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Taylor series expansion

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Need to calculate fractional uncertainty f, of M (mass of a star in this case), where f is much less than one. The hint i was given was all i need to know is M [tex]\alpha[/tex] d3, and use a taylor expansion to the first order in f.

    M = mass of a star, d = distance to star


    2. Relevant equations

    M [tex]\alpha[/tex] d3

    3. The attempt at a solution

    Firstly, im stuck at what the [tex]\alpha[/tex] means, ive never seen it used before in this sense, at least not that i remember. Cut and paste shows it to be a division sign, although that doesnt mean much. Im not very strong with taylor expansions at all. Assuming that the [tex]\alpha[/tex] is supposed to be a division sign...

    f(x) = M/d3 then...

    f(a) + (f`(a)/1!)(x-a) then...

    M/d3 + 1/d3(x-a) would be the first order? Im guessing you take the derivative with respect to M. Would 1/d3 be my uncertainty since that is the first order? Its been awhile since i worked on taylor expansions and im very confused.
     
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  3. Oct 1, 2008 #2

    gabbagabbahey

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    The [itex]\propto[/itex] means "directly proportional to", so that the equation [itex]M(d) \propto d^3[/itex] says that [itex]M(d)[/itex] is directly proportional to [itex]d^3[/itex]. It could mean that [itex]M(d)=2.6876d^3[/itex] or [itex]M(d)=-97d^3[/itex] for example. Since you aren't told what the constant of proportionality is, just call it [itex]\kappa[/itex] so that [itex]M(d) \propto d^3 \quad \Rightarrow M(d)=\kappa d^3[/itex]

    What is f supposed to be? Is it a function of M?
     
  4. Oct 1, 2008 #3

    Redbelly98

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    The [itex] \alpha [/itex] means "is proportional to". Alternatively, you could say

    M = a d3
    where "a" is a constant.

    Do you know what "d" is?

    I'll take an educated guess that you are to write f in terms of the fractional error in d, which you could call fd or maybe Δd/d

    Another educated guess: you didn't quite copy the problem statement properly. If you are really to find f "to the first order in f", the answer is simply

    f=f
     
  5. Oct 1, 2008 #4
    Here is the exact question:

    Lets call the fractional uncertainty on the distance to the Galactic center f, meaning that we think the true distance is with in d0 +/- fd0 (where D0 is our best guess and f is much smaller than 1). What is the fractional uncertainty on M? Hint: All you need to know is M [tex]\alpha[/tex] d3 and then use a Taylor expansion to first order in f.

    If M is directly proportional to d3, then wouldnt the fractional uncertainty be proportional to f? So if fd = f, and M is proportional to d3, then would fM = f1/3?

    f is the fractional uncertainty, not a function of mass or distance i dont believe.
     
    Last edited: Oct 1, 2008
  6. Oct 1, 2008 #5

    Redbelly98

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    Okay, so f is the fractional uncertainty of d, not of M as you suggested originally.

    It is. But that does not mean they are equal. Proportional does not mean equal.

    Since M α d3, see what happens when you expand

    (d0 +/- fd0)3
     
  7. Oct 1, 2008 #6
    So is my f(a) = (d0+/- fd0)3, then i take the first derivative with respect to f? f`(a) = +/-d0(x-a)? I am so confused by series expansion.
     
  8. Oct 1, 2008 #7

    gabbagabbahey

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    Basically you have [itex]M(d)=\kappa d^3= \kappa (d_0 \pm fd_0)^3 =\kappa d_0^3 (1 \pm f)^3[/itex]
    You are given that f is much smaller than 1, so you want to do a Taylor series expansion of [itex]g(f)=(1 \pm f)^3[/itex] about the point f=0. (I'm using g here so that you don't confuse your f's)

    The Taylor series of g(f) about the point f=0 is:

    [tex] g(f)=g(0)+\frac{g'(0)f}{1!}+\frac{g''(0)f
    ^2}{2!}+\ldots[/tex]
     
    Last edited: Oct 1, 2008
  9. Oct 1, 2008 #8
    Alright, and then the expansion to the first order would be g(0) + g'(0)f?

    Tell me if im getting this...

    g(f) = (1+f)3 = (1 + 3f + 3f2 + f3)
    g'(f) = (3 + 6f + 3f2)

    g(f) = g(0) + g'(0)f /1! = 1 + 3f

    That would be expanded to the first order f? 1 + 3f?

    So whatever my fractional uncertainty is for distance, say fd, the fractional uncertainty of the mass, fm = 1 + 3fd?

    BTW thank you very much for your help.
     
  10. Oct 1, 2008 #9

    gabbagabbahey

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    Your expansion is correct, however I meant to say [itex]g(f)=(1 \pm f)^3[/itex]. As you can check for yourself, the expansion of that to first order in f about f=0 is [itex]1 \pm 3f[/itex]. So substitute this into the formula for [itex]M[/itex] what do you get?
     
  11. Oct 1, 2008 #10
    The fractional uncertainty of the Mass = 1+/- 3f, where f is the fractional uncertainty of the distance? Im not sure which equation for mass your referring to since i didnt provide one initially, only the proportion, or were you talking about M(d) = a d3
     
  12. Oct 1, 2008 #11

    gabbagabbahey

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    Use the equation [itex]M(d)=\kappa d^3= \kappa (d_0 \pm fd_0)^3 =\kappa d_0^3 (1 \pm f)^3[/itex].
    Remember this comes from the fact that [itex]M[/itex] is proportional to [itex]d^3[/itex] and that [itex]d=d_0 \pm fd_0[/itex].

    All that the taylor series has shown is that [itex](1\pm f)^3 \approx 1 \pm 3f[/itex]
     
  13. Oct 1, 2008 #12
    M(d) = ad3(1+/-3f), M(d) = ad3 +/- 3fad3, now do i solve for f or something?
     
  14. Oct 1, 2008 #13

    gabbagabbahey

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    Well, you have:

    [tex]M(d)=\kappa d_0^3 \pm 3f\kappa d_0^3[/tex]

    Or, using [itex]M_0 \equiv M(d_o)=\kappa d_0^3[/itex]

    [tex]M(d)=M_0 \pm 3fM_0[/tex]

    Now, can you look at that equation and tell me what [itex]f_M[/itex] is?:wink:
     
  15. Oct 1, 2008 #14
    I hope so...its getting late here and i was up early...so here it goes...

    fM= +/-3fd
     
  16. Oct 1, 2008 #15

    gabbagabbahey

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    Yep, you got it. :)
     
  17. Oct 1, 2008 #16
    Great, thank you very much for your help.
     
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