Taylor series expansion

  • #1

Homework Statement



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


Homework Equations



M [tex]\alpha[/tex] d3

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.
 

Answers and Replies

  • #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?
 
  • #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
 
  • #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.
 
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  • #5
Redbelly98
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Okay, so f is the fractional uncertainty of d, not of M as you suggested originally.

If M is directly proportional to d3, then wouldnt the fractional uncertainty be proportional to f? .
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
 
  • #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.
 
  • #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]
 
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  • #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.
 
  • #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?
 
  • #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
 
  • #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]
 
  • #12
M(d) = ad3(1+/-3f), M(d) = ad3 +/- 3fad3, now do i solve for f or something?
 
  • #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:
 
  • #14
I hope so...its getting late here and i was up early...so here it goes...

fM= +/-3fd
 
  • #15
gabbagabbahey
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Yep, you got it. :)
 
  • #16
Great, thank you very much for your help.
 

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