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Maclaurin series for integrals

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Heres the question: write the first three nonzero terms (Maclaurin Series)


    2. Relevant equations

    3. The attempt at a solution
    so for the similar questions, i use maclaurin series for common functions (you can see it from this website http://mathworld.wolfram.com/MaclaurinSeries.html" [Broken])
    for this question, i tried to start with lnx, but from what i searched online, there is no maclaurin series for lnx (because it is not defined at 0), so now i am stuck. also, i tried to inegrate just the function itself, but it seems impossible.
    if lnx does not have a maclaurin series, that function does not have a maclaurin series?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 6, 2010 #2


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    Hi bobthetomato - welcome to PF ;)

    I think the notation is a little confusing, does it help if you re-write in terms of a dummy variable?
    [tex]f(x) = \int^{1+sinx}_{1} \frac{du}{\sqrt{1+ln(u)}}[/tex]

    note the intergal is over u from 1 to 1 + sin(x), so the ln(u) doesn't blow up

    then you can differntiate w.r.t. x to find each term, which could be a little tricky...
  4. Apr 7, 2010 #3
    thank your lanedance, that function looks better now.
    i dont understand, why do i need to differentiate it? can i integrate the whole thing? but if i integrate the equation, i wont get maclaurin series (first three terms).

    this is what i tried:

    since there is a series for ln(1-x) = -x-(x^2/2)-...

    let x=1-t



    but if i continue to do it, it becomes complicated and hard to differentiate

    1+ln(t)= t-((1-t)^2/2)-...

    1/(1+ln(t))^(1/2)= 1/(t-((1-t)^2/2)-...)^(1/2)

    i stopped here because i do not know how to integrate this function.
  5. Apr 8, 2010 #4
    Can someone please help me? This is only a university first year math question, so I think it shouldn't be that hard.

    Here is the question again...
    f(x) = \int^{1+sinx}_{1} \frac{dt}{\sqrt{1+ln(t)}}

    so i think substitution is the right place to start.

    as the range of integration is near 1 (because 1+sinx can range from 1 to 2), ln(t) will not blow up.

    so i set x=(1-t)

    [tex]\ln(1-x) = -x - \frac{x^{2}}{2} - \frac{x^{3}}{3}+...[/tex]

    if i substitute 1-t into all the x

    [tex]\ln(1-(1-t)) = -(1-t) - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...[/tex]

    and then it turns out like this

    [tex]\ln(t) = -1+t - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...[/tex]

    now if I add 1 to the whole equation

    [tex]\ 1 + ln(t) = 1 -1+t - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...[/tex]

    which will equal to

    [tex]\ 1 + ln(t) = t - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...[/tex]

    and now if i square root the whole thing and 1/the whole equation, i will get the this...
    oh and i added integrals

    f(x) = \int^{1+sinx}_{1} \frac{dt}{\sqrt{t-\frac{(1-t)^{2}}{2}-\frac{(1-t)^{3}}{3}+...}}

    so know i have made it to Maclaurin Series like, i do not know what to do next. How do you simplify and integrate this?
    This question is bugging me, please help!
  6. Apr 8, 2010 #5


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    sorry if is miss something, tex doesn't display well on this computer..
    so you need to evaluate the macluarin sereis of:
    [tex]f(x) = \int^{1+sinx}_{1} \frac{dt}{\sqrt{1+ln(t)}}[/tex]

    so rather than substituting, expanding etc. why not just evaulate the derivatives
    f'(x), f''(x) and evaluate at x = 0. f(0) should be obvious

    As an example, consider:
    [tex] F(x) = \int_0^x du u^2 = \frac{u^3}{3}|_0^x = \frac{x^3}{3}[/tex]
    [tex] F'(x) = f(x) = \frac{d}{dx}\int_0^x du u^2 = \frac{d}{dx}\frac{x^3}{3}=x^2[/tex]
    [tex] F''(x) = f'(x) = ...[/tex]
    now all these should be easy to evaluate at x=0, how can you write the maclaurin expansion using those terms? (though its a very trivial expnasion for a single polynomial, but its the method that is important here)
  7. Apr 8, 2010 #6


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    PS if its still not clear have a look at the definition of maclaurin series in terms of derivatives of the function
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