Cosine

[Farsight]

Thanks Leo. It's all to do with conic sections, ronthepon. Slicing cones at different angles.

[Qfwfq]

That way, we could imagine a circle for the [math]\theta[/math]... and assosicate a triangle for the [math]Cos\theta[/math].

Getting there... :) (sure that's the right way around?)

 

Think of the graph for the fixed point of [math]\cos x = x[/math] and... apply a bit of force somewhere... :D

[ronthepon]

It's the point where the [math]x = y[/math] line cuts the cosine curve...

[Qfwfq]

It's the point where the [math]x = y[/math] line cuts the cosine curve...

...or the arc of circle!

 

Letting the cat out of the bag, as perhaps it's a bit tricky to get through the whole thing exactly, let's first change the equation I gave (which is a parametrized curve) into the form f(x) = x (at least for the bit of this circle that we need). To do this we must write:

 

[math]y=\cos\theta[/math]

 

[math]x=\sin\theta[/math]

 

So y = x is the given equation, from which one of its two solutions is the fixed point of:

 

[math]\cos\;\arcsin x = x[/math]

 

which of course is exactly [math]\norm\frac{\sqrt2}{2}[/math], but now let's make the comparison with your original equation:

 

[math]\cos x = x[/math]

 

It's easy to see, and not difficult to show, that the two graphs are tangent to each other at x = 0 and that the arc of circle is lower than the cosine curve for 1 > x > 0. This is why the fixed point of cosine is greater than the root of 2, though not all that much different because the two curves don't get so terribly far from each other before crossing the x = y line.

[CraigD]

…

[math]\cos x = x[/math]

 

It's easy to see, and not difficult to show, that the two graphs are tangent to each other at x = 0 and that the arc of circle is lower than the cosine curve for 1 > x > 0.

The graph of [math]y = \cos x[/math] and [math]y = x[/math] obviously are not tangent at [math]x = 0[/math].

 

What I believe ron is looking for is a constant-value expression for [math]x[/math] such that [math]\cos x = x[/math] (eg: [math]x=\frac{\sqrt{\pi}}2[/math] – not correct, but an example of a constant expression). An approximate value of [math]x = 0.7391[/math] is well known, appearing in a linked-to textbook page as well as post #1.

 

The solution is not obvious (at least to me), nor is a proof that such a solution exists or not. A geometric sketch is simple but unhelpful, while a Taylor series expansion of [math]\cos x - x[/math]doesn’t “cancel” to give a polynomial with a finite number of terms.

 

So, does anyone actually have a constant-value expression for [math]x[/math]? I’ve been keeping the problem at the back of my mind for the past few days, but haven’t yet had any flashes of intuition or recollections of an approach. I even cheated and read a similar discussion at another forum (dated 1/30, which makes me suspect ron frequents this other forum), but found that thread no closer to a solution than this one.

[Qfwfq]

The graph of [math]y = \cos x[/math] and [math]y = x[/math] obviously are not tangent at [math]x = 0[/math].

Obviously. I hadn't said they were. Not those two curves.

 

So, does anyone actually have a constant-value expression for [math]x[/math]? I’ve been keeping the problem at the back of my mind for the past few days, but haven’t yet had any flashes of intuition or recollections of an approach.

I seriously doubt you might find what you're seeking. I say (again) it's almost certainly transcendental, though I'm no expert at proving such things.

[CraigD]

So, does anyone actually have a constant-value expression for [math]x[/math] [such that [math]\cos x = x[/math]]? I’ve been keeping the problem at the back of my mind for the past few days, but haven’t yet had any flashes of intuition or recollections of an approach. I even cheated and read a similar discussion at another forum (dated 1/30, which makes me suspect ron frequents this other forum), but found that thread no closer to a solution than this one.

I seriously doubt you might find what you're seeking.

I share Qfwfq’s doubt. I’ve no more idea how to prove that a solution doesn’t exist, though, than how to find it, so the search, either for a solution or a proof that none exists, continues! :)

I say (again) it's almost certainly transcendental, though I'm no expert at proving such things.

