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

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  • #1
ehj
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I'm wondering why it is that some simple equations can't be solved for an exact answer, but must be calculated numerically. For instance solving the following for x:
1 = sin(x) + 2x
And does it imply that an exact answer doesn't even exist? (Or maybe there is an exact answer but my current math skills just don't allow for solving it :)?)
 

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  • #2
daniel_i_l
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Usually (as in your example) you can prove that an exact answer exists even if you don't know how to find it.
 
  • #3
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The equation 1 = sin(x) + 2x has as solutions the abscissas of the points of intersection of the graphs of the functions f(x) = -2x + 1 and g(x) = sin(x). By logic, these graphs intersect exactly once. I do not see a way of writing x = h where h is a finite numerical expression composed of numbers and elementary functions of numbers, but that has nothing to do with the fact that the two curves intersect at a unique point.
 
  • #4
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If the answer exists, then there's an exact solution (or several). Unfortunately, many times, the exact answer is not known, but when the function is continuous, an approximate answer can be good enough. In the case of the function you're asking about, the solution is usually a transcendental (like pi or e), which is to say that it's not a fraction, and it's not the root to any polynomial. Therefore, there is no common way of writing it other than as the inverse to your function (though you might be able to write it in terms of a complex logarithm, but that doesn't sounds like what you'd want)
 
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  • #5
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try .335418 that should get you pretty damn close :D. (argument in sine is in radians)
 
  • #6
lurflurf
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try
3x-1=0
or
x^3+3x^2-180x+60
for a first approximation.
 
  • #7
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Going back to the original question regarding simple equations, is sin(x) simple to represent algebraically? If an equation has a solution, it has an exact solution. We may not be able to find it algebraically, which means we do not know of a way to represent it using any of the algebraic symbols or special mathematical numbers like [itex]\pi[/itex] or [itex]e[/itex]. It could, for example, be a very weird irrational.
 
  • #8
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The great mathematician Gauss was known to say that he did not consider it very important to find a symbol to represent the solution to an equation.

Historians have later considered this a kind of "sour grapes" attitude (wrt the results of Galois theory), much like the atittude that Gauss had towards Fermat's last theorem: "I could easily come up with a number of similar propositions that would likewise be exceeedingly difficult to prove; the only interest I have in (FLT) is if it should appear as a corollary of some more general result."
 
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  • #9
CRGreathouse
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To 500 places:
0.3354180323849400594578624117492548397118513151006308650233458056997707969
93488867326823297805099655941760946007417735305449140296885174971232582467743747
67817725202600726615999809586989494215709825435975337739655562116739626518112856
23055310151600910291959720327473494910708608367675240373868492976882512228069498
07934021595150112613380719098738548853419012578533010207808040687825471704745279
60925128393175327764000215850832461166054705141754704841341722301915669142357206
871814814939461336135943871

:approve:
 
  • #10
Hurkyl
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If the answer exists, then there's an exact solution (or several). Unfortunately, many times, the exact answer is not known
What you meant to say was
Unfortunately, many times, we want to express the exact answer in a specific form, and it is not known how to do so (or even impossible)​
 
  • #11
CRGreathouse
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Given the integers and the operators + - / *, we can't give a closed-form solution to [itex]x^2=2[/itex]. With the exponent operator, this becomes possible, but [itex]2^x=3[/itex] still has no closed form. Adding in the logarithm function, this has a closed form; but [itex]xe^x=3[/itex] doesn't. Adding in the Lambert W-function, this has a closed form, but [itex]1 = \sin(x) + 2x[/itex] doesn't. If you keep on adding in function symbols you can get 'closed forms' for anything -- but the essential problem of calculating the root is unchanged. Even finding the solution to [itex]7x=1[/itex] requires an algorithm, as does [itex]x^2=7[/itex]; computing them is quite different from writing down a symbolic form.

If you had a function symbol
[tex]\mathcal{Q}(a,b)[/tex] that gave the zero in x (least absolute value, positive if two) for [tex]\sin(x)+ax+b[/tex]
then would you feel better to know that the question has answer [itex]\mathcal{Q}(2,-1)[/itex]?
 
  • #12
If you've got the free time https://www.amazon.com/dp/1402021860/?tag=pfamazon01-20 is a great way to learn about the "unsolvability" of all but the most simple equations.

It should be noted though that methods like newtons method, etc, are often faster at finding answers to third and fourth order problems than exact formulae. With this in mind, using them to find accurate approximate answers to equations should not be viewed as a crippling setback. We often forget that even "exact" answers like [tex]\sqrt{2}[/tex] cannot be represented exactly in an arithmetical calculation, so approximating even the answers to "solvable" equations is something we inevitably have to do anyway.
 
