Q. 16 - Calculus

Question

How may we find the Maclaurin series for f(x)g(x) if we already know the Maclaurin series for the functions f(x) and g(x)? How do you find the interval of convergence for the new series?

Step-by-Step Solution

Answer

In order to find the Maclaurin series for f(x)g(x)  when the maclaurin series for individual function is known then we simply multiply the maclaurin series term by term.

Step 1. Given Information

The given data is to find the Maclaurin series for f(x)g(x) if we already know the Maclaurin series for the functions f(x) and g(x).

Step 2. Explanation

In order to find the Maclaurin series for f(x)g(x)  when the maclaurin series for individual function is known then we simply multiply the maclaurin series term by term.

For example, $f\left(x\right)=e^{x}cosx$.

We know that maclaurin series for the function $g\left(x\right)=e^x \mathrm{is} e^x=\sum _{k=0}^{\infty }\frac{x^k}{k!}$

That is, $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....$

And maclaurin series for the function $h\left(x\right)=\cos x$is $\cos x=\sum _{k=0}^{\infty }\frac{{(-1)}^k}{\left(2k\right)!}{x}^{2k}$

That is, $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$

Step 3. Calculation

There are constant terms in both the series as 1. Thus, the new series will also have 1 as its constant term.

Now, Coefficient of x term is $1·1=1$

The coefficient of $x^2$ term is $$\begin{gathered} 1·-\frac{1}{2}+\frac{1}{2}·1=-\frac{1}{2}+\frac{1}{2} \\ =0 \end{gathered}$$

The coefficient of $x^3$ term is as follows,

$$\begin{gathered} 1·\left(\frac{1}{2}\right)+\frac{1}{6}·1=-\frac{1}{2}+\frac{1}{6} \\ =\frac{-3+2}{6} \\ =-\frac{1}{6} \end{gathered}$$

Also, the coefficient of $x^4$term is below,

$$\begin{gathered} 1·\frac{1}{24}+\frac{1}{24}·1=\frac{1}{24}+\frac{1}{24} \\ =\frac{1}{12} \end{gathered}$$

As the coefficient of quadratic term is 0, so we move forward and find the coefficient of $x^5$term that is,

$$\begin{gathered} 1·\left(\frac{1}{4!}\right)+\frac{1}{3!}·\left(-\frac{1}{2!}\right)+\frac{1}{5!}·1=\frac{1}{24}-\frac{1}{12}+\frac{1}{120} \\ =-\frac{1}{30} \end{gathered}$$

Thus, we get,

$$\begin{gathered} e^x\cos x=1+1·x+0x^2-\frac{1}{6}{x}^3+\frac{1}{12}{x}^4-\frac{1}{30}{x}^5 \\ {e}^x\cos x=1+x-\frac{1}{6}{x}^3+\frac{1}{12}{x}^4-\frac{1}{30}{x}^5 \end{gathered}$$