How do you find the first three terms of a Maclaurin series for f(t) = (e^t - 1)/t using the Maclaurin series of e^x?

Question

How do you find the first three terms of a Maclaurin series for f(t) = (e^t - 1)/t using the Maclaurin series of e^x?

1 Answer

Impact of this question

5143 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-20T03:45:15

We know that the Maclaurin series of $e^{x}$ $e^x$ is

$\infty \sum n = 0 \frac{x^{n}}{n!}$ $sum_(n=0)^oox^n/(n!)$

We can also derive this series by using the Maclaurin expansion of $f (x) = \infty \sum n = 0 f^{(n)} (0) \frac{x^{n}}{n!}$ $f(x)=sum_(n=0)^oof^((n))(0)x^n/(n!)$ and the fact that all derivatives of $e^{x}$ $e^x$ is still $e^{x}$ $e^x$ and $e^{0} = 1$ $e^0=1$ .

Now, just substitute the above series into
$\frac{e^{x} - 1}{x}$ $(e^x-1)/x$

$= \frac{\sum_{n = 0}^{\infty} (\frac{x^{n}}{n!}) - 1}{x}$ $=(sum_(n=0)^oo(x^n/(n!))-1)/x$

$= \frac{1 + \sum_{n = 1}^{\infty} (\frac{x^{n}}{n!}) - 1}{x}$ $=(1+sum_(n=1)^oo(x^n/(n!))-1)/x$

$= \frac{\sum_{n = 1}^{\infty} (\frac{x^{n}}{n!})}{x}$ $=(sum_(n=1)^oo(x^n/(n!)))/x$

$= \infty \sum n = 1 \frac{x^{n - 1}}{n!}$ $=sum_(n=1)^oox^(n-1)/(n!)$

If you want the index to start at $i = 0$ $i=0$ , simply substitute $n = i + 1$ $n=i+1$ :

$= \infty \sum i = 0 \frac{x^{i}}{(i + 1)!}$ $=sum_(i=0)^oox^i/((i+1)!)$

Now, just evaluate the first three terms to get

$\approx 1 + \frac{x}{2} + \frac{x^{2}}{6}$ $~~1+x/2+x^2/6$

Science

Math

Humanities

... and beyond

How do you find the first three terms of a Maclaurin series for f(t) = (e^t - 1)/t using the Maclaurin series of e^x?

1 Answer

Related questions

Impact of this question