Thread View: alt.uu.math.misc
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Started by Stefano Baraldi
Tue, 29 Apr 1997 00:00
HELP REQUEST: Need good method to solve equations of (exp(x)>x) type
Author: Stefano Baraldi
Date: Tue, 29 Apr 1997 00:00
Date: Tue, 29 Apr 1997 00:00
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Hi all!
I'm going to give my Calculus exam next month, i still have real
doubts about solving equations of this type (i don't knoe their exact
name)
exp(x)>x, ln(x+a)<tan(x), ....
and so on...
I know the solution should be found graphically, please could you
point me to books (available online) or tell me the method to do it?
Up to now i made a simple comparing algorithm on my sci-calculator,
but that's not so ortodox.. :))
thank you and bye!
See ya,
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Re: HELP REQUEST: Need good method to solve equations of (exp(x)>x) type
Author: Warren
Date: Wed, 30 Apr 1997 00:00
Date: Wed, 30 Apr 1997 00:00
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Stefano, A good book on solving these types of equations iteratively is Burden and Faires, Numerical Analysis. I have the 2nd and 3rd editions, but I think there may be a later edition. They discuss several methods. The method you are using might fit into something called the bisection method. Nothing at all wrong with the method except it could take some time. There is also a method called the Newton-Raphson method to find the solution of f(x)=0. For Calculus students, this would be a great example to tie some of the concepts together. Burden and Faires introduce the method using a Taylor series expansion so that f(x)+(x-x_bar)f'(x_bar) ~~ 0, assuming the function is twice differentiable on some closed set and ignoring the error term. Guess at the solution then iterate until some convergence criteria is met. The iterative relationship is p_n = p_n-1 - f(p_n-1)/f'(p_n-1). baro@pt.tizeta.it (Stefano Baraldi) wrote: >Hi all! > >I'm going to give my Calculus exam next month, i still have real >doubts about solving equations of this type (i don't knoe their exact >name) > >exp(x)>x, ln(x+a)<tan(x), .... > >and so on... >I know the solution should be found graphically, please could you >point me to books (available online) or tell me the method to do it? > >Up to now i made a simple comparing algorithm on my sci-calculator, >but that's not so ortodox.. :)) > >thank you and bye! > >See ya, >--------------------------------------------------------------------------------- > ------<<<<<< T H E B A R D >>>>>>------ >--------------------------------------------------------------------------------- >
Re: HELP REQUEST: Need good method to solve equations of (exp(x)>x) type
Author: Greg Heath
Date: Wed, 30 Apr 1997 00:00
Date: Wed, 30 Apr 1997 00:00
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Date: 30 Apr 1997 14:19:37 GMT Message-ID: <5k7kdp$6c8$1@www.umsmed.edu> From Warren <wlmay@umsmed.edu>: > A good book on solving these types of equations iteratively is > Burden and Faires, Numerical Analysis. I have the 2nd and 3rd editions, > but I think there may be a later edition. > Guess at the solution then iterate until some convergence criteria is > met. The iterative relationship is > > p_n = p_n-1 - f(p_n-1)/f'(p_n-1). In very simple cases, the equation can be rearranged to yield x = f(x) where |f'(x*)| < 1 at the solution x*. Then the solution can be found by simply iterating x(n) = f(x(n-1)). For example, you can solve x - invcos(x) = 0 by rearranging to get x = cos(x) which obviously has a solution between 0 and PI/2 ~ 1.57. Arbitrarily choosing x(0) = 0.8, just keep pushing the cos button on your calculator until the display converges. Vary x(0) and see what happens. Also try x(n) = invcos(x(n-1)). Similarly, solve the equation x + ln(x) = 0. In general, the technique is inferior. However, it is easy and, for the inexperienced, can be both fun and instructive. P.S. I don't think you'll have much fun with x - exp(x) = 0. Greg Hope this helps. Gregory E. Heath heath@ll.mit.edu The views expressed here are M.I.T. Lincoln Lab (617) 981-2815 not necessarily shared by Lexington, MA (617) 981-0908(FAX) M.I.T./LL or its sponsors 02173-9185, USA
Re: HELP REQUEST: Need good method to solve equations of (exp(x)>x) type
Author: Zdislav V. Kovarik
Date: Thu, 01 May 1997 00:00
Date: Thu, 01 May 1997 00:00
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Unless something got distorted during transmission, I saw that an
inequality (namely exp(x)>x -- see below)
was called an
equation.
For your students' sake, get the terminology straight. (Some of my
students do that, too; even calling expressions such as 3*x-5 equations.
I wonder why...)
And: some inequalities are satisfied everywhere (such as exp(x)>x for
real x), some on proper subsets of the declared or assumed domain.
How does one prove that exp(x)>x everywhere? Not merely by plotting many
points of the graph (that is good as a hint), but in this case, by
finding (or estimating from below) the minimum of exp(x)-x. This is a
good place for using Calculus. (BTW, the minimum is 1. Why?)
And yes, solving inequalities such as exp(x)>5*x is helped by finding the
solutions of exp(x)=5*x first, but the students should have heard
at least something about continuity, differentiability and their
consequences (Rolle's Theorem to show that there are no more than 2
solutions, etc.)
How much to put into the exam depends, of course, on how much Calculus
information and skills the students have acquired. Is it really Calculus,
or are the students still in the (often regrettably underpracticed)
pre-Calculus part?
Good luck, ZVK (Slavek).
In article <5k5u04$kv2$1@qualcuno.nettuno.it>,
Stefano Baraldi <baro@pt.tizeta.it> wrote:
:Hi all!
:
:I'm going to give my Calculus exam next month, i still have real
:doubts about solving equations of this type (i don't knoe their exact
:name)
:
:exp(x)>x, ln(x+a)<tan(x), ....
:
:and so on...
:I know the solution should be found graphically, please could you
:point me to books (available online) or tell me the method to do it?
:
:Up to now i made a simple comparing algorithm on my sci-calculator,
:but that's not so ortodox.. :))
:
:thank you and bye!
[...]
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