# Taylor series

## Homework Statement

find the taylor series for the function

f(x) = $$\frac{x^2+1}{4x+5}$$

N/A

## The Attempt at a Solution

how to do this?

1st attempt.

i did turn it this term
$$\frac{x}{4}$$ + $$\frac{-5x+4}{16x+20}$$ can i turn this to taylor series?

maybe i know how to make $$\frac{-5x+4}{16x+20}$$ to taylor but
$$\frac{x}{4}$$ + $$\frac{-5x+4}{16x+20}$$??

2nd attempt.

should i differentiate it until i get fn(x) form??

what to do what to do?

Filip Larsen
Gold Member
If you can partition your function into two functions like f(x) = f1(x) + f2(x) and you have the Taylor series for both f1 and f2 then the sum of these is the Taylor series for f.

You can also differentiate f directly and look for a pattern that will allow you to write the series as

$$f(x) = f(x_0) + f'(x_0)(x-x_0) + \sum_{i=2}^{\infty}{g_i(x_0) (x-x_0)^i}$$

where gi is a fairly simple function having parameter i.

HallsofIvy
Homework Helper
Certainly, the basic definition of "Taylor series" is just what Filip Larsen says, but you can also do what you were trying- you just stopped too soon. Since
$$\frac{-5x+ 4}{16x+ 20}$$
has the same order in numerator and denominator you could also divide that. In fact,
$$\frac{x^2+ 1}{4x+ 5}= \frac{x}{4}- \frac{5}{16}+ \frac{11}{16}\frac{1}{4x+ 5}$$

Now, what you can do is write that last fraction as
$$\frac{1}{5- (-4x)}= \frac{1}{5}\left(\frac{1}{1- \frac{-4x}{5}}\right)}$$

Recall that the sum of a geometric series is given by
$$\sum_{n=0}^\infty r^n= \frac{1}{1- r}$$
so that can be written as a power series in
$$\frac{-4x}{5}$$

Filip Larsen
Gold Member
Neat. I guess that would "correspond" to Taylor series evolved around x0 = 0 and with -1 < x < 1. Is is possible to make something equally neat for x outside this range, or are you then "stuck" with the general Taylor series?