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February 27

Logarithms

I'm having trouble with this problem. I am pretty sure logs are used to solve it. 40^(3x) = 5^(2x+1). Any help would be appreciated. 70.12.91.235 (talk) 04:09, 27 February 2012 (UTC)[reply]

Take the logarithm of each side

3x\log 40=(2x+1)\log 5

Widener (talk) 04:43, 27 February 2012 (UTC)[reply]

Or rewrite the equation as (40³/5²)^x = 5. 96.46.204.126 (talk) 09:20, 27 February 2012 (UTC)[reply]

How to reverse XORs

m(0..79) are 80 binary bits

Given m(0..15) we calculate m(16..79) using

  m(i) = m(i-16) xor m(i-14) xor m(i-8) xor m(i-13)

Alternatively if given m(64..79) please explain how to calculate m(0..15).

84.209.89.214 (talk) 12:11, 27 February 2012 (UTC)[reply]

Try and get an equation with m(i-16) on the left instead of m(i) and work backwards. Dmcq (talk) 12:53, 27 February 2012 (UTC)[reply]

You may also need to know that bitwise xor is its own inverse. In other words, if a xor b = c then a = c xor b. Gandalf61 (talk) 13:29, 27 February 2012 (UTC)[reply]

OP here. I have been able to reduce the problem somewhat but here is where I am stuck:

m(0..79) are 80 binary bits

Given m(0..15) we calculate m(16..31) using

  m(i+16) = m(i) xor m(i+2) xor m(i+8) xor m(i+13)

Alternatively if given m(16..31) please explain how to calculate m(0..15).

84.209.89.214 (talk) 08:34, 28 February 2012 (UTC)[reply]

You needed a minus on the left, not a plus. And you can do it without adjusting the exponents up and down like that, just by manipulating the equation using the business about xor being its own inverse. What happens if you xor both sides of the original with m(i-16)? Dmcq (talk) 09:17, 28 February 2012 (UTC)[reply]

You have

m(i) = m(i-16) xor m(i-14) xor m(i-8) xor m(i-13)

but you want to turn this into an expression that has "+" offsets rather than "-" offsets, so you can calculate bits from right to left instead of from left to right. So

Replace i with j+16 throughout the expression. This makes all your offsets positive, without getting confused over which i you are using.
Use the self-inverse property of xor to move terms from one side of the "=" sign to the other until you are left with m(j) on one side and a bunch of terms in m(j + something) on the other side. Gandalf61 (talk) 11:21, 28 February 2012 (UTC)[reply]

And if that still isn't enough you can always keep generating a sequence using that rule until you get 16 just like you started with - that'll give you the sequence before that point. You can't have two different patterns before the sequence but perhaps that's harder to show :) See linear feedback shift register. Dmcq (talk) 13:48, 28 February 2012 (UTC)[reply]

U.S. income tax revenue

how much more revenue would be created if the income tax rate was increase back to 39.6 % this seems to be the rate people refer to when they talk about raising taxes ? — Preceding unsigned comment added by Poisonpaws (talk • contribs) 19:19, 27 February 2012 (UTC)[reply]

39.6% would be the top tax rate if the Bush tax cuts are allowed to expire. The article mentions that the CBO estimated in 2010 that letting the tax cuts expire in 2011 would have raised $3.3 trillion over 10 years over the scenario where they were extended. Letting the tax cuts expire only for individuals making over $200,000 per year and couples making over $250,000 (a plan often suggested by Democrats) would have raised $1.1 trillion over 10 years. They were temporarily extended instead of being allowed to expire at the end of 2010, so now they'll expire at the end of 2012 unless they're extended again. I would guess the 10 year outlook on letting them expire now is probably pretty similar to the 2010 numbers. I can't find newer numbers, though I'm sure the CBO has done more recent forecasts. Rckrone (talk) 07:27, 28 February 2012 (UTC)[reply]

Diophantine equations

Given any Diophantine equation with no solutions, does there exist a proof that it has no solutions? Widener (talk) 21:46, 27 February 2012 (UTC)[reply]

We don't know for sure that a given Diophantine equation has no solutions unless we can prove it. If the question whether all Diophantine equations are known to have solutions or not, I would think the answer is "no". Consider that a rather famous family of Diophantine equations in Fermat's last theorem were only recently proved to have no solutions. I'm pretty sure you could find specific Diophantine equations or families that are conjectured to have no solutions, but as of yet have no proof. Rckrone (talk) 23:18, 27 February 2012 (UTC)[reply]

Is it possible that, given a Diophantine equation that has no solutions, there exists no proof that it has no solutions? A bit like the continuum hypothesis? Widener (talk) 23:32, 27 February 2012 (UTC)[reply]

If a Diophantine equation does have a solution, then there always exists a proof that it has a solution. Therefore, any proof that a Diophantine equation having solutions is undecidable would imply that there are no solutions, therefore it is not possible that the question of a Diophantine equation having solutions is provably undecidable. However, the question as to whether that problem is undecidable could itself be undecidable (and you could extend that back infinitely). Widener (talk) 23:35, 27 February 2012 (UTC)[reply]

This is mistaken. Unfortunately, you're confusing two different notions of "have solutions". One is "have solutions in the standard model", and one is "have solutions in any model". As noted below, Godel's incompleteness theorem can be used to explicitly construct such equations. — Preceding unsigned comment added by 121.74.97.164 (talk) 03:35, 28 February 2012 (UTC)[reply]

Our Diophantine set article discusses this question in some detail. Let me quote the last section of the article: "Matiyasevich's theorem has since been used to prove that many problems from calculus and differential equations are unsolvable. One can also derive the following stronger form of Gödel's first incompleteness theorem from Matiyasevich's result: Corresponding to any given consistent axiomatization of number theory,[5] one can explicitly construct a Diophantine equation which has no solutions, but such that this fact cannot be proved within the given axiomatization." Looie496 (talk) 00:11, 28 February 2012 (UTC)[reply]

Calculating the equation of a line tangent to a circle that passes through a particular point

What would be the generalized equations that can be used to find the $A$ , $B$ , and $C$ constants that describe the lines tangent to a circle that pass through a given point? The variables are as follows:

$P_{x}$ is the x coordinate of the point.
$P_{y}$ is the y coordinate of the point.
$C_{x}$ is the x coordinate of the circle.
$C_{y}$ is the y coordinate of the circle.
$r$ is the radius of the circle.

--Melab±1 ☎ 22:51, 27 February 2012 (UTC)[reply]

Let Q be the point of tangency, so Q-C is a radius. We have that |Q-C| = r², and that Q-C is perpendicular to the line, which goes in the direction of Q-P, so (Q-P).(Q-C) = 0. So then we can solve for Q = (Q_x,Q_y) using these constraints, namely (Q_x-C_x)² + (Q_y-C_y)² = r² and (Q_x-P_x)(Q_x-C_x) + (Q_y-P_y)(Q_y-C_y) = 0. Once you find Q, the equation for the line through P and Q can be found with Linear_equation#Two-point_form. Rckrone (talk) 23:33, 27 February 2012 (UTC)[reply]

I was thinking of expressing it like:

A=...

B=...

C=...

and then making

Ax+By+C=0

--Melab±1 ☎ 02:12, 28 February 2012 (UTC)[reply]

In the coordinate system where P=(0,0) and C=(1,0) the tangent equations are $y=\pm (r^{-2}-1)^{-{\frac {1}{2}}}x$ . Bo Jacoby (talk) 00:18, 28 February 2012 (UTC).[reply]