The undecided multiplier method of Lagrange

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The undecided multiplier method of Lagrange (Lagrange のみていじょうすうほう British: method of Lagrange multiplier) is a mathematics (analytics)-like method to optimize it under the restraint condition. For some variables, I think about a problem to demand the extreme value of one function with the distinction under the restraint condition to fix the value of some functions. I prepare for the fixed number (undecided multiplier, Lagrange multiplier) for each restraint condition and am the method that can untie the issue of restraint as a normal extreme value problem by considering linear combination to assume these a coefficient as a new function (I assume the undecided multiplier a new variable).

Table of contents

Theorem

The undecided multiplier method of Lagrange is described as the following theorem.

In the case of two dimensions, I it

Under restraint condition g (x, y) = 0, it is clogged up the problem for the point (a, b) where f (x, y) becomes the maximum

maximize ${\displaystyle f(x,y)}$ 

Subject to ${\displaystyle g(x,y)=0}$ 

I think about the issue of という. I assume Lagrange multiplier λ,

${\displaystyle F(x,y,\lambda) =f(x,y)-\lambda g(x,y)}$ 

Distantly. If neither ∂ g/ ∂ x nor ∂ g/ ∂ y is 0 at a point (a, b), α exists and lights it (a, b, α)

${\displaystyle {\frac {\partial F}{\partial x}}={\frac {\partial F}{\partial y}}={\frac {\partial F}{\partial \lambda}} = 0}$ 

But, it is managed [¹].

In the case of general hyperspace, I it

Point x = (x₁, of the n dimension space ... The m dimension vector function that cover evaluation function z = f (x) which assumes domain R with ,xn) a domain assumes the same domain a domain

${\displaystyle {\boldsymbol {G}}({\boldsymbol {x}})={\begin{pmatrix}g_{1}(x_{1},\dots ,x_{n})\\\vdots \\g_{m}(x_{1},\dots ,x_{n})\end{pmatrix}}={\boldsymbol {0}}\qquad (1)}$ 

The requirement to take extreme value in のもとで, point x in R is a gradient vector of f in the point

${\displaystyle \nabla f=^{t}{\begin{pmatrix}{\frac {\partial f}{\partial x_{1}}},\dots ,{\frac {\partial f}{\partial x_{n}}}\end{pmatrix}}}$ 

But, group λ = (λ ₁ of a thing namely the scalar to be included in the m dimension alignment subspace where a gradient vector of each g_i of the m unit forms on at the point ... Using ,λm),

${\displaystyle \nabla f=\sum _{i=1}^{m}\lambda _{i}\nabla g_{i}\qquad (2)}$ 

But, it is to be managed. If I transpose it and take ∇,

${\displaystyle f({\boldsymbol {x}})-\sum _{i=1}^{m}\lambda _{i}g_{i}({\boldsymbol {x}})}$ 

But, it is to get a stoppage point. But it is {∇ g₁, ... Independence primary as for, ∇ gm}, in other words

${\displaystyle \dim(\nabla g_{1},\dots ,\nabla g_{m})=m}$ 

You must appear. Let the expression of m book of expression (1) and the n book of expression (2) unite; and of x and λ if untie it about the unknown quantity of the unit (n +m), a candidate point to give extreme value of f is provided [²].

Interpretation

Geometric explanation

 

When figure 1 maximizes function f (x,y) for restraint condition g (x,y) = c.

 

It is a contour map of figure 1 figure 2. The red flight is a restraint condition${\displaystyle g(x,y)=c}$ I show を. The blue line${\displaystyle f(x,y)}$ の contour line. A point close against the contour line where a red line is blue is a solution.

Let's think about the case of simple のため two dimensions. For g (x,y) = c (c is the given fixed number here), it will be said that I maximize function f (x,y). I think about the graph which persuaded a value of f high. Orbit of f given for various values of d in f (x,y) = d is thought about. It is that the restraint condition fixes a value of g to c, and there is g in one orbit. Generally this orbit will cross a lot of f orbit when I walk along orbit of g = c because the orbit of g is different from f. Therefore let's pay attention to the orbit of various f =d to cross for a different value of d. The prices of f increase if they sail across the orbit if they go up the slope (they decrease if they withdraw).

