# Taylor series expansion

1. Oct 1, 2008

### nissanztt90

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 $$\alpha$$ 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 $$\alpha$$ d3

3. The attempt at a solution

Firstly, im stuck at what the $$\alpha$$ 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 $$\alpha$$ 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.

2. Oct 1, 2008

### gabbagabbahey

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

What is f supposed to be? Is it a function of M?

3. Oct 1, 2008

### Redbelly98

Staff Emeritus
The $\alpha$ 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. Oct 1, 2008

### nissanztt90

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 $$\alpha$$ 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
5. Oct 1, 2008

### Redbelly98

Staff Emeritus
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

6. Oct 1, 2008

### nissanztt90

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. Oct 1, 2008

### gabbagabbahey

Basically you have $M(d)=\kappa d^3= \kappa (d_0 \pm fd_0)^3 =\kappa d_0^3 (1 \pm f)^3$
You are given that f is much smaller than 1, so you want to do a Taylor series expansion of $g(f)=(1 \pm f)^3$ 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:

$$g(f)=g(0)+\frac{g'(0)f}{1!}+\frac{g''(0)f ^2}{2!}+\ldots$$

Last edited: Oct 1, 2008
8. Oct 1, 2008

### nissanztt90

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. Oct 1, 2008

### gabbagabbahey

Your expansion is correct, however I meant to say $g(f)=(1 \pm f)^3$. As you can check for yourself, the expansion of that to first order in f about f=0 is $1 \pm 3f$. So substitute this into the formula for $M$ what do you get?

10. Oct 1, 2008

### nissanztt90

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. Oct 1, 2008

### gabbagabbahey

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

All that the taylor series has shown is that $(1\pm f)^3 \approx 1 \pm 3f$

12. Oct 1, 2008

### nissanztt90

13. Oct 1, 2008

### gabbagabbahey

Well, you have:

$$M(d)=\kappa d_0^3 \pm 3f\kappa d_0^3$$

Or, using $M_0 \equiv M(d_o)=\kappa d_0^3$

$$M(d)=M_0 \pm 3fM_0$$

Now, can you look at that equation and tell me what $f_M$ is?

14. Oct 1, 2008

### nissanztt90

I hope so...its getting late here and i was up early...so here it goes...

fM= +/-3fd

15. Oct 1, 2008

### gabbagabbahey

Yep, you got it. :)

16. Oct 1, 2008

### nissanztt90

Great, thank you very much for your help.