Hey that's great! i never thought of transforming it that way.. i guess what i was attempting to do at first before was nonsensical although i'm not sure why, it seems they should have revealed the same? i'm sure i'm on the right track with this now. Thanks for going to the time to show me where...
jtbell i apologise for any confusion here but I'm not referring to the normalisation process, it's the constant on the exponential. The equation shows a rationalisation of the numerator to give 1 /(2∏(hbar)^2 / 4σ^2)^1/4.
From what i understand if one wants to move a quantity from the...
Hey SteamKing, thanks for stopping by to help. I'm not referring to the normalisation process here, it's the constant on the exponential i'm having trouble with.. The equation shows a rationalisation of the numerator to give (2∏(hbar)^2 / 4σ^2)^-1/4.
From what i understand if one wants to...
Consider the following commutator for the product of the creation/annihilation operators;
[A*,A] = (2m(h/2∏)ω)^1 [mωx - ip, mωx + ip] = (2m(h/2∏)ω)^1 {m^2ω^2 [x,x] + imω ([x,p] - [p,x]) + [p,p]}
Since we have the identity;
[x,p] = -[p,x]
can one assume that..
[x,p] - [p,x] =...
There is a great deal of uncertainty when it comes to knowledge. The problem with the 'learning hysteria' arises essentially from a common misunderstanding about how cognition and the brain operates..
The truth is graduates are not really masters of the information they have assimilated, or at...
Yes i understand what you are saying. That's a valid point you have made. So how would one go about that Micro? expand the complex exponential into the trigonometric form, separate the terms and integrate?
I'm not sure you can for anything but the most vigorous proof, but one will often know from experience that the exponential function that which is a weighted sum of the sine and cosine, equals it's own derivative. Maybe someone else could provide an answer to this..
Yes okay i understand. That's pretty cool. So the bar is inserted at the end with the underscore to indicate the limits themselves?
I have bookmarked the Latex page for future reference. The sandbox is very helpful for getting started. Might take some getting used to but i'm happy to learn it...
Brilliant :) thanks for the link. It doesn't look too complicated i shall begin using it right away. I found some more latex under the sigma tab on the advanced page too. thanks again Mark ;)
Yes. i understand what you are saying. i just integrate the function e^t like an indefinite case and than plug in the upper and lower limits of integration and obtain the difference. sorry for the confusion there and thank you for pointing this out.
Where can i find the tex commands for the...
Hey Mark, thanks for sharing your insight, i guess i will have to review the topic a little and practice a bunch of examples. Much more legible with the use of dummy variable too..
with regards to the integral when i evaluate at zero, I think i understand where i'm going wrong here..
I...
Thanks for popping by Voko. That's the definite integral from zero to x.. i.e. the indefinite integral evaluated at x minus the indefinite integral evaluated at 0 right?
So if we evaluate at x we have;
∫ e^x dx = e^x +c1,
and at zero;
∫ e^0 dx = ∫ (1) dx = x +c2
Thus the...
I ask members here kindly for their assistance. I'm having some confusion over the process of integrating inequalities, in particular for obtaining the series expansion for the exponential function by integration. The text by Backhouse and Holdsworth (Pure Mathematics 2), shows the expansion of...
Thanks for your explanation, i understand what you are saying here. I can see now how is arises naturally from applying the chain rule..
You're absolutely right, i was using this method the whole time. it was just not obvious at the time since i was only computing functions up to a constant...
Thanks for pointing that out CompuChip! Your explanation is sound. i guess i was solving it as a differential when i really didn't need to, just was not obvious to me at first. Silly mistake really! What variation of u-substitution is being used here? i have not seen the use of [dx/du]du before...
i have reached a chapter in my pure mathematics text that introduces u-substitution as a method for solving integrals where in general if f(x) = dy/dx, an integral with respect to x may be transformed into an integral with respect to a related variable u such that: ∫ f(x) dx = ∫ [f(x) dx/du]du =...
I see where i'm going wrong, thanks to you both for pointing this out for me.. clearly need more practice with identifying when to use the chain rule! Thanks again!
Hi everyone, any help with this would be greatly appreciated..
I have been practicing simple differential equations for a couple months now and kinda just taking it easy and enjoying building my intuition. i have encountered a chapter in my text by Backhouse (pure mathematics 2) involving...