Wikipedia:Reference desk/Archives/Mathematics/2010 October 9

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October 9

angle trisection

Degree 4th polynomial equation solving

Please verify the following ideas:

Polynomial degree 4th equation can b easily transformed by Viete relations aiding and being put under the form:

X^4 + C*X + D =0

The root will be of the form:

X1=x0+a

X2=x0-a

X3= -x0+b

X4= -x0-b

The relations to be satisfied are:

X0*x0+ (a*a+b*b)/2=0 (1)

X0*(a*a-b*b)=k, k constant (2)

(x0*x0-a*a)*(x0*x0-b*b)=D (3)

From (1) and (3) => (a*a-b*b)^2 => a*a- b*b (4)

From (2) and (4) => x0….. then => a… and b

Thank You, Florin

Schitarea rezolvarii “de tip numai cu rigla si compasul” a ecuatiei polinomiale de gradul al 4-lea ( Incercare de Matei Florin )

Ecuatia polinomiala de gradul al patrulea este de forma:

P(x)=0, P(x)= (x^4) +A*(x^3) + B*(x^2) + C*x + D

Ideea rezolvarii este ca, in momentul in care se reuseste scrierea ecuatiei date sub forma:

(x*x+M*x+N)*(x*x+P*x+Q)+T=0,

in coditiile in care cele doua trinoame factor ar avea acelasi discriminant, ecuatia ar admite relativ usor scrierea sub forma bipatrata, dupa care se rezolva sub aceasta forma , iar dupa aceea se obtin si… radacinile lui P.

Conditiile sunt relativ trivial de clarificat, cheia obtinerii conditiei cheie, cea in legatura cu discriminantii, pare a fi notarea urmatoare:

M=A/2+z ,z grad de libertate,necunoscuta

Rezulta N=A/2-z

Ecuatia cheie este de gradul al doilea, in z, de cele mai multe ori nu ridica probleme speciale spre a fi rezolvata iar daca totusi , acest lucru se intampla, se poate lucra la transformarea lui P cu ajutorul tehnicilor de prelucrare a radacinilor (sale).

Multumesc frumos, Florin Matei —Preceding unsigned comment added by 93.113.176.190 (talk) 05:30, 9 October 2010 (UTC)

English only, not Romanian, should be used in this reference desk. Please see Angle trisection. The problem can't be solved with ruler and compass and the reason is you can't solve general cubics using ruler and compass. Dmcq (talk) 11:08, 9 October 2010 (UTC)

Second Step Analysis for Markov Chains

Hello, the following problem comes from a study guide for a test but I have no idea where to begin. Consider the Markov chain on state space S={0,1,2} with the probability transition matrix {{1/3,1/3,1/3},{0,1/2,1/2},{3/4,1/4,0}} where I have written the matrix as {row1,row2,row3}. Now I want to find the mean time spent in state one given that I start in state zero and prior to returning to state zero again for the second time. I know how to do the first step analysis but how to do this "second step analysis"? How do I take care of the fact that the chain started in state zero, spends time elsewhere, and then returns to zero? How do I define the variables and how do I derive the system? Thanks! 174.29.63.159 (talk) 06:30, 9 October 2010 (UTC)

Set up three unknowns, denoting "mean time spent in state 1 before reaching state 0, given that we start in state n". One of these (for n=0) is the answer you're looking for. You can use your transition matrix to express each of the unknown in terms of the same three unknowns (simulating a single step). This gives you three equations in three unknowns. Solve. –Henning Makholm (talk) 04:51, 10 October 2010 (UTC)

Also, try making contact with User talk:A Real Kaiser who asked a very similar question further below but seems to have gotten further than you. Chances are that you're working from the same text; you could form a study group. –Henning Makholm (talk) 14:12, 11 October 2010 (UTC)

Biostatistics / math question

It's sad I don't know how to do this, but I don't, so here I am, I'd appreciate your help:

20 years ago, a specific disease caused the deaths of 117 people every day. Today the disease causes 110 deaths per day. Assuming this annual rate of decline remains steady, how many years will pass before the deaths per day reaches zero? My own primitive math skills reaches the figure of about 313, but I don't trust the method I used. Wolfgangus (talk) 06:55, 9 October 2010 (UTC)

I get about 314. In 20 years, it dropped by 7. So it drops by 7/20 = .35 per year. 110/.35 is roughly 314, so it takes 314 years to drop to 0.--203.97.79.114 (talk) 07:23, 9 October 2010 (UTC)

Excellent, thank you so much for the input.Wolfgangus (talk) 09:05, 9 October 2010 (UTC)

"Assuming this annual rate of decline remains steady" is an awfully big assumption though. Qwfp (talk) 10:02, 9 October 2010 (UTC)

Slope of a linear equation

Can someone who understands more about algebra than I do please take a look at this diff and confirm whether or not it's an accurate change? Thanks! —Angr (talk) 13:17, 9 October 2010 (UTC)

Well, it's not wrong. I believe "ax+by+c=0" is a more common standard form for linear equations, but that doesn't make other forms erroneous. More important is the suggestion that the equation itself has a slope, when really the slope is of the line defined by the equation. I've attempted to fix this. Algebraist 13:51, 9 October 2010 (UTC)

But isn't ax+by+c=0 equivalent to ax+by=−c rather than ax+by=c? Or is this one of those times when the difference between positive and negative numbers doesn't matter? —Angr (talk) 14:54, 9 October 2010 (UTC)

c is arbitrary anyway, and replacing it with −c doesn't change anything here. Algebraist 15:02, 9 October 2010 (UTC)

Probability

I did a math league-type thingy yesterday at school, and there was one problem I had a question about. It went like: If 3 dice (6 sided cubes with the numbers 1-6 written on each) are thrown, what is the probability of the sum of the results being 17 or 18. I reasoned this is P(17)+P(18). P(18) is unequivocally 1/216, but I'm not sure I evaluated P(17) right. I thought P(17)=1/216 as well, making P(17 or 18)=1/108. However, even though you need the dice to be 6-6-6 for 18, you can have 5-6-6 or 6-6-5 or 6-5-6, which increases P(17), doesn't it? What would eb the write answer? Thanks. —Preceding unsigned comment added by 24.92.78.167 (talk) 18:19, 9 October 2010 (UTC)

You're pretty much there. As you said, of the 216 outcomes of the three dice, exactly 3 of them have a sum of 17 (5-6-6, 6-6-5, 6-5-6), so the probability of rolling a 17 is 3/216. The probability of rolling a 17 or an 18 would be 4/216 (5-6-6, 6-6-5, 6-5-6, 6-6-6). Rckrone (talk) 19:18, 9 October 2010 (UTC)

....and 3/216 = 1/72. Michael Hardy (talk) 20:22, 9 October 2010 (UTC)

I believe that it is asked for both 17 and 18, so it is 4/216=1/54.Jijo (talk) 19:07, 15 October 2010 (UTC)