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  • #13
Hurkyl
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We often forget that even "exact" answers like [tex]\sqrt{2}[/tex] cannot be represented exactly in an arithmetical calculation,
Yes it can. In fact, you did so in this every sentence. :tongue: I think what you meant to say is
[itex]\sqrt{2}[/itex] cannot be represented exactly by a terminating decimal number​
 
  • #14
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...
If you had a function symbol
[tex]\mathcal{Q}(a,b)[/tex] that gave the zero in x (least absolute value, positive if two) for [tex]\sin(x)+ax+b[/tex]
then would you feel better to know that the question has answer [itex]\mathcal{Q}(2,-1)[/itex]?
I find this to be a clever response :cool:
 
  • #15
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If you had a function symbol
[tex]\mathcal{Q}(a,b)[/tex] that gave the zero in x (least absolute value, positive if two) for [tex]\sin(x)+ax+b[/tex]
then would you feel better to know that the question has answer [itex]\mathcal{Q}(2,-1)[/itex]?
Yes, in fact it makes me feel smug and superior that I can go into a room and be confident that most likely only I have enough mathematical training to know what the Q function is. :wink:
 
  • #16
CRGreathouse
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Yes, in fact it makes me feel smug and superior that I can go into a room and be confident that most likely only I have enough mathematical training to know what the Q function is. :wink:
Ooh, this tempts me to write an article for the PF Library defining the Q function so your post gets auto-linked.
 
  • #17
mathwonk
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what do you mean by an "exact answer" what is the exact solution to X^2 = 2? and i do not mean a trivial answer like sqrt(2), which translates as "the exact number which solves this problem".

for that matter what do you consider a number? i.e., is it a finite decimal? a sequence of approximations? so would a newton's process for approximating an answer satisfy you?

if so, then one can solve most equations. if not, then one cannot even solve the one above.
 
  • #18
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But you can express many answers as an infinte sum of polynomials. Is this possible for all solutions?
 
  • #19
mathwonk
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what do you mean? all real numbers, hence all solutions, can be expressed as an infinite sum of decimals.

and i have already alluded to the process of using newtons approximations, which "solve" a wide variety of equations.
 
  • #20
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Sorry I wasnt clear enough. I meant if there is possible to find a solution on the form of an infinite series to all eqations, since that would then be an exact answer, not an approximation. I.e. is there possible to find a solution to all equations on the form:
[tex]\sum_{k=0}^{\infty} f(k)[/tex]
 
  • #21
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Sorry I wasnt clear enough. I meant if there is possible to find a solution on the form of an infinite series to all eqations, since that would then be an exact answer, not an approximation. I.e. is there possible to find a solution to all equations on the form:
[tex]\sum_{k=0}^{\infty} f(k)[/tex]
Any solution to an equation of the type you are asking about is a real (or maybe complex) number, and every real (and complex) number can be written in that form because the real and complex numbers are complete
 
  • #22
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Any solution to an equation of the type you are asking about is a real (or maybe complex) number, and every real (and complex) number can be written in that form because the real and complex numbers are complete
Yes I can see that, but is it always possible to find that form? I mean as stated earlier there is ofcourse an exact number that is the solution to for example 1 = sin(x) + 2x but it isnt possible to express it in terms of elementary functions. But if we allow the use of an infinite sum of elementary functions, is it then always possible to solve any equation and get an exact answer?
 
  • #23
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I find this to be a clever response :cool:
Agree very much in this. Maybe it was this Hermann Weyl meant by "We are left with
our symbols" - although he of course meant symbols representing more advanced things.


By using symbols everything can be solved exactly and explicitely. Define a symbol for the problem and the problem is solved using this symbol. Maybe even integers can be regarded as nondefined symbols in some way? I.e a numerical exact answer may be regarded as fundamentally resting on the integer symbols (1, 2, 3, ...) contained?

Symbols maybe the true clue to progress of mankind. :bugeye:
 
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  • #24
CRGreathouse
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But if we allow the use of an infinite sum of elementary functions, is it then always possible to solve any equation and get an exact answer?
Sure, in the sense that you can express any real number as its decimal expansion.

If the equation uses only elementary functions, then you could even encode (presumably) the bisection formula.
 
  • #25
mathwonk
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i am puzzled. you seem to think that writing down an infinite series, represents an "exact" answer, when to me it is only a (convergent) sequence of approximations. any convergent sequence of approximations is as exact as that, no? such as a convergent newtons process.

if what you say were true, it should be possible to say whether the sum of a series were say rational or not. so if you believe the values of the zeta function are exact expressions, are they rational?
 

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