Recent content by Jbreezy

Actually it ended up being V/4 not (2/4)V like you guys had suggested. I think that this was a poorly written question. I asked my tutor and he agreed with you guys but we tried the answer and it was wrong. I'm going to say this question is poor. Thanks for the help

So I'm right. It means divide but The change is what part is confusing me. Why do you want the change and not simply x/n ? Thank you.

Right you are. My bad. But again, the reduction is by 3/4 of the initial speed, not 1/4. Its Ok. Keeps me thinking. Can you please explain to me why you reduce by 3/4? I feel confused because first you say Which is what I thought so I reduced the velocity in part b by 1/4. Vi = v/4. But...

Thanks for the reply. I'm sure I understand you though. Why am I reducing it by (3/4)? And why would I solve for the new V? I want the potential difference.

Homework Statement A proton has an initial speed of 4.9 105 m/s. (a) What potential difference is required to bring the proton to rest? (b) What potential difference is required to reduce the initial speed of the proton by a factor of 4? (c) What potential difference is required to...

OK so this Ʃ (-1)^(n-1) (x^n)/n from n = 1 to ∞ Should be Ʃ (-1)^(n-2) (x^(n-1))/(n-1) from n = 1 to ∞ Right?

Homework Statement I have to find the power series representation for integral (1/x) dx Homework Equations ln (1+x) The Attempt at a Solution This is very similar to ln(1+x) but I don't know if this helps me. Is this ln(x) shifted one to the right? So maybe I can use what is...

I;m lost. This is $$\lim_{n \to \infty}e^n = ∞ $$ This is $$|x| < 1 $$ for the series to converge? Yes. I swear my book has this. I don't understand why this is wrong. Is it not true?

Yes.

This is what I did (A_n)^(1/n) = (e^(n^2) x^n)^(1/n) = e^n (x) |(A_n)^(1/n)| = |x| lim n---> ∞ e^n = ∞ = L OK the series diverges but if I want it to converge I need|x| < 1 So that's where it came from

Didn't know it was in the quote. I got |x| <1 because by the root test it converges for those values. 0 < 1. It is what they did in my book when they got infinity for the limit. They took whatever was in abs like |x-3| or in this case |x| and did |x| <1 and found the interval of convergence...

I don't know what prose is. I just did (e^(n^2) x^n)^(1/n) = e^n (x) Took the limit as n --> infinity I got infinity. This is when I said |x| < 1 So I got interval of convergence (-1,1)

Homework Statement Question. Did I do this OK? Homework Equations The Attempt at a Solution A_n = Ʃ e^(n^2) x^n from n = 1 to ∞ So I tried the root test. After you take the nth root you have x e^n so then I took the limit of this as n-->∞ and I got infinity. I then said OK...

Where did you get your 1/L radius of convergence eq. ? I want to read this

Question when using the ratio test for power series and your limit is equal to 1 then you just...like this |x| limit n --> (n^2 +1)/(n^2 +2) = |x| (1) then you say You need |x| <1 and go from there? Even though your limit is one you can still use it although with the ratio test if you get a...