mersenneforum.org Exponential/Polynomial Hybrid Expression??
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 2009-04-27, 01:51 #1 Unregistered   23×157 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(c-d)) + ((c-d)^3)t^3 + 3((c-d)^2)t^2 + 6(c-d)t + 6 = 0 and similarly, -6e^(t(c-d)) + ((c-d)^3)t^3 + 3((c-d)^2)t^2 + 6(c-d)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
 2009-04-27, 09:13 #2 davieddy     "Lucan" Dec 2006 England 11001010010102 Posts It cries out for a successive approximation method. Do you know any? You could start by sketching the exponential and cubic expressions graphically.
 2009-04-27, 10:25 #3 J.F.     Jun 2008 4816 Posts  removed nonsense, have to look better... [edit2] I agree with davieddy's hint. Last fiddled with by J.F. on 2009-04-27 at 10:27
 2009-04-27, 10:44 #4 davieddy     "Lucan" Dec 2006 England 194A16 Posts It seems that letting (c-d)t = x might help the presentation of the problem, if not its solution.
2009-04-27, 11:00   #5
davieddy

"Lucan"
Dec 2006
England

647410 Posts

Quote:
 Originally Posted by J.F.  removed nonsense, have to look better... [edit2] I agree with davieddy's hint.
Curious to know what "nonsense" you edited

2009-04-27, 13:06   #6
J.F.

Jun 2008

23×32 Posts

Quote:
 Originally Posted by davieddy Curious to know what "nonsense" you edited
Well, I saw 2 expressions with the same exp term, so I said something like "try computing the difference" :).

2009-04-27, 15:27   #7
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by J.F. Well, I saw 2 expressions with the same exp term, so I said something like "try computing the difference" :).
Yes. If the equations were true simultaneously, one might conclude
that 6b(14-b) = 0. The same thought crossed my mind.

 2009-04-29, 10:43 #8 Kevin     Aug 2002 Ann Arbor, MI 1B116 Posts Well, the polynomial part corresponds to the first 4 terms in the Taylor expansion of 6e^(t(c-d)). 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(c-d)) + ((c-d)^3)t^3 + 3((c-d)^2)t^2 + 6(c-d)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.
2009-04-29, 13:29   #9
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by Unregistered 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(c-d)) + ((c-d)^3)t^3 + 3((c-d)^2)t^2 + 6(c-d)t + 6 = 0 and similarly, -6e^(t(c-d)) + ((c-d)^3)t^3 + 3((c-d)^2)t^2 + 6(c-d)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
There is certainly no closed form solution, even in terms of the
Lambert W function.

2009-04-30, 16:02   #11
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by ewmayer Substituting x := (c-d)t as suggested by davieddy, you have: f(x) := -6*exp(x) + x^3 + 3*x^2 + 6*x + 6 = 0
I'm a fan of giving credit where (however mildly) it's due:

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 2009-04-30 at 16:08

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