Proving One Real Root for 2x - 1 - sin x = 0

[Orion1]

Show that the equation:
2x - 1 - \sin x = 0
has exactly one real root.


\frac{d}{dx} (2x - 1 - \sin x) = 2 - \cos x = 0
2 = \cos x
x = \cos^{-1} 2

Is there a better way to approach the root?
any suggestions?

 

[snoble]

Do you mean root or critical point?

 

[James R]

Suppose f(x) = 2x - 1 - sin x

Then f'(x) = 2 - cos x

For any value of x, f'(x) is greater than 0, so f(x) is a constantly increasing function.

Note that

f(-\pi) = 2(-\pi) - 1 - \sin(-\pi) = -2\pi
f(\pi) = 2\pi - 1 - \sin(\pi) = 2\pi

Since f continually increases between x=-pi and x=pi, at exactly one value of x between -pi and pi, f(x) must be equal to zero.

Thus, the equation has exactly one solution (root), and it lies between -pi and pi.

 

[Orion1]

Critical Criterion...

The question is exactly as stated: 'exactly one real root.', and does not give any parameters for a 'critical point'.

Thus, the equation has exactly one solution (root), and it lies between -pi and pi.

OK, however, is there a real equation solution for x?

 

[James R]

There is no analytic solution for x, since the equation is transcendental.

The numerical solution is approximately x = 0.887862

 

[Galileo]

\frac{d}{dx} (2x - 1 - \sin x) = 2 - \cos x = 0
2 = \cos x

This has no solutions (cos x is always between -1 and 1).

So the graph has no 'turning point'. For large values we have f(x)>0 and for small values f(x)<0. Since f is continuous there's a point in between where it is zero (Bolzano, or the intermediate value theorem). Also, there can be only one such point by using the mean value theorem.

 

[Zurtex]

If it helps to visualize the root is approximately:

0.8878622115708660240357015114947117349741536783585225169984587779581531172428676120318808231042402676

It's quite simple, function is constantly decreasing so it's not going to cross the axis again. It's simple to show that it crosses the axis as the function is continuous and is positive for a given value on one side and negative for a given value on the other side.

 

[dextercioby]

The equation \cos x =2 can be solved easily.Make the sub e^{-ix} =t and solve for "t".

Daniel.

 

[mathwonk]

isn't it kind of obvious that the line like 2x-1 with slope 2, meets the graph of sin(x) at one poiint?

 

[James R]

isn't it kind of obvious that the line like 2x-1 with slope 2, meets the graph of sin(x) at one poiint?

Not really. You can draw lots of straight lines which meet the graph of sin(x) at more than one point.

 

[mathwonk]

not too many with slope greater than one.

 

[Orion1]

Calculus Nexus...

The equation \cos x =2 can be solved easily.Make the sub e^{-ix} =t and solve for "t".

Uncertain if this is the correct identity approach:
t = e^{-ix} = \cos (-x) + i \sin (-x) = \cos x - i \sin x
\cos x = 2
t = 2 - i \sin x

Note that it is relatively easy to write a subroutine to 'scan' the equation for solutions for 'x', which is similar to a 'trace' routine on a graphing calculator, however, the key here is to achieve an exact solution for 'x' without such methods.

An algebraic-trigonometric solution does not appear to exist, however, a calculus solution may exist.

 

[jdavel]

Orion1,

"An algebraic-trigonometric solution does not appear to exist, however, a calculus solution may exist."

2x-1 - sin(x) = 0 is a transcendental equation, and it can't be solved for x.

But your problem didn't ask for the solution, it just asked you to show that there is exactly one real solution. Mathwonk gave you a big hint.

And BTW you can't take the derivative of an equation in one variable and assume that what you get has the same roots as the the original equation.

 

[sparkster]

The standard approach to these problems is to use the intermediate value problem to show that roots exists, then, argue by contradiction, and use the mean value theorem to show that there cannot be two or more roots.

 

[dextercioby]

Orion,

\cos x=\frac{1}{2}\left(e^{ix}+e^{-ix}\right)

If you make the sub i hinted,u'll get a quadratic in "t".

Daniel.

 

[sparkster]

Orion,

\cos x=\frac{1}{2}\left(e^{ix}+e^{-ix}\right)

If you make the sub i hinted,u'll get a quadratic in "t".

Daniel.

But is any of that even necessary? Since cos x := 2 for any x, the function has no critical points, so it can cross the x-axis at most one time.

 

[dextercioby]

I was addressing other problem,namely solving the equation \cos x=2,which means finding all possible "x" for which \cos x is equal to 2.

Daniel.