Wikipedia:Reference desk/Archives/Mathematics/2009 January 19

= January 19 =

math coincidence
Ok so on the Mathematical coincidence page, the last one on the page is


 * $$1782^{12} + 1841^{12} \approx 1922^{12}$$

My question is, given that there are infinite numbers that can be taken 3 at a time, will there eventually be a combination where the difference between the sides approaches zero and will not be able to be found unequal? I mean like with really big numbers. —Preceding unsigned comment added by 24.251.110.206 (talk) 03:58, 19 January 2009 (UTC)


 * One simple test to determine if they're different is by considering the powers of the final digits only. If the sum of the 2 powers on the left is not the same as the power on the right, there's no way they're equal.  It doesn't matter if the exponent is in the billions, because there are only a small number of values (max 4, which applies only with 2, 3, 7 and 8) that the last digit of a power of a digit can ever have.  If they're the same, then you go to the next test, whatever that is.


 * To illustrate with the above example: the powers of 2 can only ever end in 2, 4, 6 or 8.  The powers of 1 are all 1, just 1.  Adding 1 to 2, 4, 6 or 8 will always give you an odd number.  So we know the left hand side of the equation is odd.  But we know the right hand side is even, because it again ends in a power of 2, which is always even.  Hence, they cannot possibly be equivalent.  --  JackofOz (talk) 04:30, 19 January 2009 (UTC)
 * "not be able to be found unequal"... Well you just need to build a better computer. Both the left and the right side are integers.  And either they are the same integer or they are different integers.  The smallest difference between the right side and left side, without being 0, is 1, so it can't "approach 0".  More importantly to answer your question please see Fermat's Last Theorem.  If all the exponents are equal than the answer is no unless the exponents are all 2. Anythingapplied (talk) 09:37, 19 January 2009 (UTC)

Let me rephrase the question, because I am curious. For given n > 2 and all &delta; > 0, is it always possible to find positive integers a, b, and c such that:

$$|{a^n+b^n \over c^n} - 1| < \delta$$

That seems to a reasonable expression of approaching each other in the limit, i.e. that the fractional difference can become arbitrarily small. My guess is that yes one can always find such integers, but it doesn't strike me as obvious. Dragons flight (talk) 10:15, 19 January 2009 (UTC)


 * Actually, I take that back, it is obvious. Simply let a = c and b = 1, and the left hand side is $$1 \over c^n$$ which can be made arbitrarily small through choice of c.  Dragons flight (talk) 10:19, 19 January 2009 (UTC)


 * An Internet search on "Fermat near-misses" finds other examples such as which is also linked in Fermat's Last Theorem. PrimeHunter (talk) 11:21, 19 January 2009 (UTC)

Level in a horizontal cylinder
I need to measure the level in a cylinder tank placed horizontal. There are a few different sizes [5.4m(h) x 2.3m diameter], [12.5m(h) x 2.855m diameter] etc. We want to use a pressure transmitter with 4-20mA output to monitor the level of Heavy Furnace Oil with SG of 0.985kg/l. What will be the formula to calculate the oil volume in liters for every cm of tank? Also because of the cylindrical shape the 4-20mA signal will not be linear and so this needs to be calibrated to the determined level at each cm 41.242.212.142 (talk) 07:05, 19 January 2009 (UTC)


 * See Circular segment which gives the area of a segment of a circle cut of by a line. WHen you have found this area you can multiply by the length of the cylinder to give the volume. --Salix (talk): 07:50, 19 January 2009 (UTC)


 * Also, I want to make sure it really is a cylindrical tank. That is, it must have a flat circle at each end.  Many tanks have hemispherical-shaped ends, instead.  In that case, you'd need to use the above method for the volume of the cylinder up to where the hemispheres begin, and use the formula for the volume of a partially filled sphere to figure the volume in the ends.  Finally, for all measurements, be sure you use interior dimensions, not exterior. StuRat (talk) 15:37, 19 January 2009 (UTC)


 * These gauges do a remarkably accurate job of indicating the quantity of fuel in the tank. I don't know how they deal with the nonlinearity but they do. hydnjo talk 22:08, 19 January 2009 (UTC)


 * Actually, the description in that article makes it clear that they don't. They only measure the height of the fuel level, and the article itself points out that it's no good for estimating the volume. — Emil J. 14:57, 22 January 2009 (UTC)


 * Another way, with no math required, is to fill the tank with known volumes, then mark those volumes on the gauge where the needle is at that time. You would end up with a non-linear scale, but it would still get the job done nicely.  This would also handle imperfections on the geometry of the tank, like dents or internal supports.  For an electronic scale, you would need to record conversions from CM on the scale to L of volume.  You could either use a table look-up program to do the conversion, or just do the conversions manually. StuRat (talk) 17:03, 24 January 2009 (UTC)


 * Another factor that sometimes needs to be considered is compressibility of the fluid. However, in this case, I don't expect that to be a significant factor. StuRat (talk) 17:03, 24 January 2009 (UTC)

Meaning of 'deviated from their sample means'
I came across the following phrase in a statitics related paper: "All [independent] variables are deviated from their sample means". What does this mean? Tom pw (talk) (review) 18:45, 19 January 2009 (UTC)


 * Can you give us some context? Usually "deviation from the sample mean" merely means what you get when you subtract the sample mean from the observation.  But the way the phrase is used here seems to be about something different. Michael Hardy (talk) 23:26, 19 January 2009 (UTC)


 * I would interpret the OP's question as meaning "the independent variables were centered" which means the sample mean of each such variable was subtracted, per MH. But that language seems awkward and unconventional, so it is not clear.  Perhaps it was a non-native english speaking author?  Baccyak4H (Yak!) 03:36, 20 January 2009 (UTC)

A question about consistent mass matrices in Finite Element Analysis
I am a bit confused about a very elementary thing in finite element analysis. For dynamic structures, we have


 * $$M \ddot{x} + K x = f.$$

where M and K are mass and stiffness matrices respectively. Now, assume we are dealing with a single bar element, and the stiffness is zero, and we are using the consistent mass matrix. So, we have


 * $$M \ddot{x} = f.$$

where


 * $$\mathbf M = \begin{bmatrix}

2 & 1\\ 1 & 2\end{bmatrix}$$

and let's have f = [0;1], which means we are pulling on one of the nodes.

This will give the acceleration as [-1/3;2/3]. Shouldn't we expect the two nodes of the bar to have accelerations in the same direction? Where does this negative acceleration for the first node come from? deeptrivia (talk) 19:59, 19 January 2009 (UTC)