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#1 2010-03-19 04:56:59

GOKILL
Member
Registered: 2010-03-19
Posts: 26

Unusual equality :))

1. Find the solution of


(i think there's no solution, but i'm not sure how to prove it analitically big_smile )

2. Find the area between the curves

and

(i feel the difficulty was in finding the point where they're crossing over dunno )


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#2 2010-03-19 11:07:27

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: Unusual equality :))

Not sure if this is rigourous enough for you, but:
e^x > 0 for all x, so clearly there is no negative solution.
d(e^x)/dx = e^x, which is >1 for all x>0. Since e^0 = 1 > 0 and dx/dx = 1, there can't be any solution greater than 0 either. Hence there is no solution to e^x = x.


Why did the vector cross the road?
It wanted to be normal.

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#3 2010-03-19 11:27:02

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Unusual equality :))

mathsyperson, I believe #1 is supposed to be a question in complex analysis (this is why z is used).

As to the answer, it's an application of Rouche's theorem.  The "hint" below is really an answer...

This proves that there is exactly one value where e^z = z.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#4 2010-03-19 13:16:53

GOKILL
Member
Registered: 2010-03-19
Posts: 26

Re: Unusual equality :))

at first i think it was in real, so there's no solution.
i don't know how it would be in case of z is in complex..
so would u explain f(z) = e^z - z, g(z) = z ?
what's the answer?


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#5 2010-03-20 10:47:10

juriguen
Member
Registered: 2009-07-05
Posts: 59

Re: Unusual equality :))

Hi there!


I actually got help from quickmath.com, and then checked what did the result mean. The idea is to use what is called the Lambert W function, which, by definition (Wikipedia) allows to express any complex number as:

Then, for your equation you have to do:

But any number can be written using the W function, so:

By equating both expressions, then:


which actually has |z| = 1.3745, so the condition is satisfied.


The other problem I cannot get it yet. I will keep trying smile

Ok, so I would say you have to do this numerically. First plot both functions, to have an idea. then use Newton-Rapson's method to obtain when they intersect. For instance, to calculate the area in the interval [0, 3pi/2], first lets find where they cross:

Say we want to find

Using N-R, from a guess xn, we obtain a better guess doing:

Choose as x0 = 4.5, and a few iterations give x = 4.4934

I have calculated only the area between [0, 4.4934], which is:


Hope this helps!
Jose

Last edited by juriguen (2010-03-20 11:07:39)


“Make everything as simple as possible, but not simpler.” -- Albert Einstein

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#6 2010-03-22 18:40:48

scientia
Member
Registered: 2009-11-13
Posts: 224

Re: Unusual equality :))

juriguen wrote:

I actually got help from quickmath.com, and then checked what did the result mean. The idea is to use what is called the Lambert W function, which, by definition (Wikipedia) allows to express any complex number as:

Then, for your equation you have to do:

But any number can be written using the W function, so:

By equating both expressions, then:


which actually has |z| = 1.3745, so the condition is satisfied.

This only proves that a solution exists. I think the OP wants to find all complex solutions of the equation.

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#7 2010-03-23 01:50:07

juriguen
Member
Registered: 2009-07-05
Posts: 59

Re: Unusual equality :))

Hi!

I think there's only 1 solution, and I wrote it down, z = -W(-1)

Right?


“Make everything as simple as possible, but not simpler.” -- Albert Einstein

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#8 2010-03-24 15:14:41

GOKILL
Member
Registered: 2010-03-19
Posts: 26

Re: Unusual equality :))

Some one has sugest me to use rouche's theorem but i don't know bout it >_<


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