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As part of derivation in a book I'm reading, it is claimed that for small $x$, $\frac{1}{(1+x)}$ is approximately equal to $1-x$. This is easy to very for specific small values of $x$.

How does one prove this?

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    $\begingroup$ How about looking at $1/(1+x)-(1-x)$? $\endgroup$ – Lord Shark the Unknown Nov 30 '17 at 4:38
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Check out this general formula

$${1\over 1-x}=\sum_{i=0}^\infty x^i$$

(Think why we can neglect powers of x greater than or equals $2$)

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$$\frac{1}{1+x}=\frac{1-x}{(1+x)(1-x)}=\frac{1-x}{1-x^2} \approx 1-x $$

$\big($ Since $x\to 0 \implies 1-x^2 \to 1 \big)$

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Here is a short answer followed by a more considered one:

Certainly the two quantities are each slightly less than $1$ for small positive $x$ and slightly greater than $1$ when $x$ is negative but small. And $$\frac1{1+x}-(1-x)=\frac{x^2}{1+x} =\frac{x^2}{1+x}$$ which is much closer to $0$ than either of the two is to $1.$ So that is a result for a while range of $x$ and not just a specific value . In other words, if you can tell me that $|x| \lt 0.01$ then I can assure you that $\frac1{1+x}-(1-x) \lt 0.000102$ However $$\frac1{1+x}-(1-x+x^2)=\frac{-x^3}{1+x}$$ so if you can tell me that $|x| \lt 0.01$ then I can assure you that $-0.000001 \leq \frac1{1+x}-(1-x+x^2) \lt 0.00000102.$


But perhaps it is important to decide what it means to say that it is a very good approximation before proving that it is. After all, $\frac1{1+x} \sim 1$ is a pretty good approximation! It is better than any other constant approximation $c \neq 1$ in that $|\frac1{1+x}-1| \lt |\frac1{1+x}-c|$ for all small enough $x.$ In the same sense, $1-x$ beats any other linear approximation $ax+b.$ You can see that graphically by drawing the curve $y= \frac1{1+x}$ and the tangent line $y=1-x$ at $(0,1)$ (then imagine drawing any other straight line.) However proving it mathematically requires some manipulation of inequalities. As remarked by others, $1-x+x^2$ is the best quadratic approximation $ax^2+bx+c.$

A number of answers essentially say that $1$,$1-x$ and $1-x+x^2$ are the Taylor series (at $x=0$) of $f(x)=\frac1{1+x}$ truncated after $n=1,2,3$ terms and hence the best polynomial approximations of their degree. That is an accurate statement of a theorem which holds for any reasonably behaved function $f(x)$ but maybe not helpful in understanding this one particular approximation. In parricular, it assumes understanding of what is meant by best. A helpful detail is that each is the only polynomial of it's degree with error below $x^{n+1}$ for all small enough $x.$

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1/1+x is the Taylor's series solved by Newton Expansion 1/1+x = 1 - x + x^2 - x^3..... As x is a small number higher powers can be neglected and the equation reduces to 1-x.

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  • $\begingroup$ It would be really helpful if you formatted your answer with MathJax. $\endgroup$ – Xander Henderson Dec 10 '17 at 3:16
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In what follows, I'll assume that $x$ is positive, but it's not hard to also deal with negative $x$ close to zero.

If $x$ is small, then so is $x^2$, and in fact so are all higher powers. To be more specific

Let $$0 < x < 1$$ Then, multiplying by $x$ we get $$0 < x^2 < x$$ Thus $$0 < x^2 < x < 1$$ Similarly, $$0 < x^3 < x^2 < x < 1$$ Etc.

Subtracting that inequality from 1 gives us $$1-0 > 1-x^3 > 1-x^2 > 1-x > 1-1$$ Or $$0 < 1-x < 1-x^2 < 1-x^3 < 1$$

So if $1 - x$ is close to 1 then $1 - x^2$ is even closer. Also note that $1 + x^3$ is even closer to 1, although (for positive $x$) it's greater than 1. So we can extend that chain to

$$0 < 1-x < 1-x^2 < 1-x^3 < 1 < 1+x^3$$ Now simple algebra shows that $$(1 + x)(1 - x) = 1 - x^2$$ and $$(1 + x)(1 - x + x^2) = 1 + x^3$$ Let's combine that with our inequality chain $$0 < 1 - x^2 < 1 < 1 + x^3$$ Dividing through by $1+x$ we get $$0 < 1 - x < \frac1{1+x} < 1 - x + x^2$$ Thus $\frac1{1+x} \approx 1 - x$, and is, in fact, between $1 - x$ and $1 - x + x^2$.

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$$1-x+x^2-x^3+\cdots=\frac{1}{1+x}$$ For small $x$ you can neglect higher order terms on L.H.S

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Well it could also be taken as a limit where it can be proved that $ \lim_{x \to 0} \frac {1}{1+x} - 1-x = 0$

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    $\begingroup$ This just gives a 1st order approximation: the $-x$ part doesn't give any information since you let $x$ go to zero. The correct way to go is via Taylor approximation. $\endgroup$ – Daniel Robert-Nicoud Nov 30 '17 at 10:12

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