# Can we have a series representation of order $5$ at $x=0$ of the inverse function of $xe^{x^{2}}$?

Well, I am a freshman, This was my homework. With a bit of googling, I found that: $$f^{-1}(x) = \pm \frac{\sqrt{W(2x^2)}}{\sqrt{2}}$$ What our professor did was: $$f(x)=xe^{x^{2}}=x+x^3+\dfrac{x^5}{2} +O(x^6)$$ Then he proved that $$f(x)$$ is a bijection then he supposed that: $$f^{-1}(x) = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+O(x^6)$$ then $$f^{-1}(f(x))=x$$ $$f^{-1}(f(x))=a_0+a_1{(x+x^3+\dfrac{x^5}{2})}+a_2{(x+x^3+\dfrac{x^5}{2})}^2+a_3{(x+x^3+\dfrac{x^5}{2})}^3+a_4{(x+x^3+\dfrac{x^5}{2})}^4+a_5{(x+x^3+\dfrac{x^5}{2})}^5$$

then solved the following for the coefficient: $$a_0+a_1{(x+x^3+\dfrac{x^5}{2})}+a_2{(x+x^3+\dfrac{x^5}{2})}^2+a_3{(x+x^3+\dfrac{x^5}{2})}^3+a_4{(x+x^3+\dfrac{x^5}{2})}^4+a_5{(x+x^3+\dfrac{x^5}{2})}^5=x$$ and found that: $$a_0=0,a_1=1,a_2=0,a_3=-1,a_5=\dfrac{5}{2}$$ and found that: $$f^{-1}(x)=x-x^3+\dfrac{5x^5}{2} +O(x^6)$$ I Think it is wrong because when you plot both of them, they are not symmetric to $$y = x$$

• shouldn't it be $f^{-1}(x)=x-x^3-\frac{x^5}{2}+O(x^6)$? – Pink Panther Jul 5 at 14:22
• yes it is, but even though still not symmetrical – ahmed ben Jul 5 at 14:30
• and why is that important? – Pink Panther Jul 5 at 14:31
• @PinkPanther because function and its inverse are symmetric to y=x – ahmed ben Jul 5 at 16:20
• – Martín-Blas Pérez Pinilla Jul 5 at 17:41

$$\text{I don't agree with} \quad f^{-1}(x)=x-x^3-\dfrac{1}{2}x^5 +O(x^6)$$ I found : $$f^{-1}(x)=x-x^3+\dfrac{5}{2}x^5 +O(x^6)$$

But my answer is mainly about the question of symmetry to $$y=x$$.

Sure, you are right. The curves $$y(x)$$ and $$f^{-1}(x)$$ must be symmetrical.

The series $$y=x-x^3+\dfrac{5}{2}x^5$$ is an approximate of $$f^{-1}(x)$$ only accurate in a range close to the origin. So, don't expect to observe symmetrical curves if you consider a too large range from the origin.

This is illustrated on the figure below.

$$y=x$$ is drawn in black.

$$x\,e^{x^2}$$ is drawn in red (solid curve).

$$f^{-1}(x)$$ is drawn in red (dashed curve), that is the symmetric curve of $$x\,e^{x^2}$$.

$$x-x^3+\dfrac{5}{2}x^5$$ is the first blue curve. One can see that the approximation is correct only on a small range, approximately $$x<0.3$$

More accuracy requires more terms for the series. For example : $$x-x^3+\dfrac{5}{2}x^5-\dfrac{49}{6}x^7+\dfrac{243}{8}x^9-\dfrac{14641}{120}x^{11}$$ is the second blue curve.

One can see that the range is not much enlarged, approximately $$x<0.4$$ . This is only a small improvement. Definitively one cannot expect a large range of validity for the inverse function with this method of limited series expansion, even with a large number of terms.