Not every numerical solution to this problem is a right triangle. When t = pi/2, or 90 degrees, the intersection is y= infinity. When t= 0, y=0. This is the tangent (sin(t)/cos(t)) function, and for that particular function, solutions go to infinity at t= pi/2 whether they are stuck on the rim of the unit circle or not. They just get larger a little faster. This problem isn't worth any more detailed analysis than that. When I was 10, my parents took me to a library book sale at my elementary school and I bought a book on plane trigonometry for ten cents, a princely sum from my 50 cent a week allowance. I loved that book. It had log and common log tables and also a table of haversines. Problems In the book involving the shadows of trees on the ground using similar triangles and finding the height of a distant tower by holding a pencil at arms length helped me understand a description my father told me about how physicists in Tuella Utah calculated fallout dispersal distances from a tower when he was stationed there during atomic testing. Not a job I would have wanted, but my dad was enlisted in the army at the time, an atomic veteran. Probably explains why I am a mutant, actually. I don't fault Einstein for this sequence of events, but some misguided people might. I know better. My father didn't understand when I tried explaining the geometry to him (he never graduated high school) but how the physicists did this bit of parlor magic always fascinated him. By age 15 I had also developed a continuing fraction for pi and put together a notebook containing everything I could find out about the elements on the periodic table, including electron configurations, which I found fascinating. None of my teachers ever saw it, and I still have it, even though the standard of atomic weights changed from hydrogen to carbon.
OK, that tells me more about the requirements of this formal language. You can't touch t in a formal proof, and you can't even look at it (arcsin y). Gotcha. I sorta follow that. It's putting t on the bottom of a fraction that makes this so, from my viewpoint. Yes, that follows from your first statement. Also, 0/0 was precisely what I was trying to avoid. Its value is undefined. You can argue it's 0, or you can argue it's infinity. I'm used to languages that are less formal (or that can be- this is treated like t and arcsin(y) are pointers to the same address, and likewise sin(t) and y; whereas I was thinking y and t and arcsin(y) and sin(t) were four independent variables, each with their own storage). Yes, same again. Right, same again. Thanks for being so patient. Yery nice. I was specifically avoiding looking at outside sources; you of course are not so constrained. This caps off a great explanation. Thanks very much.
They are not rational -- except for \(x_0(n), x_1(n)\) they are not even in the field \(\mathbb{Q}[\pi]\). Since \(\pi\) is not a Liouville number, \(x_0(n)\) is not a Liouville number, and I suspect \(x_1(n)\) can be proven not to be, while I form no opinion on \(x_2(n)\) or the non-trivial solutions to \(\sin x = x \cos x\).
I'll rephrase: whether you try rational or non-rational approximations to solutions of \( x\,cos\,x - sin\,x = 0 \), they are multiples of \( \pi \). But modulo \( \pi \), say, is there any connection between rational approaches (as I tried on a calculator) or ones like yours, and Liouville's theorem? Since it seems that the solutions must be irrational multiples of \( \pi \)? Another observation: the plot of \( x\,cos\,x \cup sin\,x \) shows there is a region around x = 0 where the two curves almost coincide. And the sin curve intersects all the extrema of x cos x where x ≠ 0 (of course). Courtesy of Wolfram: Please Register or Log in to view the hidden image! So now we can compare say the integrals of sin x and x cos x from 0 to \( \frac {\pi }{n},\, n = 1,2,3, ... \). As n gets larger and the interval smaller the integrals should converge.
Oops, meant to say the intersections are the same as where the cos curve intersects the extrema of (sin x)/x; they have the same x value.
The x_0's are all half-integer multiples of π, but they are not good approximations. The x_1's are ratios of expressions in π, so they could be viewed as rational functions (ratios of polynomials) evaluated at π. But n is the free parameter, not π. Finally, the x_2's and beyond are not in the field of numbers that can be expressed by a finite number of arithmetic operations on the rational numbers and π.
Ok, we're looking at the extrema of sinc(x) the unnormalized version. Does the error formula come into it? Or can we do some linear interpolation, or some other kind?
I have not read all Posts to this Thread & apologize if somebody came up with the following answer. When the quotient of two functions approaches 0/0, The quotient approaches the quotient of the derivatives. Hence Limit (sint/t) = cost/1 as t approaches zero. Ergo the limit is 1/1 = 1
Yes, but why? Again with the lack of rigour. Your answer is correct, but it wouldn't get full marks in a test. Not that anyone cares . . .
Here's an example of another way to think about limits Please Register or Log in to view the hidden image! the function f(x) = x cos(x), as x gets larger, starts to become a set of vertical lines because the amplitude increases without bound. In that case f(x) must intersect the extrema of sin(x) = {-1,1} at infinity, but converge quickly otherwise; once x is a few hundred say, you should be pretty close to converging.
Let's demonstrate this. Rigorously. First, we have \(\lim_{t \to 0} \frac{sin \, t}{t} = 1\) to prove. So we've got \(\lim_{t \to 0} sin \, t = 0\) for \(\lim_{x \to 0} f(x) = 0\), and \(\lim_{t \to 0} t = 0\) for \(\lim_{x \to 0} g(x)\). So the first limitation is satisfied. Second, the derivative of t is 1 for all t, which includes all t≠0. So the second limitation is satisfied. Third, we have t' = cos t, so we have \(lim_{t \to 0} \frac{cos \, t}{1} = lim_{t \to 0} cos \, t\) which exists. So the third limitation is met. Now we can apply L'Hopital's Rule, and we see that \(\lim_{t \to 0} \frac{sin \, t}{t} = \lim_{t \to 0} \frac{cos \, t}{1} = \lim_{t \to 0} cos \, t = 1\). Which is what Dinosaur said.
Well, why didn't you say so? BTW, that trigonometry and calculus are inclusive isn't my fault. Complain to the math police.
Because everybody thought you knew L'Hopital's rule. But the problem you chose is your fault. You got owned on the first one so you sandbagged me. I'll remember that about you for a very long time.
And, is that your final answer for 10? I have no idea what you think you "owned", or what exactly is your problem. Well, perhaps the world will forget . . .
Not a total loss, though: 1. I (re-)learned L'Hopital's Rule. 2. I found out that you can't use trigonometric identities in a calculus proof. But you can use calculus identities in it. 3. I identified someone I no longer wish to interact with.
Congratulations! We have a winner! My goodness, it looks like he's going all in. Are you really, really sure that's something you learned, and can you please explain "calculus identities"? Yea, I get that a lot.