But the value of f does not change only when I do only contact without, actually, orbit f =d which I am going to cross intersecting with orbit g =c (restraint condition) (f is right so under a restraint condition at the point becoming maximum). The direction of the gradient vector (I assume the partial differentiation by each variable an ingredient) with g becomes same as f here.

I introduce scalar λ which is not 0 here and think about f-λg. The condition of the upper point is equal to an incline of f-λg being 0 for a value with λ (as for λ the ratio of the incline of f and g).

Physical interpretation

When I solve a physical problem, the undecided multiplier of Lagrange often expresses a certain physical quantity not a simple means.

For example, the pressure is demanded as an undecided multiplier for a speed vector field to meet the restraint condition called consecutive expressions when I solve the Navier-Stokes' equation of the incompressible flow in hydrodynamics [³].

for irregularity

In the case of one, a restraint condition may make a simultaneous equation with a two-dimensional problem as follows:

${\displaystyle {\frac {\partial f}{\partial x}}+\lambda '{\frac {\partial f}{\partial y}}=0}$ 

${\displaystyle {\frac {\partial g}{\partial x}}+\lambda '{\frac {\partial g}{\partial y}}=0}$ 

${\displaystyle g(x,y)=0}$ 

But λ' of this case is different from λ of the original theorem.

Because all differential calculus df = 0 and direction where it is and dg = 0 and direction where it is are parallel, this irregularity version is led at a point becoming the extreme value.

Example

Let's think that entropy of the information theory finds discrete probability distribution becoming maximum. The entropy is a function to assume probability a variable then,

${\displaystyle f(p_{1},p_{2},\dots ,p_{n})=-\sum _{k=1}^{n}p_{k}\log _{2}p_{k}}$ 

となる. Of course the function that the total of these probability is equal to 1 and expresses a restraint condition

${\displaystyle g(p_{1},p_{2},\dots ,p_{n})=\sum _{k=1}^{n}p_{k}-1}$ 

となる. Let's find a point of the entropy maximum using Lagrange multiplier. The next condition is necessary for all i (I take n out of 1):

${\displaystyle {\frac {\partial }{\partial p_{i}}}(f+\lambda g)=0}$ 

Therefore

${\displaystyle {\frac {\partial }{\partial p_{i}}}\left(-\sum _{k=1}^{n}p_{k}\log _{2}p_{k}+\lambda (\sum _{k=1}^{n}p_{k}-1)\right)=0}$ 

The next expression is provided from the equation of these n units:

${\displaystyle -\left({\frac {1}{\ln 2}}+\log _{2}p_{i}\right)+\lambda = 0}$ 

This shows that all p_i is equal (because the variable is only λ).

Using restraint condition ∑ _k pk = 1,

${\displaystyle p_{i}={\frac {1}{n}}}$ 

But, I understand it. In other words, the same distribution of the equal probability is the entropy's greatest distribution all phenomena: In other words, I am that expectation of the information to be provided when random variable was really observed than a case of what kind of other probability distribution is big.

References

1. ^ Toshitsune Miyake (1992). Guide differential calculus integral calculus. It is p. 培風館 104. ISBN 4-563-00221-6. 
2. ^ Akira Shimizu ratio old "undecided how to teach decline in academic ability eras fourth ラグランジ coefficient law," it is 987-992 pages in "Japan Society of Mechanical Engineers" Vol. 112 1093rd, general corporate judicial person Japan Society of Mechanical Engineers, December, 2009.
3. ^ Joel H. Ferziger; Milovan Perić; For Toshio Kobayashi, Nobuyuki Taniguchi, Makoto Tsubokura reason "hydrodynamics シュプリンガー fair Lark Tokyo with the computer", 2,003 years, it is 195-197 pages. ISBN 4-431-70842-1. 

Allied item

• The issue of optimization
• Extreme value
• Generalization of the undecided multiplier method of カルーシュ クーン Tucker condition (KKT condition) - Lagrange
• Joseph = Louis Lagrange

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