (1,0,0)$. Show more…
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Key Concepts
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(a) Maximize $ \sum_{i = 1}^{n} x_i y_i $ subject to the constraints $ \sum_{i = 1}^{n} x_i^2 = 1 $ and $ \sum_{i = 1}^{n} y_i^2 = 1 $. (b) Put $ x_i = \dfrac{a_i}{\sqrt{\sum a_j^2}} $ and $ y_i = \dfrac{b_i}{\sqrt{\sum b_j^2}} $ to show that $$ \sum a_i b_i \leqslant \sqrt{\sum a_j^2} \sqrt{\sum b_j^2} $$ for any numbers $ a_1, \ldots, a_n, b_1, \ldots, b_n $. This inequality is known as the Cauchy-Schwarz Inequality.
Partial Derivatives
Lagrange Multipliers
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