How to get a linear plot of a power function?

Question

Imagine I have a function of x as follows:

$$y=f(x) = ax^2 + bx + c + \frac{d}{x}$$

And I am trying to plot this on a graph with y as ordinate and $x^{n}$ as abscissa. Now what value of n would give me the plot a straight line"

1. $n = 2$
2. $n = 1$
3. $n = -1$

I strongly feel $n = 2$ would give me a straight line when plotted as $y$ versus $x^2$. Is my approach correct?

Answer 1

I will assume that by $x^n$ as abscissa you mean that on the $y$-axis a distance of one unit corresponds to an increase of the $y$ value by one whereas on the $x$-axis an increase of one unit corresponds to an increase of the $x$ value that depends on how far from the origin you are, in such a way that at distance $k$ units form the origin the $x$-value is $k^n$.

In that case the best you can do to get something looking a bit like a straight line at least far away from $0$ is choose $n$ to be $\frac{1}{2}$. In that case at distance $k$ from the origin in the $x$ direction your $x$ value will be $\sqrt{k}$ and thus the value of the function will be $y=ak+b\sqrt{k}+c+\frac{d}{\sqrt{k}}$ which while not actually a straight line will be pretty close to it for large $k$.

Answer 2

To solve this question for a general $n$, define a new variable $u = x^n$. Then $x = u^{1/n}$. Now plug this into $f(x)$ to obtain

$y = au^{2/n} + bu^{1/n} + c + du^{-1/n}$

You can then see that there is no value of $n$ that makes this a linear function of $u$ over the entire $u$-axis.

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However, consider each of the proposed values of $n$.

1. $n=2$

   $y = au + b\sqrt{u} + c + \frac{\displaystyle d}{\displaystyle\sqrt{u}}$

   Here, the second term increases as $u$ increases but the last term decreases as $u$ increases.

2. $n=1$

   $y = au^2 + bu + c + \frac{\displaystyle d}{\displaystyle u}$

   Here, the first term increases as $u$ increases but the last term decreases as $u$ increases.

3. $n=-1$

   $y = \frac{\displaystyle a}{\displaystyle u^2} + \frac{\displaystyle b}{\displaystyle u} + c + du$

   Here both the first and second terms decrease as $u$ increases.

None of the proposed values of $n$ result in an asymptotically linear function of $u$ for small positive values of $u$ because they all have terms that increase as $u$ approaches 0.

But, for values of $u$ with large magnitudes, the value of $n$ from the proposed list that asymptotically produces a linear function of $u$ is $n=-1$:

$y \approx c + du$

for $|u| \gg 1$.

How much bigger than 1 does |u| need to be for the linear approximation to be good? Well, you want $|du| \gg |b/u|$ and $|du| \gg |a/u^2|$ so you need $u^2 \gg |b/d|$ and $|u|^3 \gg |a/d|$.