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May 14

how many possible soduko variations are there

how many possible soduko variations are there - I mean different solved final solutions... how much of this needs to be revealed is a different question... —Preceding unsigned comment added by 79.122.61.98 (talk) 01:21, 14 May 2009 (UTC)[reply]

6,670,903,752,021,072,936,960 total configurations and 5,472,730,538 unique configuration after eliminating rotations, reflections, permutations, and relabelings. See: Sudoku. Dragons flight (talk) 01:39, 14 May 2009 (UTC)[reply]

I didn't expect the numbers to have such big prime factors – respectively

2^20 3^8 5 7 27704267971
2 11^2 23 983243

—Tamfang (talk) 05:28, 18 May 2009 (UTC)[reply]

On the other hand there are just 2³ = 8 variant spellings of the name, all of which get Google hits, with "sodoko" being the least common at 13,200. -- BenRG (talk) 02:05, 14 May 2009 (UTC)[reply]

geometry and algebra

i need some hints to solve these geomotry questions:

1. two congruent triangles (30°, 60°, 90°) are placed so that they overlap partly and their hypotenuses coincide. if the hypotenuse is 12 cm, find the common area of the triangles.

2. a circular arch of width 24m and height 9m is to be constructed. what is the radius of the circle of which the arch is an arc?

3. ABC is a right triangle with hypotenuse AB and AC=15cm. altitude CH divides AB into segments AH and HB, with HB=6cm. what's the area of ABC?

also this algebra question:

if x² - xy + y² = a and x³ + y³ = b, then xy = ? —Preceding unsigned comment added by 122.50.138.137 (talk) 05:05, 14 May 2009 (UTC)[reply]

Some hints: 1) draw the prolongations of the shorter edges and look for symmetry. 2) and 3): draw the complete picture and look for similar right triangles. For the algebra question: factorize b. Ask for details if you need further explanation --pma (talk) 07:01, 14 May 2009 (UTC)[reply]

For the first one, you've got a triangle with two 30-degree angles and one 120-degree angle. The side opposite the 120-degree angle has length 12. You can use the law of sines to find the other lengths and then use Heron's formula to find the area.

Or you could find the height by thinking about the tangent of 30°, and then say that it's (1/2)×base×height. Michael Hardy (talk) 23:02, 15 May 2009 (UTC)[reply]

Limit of sequence

Hi all, I'm having difficulty with this problem, and I was hoping someone could give me a hint. I'm trying to show that $\lim _{n\to \infty }{\sqrt[{n}]{3^{n}+5^{n}}}=5$

Looking at the real function,

$y={\sqrt[{x}]{3^{x}+5^{x}}}$

this can be rewritten as

$ln(y)=ln(3^{x}+5^{x})/x$

But showing how the limit is determined here has me stymied. What's the right thing to do? - RedWordSmith (talk) 07:56, 14 May 2009 (UTC)[reply]

This:

5\leq {\sqrt[{n}]{3^{n}+5^{n}}}=5{\sqrt[{n}]{1+(3/5)^{n}}}\leq 5{\big (}1+(3/5)^{n}{\big )}

, and have a sandwich --pma (talk) 08:11, 14 May 2009 (UTC)[reply]

Alternative solution, using ln(y) = ln(3^x+5^x)/x:

Apply l'Hôpital once, then factor 3^x out of both the numerator and the denominator. Apply l'Hôpital a second time. Simplify. --COVIZAPIBETEFOKY (talk) 12:40, 14 May 2009 (UTC)[reply]

...but that method leaves the student with no clue as to why the result is what it is. Whereas pma's method is more instructive and can be generalised to

\lim _{n\to \infty }{\sqrt[{n}]{a^{n}+b^{n}+c^{n}+...}}

where a, b, c ... are any set of positive real numbers. Gandalf61 (talk) 12:53, 14 May 2009 (UTC)[reply]

That's odd. I seem to understand it just fine. Both methods lead one to the same conclusion: the largest number under the radical prevails, while all of the rest ultimately become so insignificant that they don't matter in the limit.

Besides, while my method may not appear as insightful to the generalization you describe (it's not that it can't be generalized; it's just a bit more difficult to do), I thought it had the advantage of being less creative thought and more computational, which is what you want ideally if, say, you have to take a test on the material. --COVIZAPIBETEFOKY (talk) 13:15, 14 May 2009 (UTC)[reply]

If I gave students the original problem in an analysis class, it would be long before l'Hôpital's rule had been covered. Moreover, relying on l'Hôpital's rule to prove that the limits of sequences exist will prevent one from developing the skill with estimates. it's a bad habit to get into. If I gave a test on the material, I would not permit the use of l'Hôpital's rule on this problem in the first place. — Carl (CBM · talk) 13:26, 14 May 2009 (UTC)[reply]

l'Hopital's rule was never covered at all in my first course in analysis (which is where this kind of question would have appeared) and its use was pretty much always forbidden. As a result, I learnt how to do limits from first principles and how to use intuition to get a good guess at what it is going to be and have thus never seen the need to use l'Hopital since. l'Hopital is basically a shortcut to working out the limit using Taylor expansions - I prefer the Taylor expansions themselves, it's generally quicker (you only need the first term). --Tango (talk) 14:53, 14 May 2009 (UTC)[reply]

