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February 21[edit]

Maximum number on calculator[edit]

Hi. On most graphing calculators there is a limit to the amount of numbers it can display before resorting to scientific notation. For example, on my TI-84 Plus, it can calculate 2³³ just fine, but then at 2³⁴, it displays the answer in scientific notation. My question is, is there a method to determine the maximum number the calculator can calculate before resorting to scientific notation? I presume it has to be a number under E10, but I'd like to know if there is a definite mathematical way to do this. Thanks. Vic93 (t/c) 00:00, 21 February 2009 (UTC)[reply]

Try reading the manual. Algebraist 00:01, 21 February 2009 (UTC)[reply]

It is entirely up to the people who designed the calculator. Usually things like that are constrained by how the number is stored in memory, but there is more than one way to keep track of that kind of number. Black Carrot (talk) 03:28, 21 February 2009 (UTC)[reply]

For an idea of how it might work, try skimming our article on floating-point numbers. Black Carrot (talk) 03:34, 21 February 2009 (UTC)[reply]

To experiment, I usually try entering something like (ten nines) 9999999999 + 1 and seeing what happens. The detailed answer is in the manual (bottom of PDF page 24, printed page 21).

Be aware there's a difference between how the calculator stores numbers and how it displays them. (See the manual PDF page 659, printed page 656.) Internally, the calculator is always calculating in scientific notation, using up to 14 digits in the significand and up to 2 digits in the exponent. --Bavi H (talk) 08:30, 22 February 2009 (UTC)[reply]

The maximum number displayed is 4. Above that it says a suffusion of yellow. --Trovatore (talk) 08:38, 22 February 2009 (UTC)[reply]

Intelligent design and the value of pi[edit]

If it's true that God can't make two plus two equal five, and that everything that statement implies is true, could God still have made the value of pi less than 3.14159 or more than 3.1416, as easily as He might have made the gravitational constant less than 6.674 m³kg^-1s² or more than 6.675 m³kg^-1s² (ignoring the anthropic principle, i.e. assuming it weren't necessary for the modified universe to support sentient life)? Neon Merlin 08:58, 21 February 2009 (UTC)[reply]

This depends somewhat on what is meant by π. If π denotes the mathematical constant, then to change its value is just as hard as changing the value of 2+2, and so impossible according to mainstream christian theology. If, on the other hand, π is taken to mean something physical, say the ratio of circumference to diameter of physically existing circles, then an omnipotent power could change its value, or make it vary from place to place, and indeed (since space is not perfectly flat) this is believed to be in fact the case. Algebraist 09:54, 21 February 2009 (UTC)[reply]

Algebraist, is π a constant beyond its definition of the ration of the circumference to diameter of a circle? If it is not defined by that, then by what is it defined? If, say, an omnipotent power changed this ratio, would not then the value of π change with it? --Aseld ^talk 10:04, 21 February 2009 (UTC)[reply]

What Algebraist means is that for circles in Euclidean geometry, defined as the set of points in a certain plane having a certain distance (Euclidean norm) to a certain point in that plane, it follows from the axioms of Euclidean geometry that 3.14159 < π < 3.1416 (by Archimedes' approach but with more than 96 sides), just as it follows from the Peano axioms that 2 + 2 = 5. As these axioms don't have anything to do with the universe per se, an omnipotent God couldn't change them any more than he could change the rules of tic-tac-toe. But reality, the universe, which clearly is not exactly modelled on Euclidean geometry, is quite another matter. —JAO • T • C 10:22, 21 February 2009 (UTC)[reply]

I think you're using the wrong Peano axioms there. Algebraist 10:25, 21 February 2009 (UTC)[reply]

Now, how on Earth should I know what 2 + 2 is? —JAO • T • C 10:31, 21 February 2009 (UTC)[reply]

Not wrong, just different! --Tango (talk) 14:00, 21 February 2009 (UTC)[reply]

There's no geometry where circumference/diameter is some constant other than π. The circumference of a circle is 2π sin r in elliptical geometry and 2π sinh r in hyperbolic geometry, where π is the same constant that shows up in Euclidean geometry. There's no constant π' that would enable you to write those formulas as 2π'r. A factor of π shows up in the field equation of general relativity. π is genuinely universal, it's not at all limited to flat geometry. (Are there weirder geometries that could genuinely be said to have a "different π"? I don't know.) -- BenRG (talk) 15:08, 21 February 2009 (UTC)[reply]

The natural geometry on a cone would be interesting from that perspective. Away from the point at the top, it's flat, so circles that don't include that point and are small enough not to wrap around the cone would have the regular value of pi. Circles centred on that point would have a different ratio of circumference to diameter, but it would be constant. Circles centred elsewhere that either include the top point or wrap around the cone would do something weird... I would have to think about it. --Tango (talk) 15:14, 21 February 2009 (UTC)[reply]

You may be interested in a discussion that took place on the Science desk a few weeks ago: http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Science/2009_January_25#Cosmic_gods --Tango (talk) 14:00, 21 February 2009 (UTC)[reply]

The article Philosophy of mathematics might also be of interest, in particular the section on mathematical realism. 207.241.239.70 (talk) 06:07, 27 February 2009 (UTC)[reply]

Which of the following does the OP mean by "God can't make two plus two equal five"?

