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I want to solve the equation

$$\frac{x^{17/6}}{a^{17/6}} - \frac{x^2}{a} -1 = 0$$

where I assume $a \gg1$ and $x$ is the unknown. How do I compute the dependence of $x$ on $a$ as $a \rightarrow \infty$.

Descartes
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1 Answers1

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If $x=a^{11/5}$ then $${x^{17/6}\over a^{17/6}}-{x^2\over a}=0$$ If $a$ is big, then both term in this difference are changing rapidly, so the value of $x$ leading to $${x^{17/6}\over a^{17/6}}-{x^2\over a}=1$$ will be very close to $x=a^{11/5}$. With a bit of attention to the derivative, no doubt you could get an estimate as to just how close.

Gerry Myerson
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  • I am having a lot of trouble understanding your reasoning. I am sure it is well-intended, but it seems extremely hand-wavy to me. sorry – Descartes Feb 08 '13 at 08:50
  • It's not my fault...i was born stupid. – Descartes Feb 08 '13 at 09:39
  • I'd have a much better shot at helping you if you would be more specific as to where the difficulty is. Anyway: suppose you've got a function $f(x)$, and you want a number $b$ such that $f(b)=1$, and you have a number $c$ such that $f(c)=0$, and also such that $f'(c)$ is very large. Just think about the graph of $f$, and you will see that you can take $b$ to be very close to $c$. Or think about using Newton's Method to solve $f(x)=1$, with initial guess $x_0=c$. Well, we have $f(x)=a^{-17/6}x^{17/6}-a^{-1}x^2$, and $c=a^{11/5}$. – Gerry Myerson Feb 08 '13 at 10:54