A number being transcendental doesn’t mean it can’t be expressed as a constant-value expression. [math]\frac{\pi}4[/math] (a solution for [math]x[/math], incidentally, to [math]\sin x = \cos x[/math]) is both transcendental and (real) constant-valued, as are [math]\sqrt2[/math], [math]e[/math], etc.

 

In the context of this question, a constant-valued infinite series (eg: [math]\sum_{n=0}^\infty \frac{-1^n}{(2n)!}[/math], AKA [math]\cos 1[/math] – not a solution, just an example) would, I think, be an acceptable solution.

[LaurieAG]

Does anyone know how to solve this?

 

[math]Cos\theta \. = \. \theta[/math]

 

In the 70's we used SOHCAHTOA (some old hags cannot always hide their old age) to remember how to calculate the geometric Sin Cos and Tan of right angled triangles.

 

Sin = Opposite/Hypotenuse

Cos = Adjacent/Hypotenuse

Tan = Opposite/Adjacent

 

So, all you're really looking for is the angle (in radians) of a right angled triangle that equals it's geometric equivalents (of Sin Cos etc).

[Qfwfq]

[math]\frac{\pi}4[/math] (a solution for [math]x[/math], incidentally, to [math]\sin x = \cos x[/math]) is both transcendental and (real) constant-valued, as are [math]\sqrt2[/math], [math]e[/math], etc.

Of course, Craig, but I doubt it can be simply expressed in terms of [math]\norm\pi[/math], e or other "already known" transcendental values. But since when is [math]\sqrt2[/math] transcendental? :xx: Not by definition, being a solution of [math]\norm x^2 = 2[/math].

 

By what you had said, I didn't reckon you counted an infinite series as a constant-valued expression. A transcendental number, however defined, can AFAIK always be given as such. What I understood thepon to be looking for was some kind of intuitive insight as to why that value (at least roughly) and not more, nor less, if you know what I mean. Something like explaining the value of [math]\norm\pi[/math] by knocking the corners off the square...

[max4236]

For [math]\pi[/math] there's a non-simple continued fraction like this:

[math]\pi=3+\frac{1^2}{\6+\frac{3^2}{\6+\frac{5^2}{\6+...}}}[/math]

 

[math]({1^2},{3^2},{5^2},{7^2},{9^2},{11^2}...)[/math]

 

Is there a similar non-simple continued fraction for x=cosx? I found a simple continued fraction but I can't see a pattern in the generated sequence. :\ x=.739085133215161= 44,252,069 /59,874,116 approx.

 

[math]x=0+\frac{1}{\1+\frac{1}{\2+\frac{1}{\1+\frac{1}{\4+\frac{1}{\1+\frac{1}{\40+...}}}}}}[/math]

 

x=[0;1,2,1,4,1,40,1,9,4,2,1,15,2,12,1,10,1,2,13,

1,2,1,4,6,1,2,1,20,3,1,1,2,1,2,32,1,11,3,1 ...]

 

 

defdbl x,a,b
x=1
'Newton's method of finding the root of f(x)=x-cos(x)=0
' x(n+1)=x(n)-f(x(n)/f'(x(n))
for i=1 to 10
 x=x-(x-cos(x))/(1+sin(x))
next
?x
a=x
'Continued fraction sequence 
for i=1 to 40 step 1
 b=int(a)
 ?b;
 a=a-b
 a=1/a
next

[CraigD]

But since when is [math]\sqrt2[/math] transcendental?

You're correct - not for the first time, not likely the last, I've confused the irrational with their subset, the transcendental numbers :(

By what you had said, I didn't reckon you counted an infinite series as a constant-valued expression.

You’ve a valid point here.

 

A trivial, infinite-length solution for [math]x = \cos(x)[/math] – just an arithmatic rewriting of the program that produced the approximation - is

 

[math]x = \cos( \cos( \cos( . . . cos(0) . . . )))[/math]

 

Any real value can be used in place of [math]0[/math]. A nomenclature for this better than the dreaded “…”s isn’t so trivial, but since this solution is so unsatisfying, doesn’t seem important.

 

Still, a solution, even an infinite length expression, not involving the cosine function, would amaze me.

