Thread: Calculate the taylor series expansion

1. Calculate the taylor series expansion

Calculate the taylor series expansion about x=0 as far as the term in $x^2$ for the function

$f(x)= { \displaystyle \frac{sinx}{1-e^{-x}}$ $x not = 0,$
$f(x)= 1$ $x=0$.

so at x=0, the function blow up to 0, which is a discontinuous at 0, how do we find the expansion?

2. Originally Posted by wopashui
Calculate the taylor series expansion about x=0 as far as the term in $x^2$ for the function

$f(x)= { \displaystyle \frac{\sin x}{1-e^{-x}} \quad x \ne 0,$
$f(x)= 1 \quad x=0$.

so at x=0, the function blow up to 0, which is a discontinuous at 0, how do we find the expansion?
I don't know why you say that the function is discontinuous at 0. In fact, it is continuous at 0, because $\displaystyle\lim_{x\to0}\frac{\sin x}{1-e^{-x}} = 1$ (as you can check by using l'Hôpital's rule, for example).

To find the first few terms of the Taylor series, first write the first few terms for the series of the numerator and denominator:

$\displaystyle \frac{\sin x}{1-e^{-x}} = \frac{x - x^3/6 + \ldots}{x - x^2/2 + x^3/6 - \ldots} = \frac{1 - x^2/6 + \ldots}{1 - (x/2 - x^2/6 + \ldots)}.$

For the denominator, use the series $\frac1{1-z} = 1+z+z^2+\ldots$, with $z = x/2-x^2/6+\ldots$. Then, ignoring all powers greater than $x^2$, we get

$\displaystyle \frac{\sin x}{1-e^{-x}} = (1 - x^2/6 + \ldots)(1 + (x/2-x^2/6 + \ldots) + (x/2 - \ldots)^2).$

Now you just have to pick out the coefficients of the first few powers of x on the right side of that equation.

3. Originally Posted by Opalg
I don't know why you say that the function is discontinuous at 0. In fact, it is continuous at 0, because $\displaystyle\lim_{x\to0}\frac{\sin x}{1-e^{-x}} = 1$ (as you can check by using l'Hôpital's rule, for example).

To find the first few terms of the Taylor series, first write the first few terms for the series of the numerator and denominator:

$\displaystyle \frac{\sin x}{1-e^{-x}} = \frac{x - x^3/6 + \ldots}{x - x^2/2 + x^3/6 - \ldots} = \frac{1 - x^2/6 + \ldots}{1 - (x/2 - x^2/6 + \ldots)}.$

For the denominator, use the series $\frac1{1-z} = 1+z+z^2+\ldots$, with $z = x/2-x^2/6+\ldots$. Then, ignoring all powers greater than $x^2$, we get

$\displaystyle \frac{\sin x}{1-e^{-x}} = (1 - x^2/6 + \ldots)(1 + (x/2-x^2/6 + \ldots) + (x/2 - \ldots)^2).$

Now you just have to pick out the coefficients of the first few powers of x on the right side of that equation.
sorry, I dun see what you are really doing here, what is the substituion of z for, can you complete the last step to show what happen?

4. Why aren't we just taking the first and second derivatives and putting them into
f(a) + f'(a)x + f''(a)x^2/2 ???