sin(integer) < E for any E > 0, possible?

mikau · 2007-05-10 20:01:45

in order for sin(x) to be zero, x must be some multiple of pi. However, because the sine function is repeating, some large integer may be terribly close to some multiple of pi and therefore its sine may be . But can we say with certainty that for any error E > 0, there is some integer n such that |sin(n)| < E?

At first I thought you could show you could get any level accuracy by taking a finite number of pi's digits, such as say 3.14, and multiplying it by some power of 10 to kill the decimal point and make it an integer. So we'd get a rough approximation for 10*pi, and sin(10pi) = 0. I thought using more digits of pi would work, and essentially, I thought using as many of pi's digits as possible, without a decimal point would work to any desired accuracy as it would just be pi * (some power of ten), the sine of which would be zero.

This however, is not the case, and I realized why.

If you take any approximation of pi, call it a, and let Error = |pi - a|.
if you let 'a' be some approximation of pi to n decimal places, there will still be some Error because pi has an infinite number of digits.

so basically a = pi + error,

if we mulitply a by 10^n to kill the decimal place, and take the sine of a, we are taking the sine of (10^n*pi + error*10^n) thus the error values is magnified and the approximation for an integer multiple of pi does not become increasingly accurate.

However, could it be shown in the infinite case I wonder, and would that question even make sense. lol.

at least at the finite level, it does seem getting an approximation of pi by using its digits doesn't work. Is it therefore, a mere matter of luck that some integer could be terribly close to a multiple of pi, or is it garaunteed that we can find an any integer n to make sin(n) as close to zero as we want?

Don't know.

mikau · 2007-05-10 20:07:05

I wrote a quick program to find the ten lowest integer sines between 1 and 1000 at intervales of 100. Here are my results:

best sine between 1 and 100 is sin(22) = -0.008851309290403876

best sine between 101 and 200 is sin(110) = -0.044242678085070965

best sine between 201 and 300 is sin(289) = -0.026521020285755953

best sine between 301 and 400 is sin(355) = -3.014435335948845E-5

best sine between 401 and 500 is sin(421) = 0.02658128773805108

best sine between 501 and 600 is sin(600) = 0.044182448331873195

best sine between 601 and 700 is sin(688) = 0.008791022929352243

best sine between 701 and 800 is sin(710) = 6.0288706691585265E-5

best sine between 801 and 900 is sin(820) = -0.044302907677458626

best sine between 901 and 1000 is sin(999) = -0.026460752737064126

any of those numbers sem familiar?
the nifty thing is we know all these integers must be fairly close to some integer multiple of pi, therefore, by dividing each of these "best" numbers by pi we can figure out about what multiple of pi it is, and we're left with a bunch of rational approximations for pi. the best value gotten above was sin(355). 355/pi, right off my windows calculator is 113.00000959524568839590747199449 so pi is about 355/113. For this, windows calculator gives 3.1415929203539823008849557522124. Thats pretty good!

so, whats the best approximation of sin(n) for n ranging from zero to a million? :-) Lets see!
And the lucky integer is..... 833719!!! Dividing by pi you get about 265381. By my windows calculator, 833719/
265381 = 3.1415926535810777712044193065819

thats correct to 11 decimal places! :-)

Anyways, my system seems to have trouble calculating the sine of anything much higher than a milion, but I would assume if we kept searching we would find better and better numbers. But could we find as good an approximation as we wanted if we kept looking? or is it a mere matter of luck?

Last edited by mikau (2007-05-10 20:27:05)

mikau · 2007-05-10 20:19:26

hey! my system managed to calculate the smallest integer sine from 1 to 10 million just now. And the lucky number is.... 5419351, the sine of which is about 3.82*10^-8. Dividing by pi we get about 1725033. But it turns out 5419351/1725033 is only correct to 12 decimal places, which is 1 more than the last entry, even by expanding the range by 9 million. So can we in fact find one as close to we want if we just keep looking?

Well, what do YOU think? :-)

mathsyperson · 2007-05-10 23:23:17

Based on nothing but intuition, I would say yes.

