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Adam Ledger

Mr. Adam Ledger

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9 years, 127 days
unemployed
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Perth, Australia

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What are the roots of unity at an index ...

Adam Ledger 360
July 25 2017
0 0

What are the roots of unity at an index of 10 and this should provide a basis on C

also studied thought to be relevant bina...

Adam Ledger 360
July 25 2017
0 0

 

HERE IS AN IMAGE

@acer also unsure of this part...

Adam Ledger 360
July 25 2017
0 0

ACER6_paramterization_i_less_than_j_Known_true.mw
Also again bijectivity of (i,j) restriants  is conserved over this implict path of computation

> 

``
> 

`implies`(`implies`((-1)^(2*i*j/(-j+2*i))+(-1)^((j+2*i*j+2*i)/(-j+2*i)) = 0, 2*i*j/(-j+2*i)+1 = (j+2*i*j+2*i)/(-j+2*i)), {i = i, j = 0})

1

 (1)
> 

"for i<j:"
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(-1)^(2*j*(j-k)/(j+2*(j-k)))+(-1)^((j+2*j*(j-k)+2*(j-k))/(-j+2*(j-k))) = 0

(-1)^(2*j*(j-k)/(3*j-2*k))+(-1)^((3*j+2*j*(j-k)-2*k)/(j-2*k)) = 0

 (2)
> 

solve({(-1)^(2*j*(j-k)/(3*j-2*k))+(-1)^((3*j+2*j*(j-k)-2*k)/(j-2*k)) = 0}, [j])

[[j = -3/4+(1/2)*k+(1/4)*(9+4*k+4*k^2)^(1/2)], [j = -3/4+(1/2)*k-(1/4)*(9+4*k+4*k^2)^(1/2)], [j = 0]]

 (3)
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solve({(-1)^(2*j*(j-k)/(3*j-2*k))+(-1)^((3*j+2*j*(j-k)-2*k)/(j-2*k)) = 0}, [k])

[[k = (1/2)*j*(3+2*j)/(j+1)]]

 (4)
> 

``


 

Download ACER6_paramterization_i_less_than_j_Known_true.mw

 

@acer also unsure of this part...

Adam Ledger 360
July 25 2017
0 0


 

> 

``
> 

`implies`(`implies`((-1)^(2*j*i/(-j+2*i))+(-1)^((2*i*j+2*i+j)/(-j+2*i)) = 0, 2*j*i/(-j+2*i)+1 = (2*i*j+2*i+j)/(-j+2*i)), {i = i, j = 0})
> 

``


 

Download ACER_5_UNCERTAIN_IMPLICIT.mw

@acer pls also see attached...

Adam Ledger 360
July 25 2017
0 0


 

> 

"Determine R."
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``
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``
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"(-1)^((2 j i)/(-j+2 i))+(-1)^((j+2 j i+2 i)/(-j+2 i))=0 implies {(i,j) in `&Zopf;`^(2):R}"


 

Download ACER4.mw

Another important note to make for the a...

Adam Ledger 360
July 25 2017
0 0

The implied expression of k in terms of i and j is then injected back into the original equation to assure bijectivity of (i,j). Thus in determining the restrictions that apply you can expect these statements must themselves too be bijective, in that you can express the restrictions of i strictly in terms of j and vice versa.

Re i,j conditions...

Adam Ledger 360
July 25 2017
0 0

@acer  i have simply encountered so many erraneous results tryng to do this part but yes the next step is to define their domains and ranges. but as such i have not been able to decicively say these conditions. but there are poles. 

in a nutshell...

Adam Ledger 360
July 25 2017
0 0

its basically imposing the conditions similar to the derivation ive seen many use for the -1/12 result actually in the treatment of the Grandis series algebra used there.

sure see attached. Will send further rel...

