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Maclaurin Series

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine the first three terms in the Maclaurin series for:
    (x2+4)-1


    2. Relevant equations
    f(x)=f(a)+f'(a)(x-a)+f''(a)[itex]\frac{(x-a)^{2}}{2!}[/itex]+f'''(a)[itex]\frac{(x-a)^{3}}{3!}[/itex]


    3. The attempt at a solution
    So I start out with getting my primes of f(x):
    f'(x)=[itex]\frac{-2x}{(x^{2}+4)^{2}}[/itex]
    f''(x)=[itex]\frac{6x^{2}-8}{(x^{2}+4)^{3}}[/itex]
    f'''(x)=[itex]\frac{24x(x^{2}-4)}{(x^{2}+4)^{4}}[/itex]

    So my issue is that when I set a=0 for the Maclaurin series then the entire series is equal to zero. Is that correct? and if it is how would I write it up as an answer for my question?
     
  2. jcsd
  3. Sep 21, 2013 #2

    Simon Bridge

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    No. To start with f(0)≠0 ... please show your working.
     
  4. Sep 21, 2013 #3
    But doesn't Maclaurin series say that a=0? So then how would it not equal zero? It would have just been a lot to show my work on the derivatives and I'm on a crappy laptop right now. So I just put down the answers I got after I did my work on paper.
     
  5. Sep 21, 2013 #4
    Plug in 0 for x and you will not get all zero coefficients. Some might be zero.
     
  6. Sep 21, 2013 #5
    I just need the first three terms in the Maclaurin series. I'm not sure what exactly that means in the question so I decided it would be safe to think that what they mean by the first three terms is to go to the third prime.

    f'(0)=0
    f''(0)=0
    f'''(0)=0

    So that Maclaurin series for the first three terms looks like:

    (x2+4)-1=1/4+0+0+0

    Is that right?
     
  7. Sep 21, 2013 #6

    vela

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    I would hope you know the difference between x=0 and f(0).

    The problem is when all you do is write down a wrong answer, it's impossible for us to see what mistakes you are making.

    We don't need to see your calculations of the derivatives, but it would help us to see how you managed to go from those expressions to all of the terms in the series being 0. For example, how did you manage to get f(0)=0 or f''(0)=0?

    No. The very first term is right though.
     
  8. Sep 21, 2013 #7

    Simon Bridge

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    No.
    You figured out why f(0)≠0?
    Did you check the others too?

    Please show your working.
    I'll show you what I mean. Showing the working for the 1st term would go like this:

    $$f(x)=\frac{1}{x^2+4} \Rightarrow f(a)=\frac{1}{a^2+4}\\ \Rightarrow f(a=0)=\frac{1}{0^2+4}=\frac{1}{4}$$ ... see? Now do that for the other terms.
     
  9. Sep 21, 2013 #8

    Ray Vickson

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    Do you really believe that 1/(4+0) = 0? That IS what you claim!
     
  10. Sep 21, 2013 #9
    f(x)=(x2+4)-1

    f'(x)=(-1)(x2+4)-2=[itex]\frac{-2x}{(x^{2}+4)^{2}}[/itex]
    f''(x)=d/dx(-2(x2+4)-2)=-2(x2+4)-2+(-2x)(-2)(x2+4)-3(2x)=-2(x2+4)-2+8x2(x2+4)-3=[itex]\frac{-2}{(x^{2}+4)^{3}}[/itex]+ [itex]\frac{8x^{2}}{(x^{2}+4)^{4}}[/itex] =[itex]\frac{6x^{2}-8}{(x^{2}+4)^{3}}[/itex]
    f'''(x)=d/dx(6x2-8)(x2+4)-3=12x(x2+4)-3-6x(6x2-8)(x2+4)-4=[itex]\frac{-24x(x^{2}-4)}{(x^{2}+4)^{4}}[/itex]

    f'(0)=0
    f''(0)=[itex]\frac{0-8}{(0+4)^{3}}[/itex]=[itex]\frac{-8}{64}[/itex]=[itex]\frac{-1}{8}[/itex]
    f'''(0)=0

    (x2+4)-1=[itex]\frac{1}{4}[/itex]+0+[itex]\frac{-1}{8}[/itex][itex]\frac{x^{2}}{2!}[/itex]+0=[itex]\frac{1}{4}[/itex]+0+[itex]\frac{-x^{2}}{8}[/itex]+0
     
  11. Sep 21, 2013 #10
    ... Most days I spend at least 10 hours solving problems... Sometimes I overlook things. It's not that I don't know what f(0) means. Near the end of a long day equations start to mash together...
     
  12. Sep 21, 2013 #11

    vela

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    Close. The coefficient of x2 should be -1/16. You dropped the factor of 2.

    The problem is asking for the first three non-vanishing terms, so you need to calculate the x4 term too.

    Do you have to use the expansion in terms of derivatives? There are faster ways to obtain the series.

    Most of us can empathize. When you start making glaring mistakes like that, it's usually a sign to go to bed — or at least take a long break.
     
  13. Sep 21, 2013 #12
    Oh shoot! Why did I forget the factorial now... Yeah I think I need to go to bed but I want to finish this assignment so I can have the day off tomorrow... Sadly I still have 4 more questions to do. If I'm making mistakes like this though I'm probably over-tired and useless. I wasn't sure by the question if I just needed to go to f'''(x) or if I needed to find three non-zero terms in the series. Do you think I should just go to three non-zero terms to be safe? Haha If that's the case then I'll be up much longer then because my other questions I just left zero as a term in the Maclaurin series.
     
  14. Sep 21, 2013 #13
    I could make a general sum formula for these but I think my teacher wants me to show all of my work in this way.
     
  15. Sep 21, 2013 #14
    Haha, on my piece of paper I was doing the problem on I wrote -x^2/16 for that term. I think it's time to sleep or take a good half hour break.
     
  16. Sep 21, 2013 #15

    Simon Bridge

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    you mean... "IF the problem is asking for..."; the stated problem doesn't.

    Looks like the 2am effect there :) Do something else for 15mins then sleep.
    Otherwise you'll have to go through in tiny steps like when you first learned algebra.
    The person setting the problem wanted you to notice that basically every other term in the series is zero ... since that's a discovery process, telling you would have been "doing your work for you" and against the rules.
    Cheers.
     
  17. Sep 21, 2013 #16
    Yeah I tried taking a bit of a break and my eyes are still fuzzy from the long day. I think it's time for sleep. Thank you all for your help. I'm sure I'm going to feel stupid in the morning when I look at the mistakes I made tonight :P
     
  18. Sep 22, 2013 #17

    Simon Bridge

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    Happens to everyone :)
    One of the more powerful tools for problem-solving is "sleep on it". ;D
     
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