How to make statistics on a formula

[matteo86bo]

Hi everyone,
I've derived a formula which computes a certain quantity. This is the equation:

[tex]

f(x,t)=a(x,t)-\frac{1}{n}b(x)f(x,t)^{(1-n)/n}\frac{\partial f(x,t)}{\partial t}

[/tex]

I need to prove that [tex] f(x,t)\sim a(x,t)[/tex].

All I have is that [tex] 0.5 < n < 2 [/tex], [tex] b(x)[/tex] is a decreasing function of x (almost exponential) and [tex] f(x,t) = 1 +(c(x)-1)t[/tex] where [tex] c(x)[/tex] is another decreasing function of x.

I tried several things but I do not want to bias you answers.

Thanks in advance

 

[AndyW]

Hello forum member,

Thank you for sharing your equation and asking for help in proving that f(x,t)\sim a(x,t). I understand the importance of being able to prove the validity of our equations and formulas.

To begin with, let us define what it means for two quantities to be similar. In mathematics, two quantities are said to be similar if they have the same order of magnitude or if one can be approximated by the other. In other words, if the ratio of the two quantities approaches 1 as the values of the variables approach infinity.

In your equation, we can see that f(x,t) is a function of both x and t, while a(x,t) is only a function of x. This means that in order to prove that f(x,t)\sim a(x,t), we need to show that the ratio of the two quantities approaches 1 as both x and t approach infinity.

To begin with, let us consider the term \frac{1}{n}b(x)f(x,t)^{(1-n)/n}. We know that 0.5 < n < 2, which means that the exponent in the term is always negative. As b(x) is a decreasing function of x, this term will approach 0 as x approaches infinity. This means that we can ignore this term in our analysis.

Next, let us look at the term \frac{\partial f(x,t)}{\partial t}. From your equation, we can see that f(x,t) = 1 +(c(x)-1)t, where c(x) is a decreasing function of x. This means that as t approaches infinity, f(x,t) will approach 1. As a result, the derivative of f(x,t) with respect to t will approach 0 as t approaches infinity.

Now, let us look at the term a(x,t). As we mentioned earlier, this term is only a function of x and does not depend on t. As x approaches infinity, a(x,t) will also approach infinity. This means that the ratio of f(x,t) and a(x,t) will approach 0 as x and t approach infinity, which proves that f(x,t)\sim a(x,t).

In conclusion, based on the given conditions of 0.5 < n < 2, b(x) being a decreasing function of x