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    can someone please help, looking for a MacLaurin series for x, cant remember if it exists or not. Know you've got sin x, cos x , e^x, etc....
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    (Original post by FionaDutton)
    can someone please help, looking for a MacLaurin series for x, cant remember if it exists or not. Know you've got sin x, cos x , e^x, etc....
    sin(x)=x-(x^3)/3!+(x^5)/5!-.................[(-1)^r][x^(2r+1)/(2r+1)!]
    cos(x)=1-(x^2)/2!+...................+[(-1)^r][(x^2r)/(2r)!]
    e^x=1+x+x^2/2!...................+x^r/r!
    dont know wot ur actually lookin for
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    to be honest i want to be able to write -x^-2 as a power series.... any ideas??
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    (Original post by FionaDutton)
    to be honest i want to be able to write -x^-2 as a power series.... any ideas??
    well............f(x)=-1/x^2 f '(x)=2/(x^3) and f ''(x)=-6/(x^4)
    know f(0)=undifined and f '(0)=undifined and f ''(0)=undifined
    so u cannot use the macularin or taylor series.to be honest i have'nt started p6 yet so maybe some1 who has can help u?
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    it is 2nd yr degree level maths, was kinda wondering if anyone who hasnt got all this rubbish info had any ideas!
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    (Original post by FionaDutton)
    to be honest i want to be able to write -x^-2 as a power series.... any ideas??
    -x^(-2) = 2x^(-2) - 3x^(-2) + 4x^(-2)

    Ben
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    (Original post by Ben.S.)
    -x^(-2) = 2x^(-2) - 3x^(-2) + 4x^(-2)

    Ben
    Ach - what crap!!!

    Sorry!

    Ben
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    from where?!!!!!!!! got to prove it....
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    (Original post by FionaDutton)
    from where?!!!!!!!! got to prove it....
    Just stick it in the usual formula for a Taylor series - not a Maclaurin. You can't expand about a = 0.

    Ben
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    (Original post by Ben.S.)
    Just stick it in the usual formula for a Taylor series - not a Maclaurin. You can't expand about a = 0.

    Ben
    f(x) = f(a) + (x - a)f'(a) + [((x - a)^2)f''(a)]/2! + [((x - a)^3)f'''(a)]/3!...

    Ben
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    cant do that, need to use a set formula
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    (Original post by FionaDutton)
    cant do that, need to use a set formula
    What do you mean, exactly? Could you give us a better explanation of what you CAN do?

    Ben
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    (Original post by FionaDutton)
    it is 2nd yr degree level maths, was kinda wondering if anyone who hasnt got all this rubbish info had any ideas!
    rubbish info? well since this is a thread were all levels of maths qu are asked maybe u should say its 2nd yr degree level..........
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    I've had an idea. It probably won't work but it might - I haven't got time to try. You need the Maclaurin series for -(1/(x^2))...it has to be about a=0 because this is what makes the series a Maclaurin.

    Try working out a series for -exp(x)and then dividing through by (x^2)exp(x). It might not be valid but you never know.
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    There is an attachment on its way with the series written out.
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    Here it is. Sorry for the slanty last line but I'm tired. Really tired. I apologise if this turns out to be complete rubbish but its the only idea I have at the moment.
    Attached Images
     
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    (Original post by chrisbphd)
    Here it is. Sorry for the slanty last line but I'm tired. Really tired. I apologise if this turns out to be complete rubbish but its the only idea I have at the moment.
    Chris assures me that that's right. After I pointed out the mistake on his first sheet, anyway...

    Ben
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    My pen selectively stopped working when I was writing the factorial signs.
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    can you also expand the e^-x? then you would have an expansion in x only.
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    (Original post by elpaw)
    can you also expand the e^-x? then you would have an expansion in x only.
    I reckon so - it gives you an expansion which sums to 1, with each term muliplied by an x. So if you take out a factor of x, you'll just get x = x(1)

    Ben
 
 
 
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