I know $\left|\frac{x}{\sqrt{1+x^2}}\right| \le 1$. But I don't know how to find $L$ that $\left|\frac{x}{\sqrt{1+x^2}}-\frac{y}{\sqrt{1+y^2}}\right| \le L|x-y|$.
Would you please explain it? Thank you.
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Hint: the function $f(x):=\sqrt{1+x^2}$ has second derivative $\dfrac{1}{(1+x^2)^{3/2}}$, so $f$ is in $\mathrm C^2$.
Clue: Lagrange's theorem will apply.
Clue: Lagrange gives here that $|f'(x) -f'(y)| \leq |f''(z_0)||x-y|$ where $z_0$ is somewhere in $(x, y)$.
Linear Christmas
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Isn't that just the mean value theorem? – Troposphere Apr 24 '21 at 08:53
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@Troposphere Yes, Lagrange's theorem is also called Lagrange's MVT or simply MVT. – Linear Christmas Apr 24 '21 at 08:55
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Does that differ from the usual mean value theorem? It looks like the ordinary one will do the job perfectly well here. – Troposphere Apr 24 '21 at 08:59
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@Troposphere No, it's just a different name for the same thing you're referring to. There are other mean value theorems as well, of course (or different ways of stating). The version here is classically called Lagrange's mean value theorem. An example of another form is Cauchy's mean value theorem. – Linear Christmas Apr 24 '21 at 09:05
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Thanks but isn't the mean value theorem an equality? – Mina Apr 24 '21 at 11:09
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@Mina Yes, it is an equality. Do you see how to get the inequality from the equality? – Linear Christmas Apr 24 '21 at 12:37