Infinite Series

[Elkay]

I need help with 2 series problems:


Question 1: (http://fantasyland.250free.com/Series.jpg) I was able to solve part a, but I don't understand part b or part c.


Question 2: (http://fantasyland.250free.com/binomial_relativity.jpg) I don't get this one at all.


Any help would be appreicated.

[turin]

I was able to solve part a, but I don't understand part bReplace the function with the series and truncate it.




or part c.Sorry, I forgot how to do this.




Question 2: (http://fantasyland.250free.com/binomial_relativity.jpg) I don't get this one at all.Binomial series is a series for expressions of the form:

(1+x)n/2

where n is an integer and x < 1 if n is odd. Sorry, I don't remember what the series is. Identify what x and n are in the problem (hint: they give the expression to you in the desired form, so all you need to do is make these two identifications). Then, look up the binomial expansion and fill in the blanks.

[Elkay]

Thanks Turin. It took a while, but I was able to figure out part b, and I think I figured out how to do part c from the first question (dunno if it's right or not).



I'm still trying to figure out Question 2 (http://fantasyland.250free.com/binomial_relativity.jpg) though... If m and c are constants, then x would have to be v. I'm not sure what n should be. :frown:

[cookiemonster]

http://mathworld.wolfram.com/BinomialSeries.html

Take only the first two terms.

cookiemonster

[Elkay]

Thanks Cookiemonster. After reviewing my textbook and that site for about an hour, I understand a little bit better, but I still have questions (sorry for being such a pest).


I don't understand why the series should be expanded to only two terms. Is there something there that says to do that?

Is it even possible to take the factoral of a negative number? If so, what is the value? (equation 6 (http://mathworld.wolfram.com/BinomialSeries.html) when k=0)

[cookiemonster]

You only keep the first two terms because those are the only significant ones for small v. Notice that each additional term increases the order of v by 2, so if v is something like .01c, v^4 or v^8 is next to nothing. That being said, in this case you're keeping only two because that's what the problem asked for. =]

Yes, it's possible to take the factorial of a negative. Well, sort of. Just not the negative integers. The Gamma function is the generalized form of the factorial. It has the property that \Gamma (n + 1) = n \Gamma (n) and \Gamma (1) = 1, so if you apply that repeatedly you see that \Gamma (n + 1) = n! for integer values of n.

http://mathworld.wolfram.com/GammaFunction.html

cookiemonster

[Elkay]

Sorry, but that gamma function just confused me even more... Would it be okay to set k=1 instead of beginning with 0 in order to avoid (-1)!! in the numerator?

Also, shouldn't equations 6 and 7 (http://mathworld.wolfram.com/BinomialSeries.html) create an alternating series? That site lists all terms in the series as being positive... When I attempted to expand the series for Question 2 (http://fantasyland.250free.com/binomial_relativity.jpg), the terms were both positive and negative. :confused:

[turin]

Those are double factorials in eqs. (6) & (7). They are not nested factorials. That probably confused the hell out of you. I know it confused the hell out of me the first time I saw them. Basically, the double factorial is like the factorial missing every other factor. If I remember correctly, it goes something like:

(2k-1)!! = (2k-1)(2k-3)(2k-5)...

but certainly NOT:

(2k-1)!! /= ((2k-1)!)! /= ((2k-1)(2k-2)(2k-3)...)!

Actually, just use the eq. (7) part, since it takes you to the required order anyway. All you have to do at that point is put in the appropriate value for x, which, incidently, is NOT v.

Here's a link to the double factorial:
http://mathworld.wolfram.com/DoubleFactorial.html

[Elkay]

Yeah, the double factoral did confuse me a bit, but I think I understand it now.


So if x is not v, then is this true? x=\frac{-v^2}{c^2}

After expanding the binomial, I'm also not sure how to show that
mc^2 \approx m_{0}c^2 + \frac{1}{2}m_{0}v^2

[turin]

So if x is not v, then is this true? x=\frac{-v^2}{c^2}Almost. Be careful when you make the identification to use eqs. (6) & (7) on that website, and not the exact form that I gave you in my first post. There is one subtle difference.




After expanding the binomial, I'm also not sure how to show that
mc^2 \approx m_{0}c^2 + \frac{1}{2}m_{0}v^2 Well, what do you get when you expand, then? Don't forget to multiply both sides by c2

[Elkay]

OMG! It just dawned on me to substitute the first two terms from the series into that total energy equation and simplify.

I feel like an idiot for not getting this sooner. Turin and Cookiemonster, Thank you so much for all your help! :biggrin: