Adam Ledger

Mr. Adam Ledger

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8 years, 149 days
Perth, Australia

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These are answers submitted by Adam Ledger

This identity will enable you to see it has a corresponding sawtooth approximation as it should on the real number line, and i would have to revise my most recent opinions on this one but there is some strangeness when you extended to the the complex plane and so also a few cases that the approximations made by the graphical plotting engine will produce what i have been reading is being termed "artifacts" in some literature about computer algebra systems, anyway from memory there will be trouble in plots either side of trancendental limits, but for the scales as @acer showed in his answer this is sufficent. Also phrasing . Also expect corrections to what i have said




`mod`(x, y) = x-floor(x/y)*x

X := (rand(1 .. 20))();





`mod`(x, y)

modp(x, y)


eval(modp(x, y), [x = X, y = Y])






eval(1, [x = X, y = Y])



`mod`(ceil(x), 2)













In most cases no closed form will exist but the Solve feature in maple will almost always return output that can be used to make numerical approximations to the roots of these equations, but i've also found it's a good idea to always check these approximations but substituting them back into the original equation as well.

It wont let me download it again. Just get maple 16 why not? I err suppose i were to ask you to sail the seven seas me heartys is that.. is that too on the nose?

Ok for quite some time i did indeed write alot of analysis code for integer sequences, and the most important thing for me was to not get caught up in the question "is this a random pattern or not?"  Because it turns out that even if it is indeed completely random, there will still be underlying principles of number theory that will allow us to have at least some kind of expectation.

But in a practically applicable sense to what the human brain is capable of doing,  a high level of complexity in terms of the number theoretical meaning of the arithmetic functions that produce a given sequence are as good as random to the person unaware of their exact nature. 

Always start with the basics is my principle. 


Also please try to present you pattern in a way that is their most simplied form. In your example it is only the exponents of the numerator and denominator that exhibit variance, so the better what to introduce the question is to explain that, and ask if we can predict the pair of sequences that are not constant:

1415, 1648, 3474, 5984, 6197, 8739

893, 1040, 2192, 3776, 3910, 5514


This makes it far easier for your computer and fair easier for you, the numbers in the original sequence are just unnecessarily large.

there probably is a means of doing so, most likely it will be found amoungst the kernelopts help page. 


But as a general rule of thumb, in my experience in using maple which has been quite a number of years now, if your calculation is taking forever it is because your code is inefficient and sluggish, your computer if executing clean code can pretty much handle the most ridiculously complex of computations in terms of its processing power.


If this turns out to be the case, which it almost always does, the only solution is to make adjustments to the manner in which your calculation is structured, what functions you use, be certain that you are using all appropriate library packages relevant to your calculation, and have a vague motivation to keep the number of symbols you use to a bare minimum.

First of all instead of using numbers to represent the balls you should be assigning them as such Ball[1],Ball[2],Ball[3]..etc, and the same manner of variable declaration should also be made for the boxes ie (Box[1],Box[2],Box[3],etc)

Secondly, because you are working with distinct objects grouped in particular ways, you should be using axioms related to sets. if you use the curly brackets ("{ }") instead of the "[ ]" the maple interface with immeadiately recognise that these axioms apply to your input.




@Thomas Richard then if you still have trouble we will look at the finding your unknowns implied by


X := [200, 210, 220, 230, 240, 250, 260, 270, 280, 290]:

Y := [.4172, .3030, .4668, .3317, .1276, .1303, .1733, .1451, .3466, .4125]:

[seq([X[j], Y[j]], j = 1 .. Count(X))]

[[200, .4172], [210, .3030], [220, .4668], [230, .3317], [240, .1276], [250, .1303], [260, .1733], [270, .1451], [280, .3466], [290, .4125]]




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