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This equation is giving me a hard time.
$$e^x(x^2+2x+1)=2$$
Can you show me how to solve this problem algebraically or exactly? I managed to solve it using my calculator with one of its graph functions. But I would like to know how one would solve this without using the calculator.
Highly appreciated,
Bowser.
Take the natural logarithm of both terms, getting
$$x + \ln(x^2 + 2x + 1) = \ln(2)$$
Now watch the log argument: you see it's a square! Indeed $x^2 + 2x + 2 = (x + 1)^2$ so, again using log property:
$$x + 2\ln(x+1) = \ln(2)$$
If we assume to expect small values as a solution, then we may use Taylor Series expansion for the logarithm, up to the second order:
$$\ln(1+x) \approx x - \frac{x^2}{2}$$
Thence:
$$x + 2x - x^2 - \ln(2) = 0 ~~~~~ \to ~~~~~ x^2 - 3x - \ln(2) = 0$$
Solving like a second degree equation gives
$$x = \frac{3\pm \sqrt{9 - 4\ln(2)}}{2}$$
$$x_1 = 2.747(..) ~~~~~~~~~~~ x_2 = 0.252(..)$$
This is a numerical method to solve it and as you see the second solution fits with your result.
That solution can be improved simple taking more terms in the log expansion, indeed:
$$\ln(x+1) \approx x - \frac{x^2}{2} + \frac{x^3}{3} \cdot $$
As you see, there are many ways to solve the equation.
I see two of them which have not been described.
the first one is Newton method which, starting from a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, let us consider $$f(x)=e^x \left(x^2+2 x+1\right)-2\qquad f'(x)=e^x (x+1) (x+3)$$ Being very lazy, I shall start using $x_0=0$. The method then generates the following iterates $$x_1=0.333333333333333$$ $$x_2=0.255772423091538$$ $$x_3=0.248843724711150$$ $$x_4=0.248792699431494$$ $$x_5=0.248792696686402$$ which is the solution for fifteen significant figures.
the second one, which will lead to approximate solutions, uses Padé approximants which, for the same number of terms, are more "accurate" than Taylor series. Since I am still lazy and do not want to solve more than linear equations, I shall restrict to degree $1$ in numerator and degree $n$ in denominator. For example, I shall get $$P_{1,1}=\frac{\frac{25 x}{6}-1}{1-\frac{7 x}{6}}$$ $$P_{1,2}=\frac{\frac{301 x}{75}-1}{\frac{23 x^2}{50}-\frac{76 x}{75}+1}$$ $$P_{1,3}=\frac{\frac{4839 x}{1204}-1}{-\frac{521 x^3}{7224}+\frac{533 x^2}{1204}-\frac{1227 x}{1204}+1}$$ So, cancelling the numerators, approximations would be $$x_1=\frac{6}{25}=0.24$$ $$x_2=\frac{75}{301}\approx 0.249169$$ $$x_3=\frac{1204}{4839}\approx 0.248812$$
Being now less lazy and accepting to solve a quadratic, I could build $$P_{2,2}=\frac{\frac{521 x^2}{828}+\frac{533 x}{138}-1}{\frac{235 x^2}{828}-\frac{119 x}{138}+1}$$ for which the acceptable solution would be $$x=\frac{3}{521} \left(107 \sqrt{29}-533\right)\approx 0.248825$$
It is sure that we could do better and faster doing the same kind of work assuming (or knowing) that the solution is close to $\frac 14$. For example, the first iterate of Newton method would be $$\frac{1}{4}-\frac{16 \left(\frac{25 \sqrt[4]{e}}{16}-2\right)}{65 \sqrt[4]{e}}\approx 0.248794$$ and the solution of the $P_{1,1}$ approximant would be $$\frac{21024-7975 \sqrt[4]{e}}{4 \left(4384+5025 \sqrt[4]{e}\right)}\approx 0.248793$$
$$ e^x(x+1)^2=2\implies e^{(x+1)/2}(x+1)/2=\sqrt{e/2}\implies(x+1)/2=\mathrm{W}\!\left(\sqrt{e/2}\right) $$ Therefore, $$ x=2\mathrm{W}\!\left(\sqrt{e/2}\right)-1 $$ where $\mathrm{W}$ is the Lambert W function.
There is an iterative algorithm given in this answer to compute $\mathrm{W}$.
Alternatively, N[2LambertW[Sqrt[E/2]]-1,20] in Mathematica yields
$$
x=0.24879269668640244047
$$
The answer given by Desmos for intersection of the two curves $y=e^x$ and $y=\frac {2}{(x+1)^2}$ is $\color{red}{x=0.249}$. Now we have $$(x+1)^2=2e^{-1}\iff x^2+2x+1=2(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-\frac{x^5}{60}+O(x^6))$$ hence $$1-4x-\frac{x^3}{3}+\frac{x^4}{12}-\frac{x^5}{60}+20\cdot O(x^6)=0$$ The first approximation $1-4x=0$ gives $\color{red}{x\approx 0.25}$
The second approximation $ 1-4x-\frac{x^3}{3}=0$ gives $\color{red}{x\approx 0.24872}$
And we can continue but we see that the first approach is already good enough.
