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$$(x+D)^{5/4} - (x)^{5/4} \sim P \qquad(x → ∞)$$

$$4/3Px^{-1/4}\sim D \qquad (x → ∞)$$

I was reading a text about asymptotic relation. How do we go from the first relation to the second relation? I tried using binomial theorem to expand the fractional power, But it still did not seem to make sense. By the way, isn't the binomial theorem for expanding fractional power the same as the taylor series expansion about x=0? What are the steps between?

Michael Hardy
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zhongyuan chen
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2 Answers2

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Consider that you look for the zero of $$f(x)=(x+D)^{5/4} - (x)^{5/4} -P $$ and rewrite it as $$f(x)=(x)^{5/4}\left(\left(1+\frac D x \right)^{5/4}-1\right)-P$$ and use the generalized binomila theorem

$$\left(1+\frac D x \right)^{5/4}=1+\frac{5 D}{4 x}+O\left(\frac{1}{x^2}\right)$$ which makes $$f(x)\sim \frac{5 D}{4} x^{1/4}-P$$ So, if $f(x)\approx 0$ then $D \sim \frac 4 5P x^{-1/4}$.

I suppose that there is a typo and that $3$ is supposed to be $5$.

Claude Leibovici
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  • Thank you so much, I finally understand it. But yet, I dont know the method for expanding $(1+D/X)^{5/4}$. – zhongyuan chen Nov 25 '17 at 00:48
  • @zhongyuanchen. Don't you know the binomial theorem ? Otherwise, just use equivalent $(1+\epsilon)^n \sim 1+n\epsilon$. Is this easier ? Cheers. – Claude Leibovici Nov 25 '17 at 01:54
  • I only knew how to use the binomial theorem to expan integer power. But I just looked it up, and I know the generlized binomial thoerem now. THnak you again. – zhongyuan chen Nov 25 '17 at 15:48
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More generally, if $(x+D)^a - x^a \sim P $ as $x \to \infty$, then, by the mean value theorem, if $f(x) =x^a $, then $f(x+D)-f(x) =Df'(z) $ where $x \le z \le x+D$.

To see how much $f'(z)$ can vary, by the same argument, $f'(x+D)-f'(x) =Df''(w) $ where $x \le w \le x+D$.

In this case, since $1 < a < 2$, $f'(x) = ax^{a-1}$ and $f''(x) = a(a-1)x^{a-2}$ so $f''(x) \to 0$ as $x \to \infty$.

Therefore $f'(x)$ is approximately constant for the fixed range of size $D$, so $f'(z) \approx f'(x) $ for $x \le w \le x+D$, so $f(x+D)-f(x) \approx Df'(x) =Dax^{a-1} $.

If $P =Dax^{a-1}$, then $x \approx (P/(Da))^{1/(a-1)} $ so that $D =P/(ax^{a-1}) =x^{1-a}P/a $.

In your case, $a = 5/4$, so $x \approx (P/(D(5/4)))^{1/(1/4)} =(4P/(5D))^{4} $ and $D =x^{-1/4}4P/5 $.

marty cohen
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