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

AI Thread Summary
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 \( t \).
solakis1
Messages
407
Reaction score
0
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 $$|(\sqrt sinx +1)^2-(\sqrt sin a +1)^2|<b$$
 
Mathematics news on Phys.org
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]
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top