break the Christmas dry spell

[galactus]

The forum sure has been dead the last few days.

I reckon I'll post something just to break the dry spell.

Here's a problem I seen on another forum. There are many ways, but what are some ideas besides just graphing it?. Something rigorous.

"Prove \(\displaystyle \frac{sin(x)}{x}\) is strictly decreasing on \(\displaystyle [0,\frac{\pi}{2}]"\)

I thought of using a Taylor series for sin(x)/x:

\(\displaystyle 1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\frac{x^{6}}{7!}+\frac{x^{8}}{9!}-........\)

=\(\displaystyle \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n+1)!}\)

\(\displaystyle \L\\a_{n}=\frac{x^{2n}}{(2n+1)!}\), for any \(\displaystyle n\geq{0}\)

Therefore, \(\displaystyle \frac{a_{n+1}}{a_{n}}=\frac{x^{2}}{(2n+3)(2n+2)}\)

\(\displaystyle \frac{(\frac{\pi}{2})^{2}}{(2n+3)(2n+2)}<1\)

Because if \(\displaystyle a_{n+1}<a_{n}\), then the ratio \(\displaystyle \frac{a_{n+1}}{a_{n}}\) would be < 1.

This seems like I am onto something, but something doesn't appear quite right.

Any ideas?. Just for fun.

 

[daon]

Just a thought, but since x and sinx are both positive on [0, \(\displaystyle \frac{\pi}{2}\)], it would be sufficient to prove that for all x in given interval, and all epsilon \(\displaystyle x \,\, < \,\, \epsilon \,\, \leq \,\, \frac{\pi}{2}\)... that \(\displaystyle \,\, \frac{sin(\epsilon)}{\epsilon} \,\, < \,\, \frac{sin(x)}{x}\).

I haven't given any thought to this problem yet, but I thought I'd just throw an idea your way. Maybe you can incorporate Taylor's theorem with this.

-Daon

 

[daon]

I would try one of these:

1) Assume BWOC that \(\displaystyle \exists x \in [0, \pi/2)\) such that for some \(\displaystyle r \in (x, \pi/2]\), it is true that \(\displaystyle \frac{sin(r)}{r} \ge \frac{sin(x)}{x}\). Then try to get a contradiction.

2) Try to prove that \(\displaystyle \forall x \in [0, \pi/2]\), we have that \(\displaystyle f'(x) \,\, < \,\, 0\). I.e. that \(\displaystyle \frac{-sin(x) + xcos(x)}{x^2}\,\, < \,\, 0\) for all x in that interval.

The second would probably be easier (maybe by contradiction?). You may also find it beneficial to use the facts that \(\displaystyle sin(x)<x\) for all x>0, and that sin(x) is strictly increasing on that interval. Sin(x)<x is easily proved with a simple "grouping" trick with the taylor series expansion of sin(x) and induction.

edit: #2 simplifies to proving that sin(x)>xcos(x) for x in that interval.

 

[daon]

If one knows about the series expansion for tan(x), #2 is proved easily...

For x > 0, and |x|<pi/2
\(\displaystyle \frac{sin(x)}{cos(x)} = tan(x) = x + \frac{2x^3}{3!} + \frac{16x^5}{5!}+ \frac{272x^7}{7!} + ... \,\, > x\)

So we have
\(\displaystyle \frac{sin(x)}{cos(x)} > x \,\, \Rightarrow \,\, sin(x) > xcos(x) \,\, \Rightarrow \,\, xcos(x) - sin(x) < 0 \\ \,\, \Rightarrow \,\, \frac{xcos(x)-sin(x)}{x^2} < 0 \,\, \Rightarrow \,\, \frac{d(sin(x)/x)}{dx} < 0\).

So \(\displaystyle \frac{sin(x)}{x}\) is decreasing on \(\displaystyle (0, \frac{\pi}{2})\).
Also, \(\displaystyle f'(\frac{\pi}{2}) = \frac{-4}{\pi ^2} <0\) we know its true for \(\displaystyle \frac{\pi}{2}\) also.
Finally, when x=0 we have a relative maximum which can be verified the normal way.
Therefore, now we know its true for the entire interval \(\displaystyle [0,\frac{\pi}{2}]\) (Although, technically the function is not decreasing at x=0, but since it is 1-1 I see no reason why we can't say it is).

-Daon

 

[galactus]

Very nice, daon. I like how you used \(\displaystyle \frac{sin(x)}{cos(x)}>x\) to get the numerator in the derivative.

I used the 2nd derivative which can be used to show also.

Anyway, nice work.