 

Max’s continuing fraction approach is interesting, even though I too am unable to see a pattern in the sequence of values in the continuing fraction.

 

Perhaps it would be fruitful to consider other fixed point problems without obvious algebraic solutions, such as [math]x = e^{\frac{x}e}[/math] , which appears to have a solution of [math]e[/math].

[ronthepon]

So, here's my idea on a possible method to find this angle out. Problem arises: I cant find a way not to use trig. functions and still do it.

 

Basic point is:

 

[math]Cos\theta = \frac{OA}{OC}[/math]

 

[math]\theta = \frac{AB(arc)}{OA}[/math]

[LaurieAG]

So, here's my idea on a possible method to find this angle out. Problem arises: I cant find a way not to use trig. functions and still do it.

 

Hello Ron,

 

Can you solve the problem with trig. functions?

 

It is much easier to use Radians because the Sin Cos etc are numbers between 0 and 1 (for this exercise anyway). You only have 2 pi in 360 degrees (0 to 2xpi, 0 to roughly 6.3, 1/6.3 of the radians are in the correct range) and will have a much less chance of getting a matching angle when the angle is in degrees (0 to 360, even though you have 4 x 90 degree cycles you must discard the negative non zero answers, only 1/180 degrees are in the correct range).

[ronthepon]

Actually, it makes no difference. I'm not trying all the possible values to find this value. (I'll never succeed)

[CraigD]

Here’s a way to “construct” the solution [math]x[/math] to [math]x = \cos x[/math]. Unlike a classical straightedge-and-compass-only construction, this one needs an armature with some pivots, a spool, and some string.

 

(See the attached thumbnail) The circle has radius 1 unit. Edges that appear the same length are. Red line indicate sliding connections, all others can pivot freely. The 2 blue strings are equal length.

 

Tensioning the purple spool and moving the armature until both strings are at tension positions the armature so that the label segment length is the solution for [math]x[/math].

 

I’ve a faint recollection of straightedge-and-compass constructions of an arc of a circle with length equal a given line segment, and vice versa, so suspect, faintly, that [math]x[/math] is a constructible number.

[Qfwfq]

Perhaps it would be fruitful to consider other fixed point problems without obvious algebraic solutions, such as [math]x = e^{\frac{x}e}[/math] , which appears to have a solution of [math]e[/math].

But... this equation does have an obvious solution. The one you say it appears to have.

 

I’ve a faint recollection of straightedge-and-compass constructions of an arc of a circle with length equal a given line segment, and vice versa

If you can show me how to do these, I'll show you a straightedge-and-compass solution of the age-old problem of trisecting a generic angle!

[CraigD]

But... this equation [[LATEX]x=e^{\frac{x}e}[/LATEX]] does have an obvious solution. The one you say it appears to have.

I was going to protest “but I found that empirically, by approximating”, then noticed

[LATEX]x=e^{\frac{x}e}[/LATEX]

Let [LATEX]x=e[/LATEX]

[LATEX]e=e^{\frac{e}e}=e^1[/LATEX] = pretty obvious. :doh:

 

The problem’s less obvious as "solve for [math]x[/math] in terms of [math]a[/math]"

[LATEX]x=e^{\frac{x}a}[/LATEX], where [LATEX]a \ge e[/LATEX],

or even more generally [LATEX]x=b^{\frac{x}a}[/LATEX], where [LATEX]a \ge b[/LATEX],

 

If you can show me how to do these [constructions], I'll show you a straightedge-and-compass solution of the age-old problem of trisecting a generic angle!

Though I can see why my recent cluelessness would lead a reader to believe otherwise, I do remember a few well-know disproofs, such as trisecting an angle by construction.

 

In addition to my faint (and often unreliable) recollections, a little summary search/surfing tantalized me with mention of a famous construction of a line with length equal to an arc on a parabola, and much talk of the difficult-to-impossibility of such a construction for a general polynomial curve. If the general case being unsolved is worth mentioning, surely some special case, perhaps the circle, has been solved.

 

Somewhere down the road, there must either be satisfying expressions for the solutions of these fixed point problems, or proofs that such solutions don’t exist. Looking for them keeps us off the streets and out of trouble. ;)