Can you help me prove $\arctan x ≥ x-x^3/3$? I have thought of taylor but I have not come up with a solution.
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2$\begingroup$ what is $x$ here a real variable? $\endgroup$– Dr. Sonnhard GraubnerJan 12, 2015 at 19:24
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2$\begingroup$ this is not true for all real $x$! $\endgroup$– Dr. Sonnhard GraubnerJan 12, 2015 at 19:26
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$\begingroup$ x is a real variable have you found anything? $\endgroup$– ArgyrisJan 12, 2015 at 19:28
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$\begingroup$ arctan(-1) < -1 +1/3 $\endgroup$– Loreno HeerJan 12, 2015 at 19:35
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3 Answers
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$$ 1 - t^4 \le 1 $$ so $$ 1-t^2 \le \frac1{1+t^2} $$ now integrate $$ \int_0^x (1-t^2)dt \le \int_0^x \frac{dx}{1+t^2} $$ i.e. $$ x-\frac{x^3}3 \le \arctan x $$
answered Jan 12, 2015 at 19:39
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$\begingroup$ @Argyris It's basically the same solution as mine, but written in a trickier way in order to look cool. $\endgroup$ Jan 12, 2015 at 19:44
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$\begingroup$ yes i realised it just now because it is from 0 to x thanks you have helped me a lot you know your math $\endgroup$– ArgyrisJan 12, 2015 at 19:45
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Have you tried derivatives? $$(\arctan x-x+x^3/3)'=\frac 1{1+x^2}-1+x^2=\frac{x^4}{1+x^2}\ge0$$ So the difference is an increasing function. This fact, together the equality when $x=0$ means that $$\arctan x\ge x-x^3/3\text{ when }x\ge 0\\\arctan x\le x-x^3/3\text{ when }x\le 0$$
answered Jan 12, 2015 at 19:28
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1$\begingroup$ thanks i thought it would be more difficult because arctan=x-x3/3+x5/5... @Dr. Sonnhard Graubner $\endgroup$– ArgyrisJan 12, 2015 at 19:37
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let $$f(x)=\arctan(x)-x+\frac{x^3}{3}$$ for $x\geq 0$ for $x=0$ we get $f(0)=0$ and for $x>0$ we get $$f'(x)=\frac{x^4}{1+x^2}>0$$ therefore we obtain $$f(x)\geq 0$$ for all real $x$ with $x\geq 0$.
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$\begingroup$ thanks i thought it would be more difficult because arctan=x-x3/3+x5/5... @Dr. Sonnhard Graubner $\endgroup$– ArgyrisJan 12, 2015 at 19:37