My view is equivalent to the idea that you should use a theorem, if and only if you can prove it yourself without reference to the proof prior to doing so. The sandwich theorem is easy to prove for anyone who has even a little understanding of the epsilon-delta definition of a limit. This is not to say that L'hôpital's rule (don't forget the circumflex on the "o"!) is difficult to prove. However, many students use it freely without knowing its proof. This all links back to the idea that you will most certainly understand something well, if you have discovered it yourself. People who apply L'hôpital's rule without checking the indeterminate, clearly have not understood the rule well, nor do they have a good intuition of it. However, there is a large number of people who do understand the rule well, and have an intuition of it, and therefore should have the right to use it at liberty. However, I would strongly encourage that one at least attempts to discover other methods of calculating limits, because there will be cases when L'hôpital's rule does not apply. Nevertheless, I firmly agree that pma's solution is the easiest and most original solution to this problem. --PS T 03:54, 15 May 2009 (UTC)[reply]

Rate of change

Water is pumped out at a rate of 3 m³min^-1 from a hemispherical water tank with radius=13 m. When the depth of water is x m, the volume of water inside the tank is ( π/3 )x²(39-x) m³.At what rate is the water level dropping when the depth of water is 8 m? —Preceding unsigned comment added by Dansonncf (talk • contribs) 11:04, 14 May 2009 (UTC)[reply]

Clue: The flow is A·dx/dt where A is the surface area and dx/dt is the rate of water level dropping. Bo Jacoby (talk) 12:21, 14 May 2009 (UTC).[reply]

Very nice clue, Bo. Some people overlook that and do it by more cumbersome methods. Michael Hardy (talk) 00:54, 17 May 2009 (UTC)[reply]

Thanks Michael! Bo Jacoby (talk) 08:39, 19 May 2009 (UTC).[reply]

The OP's calculation of the water volume is correct. The surface area of the water is π(x⁴-338x²+28561). Cuddlyable3 (talk) 15:30, 14 May 2009 (UTC)[reply]

Shouldn't it be the derivative of the volume? —Tamfang (talk) 05:23, 18 May 2009 (UTC)[reply]

Totally real fields

If F is a totally real number field, is it always possible to find an element which is mapped to a negative number by one real embedding and to a positive number for all the others? For real quadratic fields, this is true (for Q(sqrt(d)), take 1+sqrt(d)), but what about higher-dimensional fields? --Roentgenium111 (talk) 14:10, 14 May 2009 (UTC)[reply]

Yes. The embeddings e satisfy e(q + rx) = q + re(x) for any rational q, r, hence the question is equivalent to whether there exist an element which can be mapped to two different numbers, which is in turn equivalent to whether there exist at least two different embeddings. They do, unless F is Q. — Emil J. 14:24, 14 May 2009 (UTC)[reply]

I need an element x such that the e(x) are pairwise different. But this can be achieved by taking an x in F that is not contained in any proper subfield of F, i.e. a primitive element. Thank you for your help! --Roentgenium111 (talk) 14:48, 14 May 2009 (UTC)[reply]

Question from a mechanics past paper

A particle P of mass 0.5 kg moves upwards along a line of greatest slope of a rough plane inclined at an angle of 40◦ to the horizontal. P reaches its highest point and then moves back down the plane. The coefficient of friction between P and the plane is 0.6.

Show that the magnitude of the frictional force acting on P is 2.25N, correct to 3 significant figures.

So I know I have to use F = μR but I'm having problem getting R. R is equal to the component of the particles mass perpendicular to the plane. I know I should get R = 0.5g cos(40) but when I work it out I get R = 0.5g / cos(40). Can anyone see what I'm doing wrong? —Preceding unsigned comment added by 212.120.248.41 (talk) 22:55, 14 May 2009 (UTC)[reply]

I get 2.25. Your formula R=mg cos (40 degrees) is correct. Is μ the coefficient of friction? Are you remembering to convert degrees to radians if your computer's cos function expects radians? Are you remembering that

g=9.85m/s^{2}

? 207.241.239.70 (talk) 04:27, 15 May 2009 (UTC)[reply]

I know that R=mg cos (40 degrees) cos I've seen the answer. it's just that i don't understand why. when i resolve the forces i get a right-angled triangle like this with the top angle of 40 degrees:

|\
|  \
|    \
|      \
|        \
W       R

So cos(theta) = adjacent/hypotenuse

So cos(40) = W/R

So R = W / cos(40), not R = W cos(40)

Where am I going wrong? --212.120.248.41 (talk) 16:47, 15 May 2009 (UTC)[reply]

First, the plane was described as being at 40 degrees to the horizontal, your diagram is to the vertical. Resolving a force into two components at right angles must give each of them as less than the original, so you can't divide by sin or cos - these must be multipliers. Think of it this way - if the plane was horizontal, i.e. at zero degrees, the perpendiculat component would be equal to mg, i.e. mg cos 0.217.43.211.26 (talk) 19:51, 15 May 2009 (UTC)[reply]

Yes but what i have drawn isn't the plane itself. Its the weight of the particle and the component of the weight perpendicular to the plane. Since the plane is at 40 degrees to the horizontal, I believe then that the angle between the weight and the perpendicular component is also 40 degrees. Right? —Preceding unsigned comment added by 212.120.248.41 (talk) 15:39, 17 May 2009 (UTC)[reply]

Think of the case where the plane is flat (i.e. zero degrees from the horizontal). Then the vertical component of the force is just mg. Now tilt it a tiny bit from flat, like 0.01 degrees. The vertical component is now slightly less than mg, i.e. mg*z where z is close to 1. We can see that z is either cos theta or sin theta, but which is it? Clearly z is cos theta rather than sin theta, since theta is close to 0, and z is close to 1 instead of close to 0. 67.122.209.126 (talk) 01:54, 18 May 2009 (UTC)[reply]

The diagram should be like that

|\
|  \
|    \
|      \
|        R
W

W is along the hypotenuse. Dauto (talk) 23:10, 19 May 2009 (UTC)[reply]

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