Finding the Intersection Of 2 Equations (difficult)

  • #1

Homework Statement


Solve the following equation:

[tex]e^{2x}=3x^2[/tex]


Homework Equations





The Attempt at a Solution


I can find an approximate solution with a graphing calculator easily, but I am interested how you would find the exact solution.

I can take the natural log of both sides and wind up with.

[tex]2x=\ln{3x^2}[/tex]

I'm not sure how I could proceed from here to get it into terms of x.

Would perhaps another method be better?
 

Answers and Replies

  • #2
Mentallic
Homework Helper
3,798
94
There is no analytic solution to this problem. You can solve it in terms of the lambert W function though.

The lambert W function is as follows:
if [tex]a=ue^u[/tex] then [tex]u=W(a)[/tex] where u is some function of x and a is a constant.

[tex]3x^2=e^{2x}[/tex]

[tex]3x^2e^{-2x}=1[/tex] divided through by e2x

[tex]xe^{-x}=\frac{1}{\sqrt{3}}[/tex] divided through by 3 then square root of both sides

[tex]-xe^{-x}=\frac{-1}{\sqrt{3}}[/tex] negative of both sides.

So now we have it in the lambert W form and thus the solution is

[tex]-x=W\left(\frac{-1}{\sqrt{3}}\right)[/tex]

[tex]x=-W\left(\frac{-1}{\sqrt{3}}\right)[/tex]

But again, we can't express the solution in terms of elementary functions so you will only get a good approximation at best for the solution.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966
Mentallic, I'm not sure why you would say using the Lambert W function (which you did very nicely, by the way) would not be an "analytic" solution. Surely, it is as analytic as, say, using the function ln(x) to solve [itex]e^x= 2[/itex].
 
  • #4
Mentallic
Homework Helper
3,798
94
Ahh I think I've mixed up the meaning of analytic with 'expressible in terms of elementary functions', my mistake :biggrin:
Thanks!
 
  • #5
There is no analytic solution to this problem. You can solve it in terms of the lambert W function though.

The lambert W function is as follows:
if [tex]a=ue^u[/tex] then [tex]u=W(a)[/tex] where u is some function of x and a is a constant.

[tex]3x^2=e^{2x}[/tex]

[tex]3x^2e^{-2x}=1[/tex] divided through by e2x

[tex]xe^{-x}=\frac{1}{\sqrt{3}}[/tex] divided through by 3 then square root of both sides

[tex]-xe^{-x}=\frac{-1}{\sqrt{3}}[/tex] negative of both sides.

So now we have it in the lambert W form and thus the solution is

[tex]-x=W\left(\frac{-1}{\sqrt{3}}\right)[/tex]

[tex]x=-W\left(\frac{-1}{\sqrt{3}}\right)[/tex]

But again, we can't express the solution in terms of elementary functions so you will only get a good approximation at best for the solution.

hmm. Interesting.

I'll bet If you wanted to take this even further you could use the series expansion for the lambert w function.

[tex]w(x)=\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}x^{n}}{(n-1)!}[/tex]

[tex]-w(\frac{-1}{\sqrt{3}})=-\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}(\frac{-1}{\sqrt{3}})^{n}}{(n-1)!}[/tex]

Otherwise a computer would be required to calculate that inverse function. This would allow you to compute the solution to any degree of accuracy. Right?
 
  • #6
Mentallic
Homework Helper
3,798
94
hmm. Interesting.

I'll bet If you wanted to take this even further you could use the series expansion for the lambert w function.

[tex]w(x)=\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}x^{n}}{(n-1)!}[/tex]

[tex]-w(\frac{-1}{\sqrt{3}})=-\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}(\frac{-1}{\sqrt{3}})^{n}}{(n-1)!}[/tex]

Otherwise a computer would be required to calculate that inverse function. This would allow you to compute the solution to any degree of accuracy. Right?

I really need to learn about series expansions, they seem to exist in numerous forms for every possible function!

You could even simplify the summation,

[tex](-1)^{n-1}\left(\frac{-1}{\sqrt{3}}\right)^n=(-1)^{n-1}(-1)^n.3^{-n/2}=(-1)^{2n-1}.3^{-n/2}=-3^{-n/2}[/tex]

therefore the sum is:

[tex]
-W\left(\frac{-1}{\sqrt{3}}\right)=\sum^{\infty}_{n=1}\frac{n^{n-2}}{3^{n/2}(n-1)!}
[/tex]

And yes, you can solve it to any degree of accuracy you wish :smile:
 
  • #7
ehild
Homework Helper
15,543
1,914
x can be obtained to any degree of accuracy also with an iterative process

[tex]x_{i+1}=-\sqrt{\frac{e^{2x_i}}{3}}[/tex]

I got -0.3906....



ehild
 
Last edited:
  • #8
2,981
5
x can be obtained to any degree of accuracy also with an iterative process

[tex]x_{i+1}=-\sqrt{\frac{e{^{2x_i}}{3} }[/tex]

I got -0.3906....



ehild

It is better to say that the solution of the equation is negative real number [itex]x = -y, \; y > 0[/itex] and the iteration should be:

[tex]
y_{i + 1} = \frac{e^{-y_{i}}}{\sqrt{3}}
[/tex]
 

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