Taylor series

  • Thread starter annoymage
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Homework Statement



find the taylor series for the function

f(x) = [tex]\frac{x^2+1}{4x+5}[/tex]

Homework Equations



N/A

The Attempt at a Solution



how to do this?

1st attempt.

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

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


2nd attempt.

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

what to do what to do?
 

Answers and Replies

  • #2
Filip Larsen
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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

[tex]f(x) = f(x_0) + f'(x_0)(x-x_0) + \sum_{i=2}^{\infty}{g_i(x_0) (x-x_0)^i} [/tex]

where gi is a fairly simple function having parameter i.
 
  • #3
HallsofIvy
Science Advisor
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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
[tex]\frac{-5x+ 4}{16x+ 20}[/tex]
has the same order in numerator and denominator you could also divide that. In fact,
[tex]\frac{x^2+ 1}{4x+ 5}= \frac{x}{4}- \frac{5}{16}+ \frac{11}{16}\frac{1}{4x+ 5}[/tex]

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

Recall that the sum of a geometric series is given by
[tex]\sum_{n=0}^\infty r^n= \frac{1}{1- r}[/tex]
so that can be written as a power series in
[tex]\frac{-4x}{5}[/tex]
 
  • #4
Filip Larsen
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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?
 

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