Check out some similar questions!

galactus · Dec 27, 2006, 04:42 AM

Here's one for my fellow 'mathies' if they want to tackle it.

It'de be fun to see some different approaches.

Prove $\frac{sin(x)}{x}$ is strictly decreasing on the interval $[0,\frac{{\pi}}{2}]$

asterisk_man · Dec 27, 2006, 07:15 AM

Let me work through it.

$\frac {\sin x} x$ is strictly decreasing in the range $\left[0,\frac\pi 2\right]$ if the first derivative is strictly negative in that range

$\frac {d \left(\frac {\sin x} x\right)} {dx} = \frac {\cos x} x - \frac {\sin x} {x^2}$

now I have to show that it is less than 0 in the range
I think I'll try showing that it is less than 0 @ x=0 and the original equation has no points of inflection in the range.
(using L'Hôpital's rule since we have 1/x situation)
$\lim_{x\to0} \frac {\cos x} x - \frac {\sin x} {x^2}=<br /> \lim_{x\to0} \frac {-\sin x} 1 - \lim_{x\to0} \frac {\cos x} {2x}=0-\lim_{x\to0}\frac{-\sin x} 2 =<br /> 0-0 = 0<br />$

hmm... the first derivative is zero @ x=0. This point is in the range and it is not negative. I don't think this fits the definition of "strictly decreasing" since at this point it is neither increasing nor decreasing. Lets clear this one up and then I can try to continue from there. Maybe the range you mean is $\left(0,\frac\pi2\right]$

Anyway, this was a fun use of the new math markup!

galactus · Dec 27, 2006, 08:50 AM

Thanks for your input. BTW, I like the Koch snowflake avatar.

Anyway, I thought maybe try the 2nd derivative to find the inflection pts.

Finding the 2nd derivative, setting it to 0 and solving for x, we find the point

2.0815789772 to the right of Pi/2. We are concave down from 0 to 2.08. Therefore, have

negative slope from 0 to Pi/2. This may not be rigorous enough for some.

Another thought I had was to use the Taylor series for sinx/x

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

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

$a_{n}=\frac{x^{2n}}{(2n+1)!}$

$\frac{a_{n+1}}{a_{n}}=\frac{x^{2}}{(2n+3)(2n+2)} < \frac{(\frac{{\pi}}{2})^{2}}{(2n+3)(2n+2)} < 1$

Therefore, the sequence is decreasing.

This may need more tweaking. Whatcha think?

asterisk_man · Dec 27, 2006, 09:17 PM

Yeah. I'm on board with the 2nd derivative but that still doesn't help with the fact that $\frac {\sin x} x$ is not strictly decreasing in the range given since the 1st derivative is not negative at x=0. If it were negative @ x=0 then I think you've just finished it.

As far as the Taylor series goes, I think that you're just showing that the terms of the series are strictly decreasing in absolute value.
You need to prove something like $\sum_{n=0}^{\infty}a_n\left(x_i\right) < \sum_{n=0}^{\infty}a_n\left(x_j\right)\left{{x_i,x _j \text{in} \left[0,\frac\pi2\right]\\x_i<x_j}\right.$

And I'm pretty sure that you can not remove the summation and compare the $a_n$ terms so I'm not sure how you would do that.

This is a good discussion so be sure to add your input.

As an aside, are you using any application to help with some of your algebra? I have been looking for good free applications but I haven't been overly thrilled by what I've found yet and was hoping to hear your experiences.

As another aside, the first time that I viewed your newest post I didn't see all of the final set of equations and none of the attached image. Keep it in mind if it looks like I may have lost my mind at some point!

galactus · Dec 28, 2006, 05:28 AM

Hey Asterisk Man:

Thanks for the input. The Taylor series may be OK at a specific point, but it doesn't show

anything over the domain 0 to Pi/2. I like the 2nd derivative thing better. BTW, I had a typo. It should, indeed, be

$(0,\frac{\pi}{2}]$ . The slope is 0 at x=0.

Anyway, I do the algebra myself. If I ever need any help, I use my TI-92

or Maple. I am sorry, I do not know of any freeware algebra programs. I will scout

around, though. As far as the images not displaying, I don't know.

Elisha Grey · Jan 17, 2007, 10:52 AM

d/dx(sinx/x) = (sin x - x cos x)/x^2 < 0 iff sin x - x cox s < 0 iff sin x < x cos x iff tan x < x, which is not true in this range. In fact the opposite is true: tanx >= x in this range, so sin x/x is strictly increasing in this range.

Also at 0 lim d/dx(sin x/x) = lim((sinx-xcos x)/x^2) =
(by L'Hospital's rule) lim((cos x - cos x + x sin x)/2x) = lim(sin x/2x) = 1/2.

Whoa, baby! Mistake above! For the limit above everything is OK until the next to last form, which should be
lim(sin x/2), and NOT lim(sin x/2x). So this limit is 0.