- #1
rugos
- 2
- 0
I need to find the maximum mean value of the function
F=[tex]\int[/tex][tex]\left|\alpha^{''}(f)\right|[/tex][tex]^{2}df[/tex]
I thought i could get an answer finding a constant K to normalize the function so as
K[tex]\int[/tex][tex]\left|\alpha^{''}(f)\right|[/tex][tex]^{2}df=1[/tex]
The boundaries could be f[tex]_{1}[/tex] and f[tex]_{2}[/tex]
as [tex]\alpha=1-\left|R\right|^{2}[/tex] i tried to find first the second derivative like this:
[tex]\frac{d\alpha}{df}=-2\left|R\right|\frac{d\left|R\right|}{df}[/tex]
[tex]\frac{d^{2}\alpha}{df}=-2\left[\frac{d\left|R\right|}{df}*\frac{d\left|R\right|}{df}+\left|R\right|\frac{d^{2}\left|R\right|}{df}\right][/tex]
[tex]\frac{d^{2}\alpha}{df}=-2\left[\left|\frac{d\left|R\right|}{df}\right|^{2}+\left|R\right|\frac{d^{2}\left|R\right|}{df}\right][/tex]
but know, replacing that into the integral seems harder.
so how can i find the normalization constant and solve the integral?
F=[tex]\int[/tex][tex]\left|\alpha^{''}(f)\right|[/tex][tex]^{2}df[/tex]
I thought i could get an answer finding a constant K to normalize the function so as
K[tex]\int[/tex][tex]\left|\alpha^{''}(f)\right|[/tex][tex]^{2}df=1[/tex]
The boundaries could be f[tex]_{1}[/tex] and f[tex]_{2}[/tex]
as [tex]\alpha=1-\left|R\right|^{2}[/tex] i tried to find first the second derivative like this:
[tex]\frac{d\alpha}{df}=-2\left|R\right|\frac{d\left|R\right|}{df}[/tex]
[tex]\frac{d^{2}\alpha}{df}=-2\left[\frac{d\left|R\right|}{df}*\frac{d\left|R\right|}{df}+\left|R\right|\frac{d^{2}\left|R\right|}{df}\right][/tex]
[tex]\frac{d^{2}\alpha}{df}=-2\left[\left|\frac{d\left|R\right|}{df}\right|^{2}+\left|R\right|\frac{d^{2}\left|R\right|}{df}\right][/tex]
but know, replacing that into the integral seems harder.
so how can i find the normalization constant and solve the integral?