Wikipedia:Reference desk/Archives/Mathematics/2013 May 9

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May 9

quick math help

y = -3x² - 7x + 8

a) state the y-intercept

b) determine the zeroes, to 1 decimal place (show your process)

a) the y-intercept is (0, 8)

b)

a = 3, b = 7, c = 8

D = b² - 4ac D = 7² - 4(3)(8)

D = 49 - 96

D = -47

since the discriminant is lower than zero there are no roots

yah im thinkin i did something wrong but im not sure — Preceding unsigned comment added by 99.250.250.65 (talk) 02:08, 9 May 2013 (UTC)

a is -3, not 3, and your Delta is 49 + 96 = 145 = 144 + 1 = 12² + 1, whose square root is therefore just a bit above 12, namely 12.041(6). — 79.113.240.179 (talk) 02:45, 9 May 2013 (UTC)

'Show your process'? Does it mean solve 'y = 0' step-by-step? Well, this is how it goes:
0 = −3x² − 7x + 8 = −3(x² + 7/3×x − 8/3) iff
0 = x² + 7/3×x − 8/3 = x² + 2×1×7/6×x + (7/6)² − (7/6)² − 8/3 = (x + 7/6)² − 145/36 iff
(x + 7/6)² = 145/36 iff
x + 7/6 = ±√145/6 iff
x = (−7 ± √145)/6
Now you only need to find (and show) an approximation of √145 precise enough to have one decimal digit when divided by 6... --CiaPan (talk) 05:48, 9 May 2013 (UTC)

Random walk collision

Two agents start at random positions on a finite square grid (of dimension 2 but extended to n dimensions if that's feasible). Given the grid size what is the expected number of steps before they collide (occupy the same space)? What is the expected number of steps if one of the agents doesn't move? 70.162.4.242 (talk) 03:59, 9 May 2013 (UTC)

That is practically identical to one staying at the origin and the other moving two on each move. If they start on points an odd distance apart then they'll never collide (if you don't count moving through each other). Unfortunately the grid of even distances is no longer a square type one in more than 2 dimensions. See the walks on a lattice in Random walk. For two dimensions they will always meet but not necessarily in higher dimensions, but as to actual expected number of steps for an arbitrary distance apart - that is related to the nerd sniping problem I believe ;-) Dmcq (talk) 13:05, 9 May 2013 (UTC)

Oops I didn't read the last bit. If one doesn't move it is exactly the problem in the article about random walks. Dmcq (talk) 13:09, 9 May 2013 (UTC)

Actually OP asks about a finite square grid, which is a case different from the infinite grid, as on a finite square grid the time the two walks first meet has a finite mean. Sadly I'm not really good in random walks so I won't be of much help here. – b_jonas 17:29, 11 May 2013 (UTC)

Solving harmonic sin waves

I'm trying to find a generalised approach that can be used to solve the following sets of equations: For a given N, find p and q satisfying the following:

 sin(2.pi.p.N/2) = 0
 sin(2.pi.p.N) = 0

 sin(2.pi.q.N/2) = 0
 sin(2.pi.q.N) = 0

 sin(2.pi.p.N/4) = 1
 sin(2.pi.q.N/4) = -1

where

 p.q = N 

but

 p or q ≠ 1
 p or q ≠ N
 p or q ≠ 0

Ideally I'm trying to find a method that solves for p and q that are prime integers, and the method should work for really big integers! Hwebster au (talk) 04:19, 9 May 2013 (UTC)

First note that if sin(pi x)=0 then x is an integer, if sin(pi x)=1 then x=..., 0.5, 2.5, 4.5,..., if sin(pi x)=-1 then x=..., 1.5, 3.5, 5.5,... So your equations are equivalent to saying

p N is an integer
p N is ..., 0.5, 1, 1.5, ...
q N is an integer
q N is ..., 0.5, 1, 1.5, ...
p N is 1, 5, 9, ...
q N is 3, 7, 11, ...

If p, q, and N are all integers the first four equations tell us nothing new. Using modular arithmetic the last two are equivalent to saying $p^{2}q\equiv 1{\pmod {4}}$ and $pq^{2}\equiv 3{\pmod {4}}$. If $k$ is odd then $k^{2}\equiv 1{\pmod {4}}$ so the equations simplify to $q\equiv 1{\pmod {4}}$ and $p\equiv 3{\pmod {4}}$, that is q is one of 1,5,9,13,.. and p is one of 3,7,11,15,... There are plenty of candidate solutions which are prime, say 3 and 5. --Salix (talk): 07:37, 9 May 2013 (UTC)

What does this example mean?

