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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.

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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.

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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. $$ b(x_0 = 0) = \lim_{x \to 0} \frac{\ln f(x_1) - \ln f(x)}{\ln x_1 - \ln x} = \lim_{x \to 0} \frac{x f'(x)}{f(x)} = k $$

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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.

Confounded
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  • You'd definitely need $ e > 0$ when $d > 0 $. Otherwise, you can show that $ f'(0) \rightarrow \infty$ and $ g'(0) = 0 $, and $ f(0) = g(0)$, so there's some $x$ where $f(x) > g(x)$. Conversely, with $ e > 0$, fix any $b$, and you should be able to find a corresponding $a$. At the most, find when these 2 graphs are tangential (given that one is convex and another is concave). – Calvin Lin Jul 14 '21 at 22:20
  • @CalvinLin Thank you for your comment. Not sure if I followed all of your points, but I think we should assume that $x_0 > 0$, as otherwise we will probably encounter $\infty$ ' s. – Confounded Jul 14 '21 at 22:29
  • There isn't a need to assume $x_0 > 0$. $x^k$ is well defined for $ k > 0$ and $ x \in [ 0, \infty ) $. In particular, $ 0^k = 0 $, which is why $ f(0) = g(0)$. – Calvin Lin Jul 14 '21 at 22:57
  • @CalvinLin Well, if $x_0 =0$ then we can't use the method to calculate $b$ using the formula in my post as both the denominator and numerator go to $\infty$ then. Maybe taking a limit using L'Hospital's rule in this case can help. Of course, it still needs to be shown that such $a$ and $b$ guarantee that $g(x) \ge f(x)$ on $[x_0, x_1]$... – Confounded Jul 15 '21 at 07:31

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