Adam Ledger

Mr. Adam Ledger

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9 years, 149 days
unemployed
hobo
Perth, Australia

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These are replies submitted by Adam Ledger

@Carl Love thanks yes its very core to the learning issues i experience i began learning this under the strict assumption of mod being applicable to only integers under the definition of knuth's floored division and so now hearing you mention trancendentals along with every other time ive had people refer to mod for non integers has me completely bush whacked.

@Axel Vogt see attached]


 

is(`mod`(`mod`(n^k, n^m)+n^m, n^k) = modp(modp(n^k, n^m)+n^m, n^k))

is(`mod`(n^k, n^m) = modp(n^k, n^m))

is(`mod`(n^m, n^k) = modp(n^m, n^k))

is(modp(n^k, n^m)+modp(n^m, n^k) = `mod`(`mod`(n^m, n^k)+n^k, n^m))

``


 

Download axelMODconfuse2.mw

@Axel Vogt  Is this in agreeance? 
 

FD := proc (a, b) options operator, arrow; a-b*floor(a/b) end proc

NULL

MOD := proc (a, b) options operator, arrow; `mod`(a, b) end proc

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

omega(n, m) = `mod`(n, m) and `mod`(n, m) = n-m*floor(n/m)


 

Download modAXEL.mw

@Axel Vogt  but so is the case for the evaluation of (2n+1)mod(3)=2n+1 no?

@acer ok thanks ive got that it's labelling some solutions and that _L is just short for label to assign a number for an index, it just seems as if solve does this everytime i consider the results to be significant, and this gives me a slightly disconcerting feeling for some reason

@ecterrab  as a means of communicating the origins of how my interest tonight came to be centered upon this particular function built into maple (the hypergeom function) ,please see the attached worksheet.

as a side note it is groups of indexed function sets exhibiting idempotence*, meaning as the number as labelled below "alpha" tends to an arbitarily large number the cardinality of the function set has a finite upper bound which is as can be seen from the graphical figures, reached by alpha>3  to an accurate extent as the hypergeometric functions added are increasingly similar to the last ones added as alpha is increased.

Or to put it a more simple way, i am asking the user to make note of the fact that consecutive graphical figures become identical beyond alpha=3.
 

 

R := sum((-1)^n*factorial(3*n)/(factorial(n)*factorial(3*n))^3, n = 0 .. infinity)-(sum(z^n*(sum((-1)^n*z^n/factorial(n)^3, n = 0 .. infinity))/factorial(n)^3, n = 0 .. infinity))

RAMANUJUAN1

hypergeom([], [1/3, 1/3, 2/3, 2/3, 1, 1, 1, 1], -1/729)-hypergeom([], [1, 1], -z)*hypergeom([], [1, 1], z)

(1)

alpha = 3*EXAMPLE

 

sum(z^n*(sum((-1)^n*z^n/factorial(n)^3, n = 0 .. infinity))/factorial(n)^3, n = 0 .. infinity) = hypergeom([1/3, 2/3], [1/2, 1/2, 1/2, 1, 1, 1, 1], -(27/64)*z^2)

hypergeom([], [1, 1], -z)*hypergeom([], [1, 1], z) = hypergeom([1/3, 2/3], [1/2, 1/2, 1/2, 1, 1, 1, 1], -(27/64)*z^2)

(1.1)

sum(z^n/factorial(n)^3, n = 0 .. infinity) = hypergeom([], [1, 1], z)

hypergeom([], [1, 1], z) = hypergeom([], [1, 1], z)

(1.2)

sum((-1)^n*z^n/factorial(n)^3, n = 0 .. infinity) = hypergeom([], [1, 1], -z)

hypergeom([], [1, 1], -z) = hypergeom([], [1, 1], -z)

(1.3)

NULL

NULL

α -GENERALISED CASE GRAPHICAL REPRESENTATION

 

