Find $a$ and $b$ such that $ax^b \ge cx + dx^k$ on $ x \in [x_0, x_1]$

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I am looking for an upper bound $g(x)$ of $f(x) = cx + dx^k$ with $c>0$, $d>0$, $1>k>0$ on $x \in [x_0, x_1]$ with $x \ge 0$ in a form $g(x) = e + ax^b$. If $e=0$, then

$$ f(x) = dx^k(1 + \frac{c}{d} x^{1-k}) < dx^k(1 + \frac{c}{d} x_1^{1-k}) $$

so $a = d + cx_1^{1-k}$ and $b = k$. But I expect that there must be a better / tighter bound then this, but I am not sure how to move forward to find it.

Add 1

It seems to me that, if $e=0$, then "the closest" $g(x)$ is the one that is equal to $f(x)$ at the boundaries of $[x_0, x1]$. In that case, for $x_0>0$, we can solve to find

$$ b = \frac{\ln f(x_1) - \ln f(x_0)}{\ln x_1 - \ln x_0} $$

and $$ a = \frac{f(x_1)}{x_1^b} $$ But I am not sure if these coefficients will guarantee that $g(x) \ge f(x)$ on the interval.

Add 2

It seems that the limit of the expression for $b$ above as $x_0 \to 0$ is equal to $k$ if $0<k<1$, i.e. $$ b(x_0 = 0) = \lim_{x \to 0} \frac{\ln f(x_1) - \ln f(x)}{\ln x_1 - \ln x} = \lim_{x \to 0} \frac{x f'(x)}{f(x)} = k $$

Add 3

I thought that maybe one can define a measure of "close-nest" between two functions and minimize this to find the parameters. For example, defining a relative error as

$$ \delta(x,a,b) = \frac{cx^{k_1} + dx^{k_2}}{ax^b} - 1 = \frac{c}{a}x^{k_1-b} + \frac{d}{a}x^{k_2-b} - 1 $$

and then minimize the total absolute relative error

$$ \Delta(a,b) = \int_{x_0}^{x_1} |\delta(x)| dx $$

with respect to $a$ and $b$, but not clear to me how to find the integral of an absolute value of a function, since there is no expression for the roots of the integrand and so can't split it into positive/negative intervals to integrate.