6.4 Working with taylor series  (Page 3/11)

1. Using the Maclaurin series for $\text{cos}x$ we find that the Maclaurin series for $\text{cos}\sqrt{x}$ is given by
   
   $\begin{array}{cc}\hfill {\displaystyle \sum _{n=0}^{\infty }\frac{{\left(\mathrm{-1}\right)}^n{\left(\sqrt{x}\right)}^{2n}}{\left(2n\right)\text{!}}}& ={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(\mathrm{-1}\right)}^n{x}^n}{\left(2n\right)\text{!}}}\hfill \\ & =1-\frac{x}{2\text{!}}+\frac{x^2}{4\text{!}}-\frac{x^3}{6\text{!}}+\frac{x^4}{8\text{!}}-\text{⋯}.\hfill \end{array}$
   
   This series converges to $\text{cos}\sqrt{x}$ for all $x$ in the domain of $\text{cos}\sqrt{x};$ that is, for all $x\ge 0.$
2. To find the Maclaurin series for $\text{sinh}x,$ we use the fact that
   
   $\text{sinh}x=\frac{e^x-e^{\text{−}x}}{2}.$
   
   Using the Maclaurin series for $e^x,$ we see that the $n\text{th}$ term in the Maclaurin series for $\text{sinh}x$ is given by
   
   $\frac{x^n}{n\text{!}}-\frac{{\left(\text{−}x\right)}^n}{n\text{!}}.$
   
   For $n$ even, this term is zero. For $n$ odd, this term is $\frac{2x^n}{n\text{!}}.$ Therefore, the Maclaurin series for $\text{sinh}x$ has only odd-order terms and is given by
   
   ${\displaystyle \sum _{n=0}^{\infty }\frac{x^{2n+1}}{\left(2n+1\right)\text{!}}}=x+\frac{x^3}{3\text{!}}+\frac{x^5}{5\text{!}}+\text{⋯}.$

The two functions are related by

so the binomial series for $\frac{1}{\sqrt{1+x}}$ is given by

Suppose that there exists a power series solution

Differentiating this series term by term, we obtain

If y satisfies the differential equation, then

Using [link] on the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,

Using the initial condition $y\left(0\right)=3$ combined with the power series representation

we find that $c_0=3.$ We are now ready to solve for the rest of the coefficients. Using the fact that $c_0=3,$ we have

Therefore,

You might recognize

as the Taylor series for $e^x.$ Therefore, the solution is $y=3e^x.$