[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 $x$.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 \(\displaystyle |(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2|<b\).
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 \(\displaystyle |(\sqrt sinx +1)^2-(\sqrt sin a +1)^2|<b\)
This is a challenge questionJameson 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.
[sp] This supose to be a high school inequality :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 \(\displaystyle |\sqrt{\sin x} - \sqrt{\sin a}| < \frac{|x - a|}{ \sqrt{\sin a}}.\)
Then \(\displaystyle |(\sqrt {\sin x} +1)^2-(\sqrt {\sin a} +1)^2| \leqslant |(\sin x -\sin a)| + 2|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right).\)
We want this to be less than $b$. So take \(\displaystyle c = \frac b{1 + \frac1{ \sqrt{\sin a}}}\). Then $$|\sqrt{\sin x} - \sqrt{\sin a}| < |x-a|\left(1 + \frac1{ \sqrt{\sin a}}\right) < c\left(1 + \frac1{ \sqrt{\sin a}}\right) = b.$$
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The inequality equation (√(sinx)+1)^2−(√(sina)+1)^2|2. How does this inequality relate to high school inequality?
This inequality is a mathematical representation of the concept of high school inequality, which refers to the unequal distribution of resources, opportunities, and outcomes among high school students. The inequality equation highlights the disparity between two expressions and serves as a reminder of the importance of addressing and reducing high school inequality.
The values of x and a in this inequality can vary depending on the context in which it is used. Generally, x and a can take on any real number values, but they may be restricted to certain ranges or intervals depending on the specific problem or situation.
This inequality can be used as a tool to analyze and quantify the level of inequality present in a high school setting. By plugging in different values for x and a, we can see how the difference between the two expressions changes and identify areas where there may be a larger disparity. This information can then be used to inform policies and interventions aimed at reducing high school inequality.
While this inequality can provide valuable insights into high school inequality, it is important to note that it is a mathematical representation and does not capture all aspects of the issue. High school inequality is a complex and multifaceted problem that requires a holistic approach to address. Additionally, the values of x and a used in this inequality may not accurately reflect the experiences and backgrounds of all high school students.