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LeeHoYeung

Mr. Ho Yeung Lee

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12 years, 234 days

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Seldom to ask question after retired math hobby Welcome August, February, July, May born girl And waited for her email to mavio@protonmail.com

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These are questions asked by LeeHoYeung

can this be a solution set?...

LeeHoYeung 535 Maple 12
solve
July 09 2017
1 1

v=u+at                      (1)
s=u*t+1/2*a*t^2        (2)

below 3 equations, can substitute  (1)  into it to form (2)
s=1/2*(u+v)*t       (3)
v^2=u^2+2*a*s    (4)
s=v*t-1/2*a*t^2    (5)

can these 5 equations be considered as a solution set of solve function?

or

is only first 2 equations be a solution set?

if so, number of equations less than 5 variables, is there something missing?
 

how to simplify this logic?...

LeeHoYeung 535 Maple 12
logic boolean
July 03 2017
1 2

https://drive.google.com/file/d/0Bxs_ao6uuBDUNmd5ZVJtX29GT3c/view?usp=sharing
https://drive.google.com/file/d/0Bxs_ao6uuBDUSUdHSTcwSEQtS3M/view?usp=sharing

would like to return K map of P1

Summation expression for logic only consider 1 but how about wildcard x ?

if consider wildcard x as 1 too, then will use below

source = [[0,0,1,0],[0,0,1,1],[0,1,0,1],[0,1,1,0],[0,1,1,1],[1,0,0,0],[1,0,0,1],[1,0,1,0],[1,0,1,1],[1,1,0,0],[1,1,0,1],[1,1,1,0],[1,1,1,1]];
i use Quine Mccluskey algorithm

got result below

wildcard is 5 or x
[[0, 5, 1, 5], [5, 0, 1, 5], [5, 5, 1, 0], [1, 0, 5, 5], [1, 5, 0, 5], [1, 5, 5, 0], [5, 5, 1, 1], [5, 1, 5, 1], [5, 1, 1, 5], [1, 5, 5, 1], [1, 5, 1, 5], [1, 1, 5, 5]]
A'C + B'C + CD' + AB' + AC' + AD' + CD + BD + BC + AD + AC + AB
 
table1 = [[0,0,0,0],
[0,0,0,1],
[0,0,1,0],
[0,0,1,1],
[0,1,0,0],
[0,1,0,1],
[0,1,1,0],
[0,1,1,1],
[1,0,0,0],
[1,0,0,1],
[1,0,1,0],
[1,0,1,1],
[1,1,0,0],
[1,1,0,1],
[1,1,1,0],
[1,1,1,1]];
 
loand(lonot(tt[0]),tt[2])
loand(lonot(tt[1]),tt[2])
loand(lonot(tt[3]),tt[2])
loand(lonot(tt[1]),tt[0])
loand(lonot(tt[2]),tt[0])
loand(lonot(tt[3]),tt[0])
loand(tt[2],tt[3])
loand(tt[1],tt[3])
loand(tt[1],tt[2])
loand(tt[0],tt[3])
loand(tt[0],tt[2])
loand(tt[0],tt[1])
 
loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loand(lonot(tt[0]),tt[2]),
loand(lonot(tt[1]),tt[2])),
loand(lonot(tt[3]),tt[2])),
loand(lonot(tt[1]),tt[0])),
loand(lonot(tt[2]),tt[0])),
loand(lonot(tt[3]),tt[0])),
loand(tt[2],tt[3])),
loand(tt[1],tt[3])),
loand(tt[1],tt[2])),
loand(tt[0],tt[3])),
loand(tt[0],tt[2])),
loand(tt[0],tt[1]));
def lonot(z):
    if z == 1:
        return 0
    else:
        return 1
def loand(a, b):
    if a == 1 and b == 1:
        return 1
    else:
        return 0
def loor(a, b):
    if a == 0 and b == 0:
        return 0
    else:
        return 1
#A'C + B'C + AB' + CD + A'BD
for tt in table1:
    print loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loor(loand(lonot(tt[0]),tt[2]),
loand(lonot(tt[1]),tt[2])),
loand(lonot(tt[3]),tt[2])),
loand(lonot(tt[1]),tt[0])),
loand(lonot(tt[2]),tt[0])),
loand(lonot(tt[3]),tt[0])),
loand(tt[2],tt[3])),
loand(tt[1],tt[3])),
loand(tt[1],tt[2])),
loand(tt[0],tt[3])),
loand(tt[0],tt[2])),
loand(tt[0],tt[1]));
 
finally i use python to verify
return
0
0
1
1
0
1
1
1
1
1
1
1
1
1
1
1
 
seems correct if wildcard is 1 too, but
can boolean simplify function simplify this
A'C + B'C + CD' + AB' + AC' + AD' + CD + BD + BC + AD + AC + AB
 
to
 
C + A + B.D   which is
P1 = D + Q0 + Q1.N in png file ?

how to compute the most simplified resul...

LeeHoYeung 535 Maple 12
solve
June 22 2017
1 0

solve(diff(-1/x,x) = (-1/x)^(b), b);

originally is 2, but it use ln(....) to express
 
if start from substitute, it seems need to replace manually.

solve(subs(a(x)=-1/x,diff(a(x),x) = (a(x))^(b)), b);

 
goal is to find b in equation below
solve(diff((x^2+x+1)/(-1+x)^2,x) = ((x^2+x+1)/(-1+x)^2)^(b), b);
(2*x+1)/(-1+x)^2-(2*(x^2+x+1))/(-1+x)^3 = ((x^2+x+1)/(-1+x)^2)^(b)
 
solve(diff((x^2+x+1)/(-1+x)^2,x) = ((x^2+x+1)/(-1+x)^2)*(b), b);

is it possible to evaluate or how to eva...

LeeHoYeung 535 Maple 12
sum summation
June 17 2017
1 4
updated

after refer from

https://en.wikipedia.org/wiki/List_of_representations_of_e

exponential1 := sum((1/n!), n=0..infinity);
exponential1 is not a decimal number, it is exp(1)
 
hoyeung1:= sum((Int(exp(LambertW(1/(-1+x))*(-1+x)), x)), x=0..infinity);
 
hoyeung2:= sum((Int(exp(LambertW(1/(-1+x!))*(-1+x!)), x)), x=0..infinity);
 
how to evalute hoyeung1 or hoyeung2 as a decimal number?
 
how to evalute hoyeung^x as a decimal number function is func1 := proc(x) return hoyeung^x end proc:
 
but i do not know whether sum((Int(exp(LambertW(1/(-1+x))*(-1+x)), x))*m^x, x=0..infinity) = hoyeung^x
 
can limit(1+(Int(exp(LambertW(1/(-1+x))*(-1+x)), x)))^x, x=infinity) = hoyeung^1 ?

how to dsolve this case and evaluate the...

LeeHoYeung 535 Maple 12
dsolve
June 17 2017
1 1

sol := dsolve(diff(ln(y(x)),x) = y(x)^(1/(1-y(x))), y(x));
x-Intat(_a^(-(-2+_a)/(-1+_a)), _a = y(x))+_C1 = 0
 

the solution is not y(x) = , but y(x) at the right hand side

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