mersenneforum.org  

Go Back   mersenneforum.org > New To GIMPS? Start Here! > Homework Help

Reply
 
Thread Tools
Old 2009-04-27, 01:51   #1
Unregistered
 

23×157 Posts
Default 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
  Reply With Quote
Old 2009-04-27, 09:13   #2
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

11001010010102 Posts
Default

It cries out for a successive approximation method.
Do you know any?
You could start by sketching the exponential and cubic
expressions graphically.
davieddy is offline   Reply With Quote
Old 2009-04-27, 10:25   #3
J.F.
 
J.F.'s Avatar
 
Jun 2008

4816 Posts
Default

[edit]
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
J.F. is offline   Reply With Quote
Old 2009-04-27, 10:44   #4
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

194A16 Posts
Default

It seems that letting (c-d)t = x might help the presentation
of the problem, if not its solution.
davieddy is offline   Reply With Quote
Old 2009-04-27, 11:00   #5
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

647410 Posts
Default

Quote:
Originally Posted by J.F. View Post
[edit]
removed nonsense, have to look better...
[edit2]
I agree with davieddy's hint.
Curious to know what "nonsense" you edited
davieddy is offline   Reply With Quote
Old 2009-04-27, 13:06   #6
J.F.
 
J.F.'s Avatar
 
Jun 2008

23×32 Posts
Default

Quote:
Originally Posted by davieddy View Post
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" :).
J.F. is offline   Reply With Quote
Old 2009-04-27, 15:27   #7
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2·3·13·83 Posts
Default

Quote:
Originally Posted by J.F. View Post
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.
davieddy is offline   Reply With Quote
Old 2009-04-29, 10:43   #8
Kevin
 
Kevin's Avatar
 
Aug 2002
Ann Arbor, MI

1B116 Posts
Default

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.
Kevin is offline   Reply With Quote
Old 2009-04-29, 13:29   #9
R.D. Silverman
 
R.D. Silverman's Avatar
 
Nov 2003

22×5×373 Posts
Default

Quote:
Originally Posted by Unregistered View Post
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.
R.D. Silverman is offline   Reply With Quote
Old 2009-04-29, 21:47   #10
ewmayer
2ω=0
 
ewmayer's Avatar
 
Sep 2002
Rep├║blica de California

2×3×1,931 Posts
Default

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
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

xk+1 = xk - f(xk)/f'(xk) .

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 zero-derivative issue should not be a problem for your 2nd equation as long as the right-hand 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 3rd-order method (cf. the bottom of this page):

xk+1 = xk - f f'/[(f')^2 - f f''/2], where all the f-terms are evaluated at the current iterate xk, 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 2009-04-29 at 21:48
ewmayer is offline   Reply With Quote
Old 2009-04-30, 16:02   #11
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2·3·13·83 Posts
Default

Quote:
Originally Posted by ewmayer View Post
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
davieddy is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
exponential growth patterns MattcAnderson Puzzles 3 2015-10-04 02:33
exponential distribution davieddy Puzzles 10 2010-05-25 03:43
Exponential Inequality Primeinator Math 13 2009-11-05 17:30
Exponential prime search Citrix Prime Sierpinski Project 37 2009-08-17 06:32
Exponential Digits ndpowell Math 18 2005-07-15 22:31

All times are UTC. The time now is 11:53.

Mon Jan 25 11:53:09 UTC 2021 up 53 days, 8:04, 0 users, load averages: 3.89, 3.06, 2.68

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.