H[1] := proc (alpha) options operator, arrow; sum(z^n*(sum((-1)^n*z^n/factorial(n)^alpha, n = 0 .. infinity))/factorial(n)^alpha, n = 0 .. infinity) end proc:

Y := {seq([H[1](k), H[2](k), H[3](k)], k = 1 .. 10)}

{[1, exp(z), 1/exp(z)], [BesselI(0, 2*(-z)^(1/2))*BesselI(0, 2*z^(1/2)), BesselI(0, 2*z^(1/2)), BesselI(0, 2*(-z)^(1/2))], [hypergeom([], [1, 1], -z)*hypergeom([], [1, 1], z), hypergeom([], [1, 1], z), hypergeom([], [1, 1], -z)], [hypergeom([], [1, 1, 1], -z)*hypergeom([], [1, 1, 1], z), hypergeom([], [1, 1, 1], z), hypergeom([], [1, 1, 1], -z)], [hypergeom([], [1, 1, 1, 1], -z)*hypergeom([], [1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1], -z)], [hypergeom([], [1, 1, 1, 1, 1], -z)*hypergeom([], [1, 1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1, 1], -z)], [hypergeom([], [1, 1, 1, 1, 1, 1], -z)*hypergeom([], [1, 1, 1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1, 1, 1], -z)], [hypergeom([], [1, 1, 1, 1, 1, 1, 1], -z)*hypergeom([], [1, 1, 1, 1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1, 1, 1, 1], -z)], [hypergeom([], [1, 1, 1, 1, 1, 1, 1, 1], -z)*hypergeom([], [1, 1, 1, 1, 1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1, 1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1, 1, 1, 1, 1], -z)], [hypergeom([], [1, 1, 1, 1, 1, 1, 1, 1, 1], -z)*hypergeom([], [1, 1, 1, 1, 1, 1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1, 1, 1, 1, 1, 1], z), hypergeom([], [1, 1, 1, 1, 1, 1, 1, 1, 1], -z)]}

(2.1)

NULL

plot(Y[1], title = typeset("A plot of Y[%4]= {%1 ,%2,%3}.", Y[1][1], Y[1][2], Y[1][3], 1), titlefont = ["ROMAN", 15]);

 

 

 

 

 

 

 

 

NULL

NULL

NULL

``


 

Download HYPERGEOM_RAMANUJUAN_SERIES_001_MAPLE_PRIMES.mw

@Axel Vogt it was the convention used for the empty list values of n,d to be honest so my flaw in understanding is really concerned with the iterated product form in a general sense rather than specifically the hypergeom. A bit imbarassing when the flaw has such generality but what can i do.

 

 

 

 

 

 

 

its not a very popular subject in some ways yes.

@Axel Vogt  thankyou ok well for the student part, all i really need is a text/publication referal that is best fitting and  well as far as advisors are concerned asking for help on this site is the limit of my budget so... sure if you know of any good publications ill start there.

@ecterrab  ok thanks much appreciated after further consideration and having noted the convention implied by the Canonical representation of a positive integer under the fundamental theorem of arithmetic also being that the empty product is equal to 1, so in as much as the choice may or may not be entirely unrelated to choice of doing the same for the pochhammer representation of the hypergeometric function  at least now  i have another scenario where the convention is used which will come in handy when the "ït's not real mathematics because mathmatical functions dont accept lists as arguments" debate initiation is used on me again by someone

@Markiyan Hirnyk ok thanks

@Markiyan Hirnyk thanks for that i still struggle to understand what this means in the context of the pochhammer expression for hypergeom. i have also spent all night trying to work out how maple is evaluating so many infinite series expressions to hypergeom(n,d,z) expressions, no luck there either.

@Carl Love  much appreciated thankyou

@vv yep thats great so far ive just been converting all my factorials into gamma expressions but i dont know how much good thats doing for me.

@vv  sorry can you just explain to me the general jist of how you knew to select that range of 33..127 ive tried the code and its perfect i just dont understand that part

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