Finding Max Mean Value of Function F: Integral Solution

• rugos
In summary, to find the maximum mean value of the function F, we can either solve it analytically by taking the average of F between the boundaries f1 and f2, or numerically using the MATLAB code provided. The code involves taking the second derivative of alpha with respect to the variable of interest and then integrating it with respect to f. However, it is unclear what the variable a'' represents, so further information is needed for a more accurate solution.

rugos

I need to find the maximum mean value of the function
F=$$\int$$$$\left|\alpha^{''}(f)\right|$$$$^{2}df$$

I thought i could get an answer finding a constant K to normalize the function so as

K$$\int$$$$\left|\alpha^{''}(f)\right|$$$$^{2}df=1$$

The boundaries could be f$$_{1}$$ and f$$_{2}$$

as $$\alpha=1-\left|R\right|^{2}$$ i tried to find first the second derivative like this:

$$\frac{d\alpha}{df}=-2\left|R\right|\frac{d\left|R\right|}{df}$$

$$\frac{d^{2}\alpha}{df}=-2\left[\frac{d\left|R\right|}{df}*\frac{d\left|R\right|}{df}+\left|R\right|\frac{d^{2}\left|R\right|}{df}\right]$$

$$\frac{d^{2}\alpha}{df}=-2\left[\left|\frac{d\left|R\right|}{df}\right|^{2}+\left|R\right|\frac{d^{2}\left|R\right|}{df}\right]$$

but know, replacing that into the integral seems harder.

so how can i find the normalization constant and solve the integral?

Esto basicamente de puede resolver de dos maneras dependiendo de lo que te sirva como resultado. De esta manera la solucion podra ser analitica es decir

El valor medio de una funcion F entre f1 y f2 esta dado por (1/(f2-f1)) por la integral entre (f1:f2) de F con respecto a f o la que sea la variable independiente de interes.

me queda la duda que significa a", si es o no la segunda derivada de alfa con respecto al tiempo o con respecto a que variable independiente es.

Sea como sea tendrias que derivar alfa dos veces con respecto a esa variable luego calcular el valor absoluto y elavear al cuadrado para luego integrala con respecto a f. Eso te da com resultado F. Posteriormente tendrias que volver a calcular la integral de F con respecto a f y multiplica por (1/(f2-f1)).

tambien se puede resolver numericamente, en pseudocodigo de MATLAB seria

Fmean_max= max((1/(f2-f1))*trapz(trapz(abs(alfa")^2)));

donde

alfa" es la segunda derivada de alfa con respecto a la variable

si me das informacion acerca de alfa y de la variable con que se debe derivar

no se si aun te sirve esta informacion disculpa si no te escribi antes

Saludos

juan

o sea

Fmean_max= max((1/(f2-f1))*trapz(trapz(abs(diff(diff(alfa)))^2));

1. What is the meaning of "Max Mean Value" in this context?

In this context, "Max Mean Value" refers to the highest average value that a function takes on over a given interval. It is the maximum value of the average of the function over the entire interval.

2. How is the Max Mean Value of a function calculated?

The Max Mean Value of a function is calculated by finding the integral of the function over the given interval and dividing it by the length of the interval. This gives the average value of the function over the interval. The maximum of these average values is then considered the Max Mean Value.

3. Why is it important to find the Max Mean Value of a function?

Finding the Max Mean Value of a function can provide valuable information about the behavior of the function over a given interval. It can help identify the most extreme values that the function takes on and give insight into the overall shape and trends of the function.

4. Can the Max Mean Value of a function be negative?

Yes, the Max Mean Value of a function can be negative if the function takes on negative values over the given interval. This indicates that the function is decreasing over the interval and the average of its values is negative.

5. How is the Max Mean Value of a function used in real-life applications?

The Max Mean Value of a function is used in various fields of science, such as physics, economics, and engineering. It can help in optimization problems, such as maximizing profits or minimizing costs, by identifying the most favorable values of a function. It can also be used to analyze data and make predictions based on average values.