Finding the Intersection Of 2 Equations (difficult)

[themadhatter1]

Homework Statement

Solve the following equation:

$$e^{2x}=3x^2$$

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.

$$2x=\ln{3x^2}$$

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

Would perhaps another method be better?

 

[Mentallic]

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 $$a=ue^u$$ then $$u=W(a)$$ where u is some function of x and a is a constant.

$$3x^2=e^{2x}$$

$$3x^2e^{-2x}=1$$ divided through by e^2x

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

$$-xe^{-x}=\frac{-1}{\sqrt{3}}$$ negative of both sides.

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

$$-x=W\left(\frac{-1}{\sqrt{3}}\right)$$

$$x=-W\left(\frac{-1}{\sqrt{3}}\right)$$

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.

 

[HallsofIvy]

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 $e^x= 2$.

 

[Mentallic]

Ahh I think I've mixed up the meaning of analytic with 'expressible in terms of elementary functions', my mistake
Thanks!

 

[themadhatter1]

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 $$a=ue^u$$ then $$u=W(a)$$ where u is some function of x and a is a constant.

$$3x^2=e^{2x}$$

$$3x^2e^{-2x}=1$$ divided through by e^2x

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

$$-xe^{-x}=\frac{-1}{\sqrt{3}}$$ negative of both sides.

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

$$-x=W\left(\frac{-1}{\sqrt{3}}\right)$$

$$x=-W\left(\frac{-1}{\sqrt{3}}\right)$$

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.

$$w(x)=\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}x^{n}}{(n-1)!}$$

$$-w(\frac{-1}{\sqrt{3}})=-\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}(\frac{-1}{\sqrt{3}})^{n}}{(n-1)!}$$

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?

 

[Mentallic]

hmm. Interesting.

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

$$w(x)=\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}x^{n}}{(n-1)!}$$

$$-w(\frac{-1}{\sqrt{3}})=-\sum^{\infty}_{n=1}\frac{(-1)^{n-1}n^{n-2}(\frac{-1}{\sqrt{3}})^{n}}{(n-1)!}$$

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,

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

therefore the sum is:

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

And yes, you can solve it to any degree of accuracy you wish

 

[ehild]

x can be obtained to any degree of accuracy also with an iterative process

$$x_{i+1}=-\sqrt{\frac{e^{2x_i}}{3}}$$

I got -0.3906...
ehild

 

[Dickfore]

x can be obtained to any degree of accuracy also with an iterative process

$$x_{i+1}=-\sqrt{\frac{e{^{2x_i}}{3} }$$

I got -0.3906...
ehild

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

$$y_{i + 1} = \frac{e^{-y_{i}}}{\sqrt{3}}$$