How can one show that for small values of $x$, $\sqrt[3]{x+1}\approx1+\frac{x}{3}$?
Approximating cube root function for small values of $x$
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0yes, but I still don't see the connection. – 2012-11-09
3 Answers
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}.$
$(1+ \frac{x}{3})^3 = 1 + 3 \frac{x}{3} (1) ( \frac{x}{3} +1 ) + \frac{x^3}{3^3} = 1+ x+ \frac{x^2}{3} + \frac{x^3 }{27}$ So since $x$ is small $1+ \frac{x}{3}$ is approximately same to $(1+x)^\frac{1}{3}$
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.