Wikipedia:Reference desk/Archives/Mathematics/2012 September 26

= September 26 =

Limits
How can the following limit problem be solved: $$\lim_{x \rightarrow 2}\frac{2x^2 + \cos x}{3x^2 + x - 1 + e^{-x}}$$

Neither 0.577 nor O is correct. What would be the proper way to evaluate this limit. Thank you — Preceding unsigned comment added by 24.86.170.175 (talk) 06:32, 26 September 2012 (UTC)


 * Sure looks like 0.577 to me. Since the function is continuous across x = 2, you can just plug 2 in for x to calculate it.  What makes you think 0.577 is wrong ?  Could it be you just need more decimal places ?  StuRat (talk) 06:43, 26 September 2012 (UTC)


 * The program in which I have to enter the answer in states that 0.577 is incorrect. — Preceding unsigned comment added by 24.86.170.175 (talk) 06:56, 26 September 2012 (UTC)


 * Well, try more decimal places, I bet that will fix it. StuRat (talk) 07:46, 26 September 2012 (UTC)
 * Gives 0.685 with angle in degrees (which wouldn't make a lot of sense), 0.5773627.. if radians are used. What program is it, does it have specific instructions regarding results that have to be rounded off? Ssscienccce (talk) 09:00, 26 September 2012 (UTC)

It is still incorrect — Preceding unsigned comment added by 142.58.95.144 (talk) 17:14, 26 September 2012 (UTC)
 * Note: I have taken the liberty of reformatting the expression in math mode to make it easier to read. Looie496 (talk) 16:31, 26 September 2012 (UTC)


 * Another possibility is that it doesn't want a numeric answer, just a simplified expression:

$$\frac{8 + \cos(2)}{13 + e^{-2}}$$


 * On one line, that would be:

(8 + cos(2))/(13 + e^(-2))


 * Try that. StuRat (talk) 18:55, 26 September 2012 (UTC)


 * If you've tried the formula answer and the degree answer and radian answer, with various decimal places of precision, and it rejects them all, then it looks like somebody entered the wrong answer into the program which checks your answer. It's time to contact the instructor and complain about it. StuRat (talk) 18:48, 26 September 2012 (UTC)

The simplified answer worked. Thank you everyone for your help. — Preceding unsigned comment added by 24.86.170.175 (talk) 05:30, 27 September 2012 (UTC)


 * You're quite welcome. StuRat (talk) 05:54, 27 September 2012 (UTC)


 * Nice one Stu! Would never have thought of that. Ssscienccce (talk) 06:47, 27 September 2012 (UTC)


 * Thanks ! StuRat (talk) 07:03, 27 September 2012 (UTC)

Representations of so(3;1) = Lorentz algebra
Is the following correct?

The irreducible finite-dimensional (m,n)-representations of so(3;1) are complexified tensor products of irreducible real linear representations of su(2).

How is it shown that the (m,n)-representations are, in fact, irreducible (whether the above is correct or not)?

As far as I can see, complexification (both of the algebra [so(3;1)->so(3;1)C] and vector space of the rep [V->VC]) is necessary. Tensor products of su(2) reps are not generally irreducible. Moreover, tensor products of reps of sl(2;C) are not irreducible either (by the same formalism). So complexification first and then tensor product wouldn't work. It seems clear that the above recipe does yield a representation of sl(2;C). It's not immediate (to me) that it is irreducible.

What I'm asking about here is very carefully avoided in every math and physics book I've seen, and I've seen a few. Well, it's probably obvious to everybody... To me it's just plausible that the (m,n) reps are irreducible.

Ironically, when I try to Google (with sensible keywords) for a proof of irreducibility of the (m,n)-reps, the first hit is the thing I'm personally writing on at the moment. YohanN7 (talk) 16:59, 26 September 2012 (UTC)