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I'm trying to solve equations like the following one:

$$5 + 3x - 4x^3 = e^{x^2}$$

I've tried using the Lambert W function, but I didn't get any success. I must admit I'm relatively new to Lambert W function.

I've managed to bring the equation to:

$$e^{-x^2}(5 + 3x - 4x^3) = 1$$

But I can't bring the exponent and the second term on the LHS to the same value, in order to apply Lambert W function.

Also I tried to search on the Internet, but I never found a equation from a simular type, all of them were from the type:

$$5^x = 7x$$

This type of equation is fairly easy to solve using Lambert W function, but it doesn't help me solving an equation from the first type.

Also I couldn't come up with another idea how to get solution, except for Lambert W function. Can you please help me?

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    $\begingroup$ May be you can define the function $f(x) = 5 + 3x - 4x^3 - e^{x^2}$. Then find all the intervals where $f$ is strictly increasing or decreasing (so its one to one in that interval) and define inverses $g_1, \dotsc, g_n$ on those intervals. Then may be you can try to find $g_i(0)$ using Taylor series. Also, using the Lambert $W$ function solves the example but the problem of computing the number remains (which is essentially the problem you have in the first place). Same with using any "special" function. So in that case you can just consider the functions $g_i$ and say the answer is $g_i(0)$. $\endgroup$ Commented Sep 28, 2013 at 0:03
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    $\begingroup$ I've found out that $f(x) = 5 + 3x - 4x^3 - e^{x^2}$ is strictly increasing on the intervals: $$(-\infty, - 1.411671)$$ and $$(- 0.64129,0.410975)$$ And strictrly decreasing on the intervals: $$(- 1.411671,- 0.64129)$$ and $$(0.410975, \infty)$$ But I have problems with the inverses, how to find them on specific interval. $\endgroup$
    – Stefan4024
    Commented Sep 28, 2013 at 0:29
  • $\begingroup$ Yeah, you won't be able to find them explicitly. If you could you would have been able to solve the problem easily in the beginning. What I suggested is find $g_i(0)$ using Taylor series. You don't know $g_i$ explicitly but you can find the Taylor series since you can find the derivatives of $g_i$ in terms of derivatives of $f$. Note that you need to Taylor expand about a point that you do know. For example you know $f(0) = 4$ so there is a $g_i$ with $g_i(4) = 0$ so now Taylor expand about that point. Then you will get a series for the value $g_i(0)$ which is a solution. $\endgroup$ Commented Sep 28, 2013 at 0:39
  • $\begingroup$ Also note that you need to find max and mins (or find bounds for them) of $f$ to see which intervals actually contain a root. There is no point in looking at intervals where the function is never $0$. $\endgroup$ Commented Sep 28, 2013 at 0:42
  • $\begingroup$ I'm little confused, should I Taylor expand $f(x)$ at $x=4$ and set it to $0$ or should I Taylor expand $f(x)$ at $x=0$ and set it to $4$ $\endgroup$
    – Stefan4024
    Commented Sep 28, 2013 at 1:03

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There are a few problems with addition.

$$W(x+a)=?$$

$$\log(x+a)=?$$

The inverse functions of exponential related functions don't like addition on the inside. Addition on the outside is perfectly fine though.

$$\log(x)+a=\log(e^ax)$$

Except even then, the Lambert W Function won't simplify if you have addition on the outside... most of the time. And this question isn't one of the ones that simplifies.

So the best is to go with linear approximation methods.

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$$5+3x-4x^3=e^{x^2}$$

We see, this equation is a polynomial equation of more than one algebraically independent monomials ($x,e^{x^2}$) and with no univariate factor. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (operations) we can read from the equation.

Because your equation can be rearranged to a polynomial exponential equation over the algebraic numbers, it cannot have solutions that are elementary numbers.

Because your exponential polynomial equation has a polynomial term of more than one non-constant monomials of different degree, Lambert W cannot applied here.

But the equation can possibly be solved by Generalized Lambert W - see the references below. $\ $

[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function with applications in theoretical physics. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

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There is no really fundamental difference between the function $$ f(x) = x e^x$$ and your $$g(x) = (5+3x-4x^3)e^{-x^2}$$.

The only point is that the first function is famous that the inverse $f^{-1}$ has an name ("Lambert W function"). But its just a name. Mathematically, the situation is essentially the same:

  • There is no nice closed expression (say, using only $+,-,\times,/,\sqrt{},\sin,\cos,\exp$, or whatever you would consider a simple expression).
  • But its easy enough to find any value numerically.
  • Also you can derive Taylor-expansions, various limits, and plenty of properties of (for example) the derivatives of $g^{-1}$.

This situation is actually very common in math. There are plenty of functions (actually most) that exist perfectly well, but cant be written down in a nice closed form. Some are famous enough to get a name ("Gamma function", "Logarithmic integral function", "Pochhammer-symbol", "hypergeometric functions", "elliptic integral", and many many more), these are usually called "special functions". But just giving something a name doesnt really do anything mathematically. And just because you cant write down a nice equation for a function doesnt mean it is any less real.

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