20090427, 01:51  #1 
2^{3}×157 Posts 
Exponential/Polynomial Hybrid Expression??
Hello, I am taking a differential equations class and I am a little stuck on a particular step. I am trying to solve for t. I have the following expressions:
6e^(t(cd)) + ((cd)^3)t^3 + 3((cd)^2)t^2 + 6(cd)t + 6 = 0 and similarly, 6e^(t(cd)) + ((cd)^3)t^3 + 3((cd)^2)t^2 + 6(cd)t + 6 = 6b(14 + b) Note that b, c and d are constants, and their average values are known (they are probabilities). This problem seems extremely difficult because it has both an exponential and polynomial that include the independent variable. I have tried a couple methods but I seem unable to figure out how to solve for t. In fact, I do not believe there is an elementary method of solving for t unless I am missing something obvious. It should also be noted that my professor has a habit of assigning problems that are beyond this class. Can anyone provide some hints that would prove beneficial? Any help would be greatly appreciated. I assume that both expressions have very similar methods of solving. Thanks, a puzzled math student 
20090427, 09:13  #2 
"Lucan"
Dec 2006
England
1100101001010_{2} Posts 
It cries out for a successive approximation method.
Do you know any? You could start by sketching the exponential and cubic expressions graphically. 
20090427, 10:25  #3 
Jun 2008
48_{16} Posts 
[edit]
removed nonsense, have to look better... [edit2] I agree with davieddy's hint. Last fiddled with by J.F. on 20090427 at 10:27 
20090427, 10:44  #4 
"Lucan"
Dec 2006
England
194A_{16} Posts 
It seems that letting (cd)t = x might help the presentation
of the problem, if not its solution. 
20090427, 11:00  #5 
"Lucan"
Dec 2006
England
6474_{10} Posts 

20090427, 13:06  #6 
Jun 2008
2^{3}×3^{2} Posts 

20090427, 15:27  #7 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 

20090429, 10:43  #8 
Aug 2002
Ann Arbor, MI
1B1_{16} Posts 
Well, the polynomial part corresponds to the first 4 terms in the Taylor expansion of 6e^(t(cd)). You might be able to get some information by looking at the Taylor expansion of the function involved and get some information out of that. For example, if you set f(t)=6e^(t(cd)) + ((cd)^3)t^3 + 3((cd)^2)t^2 + 6(cd)t + 6, then f(0)=0 and looking at the Taylor expansion makes it obvious that f'(t)>0 for t>0, so you don't have any positive solutions.

20090429, 13:29  #9  
Nov 2003
2^{2}×5×373 Posts 
Quote:
Lambert W function. 

20090429, 21:47  #10 
∂^{2}ω=0
Sep 2002
República de California
2×3×1,931 Posts 
Substituting x := (cd)t as suggested by davieddy, you have:
f(x) := 6*exp(x) + x^3 + 3*x^2 + 6*x + 6 = 0 and g(x) := 6*exp(x) + x^3 + 3*x^2 + 6*x + 6  6*b*(14 + b) = 0. The first one is easy: Rewriting as 6*exp(x) = x^3 + 3*x^2 + 6*x + 6, we see by inspection (a.k.a. "lucky guess") that x = 0 is a solution. (In fact the unique solution). More generally, we could use a standard Newton iteration scheme, i.e. pick a starting guess for a solution of the first equation, x0, then iterate as x_{k+1} = x_{k}  f(x_{k})/f'(x_{k}) . In your case f' = 6*exp(x) + 3*x^2 + 6*x + 6; Starting with (say) x = 1 (x = 0 won't work because f'(0) = 0), here is the sequence of iterates: x0 = 1 x1 = 0.7635388975951208073691778049 x2 = 0.5804051861431977391558499448 x3 = 0.4397186311643678335652810243 x4 = 0.3322945610689959985673151052 x5 = 0.2506394392087383621381660155 x6 = 0.1887812521961736894907693542 x7 = 0.1420384340213115156125680594 x8 = 0.1067839994724474469983559004 x9 = 0.08023180409699953043063747583 x10= 0.06025485561260824648628371975, etc. To anyone familiar with Newton`s method this seems odd, because the method is supposed to converge quickly (specifically, quadratically fast) once we are close to a root of the equation in question, but since we already found x = 0, we see that the problem is that f' = 0 at the true root, which is what causes the convergence to become subquadratic. The same zeroderivative issue should not be a problem for your 2nd equation as long as the righthand side is nonzero, i.e. b is not equal to 0 or 14, but only certain ranges of b admit of a solution. For instance, letting B := 6*b*(14 + b), for B = 1 we get a solution x = 1.315946600918394508719581079... You can usually get faster and more reliable convergence by using Halley's 3rdorder method (cf. the bottom of this page): x_{k+1} = x_{k}  f f'/[(f')^2  f f''/2], where all the fterms are evaluated at the current iterate x_{k}, obviously. That gives quadratic convergence to zero for your first equation (again, this is one order slower than the cubic convergence expected for Halley applied to "nice" functions), and should allow you to investigate the second question, that is, which values of B make the 2nd equation admit of a solution. Last fiddled with by ewmayer on 20090429 at 21:48 
20090430, 16:02  #11  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
If Kevin's astute observation was correct, we get (on expanding exp(x)) a series whose lowest power is x^4. Although we might have taken a homework thread and run with it, it would be nice to hear from the OP again, if only to say "THX but I don't understand a word you're saying". David PS I think OP's "professor" may well be guilty as charged. Last fiddled with by davieddy on 20090430 at 16:08 

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