There exists a mighty being who needs to be identified as "God". Any combination of inabilities and abilities can be proposed for such a being and none is verifiable, so

the only answer to the question is "perhaps". OR

There is no God, hence no godly abilities exist. The mathematical definitions of addition and pi are understood and agreed upon by all reputable sources. Consensus wins and the answer is No. Cuddlyable3 (talk) 14:17, 27 February 2009 (UTC)[reply]

Quick vector calculus question[edit]

Hi there - I'm a little confused about a vector calculus question, or rather the hint they're giving me for the question, was wondering if I could grab a little help!

If $\nabla \cdot {\textbf {J}}=0$ in volume V and ${\textbf {J}}\cdot {\textbf {n}}=0$ on the surface S enclosing V, show that

$\int _{V}{\textbf {J}}dV=0$ . Hint: use ${\frac {\partial {(x_{i}J_{j})}}{\partial {x_{j}}}}$ .

I'm really not sure where the hint is pointing me! Differentiating gives $(\nabla \cdot J)x+(J\cdot \nabla )x=0+(J\cdot \nabla )x=(J\cdot \nabla )x$ but what then?

Thanks a lot for any help, Spamalert101 (talk) 11:12, 21 February 2009 (UTC)spamalert[reply]

I reckon you were pretty close but I think we can take your identity a little further,

{\frac {\partial {(x_{i}J_{j})}}{\partial {x_{j}}}}=(J\cdot \nabla )x_{i}=\sum _{j}J_{j}{\frac {\partial {x_{i}}}{\partial {x_{j}}}}=J_{i}

This lets us express their integral in terms of the derivative they suggested.

\int _{V}J_{i}dV=\int _{V}\nabla \cdot (x_{i}{\textbf {J}})dV

Do you see how to finish the proof now? Vespertine1215 (talk) 16:43, 21 February 2009 (UTC)[reply]

Permutation and Combination[edit]

I just need to know what this question wants me to do

Find total no. of distinct ways the product " XY³Z³ " can be written without using exponent.

I haven't understood what the sum means so i'd be glad if anyone can explain what this question means. Vineeth h (talk) 13:27, 21 February 2009 (UTC)[reply]

Does it mean how many distinct ways are there to order an X, three Y's and three Z's? (eg. XYYYZZZ, XYYZZZY, XYZZZYY, etc.) --Tango (talk) 14:04, 21 February 2009 (UTC)[reply]

Well Tango according to that the answer should come 140 but it isn't 140!!! So i guess thats not it!

And what is the given answer then? 18? --pma (talk) 16:37, 21 February 2009 (UTC)[reply]

Well, you have 2XYYYZZZ / 2, 3XYYYZZZ / 3 etc. so you could argue there are a countably infinite number of ways of writing this product with using an exponent. Pretty meaningless question without clearer constraints on which expressions are allowed. Gandalf61 (talk) 18:11, 21 February 2009 (UTC)[reply]

Even if you take "product" as meaning you can only multiply, not divide, you still have things like (-X)(-Y)YYZZZ. That keeps it finite, but still quite a bit bigger - 140*(7*6+7*6*5*4+7*6*5*4*3*2)=829080. --Tango (talk) 18:36, 21 February 2009 (UTC)[reply]

Level Curve Interpolation[edit]

If I've got a surface defined as a function of R2, and several level curves of that surface, is there a computationally simple way to approximately recreate the original surface from its level sets? I'm trying to find a way to turn a cell-shading of a drawing into a smooth gradient. Black Carrot (talk) 16:55, 21 February 2009 (UTC)[reply]

Presumably you mean a function "on", rather than "of", R2, and you mean R². If ƒ is any real-valued function on R² and g is any strictly increasing function from R to R, then ƒ and the composition g o ƒ will both have exactly the same level sets. Thus many different functions can share the same level sets. Michael Hardy (talk) 20:51, 21 February 2009 (UTC)[reply]

I think this a classic computer graphics problem and there is extensive literature on such matters. I'm guessing that do don't actually have the level sets in their entirity, just some points lying on them. You will probably want some sort of surface fitting algoritm, say fitting a Bézier surface to the data. Inverse distance weighting seems to be one technique. Kriging is one well known method used in geostatistics with a strong statistical basis. --Salix (talk): 23:50, 21 February 2009 (UTC)[reply]

That's a good start. I just tried out inverse distance weighting. It hasn't quite worked yet, but I think I can fix it. The primary difficulty is that it's designed to interpolate a small, sparse set, whereas I'd like to interpolate hundreds of pixels along a sequence of curves. Some of these curves are different lengths (imagine two concentric circles) so the number of pixels along them are different, and that throws it off a bit. BTW, is it just me or does that formula look a lot like gravity when p=2? Black Carrot (talk) 04:23, 22 February 2009 (UTC)[reply]

Try googling for "surface reconstruction contour" there are many methods for this, I don't really know enough on the subject to advise on the best. Although there does seem some interesting connections with Voronoi diagrams. --Salix (talk): 08:19, 22 February 2009 (UTC)[reply]