As you say in your first post, it's basically just luck whether or not an integer will be close to a multiple of pi, but as we have infinite integers to chose from, then there must exist one that is closer than whatever boundary we may specify.

We can't necessarily find what it is, but we know it's got to be there.

Edit: I just realised something important. It probably doesn't matter much as far as you've gone, but if you try to refine the pi fraction much further, then the approximation of pi that your computer uses will start to become an issue.

mikau · 2007-05-11 05:04:19

I agree. Plus the sine calculations are only so accurate as well. The approximations I determined thus far yielded up to 12 digits of pi, but yeah there's only so far I can go.

But do we necessarily know its got to be there? As we've said before, pi's numbers have no known pattern but that doesn't mean its random, so is it, in fact, a mere matter of chance for some integer to fall close to it? Intuitively, it makes sense that we should keep findind better and better ones but I think it would be impossible to proove formally. Or would it?

Last edited by mikau (2007-05-11 05:12:35)

mathsyperson · 2007-05-11 06:36:05

Well, there are quite a few infinite sequences where all the terms add up to give pi. The best values of pi are calculated by huge computers summing the first few squillion terms of those.

The simplest one is probably that π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - ...
Unfortunately that one converges very slowly.

A better one is that π/6 = 1/2 + 1/(2*3*2^3) + (1*3)/(2*4*5*2^5) + (1*3*5)/(2*4*6*7*2^7) + ..., which gives slightly less than one decimal place per term.

And there are also ridiculously complicated ones that can give around 10 places with each term.

So if we got one of these sequences and summed the first n terms, we'd have an approximation for the pi fraction. The problem is that we can easily approximate a pi fraction just by doing something like 31416/10000, but as your first post says, that's not good enough.

So we'd need to find a sequence that gives decimal places of pi faster than it adds digits to the denominator of whatever fraction it's generating, and I'm not sure that one like that exists.

mikau · 2007-05-11 08:15:59

very interesting..

if you do as you say, and add a finite set of terms, 1 - 1/3 + 1/5 - 1/7 + 1/9, etc. You can combine them into a rational expression by adding the fractions. So you'd end up with two integers num and den such that pi/4 = num/dem + error. So pi = 4*num/dem + 4*error, since we can make error as small as we want by using more terms of the series, we should be able to find a rational expression for pi to any degree of accuracy.

However! If we were to find multiples of pi by the relationship, den*pi = num + error*den, for ratios of pi that have very large denominators, such as 5419351/1725033, the innacruacy would be 1725033*error, which is bad news.

So while (I think) we can proove that we can approximate pi with a rational expression to any desired level of accuracy, it does not mean that the numerator of the fraction is terribly close to some integer multiple of pi, therefore, we really can't conclude from this that we can make sin(integer) as small as we want.

Last edited by mikau (2007-05-11 08:17:39)

mathsyperson · 2007-05-11 10:17:16

Yes, I noticed that as well. Still, that doesn't disprove it, it just means that we can't use that way to prove it, unless we find a better approximating sequence.

I still think that if N can go as high as we need, then sin(N) can go as low as we want.

mikau · 2007-05-11 10:34:01

yeah, naturally that doesn't disprove it.

I agree it seems likely, but I can't say it with certainty.

Math Is Fun Forum

#1 2007-05-10 20:01:45

sin(integer) < E for any E > 0, possible?

#2 2007-05-10 20:07:05

Re: sin(integer) < E for any E > 0, possible?

#3 2007-05-10 20:19:26

Re: sin(integer) < E for any E > 0, possible?

#4 2007-05-10 23:23:17

Re: sin(integer) < E for any E > 0, possible?

#5 2007-05-11 05:04:19

Re: sin(integer) < E for any E > 0, possible?

#6 2007-05-11 06:36:05

Re: sin(integer) < E for any E > 0, possible?

#7 2007-05-11 08:15:59

Re: sin(integer) < E for any E > 0, possible?

#8 2007-05-11 10:17:16

Re: sin(integer) < E for any E > 0, possible?

#9 2007-05-11 10:34:01

Re: sin(integer) < E for any E > 0, possible?

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