MHB High school inequality |(√(sinx)+1)^2−(√(sina)+1)^2|<b

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The discussion revolves around finding a positive constant \( c \) such that for all \( x \) in the interval \( (0, \pi/2) \) and \( |x-a| < c \), the inequality \( |(\sqrt{\sin x} + 1)^2 - (\sqrt{\sin a} + 1)^2| < b \) holds. The Mean Value Theorem is used to establish that \( |\sin x - \sin a| < |x - a| \). The analysis shows that \( |(\sqrt{\sin x} + 1)^2 - (\sqrt{\sin a} + 1)^2| \) can be bounded by \( |x - a| \left(1 + \frac{1}{\sqrt{\sin a}}\right) \). To ensure this is less than \( b \), \( c \) is defined as \( \frac{b}{1 + \frac{1}{\sqrt{\sin a}}} \). The discussion also questions whether high school students learn the inequality \( |\sin t| \leq |t| \) for all \( t \).
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Given 0<a<π/2 , b>0 find a c>0 such that :

for all ,x : if 0<x<π/2 and |x-a|<c ,then $$|(\sqrt sinx +1)^2-(\sqrt sin a +1)^2|<b$$
 
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solakis said:
Given $0<a<\pi/2$, $b>0$, find a $c>0$ such that :

for all $x$ : if $0<x<\pi/2$ and $|x-a|<c $, then $$|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2|<b$$.
[sp]First step: $|\sin x - \sin a| < |x-a|$. That is essentially the Mean Value theorem, which says that $\sin x - \sin a = (x-a)\cos y$ for some $y$ between $a$ and $x$.

Next, $|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2| = |\sin x + 2\sqrt{\sin x} + 1 - (\sin a + 2\sqrt{\sin a} + 1)| = |(\sin x -\sin a) + 2(\sqrt{\sin x} - \sqrt{\sin a})| \leqslant |(\sin x -\sin a)| + 2|\sqrt{\sin x} - \sqrt{\sin a}|.$

To get an estimate for $\sqrt{\sin x} - \sqrt{\sin a}$, multiply top and bottom by $\sqrt{\sin x} + \sqrt{\sin a}$: $$\sqrt{\sin x} - \sqrt{\sin a} = \frac{(\sqrt{\sin x} - \sqrt{\sin a})(\sqrt{\sin x} + \sqrt{\sin a})}{\sqrt{\sin x} + \sqrt{\sin a}} = \frac{\sin x - \sin a}{\sqrt{\sin x} + \sqrt{\sin a}},$$ from which $$|\sqrt{\sin x} - \sqrt{\sin a}| < \frac{|x - a|}{ \sqrt{\sin a}}.$$

Then $$|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2| \leqslant |(\sin x -\sin a)| + 2|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right).$$

We want this to be less than $b$. So take $$c = \frac b{1 + \frac1{ \sqrt{\sin a}}}$$. Then $$|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right) < c\left(1 + \frac1{ \sqrt{\sin a}}\right) = b.$$
[/sp]
 
solakis said:
Given 0<a<π/2 , b>0 find a c>0 such that :

for all ,x : if 0<x<π/2 and |x-a|<c ,then $$|(\sqrt sinx +1)^2-(\sqrt sin a +1)^2|<b$$

Hi solakis,

Is this a question that you want help with or know the answer to already and want to challenge others? Either way is fine but if it's the former then I'll move the thread to a different part of the forum.
 
Jameson said:
Hi solakis,

Is this a question that you want help with or know the answer to already and want to challenge others? Either way is fine but if it's the former then I'll move the thread to a different part of the forum.
This is a challenge question
 
Opalg said:
[sp]First step: $|\sin x - \sin a| < |x-a|$. That is essentially the Mean Value theorem, which says that $\sin x - \sin a = (x-a)\cos y$ for some $y$ between $a$ and $x$.

Next, $|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2| = |\sin x + 2\sqrt{\sin x} + 1 - (\sin a + 2\sqrt{\sin a} + 1)| = |(\sin x -\sin a) + 2(\sqrt{\sin x} - \sqrt{\sin a})| \leqslant |(\sin x -\sin a)| + 2|\sqrt{\sin x} - \sqrt{\sin a}|.$

To get an estimate for $\sqrt{\sin x} - \sqrt{\sin a}$, multiply top and bottom by $\sqrt{\sin x} + \sqrt{\sin a}$: $$\sqrt{\sin x} - \sqrt{\sin a} = \frac{(\sqrt{\sin x} - \sqrt{\sin a})(\sqrt{\sin x} + \sqrt{\sin a})}{\sqrt{\sin x} + \sqrt{\sin a}} = \frac{\sin x - \sin a}{\sqrt{\sin x} + \sqrt{\sin a}},$$ from which $$|\sqrt{\sin x} - \sqrt{\sin a}| < \frac{|x - a|}{ \sqrt{\sin a}}.$$

Then $$|(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2| \leqslant |(\sin x -\sin a)| + 2|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right).$$

We want this to be less than $b$. So take $$c = \frac b{1 + \frac1{ \sqrt{\sin a}}}$$. Then $$|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right) < c\left(1 + \frac1{ \sqrt{\sin a}}\right) = b.$$
[/sp]
[sp] This supose to be a high school inequality :

|sinx-sina|=$$2|cos\frac{x+a}{2}|.|sin\frac{x-a}{2}|
\leq 2|sin\frac{x-a}{2}|$$....because $$|cost|\leq 1\forall t$$$$\leq 2|\frac{x-a}{2}|=|x-a|$$...since $$|sint|\leq |t|\forall t$$.But i am wondering do they learn the inequality $$|sint|\leq | t|\forall t $$ at high school ?[/sp]
 
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