How to compute the first three terms of the Taylor expansion?

Question

Consider the equation ${\epsilon}(x^{5} + \cos(x)) + x^{2} - 1 = 0$. When $\epsilon = 0, x_{0} = 1$ is a solution. Compute the first three terms of the Taylor expansion $$x = x_{0} + {\epsilon}x_{1} + {\epsilon}^{2}x_{2} + ... $$ centered at $\epsilon = 0$ with $x(0) = 1$.

how to compute this expansion?

Thank you in advance.

Answer 1

We need to do some inserting. First of all, insert the Taylor expansion of $\cos(x)$ with respect to $x$: $$ \epsilon(x^5 + 1 -\frac{x^2}{2} + \frac{x^4}{24} - \cdots) + x^2 - 1 = 0 $$ Next, we insert the Taylor expansion of $x$ in terms of $\epsilon$: $$ \epsilon\left((x_0 + \epsilon x_1 + \epsilon^2x_2 + \cdots)^5\vphantom{\frac{2^2}{2}} \right.\\+ 1 - \frac{(x_0 + \epsilon x_1 + \epsilon^2x_2 + \cdots)^2}{2} \left. + \frac{(x_0 + \epsilon x_1 + \epsilon^2x_2 + \cdots)^4}{24} - \cdots\right)\\ + (x_0 + \epsilon x_1 + \epsilon^2x_2 + \cdots)^2 - 1 = 0 $$ Now sort things by order of $\epsilon$ (note that since the parenthesis is multiplied by $\epsilon$, everything inside it has its degree upped by one): $$ \epsilon^0: \quad x_0^2 - 1 = 0 \implies x = \pm 1 $$ We're told specifically to use the solution $x_0 = 1$, so we'll do that. Moving on: $$ \epsilon^1:\quad \epsilon(x_0^5 + 1 - \frac{x_0^2}{2} + \frac{x_0^4}{24} - \cdots) + 2\epsilon x_0x_1 = 0\\ 1 + \cos(1) + 2x_1 = 0\\ x_1 = -\frac{1+\cos(1)}{2} \approx -0.77 $$ and, finally $$ \epsilon^2: \quad\epsilon(5\epsilon x_0x_1 - \frac{2\epsilon x_0x_1}{2} + \frac{4\epsilon x_0x_1}{24} - \cdots) + \epsilon^2x_1^2 + 2\epsilon^2x_0x_2 = 0\\ 5x_1 - \sin(x_1) + x_1^2 + 2x_2 = 0\\ x_2 = \frac{\sin(x_1) - x_1(5+x_1)}{2} \approx 1.28 $$

Answer 2

First of all you must expand $\cos x$ around $x=1$. As we are interested only to terms up to $\epsilon^2$ we need only retain the first term (because $\cos x$ is multiplied by $\epsilon$): $$ \cos x=\cos1-\sin1(x-1)+\dots $$ Substitute that into your equation and substitute there also $x = 1 + {\epsilon}x_{1} + {\epsilon}^{2}x_{2} + ... $, retaining only terms up to $\epsilon^2$. Your equation then becomes: $$ \epsilon (1 + 2 x_1 + \cos1 ) + \epsilon^2 (5 x_1 + x_1^2 + 2 x_2 - x_1 \sin1)+\dots=0. $$ To find $x_1$ and $x_2$ just equate to zero the coefficients of $\epsilon$ and $\epsilon^2$: $$ 1 + 2 x_1 + \cos1 =0 \quad\implies\quad x_1=-{\cos1+1\over2} $$ $$ 5 x_1 + x_1^2 + 2 x_2 - x_1 \sin1 =0; \quad\implies\quad x_2=-{\cos1+1\over8}(2\sin1+\cos1-9). $$