Recent content by ritwik06

Never mind. Take your time. I am in no hurry. Well, I got this in my class for maths. And my teacher said it was a challenging problem. But I don't think this integral can be done while remaining within the limits of our course. as a matter of fact, we will be told about the solution tomorrow...

well, A, B and C do not come independent of a in there. For eg. B= 1/(2(i-a)) there. But I can express \frac{1}{(x+i)(x-i)(1+ax)} as \frac {-i}{(2 (1 + ax) (-i + x))} +\frac{i}{(2 (1 + ax) (i + x))}

Thanks a lot, I have already read that from Mathematica. Its a series. I have never done anything with PolyLog b4. Well, In all my possible attempts to solve the question, I was unsuccessful, therefore I didnt mention them. Here is a list of all the things i did: 1. Tried Integration by parts. I...

Homework Statement Integrate: \frac{Log[1+x]}{1+x^{2}} Homework Equations Integration by parts. The Attempt at a Solution If the integration is easier than what I think. Please let me know how to get started. I used integration by parts to get uptil here...

Yup! I agree with david completely. If you still think you are right, please prove that there is a potential difference between the inner plates, in the state I mentioned,which you say was not the stady state! Thanks

Yup! No objections till now. It doesn't necessarily follow that the capacitors in series always have same charge. I would have believed all this, if and only if I had no doubts regarding this particular statement below. Can you prove that when the inner surface of plate 1 of capacitor 1 has...

OK, I get what you want to say. But I guess that it would be correct only if the capacitor in series were uncharged initially. Lets shift our discussion to a 2 capacitor system in diagram given below. http://img200.imageshack.us/img200/6681/capacitor.jpg Capacitor 1 has a charge of 0.6 CE...

Sure, I am ready to wait another 12 hours. All I want is my concepts to be crystal clear. I'll be waiting for ya. Thanks.

I would still ask, why? What makes capacitor 3 to have a total charge which is the sum of the charges present on 1 and 2? See, if I take this assumption: 1== (CE/3)+2q 2== (CE/6)+q 3== 3q to be true, 3q=2CE/5 1== 0.6CE 2== 0.3 CE 3== 0.4 CE 0.6CE+ 0.3 CE is not equal to 0.4 CE...

This is for all the members at PF. I still need help on why the first method went wrong. queenofbabes and rl.bhat told me the first method was wrong. But I stillcant figure out, why? Please help me!

Actually, I don't understand, why the first method is wrong? Potential difference across 1 is E/3 across 2 is also E/3 and across 3 is 0 initially. So why won't the charge divide itself in ratio 2:1 ? Yes, please explain to me the 2nd method via the method of charge flow from the battery. I...

I wonder if you could please explain to me the reason behind it. If method 1 is wrong, then there must be something that I did wrong in it. What is that? PS. Sir, please answer me only if you are sure about it.

Homework Statement Image URL: http://img13.imageshack.us/img13/1935/capacitor.png Before the switch was closed, Capacitor 1 had a charge of magnitude CE/3 on its plates and Capacitor 2 had a charge of magnitude CE/6. The third capacitor was initially neutral. Find the charge flown through...

This a rational function. The domain will be all real numbers except for where the rational expression in 'x' is undefined. Find it out where the expression is undefined and from next time, please show some of your attempts. Good Luck and accept a hearty welcome :) Ritwik

Homework Statement If f'(x)>0 for all real positive x, where f:R+ ---> R and f(x)+(1/x)=f-1(1/(f(x))), f-1(1/(f(x)))>0 for all x>0. Find all the possible values of (i) f(2),(ii) f'(2) and (iii) Limit (x f(x)) as x ----->0 . The Attempt at a Solution Guessing from the last...