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Taylor series

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data

    find the taylor series for the function

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

    2. Relevant equations


    3. 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?
  2. jcsd
  3. Apr 28, 2010 #2

    Filip Larsen

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    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

    [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.
  4. Apr 28, 2010 #3


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    Science Advisor

    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
  5. Apr 28, 2010 #4

    Filip Larsen

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    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?
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