Adam Ledger 360
July 25 2017
0 0

UPDATED* (BIG ERROR)  
 

> 

restart
> 

solve(-(((-1)^(-j))^(1/(-j+2*i)))^(-2*i)*(-((-1)^(-j))^(2/(-j+2*i))+((-1)^(-j))^(1/(-j+2*i)))/(((-1)^(-j))^(2/(-j+2*i))-1) = (-1)^j*(((-1)^(-j))^(1/(-j+2*i)))^(-j)*((-1)^(-j))^(1/(-j+2*i))/(1+((-1)^(-j))^(1/(-j+2*i))), i)

Error, (in RootOf) expression independent of _S000006

  
> 

``
> 

``

`implies`(sum(1/k^(2*n), n = i .. infinity)-(sum(1/k^(2*n+1), n = i .. infinity)) = sum((-1)^n/k^n, n = j .. infinity), k[i, j] = ((-1)^j)^(-1/(-j+2*i)))

 

 

k[i, j] = ((-1)^j)^(-1/(-j+2*i))
" implies (&sum;)1/((((-1)^(-j))^(1/(-j+2 i)))^(2 n))-(&sum;)1/((((-1)^(-j))^(1/(-j+2 i)))^(2 n+1))=(&sum;)((-1)^(n))/((((-1)^(-j))^(1/(-j+2 i)))^(n))"

 

 

 

`implies`(sum(1/(((-1)^(-j))^(1/(-j+2*i)))^(2*n), n = i .. infinity)-(sum(1/(((-1)^(-j))^(1/(-j+2*i)))^(2*n+1), n = i .. infinity)) = sum((-1)^n/(((-1)^(-j))^(1/(-j+2*i)))^n, n = j .. infinity), ((-1)^(2*j*i/(-j+2*i))-(-1)^(j*(1+2*i)/(-j+2*i)))/(-1+(-1)^(2*j/(-j+2*i)))-(-1)^(j*(-1+2*i)/(-j+2*i))/(1+(-1)^(-j/(-j+2*i))) = 0)

 

`&therefore;`(((-1)^(2*j*i/(-j+2*i))-(-1)^(j*(1+2*i)/(-j+2*i)))/(-1+(-1)^(2*j/(-j+2*i))))-(-1)^(j*(-1+2*i)/(-j+2*i))/(1+(-1)^(-j/(-j+2*i))) = (2*((-1)^(2*j*i/(-j+2*i))+(-1)^((2*i*j+2*i+j)/(-j+2*i))))/(((-1)^(j/(-j+2*i))+1)*(-1+(-1)^(2*j/(-j+2*i)))) and (2*((-1)^(2*j*i/(-j+2*i))+(-1)^((2*i*j+2*i+j)/(-j+2*i))))/(((-1)^(j/(-j+2*i))+1)*(-1+(-1)^(2*j/(-j+2*i)))) = 0

> 

``
> 

``


 

Download ACER3.mw

see attached (thanks in advance)...

Adam Ledger 360
July 25 2017
0 0


 

> 

solve(-(((-1)^(-j))^(1/(-j+2*i)))^(-2*i)*(-((-1)^(-j))^(2/(-j+2*i))+((-1)^(-j))^(1/(-j+2*i)))/(((-1)^(-j))^(2/(-j+2*i))-1) = (-1)^j*(((-1)^(-j))^(1/(-j+2*i)))^(-j)*((-1)^(-j))^(1/(-j+2*i))/(1+((-1)^(-j))^(1/(-j+2*i))), i)

Error, (in RootOf) expression independent of _S000006

  
> 

``


 

Download ACER2.mw

@acer 

or.......

Adam Ledger 360
July 25 2017
0 0

@tomleslie  no that would actually make sense so i can only assume there is something in interface i can write to allow me to view the globals defined by the library requested for

yes, meaningless is my point entirely...

Adam Ledger 360
July 25 2017
0 0

Thats why this is in front of you, im saying its just not very worth while error code if maple prefers to keep its context to itself while knowing all of mine lol

fine...

Adam Ledger 360
July 25 2017
0 0

@tomleslie well alright then i i guess my point, is that the error message i get has no context to anyone other than someone that knows what _S000100 is referring to so the CAS may as well tell me nothing.

context...

Adam Ledger 360
July 24 2017
0 0

@tomleslie 

just in case it was lost in transit on t...

Adam Ledger 360
July 24 2017
0 0

@tomleslie come on you know how DHL etc is i just wanted to remind engine that expression is independent of _S000100 just in case we have a logistics issue in the dispatch

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