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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 $g_i(0)$. $\endgroup$ – Pratyush Sarkar Sep 28 '13 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 Sep 28 '13 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. $\endgroup$ – Pratyush Sarkar Sep 28 '13 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 $0$. $\endgroup$ – Pratyush Sarkar Sep 28 '13 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 Sep 28 '13 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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