How can I prove that this 2D expression is always less than 1 for n≠β?

Karthiksrao
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Hello,

While analysing the asymptotic value of a ratio of a bessel and a hankel function, I reduced it to something of the form

[(1 + β/n)^ n * (1 + n/β)^ β] / 2^(n+β) ; n and β are integers and greater than 1

how do I show that the above expression is always less than 1, for n≠β. When n=β, the above expression becomes equal to 1.

Or relatedly, if I have to find the line of maximum for a 2D expression given above (for varying n and β), how do I go about ?

Thanks!
 
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Karthiksrao said:
Hello,

While analysing the asymptotic value of a ratio of a bessel and a hankel function, I reduced it to something of the form

[(1 + β/n)^ n * (1 + n/β)^ β] / 2^(n+β) ; n and β are integers and greater than 1

how do I show that the above expression is always less than 1, for n≠β. When n=β, the above expression becomes equal to 1.

Or relatedly, if I have to find the line of maximum for a 2D expression given above (for varying n and β), how do I go about ?

Thanks!

I've played around with this a little. This formula is symmetric in β,n with two variables but as mentioned, β = n is enough to give the maximum value.

Let x = 1 + n/β, y = n + β.
Thus ##\frac{y}{x} = β, \frac{y(x-1)}{x} = n## and x > 1, y ≥ 2.

The formula simplifies to
##[ \frac{1}{2} x (x-1)^{\frac{1}{x} - 1} ]^y##

Y is irrelevant here, it won't affect which x gives the maximum. Discarding y, the derivative of what remains has numerator
##-(x-1)^{\frac{1}{x}} ln(x-1)##

The exponential part is never 0, therefore x = 2 is the only stationary point. I hope this is not the best way to show this.
 
Your question is from calculus, or mathematical analysis if you prefer. Functional analysis is built on point set topology and is an abstractization of calculus.
 
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