Showing that there exists a solution to an equation and writing its Taylor polynomial.

Question

I need to show that for the equation $e^{y}+2y+x=1$, there exists a solution to $y=f(x)$ near $x=y=0$, and then I need to write $f(x)$ as its 3rd degree Taylor polynomial expanded at $x=0$. Is there any hint I could have to figure this out?

Answer 1

$$e^y+2y+x=1 \tag 1$$ Obviously $(x=0,y=0)$ is solution. Expanding to series around this point :

$$y=a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...$$

$$e^{a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...}+2(a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+...)+x-1=0$$

Expanding to series leads to : $$(3a_1+1)x+(3a_2+\frac12 a_1^2)x^2+(3a_3+a_1a_2+\frac16 a_1^3)x^3+...=0$$ Then we have to solve $$3a_1+1=0$$ $$3a_2+\frac12 a_1^2=0$$ $$3a_3+a_1a_2+\frac16 a_1^3=0$$ $$\text{etc.}$$ The result is : $$a_1=-\frac13$$ $$a_2=-\frac{1}{54}$$ $$a_3=0$$ $$a_4=\frac{1}{8748}$$ $$\text{etc.}$$ $$y(x)=-\frac13 x -\frac{1}{54}x^2+\frac{1}{8748}x^4+\frac{1}{196830}x^5+...$$ NOTE for information :

This equation can be solved analytically, but it requires a special function.

Of course, this is not the kind of answer expected by the OP. $$\frac12 e^y=-y+\frac{1-x}{2}$$ $$(-y+\frac{1-x}{2})e^{-y}=\frac12$$ $$(-y+\frac{1-x}{2})e^{-y+\frac{1-x}{2}}=\frac12e^{\frac{1-x}{2}} \tag 2$$ The solution of an equation of this kind : $$Xe^X=Y$$ cannot be expressed with a finit number of elementary functions. It requires a special function, namely the Lambert W function : $$X=W(Y)$$ http://mathworld.wolfram.com/LambertW-Function.html

With $\quad X=(-y+\frac{1-x}{2})\quad$ and $\quad Y=\frac12e^{\frac{1-x}{2}}\quad$ Eq.$(2)$ becomes $Xe^X=Y$ which solution is : $$(-y+\frac{1-x}{2})=W\left(\frac12e^{\frac{1-x}{2}} \right)$$ The solution of Eq.$(1)$ is : $$y=\frac{1-x}{2} - W\left(\frac12e^{\frac{1-x}{2}}\right)$$ Expanding the Lambert W function to series would lead to the same result as above.

Answer 2

For the first part, this is the implicit function theorem. For the second part, I suggest the following. Differentiate your equation to get $y'(x)e^{y(x)}+2y'(x)+1=0.$ Then, at $x=0$, knowing that $y(0) = 0$), we get $y'(0)+2y'(0) +1=0.$ Hence, $y'(0) = -\frac{1}{3}.$ For $y''(0)$ and $y'''(0)$, same idea.