MHB Trigonometric inequality bounded by lines

AI Thread Summary
The discussion centers on proving the inequality $$16x\cos(8x)+4x\sin(8x)-2\sin(8x)<|17x|$$ in the context of damped motion in spring-mass systems. The approach involves expressing the left-hand side as a vector dot product and applying the Cauchy-Schwarz inequality. The key question raised is how to validate the inequality $$\sqrt{(16x)^{2}+(4x-2)^2}\leq \sqrt{(17x)^{2}}$$ without knowing the range of x. It is noted that the inequality holds true for specific intervals of x, specifically $$\left(-\infty,\frac{2\left(-4-\sqrt{33} \right)}{17} \right)\,\cup\,\left(\frac{2\left(-4+\sqrt{33} \right)}{17},\infty \right)$$. The discussion emphasizes the need for further exploration of the inequality's validity across different ranges of x.
kalish1
Messages
79
Reaction score
0
How can it be shown that $$16x\cos(8x)+4x\sin(8x)-2\sin(8x)<|17x|?$$

This problem arises from work with damped motion in spring-mass systems in Differential Equations. I have gotten to this inequality after some algebraic manipulation, but am completely stuck here.

Here is the illustrative graph provided by Wolfram Alpha:

[1]: http://i.stack.imgur.com/oWf9E.png

Thanks!
 
Mathematics news on Phys.org
I would use a linear-combination identity to obtain the amplitude $A$ of the trigonometric expression:

$$A=\sqrt{(16x)^2+(4x-2)^2}<\sqrt{(17x)^2}$$

What do you find?
 
The LHS can be expressed as a vector dot product:

\[16xcos(8x)+4xsin(8x)-2sin(8x)=16xcos(8x)+(4x-2)sin(8x)=\\\\ \binom{16x}{4x-2}\cdot \binom{cos(8x)}{sin(8x)}=\vec{a}\cdot \vec{e}\\\\ Applying\; the \; Cauchy-Schwarz \; inequality:\\\\ \left |\binom{16x}{4x-2}\cdot \binom{cos(8x)}{sin(8x)} \right |\leq \left \| \binom{16x}{4x-2} \right \|=\sqrt{(16x)^{2}+(4x-2)^2}\leq \sqrt{(17x)^{2}}\]

My question: How do you show the validity of the last inequality, if we don´t know the range of x:

\[\sqrt{(16x)^{2}+(4x-2)^2}\leq \sqrt{(17x)^{2}}\;?\]
 
We can show that this inequality is true on:

$$\left(-\infty,\frac{2\left(-4-\sqrt{33} \right)}{17} \right)\,\cup\,\left(\frac{2\left(-4+\sqrt{33} \right)}{17},\infty \right)$$
 
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

Replies
15
Views
3K
Replies
6
Views
2K
Replies
3
Views
2K
Replies
2
Views
2K
Replies
13
Views
3K
Replies
10
Views
5K
Back
Top