I have a hard time figuring out what the math looks like in this example.

"A coefficient of capitalization is defined by means of dividing of an outstanding debt under a shareholder’s loan by an equity capital in proportion to a lender’s percentage of direct or indirect shareholding" What does "in proportion to" mean in this case?

Mityuy (talk) 09:18, 9 May 2013 (UTC)

See Proportionality (mathematics). It is tricky to interpret the situation we have three main things

a=outstanding debt under a shareholder’s loan

b=equity capital

c=lender’s percentage of direct or indirect shareholding

then your statement is either saying a/b is proportional to c. Or a/(b in proportion to c). You question seems to relate to russian tax codes and [1] may help. Note we cannot answer financial questions see Wikipedia:Reference_desk/Guidelines and WP:NOLEGAL so its safest to ask a relavant professional. --Salix (talk): 10:12, 9 May 2013 (UTC)

The most probable interpretation is a/(b x c). If a shareholder holds c% of the shares issued by the company, and the company's total equity capital is b then the shareholder's "share" of the equity capital is b x c. Gandalf61 (talk) 11:02, 9 May 2013 (UTC)

The proof of the splitting lemma

Dear Wikipedia users,

I have been studying the proof of the splitting lemma. Although I agree on most of it, there is a single (but important) part omitted from the proof, and this part is not evident (to me). Look at the proof of (1) => (3) [or (2) => (3)]. The proof that B is isomorphic to the direct sum of A and C is crystal clear, but (3) also states that the maps q and r are the natural embedding and the natural projection, respectively, but this isn't proved. Is this 'obvious' (if so, how to see it?), or is it an issue with the proof given in the article? --81.170.199.71 (talk) 11:22, 9 May 2013 (UTC)

Puzzling trig problems

Two trig problems have completely puzzled me, can someone please help me. You have to out the value of these, without a calculator, of course: (angles in degrees)

1: (tan 1)^2 + (tan 2)^2 + (tan 3)^2 + . . . + (tan 89)^2

2: (sin 40 + sin 80)/sin 110

117.197.84.100 (talk) 13:00, 9 May 2013 (UTC)

Part 2 is quite easy - use the well-known identity for sin A + sin B for the numerator, and write the denominator as the sin of the sum of two angles and expand. No idea about part 1, though it looks vaguely familiar. Question slightly reformatted for readability AndrewWTaylor (talk) 13:29, 9 May 2013 (UTC)

Thanks, the formula got out of my mind. Actually its even simpler, just convert sin 110 to cos 20, and you're done.117.197.70.116 (talk) 16:28, 9 May 2013 (UTC)

Ah, found it! A generalisation of question 1 is posed and proved here. AndrewWTaylor (talk) 14:24, 9 May 2013 (UTC)

1) has a much simpler solution than the one posted on stackexchange. Count Iblis (talk) 14:46, 9 May 2013 (UTC)

OK, give us a clue.. AndrewWTaylor (talk) 12:54, 10 May 2013 (UTC)

$\tan ^{2}(x)={\frac {\sin ^{2}(x)}{\cos ^{2}(x)}}={\frac {1-\cos ^{2}(x)}{\cos ^{2}(x)}}={\frac {1}{\cos ^{2}(x)}}-1$ So, we need to know how to sum 1/cos^2(x) from 1 to 89 degrees. Since summing 1/sin^2(x) over the same values will yield the same result, we can consider summing ${\frac {1}{2}}\left[{\frac {1}{\cos ^{2}(x)}}+{\frac {1}{\sin ^{2}(x)}}\right]={\frac {\cos ^{2}(x)+\sin ^{2}(x)}{2\cos ^{2}(x)\sin ^{2}(x)}}={\frac {2}{\sin ^{2}(2x)}}$

Then the summation over 1/sin^2(2x) can be rewritten in terms of the summation of 1/sin^2(x) over only the even angles from 2 to 88 degrees, which is of course also the summation of 1/cos^2(x) over those values. In general, this means that if we sum over angles with step size delta, we can express that summation in terms of the summation with step size 2*delta. So, by iterating this relation a few times, we can relate the original summation to a trivial summation. Count Iblis (talk) 15:32, 10 May 2013 (UTC)