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

[annoymage]

Homework Statement

find the taylor series for the function

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

Homework Equations

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]

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 g_i is a fairly simple function having parameter i.

 

[HallsofIvy]

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]

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?