Adam Ledger

Mr. Adam Ledger

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8 years, 329 days
unemployed
hobo
Perth, Australia

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These are replies submitted by Adam Ledger

@tomleslie  no it's fine this is just me when  i am tired and expecting myself to be as concise as i was 24 hours ago

@Carl Love  Ok, so IF this is indeed the case, am i allowed to assume that the gamma expression is convergent to zero as the product expression is in the limit for n-> infinity? If so can you please show the working out for the gamma limit evaluation. Maple 16 refuses to even evaluate it to a float approximation with evalf. In general it would be helpful to be able to see how all the limit evaluations are dont step by step by maple, and i also want to know why it is unable to evaluate this limit.  

@Carl Love 

 

i agree that it is indeed the levels of recursion that my particular brand  crazy needs for the nested data structures required which does often accopany the error.

 

I have tried my best to think of a solution for this but still achieving the results as informative as they are when the output is produced successfully. and this is often a hit and miss scenario for the exact same execution i have found, particularly if the code involves the generation of random number parameters. If i have been extremely ridiculous, we have less that black jack odds. But i have improved to some extent at least this must be true, it is a fatal error for every 5 execution for this worksheet anyway, and i am slowly but surely gathering the desired data, but it isnt ideal of course.

 

But i was thinking, in the same manner a try catch statement picks up an error, is it possible to implement some kind of "flag" for each level of recursion for a particular command, with value that has proportion to how close i am getting to the danger zone, or a zone that is one of danger to paraphrase?

@Carl Love Cheers much appreciated

@tomleslie  well ok fine i guess it isnt so important for it to be a command line from inside maple and at least this way i don't need to fabricate a story to dad about a secret bourgeise cabal that are systematically attempting to ruin my fun in math by hacking every  OS  he passes on to me

@Carl Love  I thought as much because i usually have at least half a clue or even at least an eigth of a clue what is happening but im drawing a blank for this and well i have found quite a number of those in the past 18 months so ill have to pencil it in for beyond 2030 sometime assuming someone hasn't strangled me by then

@Carl Love ok but doesnt any program with the ability to save a file of some kind in that folder ie, any kind of editor, word, notepad, maple not have write permissions? Whatever is done when i make a folder via the "save as" clicky must bypass  the exception that causes the error im getting in mkdir that tells me no write access, it's going to be some kind of command like hocus pocus I'm sure but anyway, it has to exist right?

@Carl Love  Sure i certainly don't intend on ignoring complex analysis, but it's more that i presently have someone providing me with tutition in RA, and the fact that I have identified my ability to produce rigorous proofs as a key weakness that has been holding me back, in conjunction with my ability to explain the math i am presenting in natural language in a concise and "not rambling" manner. I suppose the number one thing that i have always thrown under rug is a deep understanding of the natural logarithm being multivalued, this is about the only part of CA i am going to be thinking about for the time being, as well as an on going study of the nth roots of unity.

@Carl Love  the ModExp part i understand in how it works, I think it's more that I am yet to study the mathematics of the BBP and am not yet at a level of competence to immeadiately see the why it works.

@Mariusz Iwaniuk Ok thankyou so much for your help that at least allows me to know it's either a branch of mathematical understanding that maple staff have reached a consensus that must be considered in their development of inbuilt features such as float approximation for particular families of complex functions, I wonder, is there a command that allows me to see a full list of the assumptions and axioms for which computation is by default based upon in the maple interface, there by allowing me to make changes in the event i want to study mathematics that are not complementary to these assumptions, and or switch back and forth between the two, allowing me to further understand the basis of the theorems and axioms the maple staff understand to be important enough to be the default, but i am yet able to grasp?

@Carl Love 

I am embarassed to admit  the fact that i do not yet understand the reasoning for assigning divergent series finite values calculated by their analytic extension onto C ( And a few have said i need to learn how to derive the functional equation for the zeta to understand this, but yet they were unable to tell me similair  things here, the only work done I have ever seen  in connecting a zeta for the floor/fractional part was done by Nyman i think the name was,) and even still, I need to finish real analysis and learn how to write a half decent proof something significant before i should even be wasting my time on complex analysis, but oh well I still get curious.

@Mariusz Iwaniuk  That's fine i am the worst with double checking things! But i guess it would be time for me to upgrade to maple 2018 i suppose, otherwise i will find myself posting about bugs that have been fixed 2 years ago again

@Carl Love  thankyou so much for sharing this sir, Quite far above what i am able to understand at this stage, but I am still keenly interested none the less!

@Mariusz Iwaniuk 

 

sorry can you show me how (5^(n-1))*floor((1/4)*5^n)=n/(sin(n)+1)^2

@Carl Love 

ok true it is as simplified as it gets. But you can expand it as follows, and the cosine terms as summands would allow the infinite series to be evaluated no?


 

floor(j*(1/j)) = (i-(sum((j-m)*(sum(cos(2*Pi*(m+i)*n/j)/j, n = 0 .. j-1)), m = 1 .. j-1)))/j

f := proc (i, j) options operator, arrow; (i-(sum((j-m)*(sum(cos(2*Pi*(m+i)*n/j)/j, n = 0 .. j-1)), m = 1 .. j-1)))/j end proc:

f(5^k, 4)

(1/4)*5^k-3/8-(3/16)*cos((1/2)*Pi*(1+5^k))-(3/16)*cos(Pi*(1+5^k))-(3/16)*cos((3/2)*Pi*(1+5^k))-(1/8)*cos((1/2)*Pi*(2+5^k))-(1/8)*cos(Pi*(2+5^k))-(1/8)*cos((3/2)*Pi*(2+5^k))-(1/16)*cos((1/2)*Pi*(3+5^k))-(1/16)*cos(Pi*(3+5^k))-(1/16)*cos((3/2)*Pi*(3+5^k))

(1)

``


 

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