3
$\begingroup$

Define if necessary the given function so as to be continuous at $x_0=0$ and find the first four nonzero terms of its MacLaurin series. $$ \frac{xe^x}{\sin x}$$

Given $f(x)= \frac{h(x)}{g(x)}$ where $h(x)= xe^x$, and $g(x)=\sin x$

We have the following MacLaurin series $$f(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x +a_2 x^2 + a_3 x^3 +a_4 x^4 + ...$$ $$g(x) = \sum_{n=0}^{\infty} b_n x^n = 0 + x+ 0 -\frac{1}{6} x^3 + 0 + \frac{1}{120} x^5 +...$$ $$h(x) = xe^x = \sum_{n=0}^{\infty} c_n x^n = \sum_{n=0}^{\infty} \frac{n}{(n)!}x^n = 0 + x + x^2 + \frac{1}{2} x^3 + \frac{1}{6} x^4 + \frac{1}{24} x^5$$ .

The Maclaurin coefficients of $f(x)g(x)$ $$ c_0 = a_0b_0 = 0$$ $$ c_1 = a_0b_1 + a_1b_0 = a_0$$ $$ c_2 = a_0b_2 + a_1b_1 +a_2b_0 = a_1 $$ $$ c_3 = a_0b_3 +\cdots a_3b_0 = - \frac{1}{6} a_0 + a_2$$ $$ c_4 = a_0b_4 +\cdots + a_4b_0 = - \frac{1}{6} a_1 + a_3 $$ $$ \implies f(x)g(x) = a_0 x^1 + a_1 x^2 + \left( - \frac{1}{6} a_0 + a_2\right)x^3 + \left(- \frac{1}{6} a_1 + a_3\right) x^4 + \cdots$$

.

Equating coefficients of $f(x)g(x)$ with coefficients of $h(x)$ $$ c_0 = 0$$ $$ c_1 = a_0 = 1 $$ $$ c_2 = a_1 = 1 $$ $$ c_3 = - \frac{1}{6} a_0 + a_2 = 1/2 \implies a_2 = 2/3 $$ $$ c_4 = - \frac{1}{6} a_1 + a_3 = 1/6 \implies a_3 = 1/3 $$ .

it follows that the first 4 nonzero terms are: $$f(x) = \frac{x e^x}{\sin(x)} = 1 + x + \frac{2}{3} x^2 + \frac{1}{3} x^3$$

.

I would like to know if I am going in the right direction?

My main question regards the original question "Define if necessary the given function so as to be continuous at $x_0=0$" . I am not sure how to handle this question, if a definition is needed how? and why? if a definition is not needed how? and why?

Thx for your input/help.

$\endgroup$
  • 1
    $\begingroup$ The definition $$f(0)=1$$ is definitely needed. $\endgroup$ – Did Aug 17 '17 at 11:43
  • 1
    $\begingroup$ Because $\lim_{x\to 0} \, \dfrac{x}{\sin (x)}=1$ you need to define $f(0)=1$ to have continuity $\endgroup$ – Raffaele Aug 17 '17 at 19:13
3
$\begingroup$

I didn't check the computations but, yes, you are doing it right.

As for the other question, note that the expression $\frac{xe^x}{\sin x}$ is undefined if $x=0$. In order to make $f$ continuous at $0$; you must define $f(0)$ as $\lim_{x\to0}\frac{xe^x}{\sin x}$, which is equal to $1$.

$\endgroup$
2
$\begingroup$

I think that your expansion is partially incorrect. After setting $f(0)=1$, note that for $x\not=0$, \begin{align*} f(x)=\frac{x e^x}{\sin(x)}&=x \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)\left(x-\frac{x^3}{6}+o(x^4)\right)^{-1}\\ &=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)\left(1-\frac{x^2}{6}+o(x^3)\right)^{-1}\\ &=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right) \cdot\left(1+\frac{x^2}{6}+o(x^3)\right)\\ &=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)\left(1+\frac{x^2}{6}+o(x^3)\right)\\ &=1+x+\frac{x^2}{2}+\frac{x^3}{6} +\frac{x^2}{6}+\frac{x^3}{6}+o(x^3)\\ &=1+x+\frac{2x^2}{3}+\frac{x^3}{3}+o(x^3) \end{align*} where we used the fact that $$(1-z)^{-1}=1+z+z^2+o(z^2).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.