Why does the approximation of $(1+x^2)^n$ work this way?

Question

I was trying to solve this problem:
Show that, for large value of $x$, $\frac{1}{y} = \frac{2 + 3x^2}{3(1+x^2)^\frac{3}{2}} + \frac{1}{3} \approx 3 -\frac{9}{x}$. So, I looked that the mark scheme, it indicated that $(1+x^2)^\frac{3}{2} \approx (x^2)^\frac{3}{2}$ And $(2+3x^2) \approx 3x^2$. Then, we would get:
$\frac{1}{y} \approx \frac{1}{3} (1+\frac{3}{x})$.
So my question is, why does $(1+x^2)^\frac{3}{2} \approx (x^2)^\frac{3}{2}$ and $ (2+3x^2) \approx 3x^2$. Is there a general rule?
For example if we want to approximate $(a+bx^2)$ we would abandon a, and keep b. Is that so? If it is, may I know what is the mechanism behind this?

Thank you so much for you guys' replies.

If you plug in large $x$, then does $a$ make much of a difference? — Qi Zhu, Feb 08 '20 at 00:03
So, the a would not matter but the coefficient of x would matter. Right ? — Henry Cai, Feb 08 '20 at 00:04
Yes. It influences the change of the part associated to $x$ after all. — Qi Zhu, Feb 08 '20 at 00:08

Mike · Accepted Answer · 2020-02-08T00:50:52.907

1

I got $\frac{2 + 3x^2}{3(1+x^2)^\frac{3}{2}} + \frac{1}{3} \approx \frac{1}{3} + \frac{1}{x}$, actually.

But the larger point is this: What you are really using is $(1+x^2)^{\frac{3}{2}} = (1+\epsilon_1(x))(x^2)^{\frac{3}{2}}$ and $2+3x^2 = (1+\epsilon_2(x))(3x^2)$, where $\epsilon_1(x),\epsilon_2(x)$ are vanishing with $x$ large. So:

$$\frac{2 + 3x^2}{{3(1+x^2)^\frac{3}{2}}} = \frac{(1+\epsilon_2(x))3x^2}{3(1+\epsilon_1(x))(x^2)^{\frac{3}{2}}}$$

which as $x$ goes to infinity and $\epsilon_1(x)$ and $\epsilon_2(x)$ vanish, gives the result.

In fact, you could even put [say] $ax^2 + bx^4 = (1+\epsilon(x))bx^4$ where $\epsilon(x)$ goes to 0 as $x$ gets large, and on the other end, $ax^2+bx^4 = (1+\epsilon(x))ax^2$ where $\epsilon(x)$ goes to 0 as $|x|$ goes to 0.

edited Feb 08 '20 at 00:50

answered Feb 08 '20 at 00:43

Mike

20,434

Thank you very much. But how do I know which term is vanishing? For example what if $(x^2 + x)$ approximate to... – Henry Cai Feb 08 '20 at 00:50
$x^2+x$ approximates to $(1+o(1))x^2$. – Mike Feb 08 '20 at 00:52
....what is that symbol sir, may I know is there any general law..? – Henry Cai Feb 08 '20 at 00:53
For a polynomial, the highest power term controls the behavior as the variable gets "large" – David P Feb 08 '20 at 00:53
So $(x^3 + x^2) \approx x^3$ ? – Henry Cai Feb 08 '20 at 00:54
Thank you very much sir . So when x gets large Highest power term would dominate , and when it gets to 0 lowest power term will dominate. – Henry Cai Feb 08 '20 at 00:57
@HenryCai sorry, $o(1)$ means, informally, that the term is vanishing i.e., $1+o(1)$ goes to 1. But it looks like you have the general idea – Mike Feb 08 '20 at 01:08
See this for yourself: Calculate $(x+x^2)/x^2$ for $x=10^6,10^7$ ect and see what happens – Mike Feb 08 '20 at 01:11

Why does the approximation of $(1+x^2)^n$ work this way?

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