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How can one show that for small values of $x$, $\sqrt[3]{x+1}\approx1+\frac{x}{3}$?

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Did you try Taylor series? –  lhf Nov 9 '12 at 13:21
1  
Check en.wikipedia.org/wiki/Binomial_series –  Jorge Nov 9 '12 at 13:25
    
yes, but I still don't see the connection. –  Vafa Nov 9 '12 at 13:25

3 Answers 3

up vote 2 down vote accepted

Well, look at the function $f(x)=\sqrt[3]{x+1}$ in the vicinity of $0$. $$ f'(x) = ((x+1)^{1/3})' = \frac{1}{3}(x+1)^{-2/3}. $$ So $f(x)$ is differentiable at $0$ and $f'(0) = 1/3$. Then $f(x)=f(0) + f'(0)\cdot x + o(x)$, that is $$ \sqrt[3]{1+x} = 1 + \frac{x}{3} + o(x). $$ I guess this is what we wanted. Unfortunately, this doesn't really give you any meaningful information in terms of percision.

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$(1+ \frac{x}{3})^3 = 1 + 3 \frac{x}{3} (1) ( \frac{x}{3} +1 ) + \frac{x^3}{3^3} = 1+ x+ \frac{x^2}{3} + \frac{x^3 }{27}$ So since $x$ is small $1+ \frac{x}{3}$ is approximately same to $(1+x)^\frac{1}{3}$

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Assume $a>b>0$. Then: $$3b^2\leq \frac{a^3-b^3}{a-b}=a^2+ab+b^2\leq 3a^2, $$ $$ (a-b)\leq\frac{a^3-b^3}{3b^2}.$$ Now plug in the last inequality $a=1+x/3$ and $b=(1+x)^{1/3}$: $$ (1+x/3)-(1+x)^{1/3} \leq \frac{x^2/3 + x^3/27}{(1+x)^{2/3}}.$$ If $x\geq 0$ you have: $$ (1+x/3)-(1+x)^{1/3} \leq \frac{1}{3}\,x^2(1+x)^{1/3}.$$

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