How come some equations can't be solved for an exact answer?

[mathwonk]

let me pose a simple question. do you think pi has an exact value? if so, what is it?let me make it easier for you, what is the exact value of sqrt(2)?

if you think about it you may realize I am trying to discern what is your definition of the phrase "exact answer".

by the way matt, i understand that you are busy, but i still convey to you the fact that many people including me miss you here a great deal, and there is a whole, thread called "where is matt grime?!"

 

[LukeD]

As to this exact question, I don't know if there is any "simple formula" that would be close to what you mean, but sometimes there are.

For instance, for pi, there is
<br /> \pi=3 + \cfrac{1}{6 + \cfrac{9}{6 + \cfrac{25}{6 + \cfrac{49}{6 + \cfrac{81}{6 + \cfrac{121}{\ddots\,}}}}}}<br />(taken from Wikipedia's article on continued fractions)

The functions in the formula you're asking about having an exact solution do have nice formulas of the sort you want, so it might be that your question might have a formula of that sort.

However, in general, no one knows if there is a "nice formula" for your problem. And really, the only reason for that is because either no one has yet figured one out or no one has been trying to find one. I don't know if it's the case that every real number will be able to written in some nice formula though.

 

[CGUE]

To 500 places:
0.3354180323849400594578624117492548397118513151006308650233458056997707969
93488867326823297805099655941760946007417735305449140296885174971232582467743747
67817725202600726615999809586989494215709825435975337739655562116739626518112856
23055310151600910291959720327473494910708608367675240373868492976882512228069498
07934021595150112613380719098738548853419012578533010207808040687825471704745279
60925128393175327764000215850832461166054705141754704841341722301915669142357206
871814814939461336135943871

Haha win, don't we all love Mathematics Computing Packages ?
(Or Biological geniuses by a slim chance if you calculated that in your head?)

Maple 11 output:

>Digits:=500;

Digits := 500

> fsolve(2*x+sin(x)=1);

0.33541803238494005945786241174925483971185131510063086502334580\
5699770796993488867326823297805099655941760946007417735305\
4491402968851749712325824677437476781772520260072661599980\
9586989494215709825435975337739655562116739626518112856230\
5531015160091029195972032747349491070860836767524037386849\
2976882512228069498079340215951501126133807190987385488534\
1901257853301020780804068782547170474527960925128393175327\
7640002158508324611660547051417547048413417223019156691423\
57206871814814939461336135943871

I should have asked my friend to go solve an equation like this in a computing package like this instead of asking me in the middle of a lecture.

 

[CRGreathouse]

However, in general, no one knows if there is a "nice formula" for your problem. And really, the only reason for that is because either no one has yet figured one out or no one has been trying to find one. I don't know if it's the case that every real number will be able to written in some nice formula though.

It's awfully hard to determine if there's a nice formula without a working definition for "nice formula". :)

But a simple counting argument will show that almost all real numbers have no formula, where for a given finite set of symbols (0, 1, +, -, *, /, etc.) a "formula" is some subset of the finite strings of those symbols.

 

[CRGreathouse]

Haha win, don't we all love Mathematics Computing Packages?

Yep. This one was a custom implementation of the secant method in Pari (v. 2.4.2; won't work in 2.3.x unless you remove ff from the variable list and predefine it):

secant_root(ff, first, second)={
	local(ff1, ff2, oldfirst, ep);
	ff2 = ff(second);
	ep = 2 * eps();
	while (abs(first - second) > ep,
		ff1 = ff(first);
		oldfirst = first;
		first = first - (first - second) / (ff1 - ff2) * ff1;
		second = oldfirst;
		ff2 = ff1;
	);
	first
}

\\ 2 ^ -(decimal_precision * log(10)/log (2) / 32) * 32 - 1)
eps()={
	precision(2. >> (32 * ceil(precision(1.) * 0.103810252965230073370947482171543443)), 9)
}

I also have bisection, Newton's method, and Halley's method, but these are slower for this problem (and probably slower is most cases). Higher-order convergence isn't all it's cracked up to be when it means you need to calculate more functions!

 

[mathwonk]

NOW THAT YOU have defined an exact answer as a limit of a nice sequence, we ought to take a vote on how many think the usually convergent sequence arising from Newton's method is nice enough to qualify.