Contour Integrals - Example 2.5, Palka, Section 2.2, Ch.4 .... ....

[Math Amateur]

I am reading Bruce P. Palka's book: An Introduction to Complex Function Theory ...

I am focused on Chapter 4: Complex Integration, Section 2.2 Properties of Contour Integrals ...

I need help with an aspect of Example 2.5,Section 2.2, Chapter 4 ...

Example 2.5, Chapter 4 reads as follows:View attachment 7435In the above text from Palka, we read the following:" ... ... But \(\displaystyle \text{ sin } u \ge 2u/ \pi \text{ whenever } 0 \le u \le \pi / 2\) ... ... "Can someone please explain exactly why/how \(\displaystyle \text{ sin } u \ge 2u/ \pi \text{ whenever } 0 \le u \le \pi / 2\) ... ... ?
Help will be much appreciated ...

Peter

 

[Euge]

In the above text from Palka, we read the following:" ... ... But \(\displaystyle \text{ sin } u \ge 2u/ \pi \text{ whenever } 0 \le u \le \pi / 2\) ... ... "Can someone please explain exactly why/how \(\displaystyle \text{ sin } u \ge 2u/ \pi \text{ whenever } 0 \le u \le \pi / 2\) ... ... ?

Consider the graphs below.

[DESMOS=-0.5496108949416503,1.9503891050583497,-0.1562499999999729,2.343750000000027]\sin\ x;\frac{2x}{\pi}[/DESMOS]

Notice the sine wave lies above the green line from $x = 0$ to $x = \pi/2$, where the sine reaches its peak. The equation of the line passing through the origin and $(\pi/2, 1)$ is indeed $y = (2/\pi)x$. So we have $\sin x \ge (2/\pi)x$ whenever $0 \le x \le \pi/2$.

 

[Math Amateur]

Consider the graphs below.
Notice the sine wave lies above the green line from $x = 0$ to $x = \pi/2$, where the sine reaches its peak. The equation of the line passing through the origin and $(\pi/2, 1)$ is indeed $y = (2/\pi)x$. So we have $\sin x \ge (2/\pi)x$ whenever $0 \le x \le \pi/2$.

OK Thanks Euge ... can see that ...

Mind you ... I still have no idea how one would prove the inequality is true ...

Peter

 

[Euge]

I was hinting at concavity of the sine function on the interval $[0, \pi/2]$. Since $(\sin x)'' = -\sin x \le 0$ over $[0, \pi/2]$, the sine is concave in that interval. Hence, the line segment joining any two points in $[0, \pi/2]$ on the sine graph must lie below the sine graph. In particular, the line segment from the origin to $(\pi/2, 1)$ must be below the graph of the sine on $[0, \pi/2]$, which means $(2/\pi)x \le \sin x$ for all $x\in [0, \pi/2]$.

 

[Math Amateur]

I was hinting at concavity of the sine function on the interval $[0, \pi/2]$. Since $(\sin x)'' = -\sin x \le 0$ over $[0, \pi/2]$, the sine is concave in that interval. Hence, the line segment joining any two points in $[0, \pi/2]$ on the sine graph must lie below the sine graph. In particular, the line segment from the origin to $(\pi/2, 1)$ must be below the graph of the sine on $[0, \pi/2]$, which means $(2/\pi)x \le \sin x$ for all $x\in [0, \pi/2]$.

Thanks for a MOST helpful post, Euge ...

Sorry that I did not pick up on your hint regarding concavity ...

Peter