- #1
xman
- 93
- 0
Show that if we have N positive numbers
[tex] \left[ p_{i}\right]_{i=1}^{N} [/tex]
such that
[tex] \sum_{i} p_{i} =1 [/tex]
then for any N numbers
[tex] \left\{x_{i}\right\}_{i=1}^{N} [/tex]
we have the inequality
[tex] \prod_{i=1}^{N} x_{i}^{2 p_{i}} \leq \sum_{i=1}^{N} p_{i}x_{i}^{2} [/tex]
So I am thinking to show the inequality is true using Lagrange multipliers first take the set
[tex] W = \sum_{i} p_{i}x_{i}^{2} [/tex]
and we want to minimize above subject to constraint
[tex] S = \prod_{i} x_{i}^{2p_{i}} [/tex]
so we form the function
[tex] f^{\star} = f + \lambda g \Rightarrow f^{\star} =\sum_{i} p_{i}x_{i}^{2}+\lambda \left(S-\prod_{i} x_{i}^{2p_{i}}\right) [/tex]
So I think everything so far is ok...my question is how do you differentiate an infinite series and an infinite product. Also in this case is the Lagrange multiplier a single value [tex]\lambda[/tex] or is there one multiplier for each value of i , that is; do I need a [tex] \lambda_{i}[/tex] Any direction or input is greatly appreciated.
[tex] \left[ p_{i}\right]_{i=1}^{N} [/tex]
such that
[tex] \sum_{i} p_{i} =1 [/tex]
then for any N numbers
[tex] \left\{x_{i}\right\}_{i=1}^{N} [/tex]
we have the inequality
[tex] \prod_{i=1}^{N} x_{i}^{2 p_{i}} \leq \sum_{i=1}^{N} p_{i}x_{i}^{2} [/tex]
So I am thinking to show the inequality is true using Lagrange multipliers first take the set
[tex] W = \sum_{i} p_{i}x_{i}^{2} [/tex]
and we want to minimize above subject to constraint
[tex] S = \prod_{i} x_{i}^{2p_{i}} [/tex]
so we form the function
[tex] f^{\star} = f + \lambda g \Rightarrow f^{\star} =\sum_{i} p_{i}x_{i}^{2}+\lambda \left(S-\prod_{i} x_{i}^{2p_{i}}\right) [/tex]
So I think everything so far is ok...my question is how do you differentiate an infinite series and an infinite product. Also in this case is the Lagrange multiplier a single value [tex]\lambda[/tex] or is there one multiplier for each value of i , that is; do I need a [tex] \lambda_{i}[/tex] Any direction or input is greatly appreciated.