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Given that the $0$ degree Taylor polynomial centered at $x=\frac{1}{2}$ for $f(x)=\arcsin(x)$ is given by $T_{0,\frac{1}{2}}(x)=\frac{\pi}{6}$ how would I use this to show that $1.2(x-\frac{1}{2})\leq R_{0,\frac{1}{2}}(x) \leq(x-\frac{1}{2})$ on $[0.4,0.5]$

I know that by Taylor's theorem $R_{0,\frac{1}{2}}(x)=\frac{1}{\sqrt{1-c^2}}(x-\frac{1}{2}$) and since $\frac{1}{\sqrt{1-x^2}}$ is an increasing function on $[0.4,0.5]$ that $R_{0,\frac{1}{2}}(x) \leq \frac{1}{\sqrt{1-0.5^2}}(x-\frac{1}{2})$

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  • $\begingroup$ The function you mention is not increasing, it is decreasing. $\endgroup$
    – daulomb
    Dec 9, 2017 at 21:20
  • $\begingroup$ @daulomb ??? It is increasing. $\endgroup$
    – A.Γ.
    Dec 9, 2017 at 21:23
  • $\begingroup$ Yes you are right I supposed just square root thanks $\endgroup$
    – daulomb
    Dec 9, 2017 at 21:25

1 Answer 1

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Since as you notice $$ \frac{1}{\sqrt{1-0.4^2}}\le\frac{1}{\sqrt{1-c^2}}\le\frac{1}{\sqrt{1-0.5^2}}, $$ all you need to prove is that $$ \color{red}{1\le\frac{1}{\sqrt{1-0.4^2}}}\le\frac{1}{\sqrt{1-c^2}}\le\color{blue}{\frac{1}{\sqrt{1-0.5^2}}\le 1.2} $$ and then multiply by $x-0.5$, which is negative for $0.4\le x\le 0.5$, so the inequalities become reversed.

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  • $\begingroup$ why the negative? when the bounds are with a positive (x-0.5) $\endgroup$ Dec 9, 2017 at 21:38
  • $\begingroup$ @Skrrrrrtttt I mean, for $0.4\le x\le 0.5$ the quantity $x-0.5$ is negative. For example. $0.4-0.5=-0.1$. $\endgroup$
    – A.Γ.
    Dec 9, 2017 at 21:40

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