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Diffstat (limited to 'source/know/concept/lagrange-multiplier')
| -rw-r--r-- | source/know/concept/lagrange-multiplier/index.md | 58 |
1 files changed, 29 insertions, 29 deletions
diff --git a/source/know/concept/lagrange-multiplier/index.md b/source/know/concept/lagrange-multiplier/index.md index 9761e75..8ee1054 100644 --- a/source/know/concept/lagrange-multiplier/index.md +++ b/source/know/concept/lagrange-multiplier/index.md @@ -10,27 +10,27 @@ layout: "concept" The method of **Lagrange multipliers** or **undetermined multipliers** is a technique for optimizing (i.e. finding the extrema of) -a function $f(x, y, z)$, -subject to a given constraint $\phi(x, y, z) = C$, -where $C$ is a constant. +a function $$f(x, y, z)$$, +subject to a given constraint $$\phi(x, y, z) = C$$, +where $$C$$ is a constant. -If we ignore the constraint $\phi$, -optimizing $f$ simply comes down to finding stationary points: +If we ignore the constraint $$\phi$$, +optimizing $$f$$ simply comes down to finding stationary points: $$\begin{aligned} 0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z} \end{aligned}$$ This problem is easy: -$\dd{x}$, $\dd{y}$, and $\dd{z}$ are independent and arbitrary, +$$\dd{x}$$, $$\dd{y}$$, and $$\dd{z}$$ are independent and arbitrary, so all we need to do is find the roots of -the partial derivatives $f_x$, $f_y$ and $f_z$, -which we respectively call $x_0$, $y_0$ and $z_0$, -and then the extremum is simply $(x_0, y_0, z_0)$. +the partial derivatives $$f_x$$, $$f_y$$ and $$f_z$$, +which we respectively call $$x_0$$, $$y_0$$ and $$z_0$$, +and then the extremum is simply $$(x_0, y_0, z_0)$$. -But the constraint $\phi$, over which we have no control, -adds a relation between $\dd{x}$, $\dd{y}$, and $\dd{z}$, -so if two are known, the third is given by $\phi = C$. +But the constraint $$\phi$$, over which we have no control, +adds a relation between $$\dd{x}$$, $$\dd{y}$$, and $$\dd{z}$$, +so if two are known, the third is given by $$\phi = C$$. The problem is then a system of equations: $$\begin{aligned} @@ -42,15 +42,15 @@ $$\begin{aligned} Solving this directly would be a delicate balancing act of all the partial derivatives. -To help us solve this, we introduce a "dummy" parameter $\lambda$, +To help us solve this, we introduce a "dummy" parameter $$\lambda$$, the so-called **Lagrange multiplier**, -and contruct a new function $L$ given by: +and contruct a new function $$L$$ given by: $$\begin{aligned} L(x, y, z) = f(x, y, z) + \lambda \phi(x, y, z) \end{aligned}$$ -At the extremum, $\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$, +At the extremum, $$\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$$, so now the problem is a "single" equation again: $$\begin{aligned} @@ -58,13 +58,13 @@ $$\begin{aligned} = (f_x + \lambda \phi_x) \dd{x} + (f_y + \lambda \phi_y) \dd{y} + (f_z + \lambda \phi_z) \dd{z} \end{aligned}$$ -Assuming $\phi_z \neq 0$, we now choose $\lambda$ such that $f_z + \lambda \phi_z = 0$. +Assuming $$\phi_z \neq 0$$, we now choose $$\lambda$$ such that $$f_z + \lambda \phi_z = 0$$. This choice represents satisfying the constraint, -so now the remaining $\dd{x}$ and $\dd{y}$ are independent again, -and we simply have to find the roots of $f_x + \lambda \phi_x$ and $f_y + \lambda \phi_y$. +so now the remaining $$\dd{x}$$ and $$\dd{y}$$ are independent again, +and we simply have to find the roots of $$f_x + \lambda \phi_x$$ and $$f_y + \lambda \phi_y$$. -In effect, after introducing $\lambda$, -we have four unknowns $(x, y, z, \lambda)$, +In effect, after introducing $$\lambda$$, +we have four unknowns $$(x, y, z, \lambda)$$, but also four equations: $$\begin{aligned} @@ -73,19 +73,19 @@ $$\begin{aligned} \phi = C \end{aligned}$$ -We are only really interested in the first three unknowns $(x, y, z)$, -so $\lambda$ is sometimes called the **undetermined multiplier**, +We are only really interested in the first three unknowns $$(x, y, z)$$, +so $$\lambda$$ is sometimes called the **undetermined multiplier**, since it is just an algebraic helper whose value is irrelevant. This method generalizes nicely to multiple constraints or more variables: -suppose that we want to find the extrema of $f(x_1, ..., x_N)$ -subject to $M < N$ conditions: +suppose that we want to find the extrema of $$f(x_1, ..., x_N)$$ +subject to $$M < N$$ conditions: $$\begin{aligned} \phi_1(x_1, ..., x_N) = C_1 \qquad \cdots \qquad \phi_M(x_1, ..., x_N) = C_M \end{aligned}$$ -This once again turns into a delicate system of $M+1$ equations to solve: +This once again turns into a delicate system of $$M+1$$ equations to solve: $$\begin{aligned} 0 &= \dd{f} = f_{x_1} \dd{x_1} + ... + f_{x_N} \dd{x_N} @@ -97,15 +97,15 @@ $$\begin{aligned} 0 &= \dd{\phi_M} = \phi_{M, x_1} \dd{x_1} + ... + \phi_{M, x_N} \dd{x_N} \end{aligned}$$ -Then we introduce $M$ Lagrange multipliers $\lambda_1, ..., \lambda_M$ -and define $L(x_1, ..., x_N)$: +Then we introduce $$M$$ Lagrange multipliers $$\lambda_1, ..., \lambda_M$$ +and define $$L(x_1, ..., x_N)$$: $$\begin{aligned} L = f + \sum_{m = 1}^M \lambda_m \phi_m \end{aligned}$$ -As before, we set $\dd{L} = 0$ and choose the multipliers $\lambda_1, ..., \lambda_M$ -to eliminate $M$ of its $N$ terms: +As before, we set $$\dd{L} = 0$$ and choose the multipliers $$\lambda_1, ..., \lambda_M$$ +to eliminate $$M$$ of its $$N$$ terms: $$\begin{aligned} 0 = \dd{L} |