Closed Form Solution | Definition & Meaning

[koolraj09]

Hi Guys,
What does the term Closed form solution mean?

 

[Mentallic]

It means the expression is not infinite in length or can be expressed in a much shorter way.
For example, for each of the following equalities, the right side is a closed form expression while the left isn't.

1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...=e^x

1+2+3+...+n=\frac{n(n+1)}{2}

1+\frac{1}{1+\frac{1}{1+\frac{1}{...}}}=\frac{1+\sqrt{5}}{2}

 

[DarthPickley]

I think it also means you can't have an integral or summation symbol of any sort, you have to have simple functions and algebraic operations and numbers. also, what would be a closed form of the roots of x^5 - x + 1 = 0 ? you can't get it by infinite process unless you define a recursive function (such as Newton's method) that converges to it, so what is an open form of it?

 

[vikas.tyagi5]

If the solution is exact, its a closed form solution. For example, a polynomial of fourth order can be solved using quadratic formulae as base(though tedious) or using simple approximate numerical methods. If u solve for exact values(using quadratic formulae), its a closed form solution. So its obvious it can not contain any infinity terms too.

 

[JJacquelin]

what would be a closed form of the roots of x^5 - x + 1 = 0 ?

The roots of the quintic equation can be expressed on closed form, thanks to the theta functions which are related to the Jacobi elliptic functions. (Very arduous)
http://mathworld.wolfram.com/QuinticEquation.html

 

[pmsrw3]

The roots of the quintic equation can be expressed on closed form, thanks to the theta functions which are related to the Jacobi elliptic functions. (Very arduous)
http://mathworld.wolfram.com/QuinticEquation.html

But this illustrates one of the difficulties in answering the question. The phrase "closed-form expression" is ambiguous -- it depends on what you define as allowable elementary functions. I mean, if I'm allowed to define new functions, I can express anything in closed form -- I just define a function whose value is the solution to my problem. This sounds like cheating, but that's basically where most of our more recondite special functions come from: elliptic functions, Bessel function, Hypergeometric functions, etc. There was no closed-form expression to the problem so some old 19th century German said, "I hereby define this function to be the solution." And presto! the problem had a closed-form solution.

 

[JJacquelin]

But this illustrates one of the difficulties in answering the question. The phrase "closed-form expression" is ambiguous -- it depends on what you define as allowable elementary functions. I mean, if I'm allowed to define new functions, I can express anything in closed form -- I just define a function whose value is the solution to my problem. This sounds like cheating, but that's basically where most of our more recondite special functions come from: elliptic functions, Bessel function, Hypergeometric functions, etc. There was no closed-form expression to the problem so some old 19th century German said, "I hereby define this function to be the solution." And presto! the problem had a closed-form solution.

I agree, but even if this way to see things is correct it is still incomplete. Of course, it should be too easy to define a new special function as the solution of a problem and then, to say : The problem as a solution which is expressed thanks to the new special function !
Special functions are more than that. For exemple, consider the Riemann zeta function.
If you say : I define a closed-form expression f(x) for the infinite series of general term 1/n^x (for n=1 to infinity), the function f(x) is defined only for x>1 since the series doesn't converges for x<1. However zeta(x) is defied for any real x (except x=1) and even for complex x. The special function zeta covers much more background : integral definition of the function, analytic continuation and much more.
In fact, when we use a closed-form expression to express the solution of a problem, we refer to a background of knowledge and we give a relationship to standard functions, i.e. functions which have been widely studied.
If the solution of a problem is only used to define à new special function, a relationship is not established to any previous background. So, this supposedly "closed-form" is useless.
A funny example is given as a preamble in the paper "Sophomores Dream Function" (pp.2-3). By the link :
http://www.scribd.com/JJacquelin/documents
This is also a main theme in the paper "Safari au pays des fonctions speciales" ( not translated yet), same link.

 

[pmsrw3]

I agree, but even if this way to see things is correct it is still incomplete.

Yes -- I was exaggerating a bit for effect. Actually, elliptic functions are an interesting case, because, although it is true that Jacobi developed them essentially as I described, as an easy-out function to solve a particular class of problems, they're more broadly applicable. I doubt he was thinking of solving quintics.