anonymous
  • anonymous
When is an equation solve-able analytically? When is it not?
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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lgbasallote
  • lgbasallote
depends...
anonymous
  • anonymous
What does it depend upon? Say you have an equation involving some trigonometric terms, how do you determine that it is not analytically solve-able? Is there a way to test these, or should it be compared to a select few that have been proven to be not solve-able? Or something else?
agreene
  • agreene
what do you mean by analytically?

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anonymous
  • anonymous
I mean, an equation is solve-able analytically if an exact solution can be found via algebraic manipulation or something similar. Some equations cannot be solved analytically because they can't be manipulated into any form for which an exact solution can be found. In that case, a numberical method is needed to find an approximate solution to the equation.
agreene
  • agreene
I cant think of something which absolutely requires a numerical approximation. I can certainly think of times when if you dont do it you can spend hours--days--even years trying to get an exact answer... however, I cannot think of anything that has been proved to be true that cannot be solved without approximation (mostly because mathematicians dont like sloppy proofs)
anonymous
  • anonymous
Suppose I have some function like 50sin^10(x) + 134sin^9(x) + ... + 578sin(x) + 498cos^7(x) + ... + 17cos(x) = 0. That to me looks like something that is not solve-able analytically. Sin and cos are functions for which we can't ever find the exact value, except for certain types of input values for x. That's because sin and cos are calculated as infinite series. To solve an equation such as the above, the equation needs to be rearranged into something similar in order to find which x will solve such a thing. Also, numerical approximation in maths is not necessarily seen as sloppy. It is well known in maths that some numbers can never be known to an exact value. That's because some values have infinite, non repeating decimal expansion. Even though those things are approximated, the error of the approximation is known, and so it's not considered sloppy.
anonymous
  • anonymous
For the example I wrote above, the coefficients I have are different... I just wrote some random numbers above. I am fairly sure that thing is not analytically solve-able. But there should be a better answer for why that is than "because it looks too hard".
agreene
  • agreene
I think you have been misinformed as to an exact answer. \[\sqrt{2}\] is exact... whereas calling it 1.414... is not. (no matter to what degree of certainty) pi is exact. \[\large \cot^{1}\frac{8\pi}{\sqrt{7}}\] is also exact, whereas calling it 0.699 radians is an approximation. in your trigonometric example, it seems to me that it is some polynomial and thus can be solved via the literally hundreds of methods we have for solving polynomials. If it is just the trig functions, remember we do have a number of identities and also the inverse trig functions that can be used to cancel out certain sections if applied to both sides of the equation, etc.
agreene
  • agreene
the cot example should be the inverse cotangent, i forgot my -
anonymous
  • anonymous
Thanks for your reply. The thing is that things like pi have a symbol attached to them, and that is isolated on its own. To find a solution to something, you need to be able to say "x=....". if x=pi then you have an exact solution. But, saying "x is whatever the solution is to sin^4509(x) + cos^39(x)=0" isn't an exact solution. First you have to isolate x. There is theory which says that sometimes it can't be done. For example, the function e^(-x)*(1/x - ln x) = 0 can't be solved analytically. Any solution you get for that equation will only ever be an approximation obtained by numerical methods. The function I have is not a polynomial. They're powers of trigonometric functions. I know about the trigonometric identities, but I think they won't be able to break it down all the way to a point where I can say "x=..." because that's not always possible (as described above). I want to know when I can know that it's not possible.
agreene
  • agreene
e^(-x)*(1/x - ln x)=0 is solvable.... Assuming you know about Lambert and his W-Function... in fact the EXACT answer to this is: \[x=e^{W(1)}\] where W is the Lambert W-Function (aka inverse omega function) where: \[z=W(z)e^{W(z)}\] proven around 1760 or so.
agreene
  • agreene
because mathematics does not allow (as a community) research that is solely based on CAS (Computer Algebra Systems) or Numerical Approximation (because it is not necessary) I cannot think of anything that has been PROVEN that would be impossible to prove--that said, the Riemann Zeta Function, and its Non-Trivial Zeros would be an example whereby we MUST use approximation methods to obtain meaningful answers--HOWEVER this is an UNPROVEN hypothesis (stating that the RZF-NTZ's are the counting function for the prime numbers)
agreene
  • agreene
I meant to say... has been PROVEN that would be impossible to RESOLVE.
anonymous
  • anonymous
Hmn, we may have a different idea about what an analytic solution is. I didn't know about the W function, but I have just spent some time reading about it. It is said that it can't be expressed in terms of elementary functions. That means that for any input value, apart from a few special cases, must be calculated by an infinite series. That means those values can only ever be approximated to some specified accuracy. I think I am supposed to determine if I can say "x= some combination of elementary functions for which an exact solution can be found" and if I can do that then I say it's analytically solve-able. I'm not sure if that's what the exact definition of analytically solve-able should be, but I think that's what I am supposed to think of it as, here.
anonymous
  • anonymous
I just got onto my professor about this and he seems to say that by "solve-able analytically" he means if there is a closed form expression: http://en.wikipedia.org/wiki/Closed_form_solution Sorry if there was some confusion! I will look into that W function more later, I'd like to know how they got that answer...
agreene
  • agreene
I haven't run in to this "closed form solution" deal before, I'll have to look into it in more depth, however--from what I've read, it depends entirely on what you define as a "well known" example/solution... and I for one, take issue with the entire premise, due to the fact that EVERY known solution falls back to a single postulate and is then ENTIRELY relying on the proof of addition. What I mean is this: All functions/expressions can be written in terms of addition. Since this is true, is not every solution (even if a trivial solution or a numerical approximation) a closed form solution?
agreene
  • agreene
I say the above and then a counter-proof arrives: http://en.wikipedia.org/wiki/Galois_theory#A_non-solvable_quintic_example can't say I'm surprised by that, lol.

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