Adam Ledger

Mr. Adam Ledger

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10 years, 36 days
unemployed
hobo
Perth, Australia

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These are replies submitted by Adam Ledger

@vv very much appreciated thankyou so much

@tomleslie  i didnt catch that engine can you redispatch that over

@Rouben Rostamian  why does the set operation not see "−" but does collect all the others as non alpha-numeric ( {"
", ", ", "+", "-", "/", "="}) ?


 

with(Statistics); with(StringTools)

S[0] := "-----BEGIN CERTIFICATE-----
MIIDdTCCAl2gAwIBAgILBAAAAAABFUtaw5QwDQYJKoZIhvcNAQEFBQAwVzELMAkG
A1UEBhMCQkUxGTAXBgNVBAoTEEdsb2JhbFNpZ24gbnYtc2ExEDAOBgNVBAsTB1Jv
b3QgQ0ExGzAZBgNVBAMTEkdsb2JhbFNpZ24gUm9vdCBDQTAeFw05ODA5MDExMjAw
MDBaFw0yODAxMjgxMjAwMDBaMFcxCzAJBgNVBAYTAkJFMRkwFwYDVQQKExBHbG9i
YWxTaWduIG52LXNhMRAwDgYDVQQLEwdSb290IENBMRswGQYDVQQDExJHbG9iYWxT
aWduIFJvb3QgQ0EwggEiMA0GCSqGSIb3DQEBAQUAA4IBDwAwggEKAoIBAQDaDuaZ
jc6j40+Kfvvxi4Mla+pIH/EqsLmVEQS98GPR4mdmzxzdzxtIK+6 NiY6arymAZavp
xy0Sy6scTHAHoT0KMM0VjU/43 dSMUBUc71DuxC73/OlS8pF94G3VNTCOXkNz8kHp
1 Wrjsok6Vjk4bwY8iGlbKk3Fp1S4bInMm/k8yuX9ifUSPJJ4ltbcdG6TRGHRjcdG
snUOhugZitVtbNV4FpWi6cgKOOvyJBNPc1STE4U6G7weNLWLBYy5d4ux2x8gkasJ
U26Qzns3dLlwR5EiUWMWea6xrkEmCMgZK9FGqkjWZCrXgzT/LCrBbBlDSgeF59N8
9 iFo7+ryUp9/k5DPAgMBAAGjQjBAMA4GA1UdDwEB/wQEAwIBBjAPBgNVHRMBAf8E
BTADAQH/MB0GA1UdDgQWBBRge2YaRQ2XyolQL30EzTSo//z9SzANBgkqhkiG9w0B
AQUFAAOCAQEA1nPnfE920I2/7 LqivjTFKDK1fPxsnCwrvQmeU79rXqoRSLblCKOz
yj1hTdNGCbM+w6DjY1Ub8rrvrTnhQ7k4o+YviiY776BQVvnGCv04zcQLcFGUl5gE
38 NflNUVyRRBnMRddWQVDf9VMOyGj/8 N7yy5Y0b2qvzfvGn9LhJIZJrglfCm7ymP
AbEVtQwdpf5pLGkkeB6zpxxxYu7KyJesF12KwvhHhm4qxFYxldBniYUr+WymXUad
DKqC5JlR3XC321Y9YeRq4VzW9v493kHMB65jUr9TU/Qr6cf9tveCX4XSQRjbgbME
HMUfpIBvFSDJ3gyICh3WZlXi/EjJKSZp4A==
 -----END CERTIFICATE-----"

THE "DELIMITER PRIMATIVE SET OF S[0]" DPS: IE THE SET OF DELIMITERS USED AT LEAST ONCE IN THE STRING S[0]

DPS := `minus`(`minus`(convert(StringTools['Explode'](S[0]), 'set'), {"A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"}), {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9"})

``


 

Download Rouban_R002.mw

@Rouben Rostamian  

@tomleslie what is the complete set of escape character symbols?

@Rouben Rostamian  ok these are print worthy cheers

@John Fredsted  well ive only encountered the works of landau and kronecker from that list so its always a gut wrencher when you know the obligated reading time is going to step quite a bit im probably the worst reader in the history of humans i think particularly when for things that use the word tensor and metric for some reason.

@Kitonum  yea thanks i pick that up and am going to be feeling pretty imbarrased about that one for eternity

To try again, i mean what the algorithm is to produce a random test point on a conformal plot. I know it has something to do with it being analytic on C but regardless, in general i want to know procedurally when each of the various plot functions are used.

I am completely noob in this area of math can you recommend a good text book for metrics and tensor physics in general?

Im still not entirely sure what im doing at this point but i can see how the one you provided encounters the same problems.
 

restart

 

 

APPROACH 1

 

sum(1/k^(2*n), n = m .. infinity)-(sum(1/k^(2*n+1), n = m .. infinity)) = sum((-1)^n/k^n, n = m .. infinity)" implies 1/((-1+k^2) k^(2 m-2))-1/((-1+k^2) k^(-1+2 m))=((-1)^m k)/(k^m (1+k))"

`implies`(1/((k^2-1)*k^(2*m-2))-1/((k^2-1)*k^(-1+2*m)) = (-1)^m*k/(k^m*(1+k)), (k^(-2*m+2)-k^(1-2*m))/(k^2-1) = -(-k)^(-m+1)/(1+k))

 

 

 

(k^(-2*m+2)-k^(1-2*m))/(k^2-1) = -(-k)^(-m+1)/(1+k)" implies ""k in {0,(e)^(-(I Pi (m+2 n m-2 n))/(m):(n,m) in `ℤ`^(2))}"

 

 

 

APPROACH 2

 

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

 

Collectively in one statement:

 

 

((-1)^j)^(-1/(-j+2*i))= exp(-I*Pi*(2*m*n+m-2*n)/m)

 

 

 

``

1/((-1+k^2)*k^(2*m-2))-1/((-1+k^2)*k^(-1+2*m)) = (-1)^m*k/(k^m*(1+k))

(1)

((-1)^j)^(-1/(-j+2*i)) = exp(-I*Pi*(m+2*n*m-2*n)/m)``

((-1)^j)^(-1/(-j+2*i)) = exp(-I*Pi*(m+2*n*m-2*n)/m)

(2)

solve({((-1)^j)^(-1/(-j+2*i)) = exp(-I*Pi*(m+2*n*m-2*n)/m)}, [j])

[[j = (m+2*n*m-2*n)*i/(m+n*m-n)]]

(3)

solve({((-1)^j)^(-1/(-j+2*i)) = exp(-I*Pi*(m+2*n*m-2*n)/m)}, [i])

[[i = -((1/2)*I)*(ln((-1)^j)*m+I*j*Pi*m+(2*I)*j*Pi*n*m-(2*I)*j*Pi*n)/(Pi*(m+2*n*m-2*n))]]

(4)

solve({((-1)^j)^(-1/(-j+2*i)) = exp(-I*Pi*(m+2*n*m-2*n)/m)}, [n])

[[n = -m*(-j+i)/((-j+2*i)*(m-1))]]

(5)

solve({((-1)^j)^(-1/(-j+2*i)) = exp(-I*Pi*(m+2*n*m-2*n)/m)}, [m])

[[m = n*(-j+2*i)/(-j+i-n*j+2*n*i)]]

(6)

 


 

Download ACER9_Unsure_if_This_helps_or_hinders.mw

N root unity N root unity N root unity

the alternative approach to initial expression previously mentioned hitherto

 

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