Solving this equation with both x, sinx and cosx

[lo2]

Homework Statement

I have this function:

f(x) = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}}

For all x \in R where x \neq n \pi, n \in Z

And then I have to solve this equation f(x)=0:

f(x) = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}}

Where I have to show that it has no solutions in the interval 0 < x < Pi and that it has got one solution in the interval Pi < x < 2Pi, and then I have to approximate that solution using Maple.

Homework Equations

The Attempt at a Solution

You get this equation:

0 = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}} ⇔ \frac{1}{x}=\frac{\cos{(x)}}{\sin{(x)}}

Or you can rewrite it to:

0 = \frac{\sin{(x)}-x \cos{(x)}}{x \sin{(x)}}

And then just show that the nominator will never be 0.

But anyhow I cannot really see how to make proper progess, so how can I solve this? And is it true that an equation with both x and sin(x) (or cos(x)) will not have an algebraic solution?

 

[jackmell]

Homework Statement

I have this function:

f(x) = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}}

For all x \in R where x \neq n \pi, n \in Z

And then I have to solve this equation f(x)=0:

f(x) = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}}

Where I have to show that it has no solutions in the interval 0 < x < Pi and that it has got one solution in the interval Pi < x < 2Pi, and then I have to approximate that solution using Maple.

Homework Equations

The Attempt at a Solution

You get this equation:

0 = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}} ⇔ \frac{1}{x}=\frac{\cos{(x)}}{\sin{(x)}}

Or you can rewrite it to:

0 = \frac{\sin{(x)}-x \cos{(x)}}{x \sin{(x)}}

And then just show that the nominator will never be 0.

But anyhow I cannot really see how to make proper progess, so how can I solve this? And is it true that an equation with both x and sin(x) (or cos(x)) will not have an algebraic solution?

When confronted with a big problem, try and learn to scale it down and just focus on a small part or a simplified version of it. So, just for now, assume you only had one problem to solve:

For f(x)=1/x-\frac{\cos(x)}{\sin(x)}, show that it doesn't have a zero in the interval (0,\pi)

Now let's cheat a little bit. Huh? Nothing's wrong with that. I mean just plot them for heaven's sake. Yeah, I did and I can see 1/x and cot(x) don't. Now when you combined them you determined that in order for it to have a solution there, the expression:

\sin(x)-x\cos(x)

would have to have a zero there. Alright, I'm not proud. Plot those two also. Yeah, they both start at the origin and one is . . . ALWAYS . . . below the other one. Now, for f(x)=sin(x) and g(x)=x cos(x) with g(x) always being lower than the other, what analytic property of g(x) would ensure it to always be lower if they start off at the same point?

 

[HallsofIvy]

There is, in general, no "algebraic" method to get an exact value of an equation in which the uknown, x, appears both inside transcendental functions (such as sin(x) and cos(x)) and outside. I suppose you might be able to write sine and cosine as complex exponentials (sin(x)= (e^{ix}- e^{-ix})/2i and cos(x)= (e^{ix}+ e^{-ix})/2) and use a complex version of the "Lambert W function" (defined as the inverse function to f(x)= xe^x) but that would be extremely complex. Best to use a numerical method like "Newton's method".

 

[lo2]

Ok thanks for the answers.

But is there an analytical way of showing that the equation has no solution in the interval ]0;Pi[ and has got one solution ]Pi;2Pi[?

 

[D H]

What's the first tool that comes to mind when you are asked to show something analytically? Have you tried using it?

 

[lo2]

What's the first tool that comes to mind when you are asked to show something analytically? Have you tried using it?

Ok well I do not know if what I mean can be called analytical.

I just wanted to be able to reach an expression from which I can show that there is no solution and one solution respectively.

 

[D H]

You posted this in the "Calculus & Beyond" section. What is one of the first things you learn when you are taking calculus, just after learning about limits?

 

[jackmell]

Ok well I do not know if what I mean can be called analytical.

I just wanted to be able to reach an expression from which I can show that there is no solution and one solution respectively.

Ok, let's do an easier one. Consider:

f(x)=x^2

g(x)=1/2 x^2

They too start at zero and never cross so the expression x^2-1/2 x^2 is never zero for x ne 0. Note also that the plot of 1/2 x^2 is always below the plot for x^2. Now, they both start at zero and one way (the simplest) for g(x) to always be lower than the other is for the rate at which g(x) is changing to be always less than the rate at which f(x) is changing. Now, how can you analytically represent that rate phrase in terms of Calculus?

 

[SammyS]

I don't think I see the following approach used in this thread so far, but I have a history of missing stuff that's been previously posted in a thread.

Finding the zeros of \displaystyle f(x)=\frac{1}{x}-\frac{\cos(x)}{\sin(x)} is equivalent to solving \displaystyle \frac{1}{x}=\cot(x)\ .

But that's equivalent to solving \displaystyle x=\tan(x)\,, for all x in the domain of f(x)\ .

The solutions to the equation \displaystyle x=\tan(x) are the same as the zeros of the function \displaystyle g(x)=\tan(x)-x\ .

It's not too difficult to show that g(x) has no zeros on the interval, (0, π).

 

[Ray Vickson]

Ok thanks for the answers.

But is there an analytical way of showing that the equation has no solution in the interval ]0;Pi[ and has got one solution ]Pi;2Pi[?

To show that g(x) = sin(x) - x*cos(x) has no root in the first interval, look at its derivative g'(x) = cos(x) - cos(x) + x*sin(x) = x*sin(x). This is > 0 in (0,π), which means that g is strictly increasing throughout the whole interval. Since g(x) --> 0 as x --> 0+, that means that the graph of y = g(x) cannot cross zero within the interval. As for the interval (π,2π): g starts off at g(π) = +π, and ends at g(2π) = -2π, so switches from + to - in the interval. Therefore, it has a root. Within the interval (π,2π) its derivative is < 0, so it is strictly decreasing throughout that interval, and that makes the root unique.

RGV

 

[Chestermiller]

Are you allowed to plot a graph of y = x and y = tan (x) on the same plot, or is that against the rule? On such a graph, you will see right away that the two plots don't cross over each other until the third quadrant.

 

[lo2]

Ah ok I think I got it now!

I just showed that in the first interval ]0;Pi[ it starts out at f(x) > 0 and that it then heads toward infinity and since it is both only growing and continuous it will not cross the x axis.

And in the other interval ]Pi;2Pi[ it starts out coming from -infinity and goes toward infinity and since it also only growing and continuous in this interval, it has to cross the x-axis just once.

Btw does anyone how to approximate this solution in Maple? Where you only get this one solution.

 

[Ray Vickson]

Ah ok I think I got it now!

I just showed that in the first interval ]0;Pi[ it starts out at f(x) > 0 and that it then heads toward infinity and since it is both only growing and continuous it will not cross the x axis.

And in the other interval ]Pi;2Pi[ it starts out coming from -infinity and goes toward infinity and since it also only growing and continuous in this interval, it has to cross the x-axis just once.

Btw does anyone how to approximate this solution in Maple? Where you only get this one solution.

In good English, are you asking how to get Maple to solve the equation in the interval (π, 2π)?
It's easy:

> eq:=1/x - cot(x)=0;
eq := 1/x - cot(x) = 0

> fsolve(eq,x=Pi..2*Pi);
4.493409458
If you want more accuracy, just increase the Digits count:
> Digits:=20;
Digits := 20

> fsolve(eq,x=Pi..2*Pi);
4.4934094579090641753

RGV