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First of all, I'm sorry to ask too simple math question.
But I have little backgroud knowledge of mathmatics so it took so long time to me.
In my lecture note it says $\frac{1}{1+x} -1 \approx -x$.
How this is possible? Is it a kind of taylor approximation? Or it uses the fact $\log(1+x) \approx x$?

Thanks for your help.

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    All you need is the formula $$\frac{1}{1-x}=\sum_{k=0}^\infty x^k$$ for $|x|<1$, replacing $x$ by $-x$ gives $$\frac{1}{1+x}=1-x+x^2-x^3\pm \cdots$$ hence $\frac{1}{1+x}-1 \approx -x$ for $x\approx 0$ – Peter Oct 15 '21 at 07:57
  • It is indeed a Taylor approximation. It is the degree $1$ Taylor approximation, also known as a linear approximation. – Theo Bendit Oct 15 '21 at 07:58
  • @Peter Thanks a lot! I completely understand it now. –  Oct 15 '21 at 08:00
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    You can see this also intuitively. The given expression is equal to $\frac{-x}{1+x}$. If $x\approx 0$, the denominator is approximately $1$, so the value is near $-x$. – Peter Oct 15 '21 at 08:01
  • @Peter Would you convert your comment into an answer? – PierreCarre Oct 15 '21 at 08:03
  • Besides the geometric series expansion that @Peter gave (of course, an analysis of the combined sum of all the remaining terms needs to be made, to show this combined sum is sufficiently small to ignore -- GreginGre just did this in his answer written while I was writing this comment), another school level method one can use is "rationalizing the denominator" by multiplying numerator and denominator by $1-x.$ --- $\frac{1}{1+x} = \frac{1}{1+x} \cdot \frac{1-x}{1-x} = \frac{1-x}{1-x^2},$ and now neglect the additive $x^2$ term. – Dave L. Renfro Oct 15 '21 at 08:03
  • $$\frac1{1+0.001}-1=-0.000999000999000\cdots\approx-0.001$$ $$\frac1{1+0.0001}-1=-0.0009999000099990000\cdots\approx-0.0001$$ –  Oct 15 '21 at 08:09

3 Answers3

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This is Taylor approximation at the first order, and it is true for $x$ small enough only. To see it without to many sophisticated tool, you can notice that $f(x)=\dfrac{1}{1+x}-(1-x)=\dfrac{x^2}{1+x}$.

When $x\to 0$, $f(x)\to 0$, so for $x$ close enough to $0$, $f(x)$ will be small, and even negligible compared to $x$. For example, if $x\geq 0$, you have $0\leq f(x)\leq x^2$ and the error term is quadratic in $x$.

Of course, you can have finer approximations by studying the variations of $f$.

GreginGre
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Multiplying by $1+x$, $$\frac1{1+x}-1\approx-x\iff-x\approx-x-x^2.$$ The approximation is good when $x\ll x^2$ or $1\ll x$.

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It is precisely Taylor series to first (linear) order.

It is valid, i.e. a good approximation, for small $x$, as you can see in the plot: enter image description here

For larger $x$, it can be arbitrarily bad (e.g. for $x\to-1$, where the original function diverges.

Toffomat
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