I am looking at an lmer model that's been coded, and I don't understand what the two -1's are doing in this growth curve model.
The code looks like lmer(y ~ -1 + time_0 + time_1 + time_2 + time_3 + (-1 + time_0 + time_1 + time_2 + time_3 | ID), data, REML = FALSE, control = lmerControl(optimizer = "bobyqa"))
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This question actually had two parts. In the first part I had to prove that a + 1/a >=2. I proved it by rearranging it to (a-1)^2 >= 0, which is always true.
So, I thought the second problem would require a similar method.
(x+y)/z + (y+z)/x + (x+z)/y >=6, where x,y,z>0
But I cant figure it out.
I've tried simplifying it and factoring it for ideas but I've got nothing.
Once you know that a + 1/a >= 2, the second part is easy. Define:
a := x/z, b := y/z, c := y/x
and now
(x+y)/z + (y+z)/x + (x+z)/y = x/z + y/z + y/x + z/x + x/y + z/y
= a + b + c + 1/a + 1/c + 1/b
>= 2 + 2 + 2
= 6
I'm working on some logic homework and I can't figure out the next step in reducing the number of literals. Any help would be greatly appreciated.
(A + B + C) (A’B’ + C)
A’B’C + AC + BC + C
C(A’B’ + A + B + C)
C((A + B)’ + A + B + C)
I'm pretty sure I use the associative law next, but I don't understand how the not operator is distributed when rearranging.
From the point where you left:
C((A + B)’ + A + B + C)
C(1 + C) ; X' + X = 1 applied to X = A + B
C(1) ; 1 + <anything> = 1
C ; <anything>1 = <anything>
Hello I am trying to simplify this expression (proving consensus expression):
(a + b)(b'+ c)(a + c) = (a + b)(b'+ c)
I was thinking of adding (a+b)(b'+ c)(a + c + b' + b), but I don't know what to do after.
Take a look at this:
(a + b)(b' + c)(a + c)
= (ab' + 0 + ac + bc)(a + c)
= (ab' + ab'c +ac + ac + abc + bc)
= (ab' + ab'c + ac + abc + bc)
= (ab'(1+c) + ac + bc(a + 1))
= (ab' + ac + bc)
= (ab' + c(a+b))
= (ab' + bb' + c (a+b))
= (a+b)(b' + c)
The key step is realising that bb' = 0 so you can safely add that term without affecting the result in the penultimate step.
(A+C')(B'+C')
AB'+AC'+BC'+C'
AB'+(A+B+1)C'
We Know ( 1+anything =1)
So required expression AB'+C' .
I'm taking a class on digital logic and I am having a hard time with boolean algebra and simplifying logic functions with it. I have tried answering this problem several times and I keep coming to the answer "1", which I feel is absolutely wrong.
The question is
Consider the logic function f(a,b,c) = abc + ab'c + a'bc + a'b'c + ab'c'. Simplify f using Boolean algebra as much as possible.
I have tried solving it several ways using the boolean identities given in my textbook and from lecture, but I keep coming to something like c + 1 which is equivalent to 1, which I don't feel is the correct answer considering the next question in the problem.
Here is my last attempt:
f(a,b,c) = abc + ab'c + a'bc + a'b'c + ab'c'
= a(bc + b'c + b'c') + a'(bc + b'c) # Distributive, took out the a and the a' separately.
= (a + a')((bc + b'c + b'c') + (bc + b'c)) # Distributive(?), took out the a and a' together (This is probably where I screwed up).
= (1)((c + b'c') + c) # a + a' = 1; bc + b'c = c (Combining).
= c + b'c' + c # cleaned up a little.
= c + b'c' # c + c = c.
= c + (b' + c') # b'c' = b' + c' (DeMorgan's Theorem).
= 1 + b' # c + c' = 1.
= 1 # 1 + b' = 1
This feels absolutely wrong to me, and the next question asks me to make the logic circuit for it, which I don't think is possible.
Can anyone help/walk me through what I am doing wrong? I would really appreciate it. :(
(P.S. I used code formatting, I apologize if this is annoying to some.)
By this table:
A 1 1 1 1 0 0 0 0
B 1 1 0 0 1 1 0 0
C 1 0 1 0 1 0 1 0
Y 1 0 1 1 1 0 1 0
Y=ab'+c
I've got it :D
f(a,b,c) = abc + ab'c + a'bc + a'b'c + ab'c'
= a(bc + b'c + b'c') + a'(bc + b'c)
= a(c(b + b') + b'c') + a'(c(b + b'))
= a(c * 1 + b'c') + a'(c * 1)
= a(c + b'c') + a'c
= a(c'(b'c')')' + a'c
= a(c'(b + c))' + a'c
= a(c'b +cc')' + a'c
= a(c'b)' + a'c
= a(c+b') + a'c
= ac + ab' + a'c
= c(a + a') + ab'
= ab' + c
I have a boolean simplification problem that's already been solved.. but I'm having a hard time understanding one basic thing about it.. the order in which it was solved.
The problem is simplifying this equation:
Y = ¬A¬B¬C + ¬AB¬C + A¬B¬C + A¬BC + ABC
The solution is:
Y = ¬A¬B¬C + ¬AB¬C + A¬B¬C + A¬BC + ABC
= ¬A¬B¬C + ¬AB¬C + A¬B¬C + A¬BC + A¬BC + ABC (idempotency for A¬BC)
= ¬A¬C(¬B + B) + A¬B(¬C + C) + AC(¬B + B)
= ¬A¬C + A¬B + AC
The way I solved it is:
Y = ¬A¬B¬C + ¬AB¬C + A¬B¬C + A¬BC + ABC
= ¬A¬B¬C + ¬AB¬C + ¬A¬B¬C + A¬B¬C + A¬BC + ABC (idempotency for ¬A¬B¬C)
= ¬A¬C(¬B + B) + ¬B¬C(¬A + A) + AC(¬B +B)
= ¬A¬C + ¬B¬C + AC
So how do I know which term to use the law of idempotency on? Thanks.
¬A¬B¬C + ¬AB¬C + A¬B¬C + A¬BC + ABC
¬A¬C(¬B + B) + A(¬B¬C + ¬BC + BC)
¬A¬C + A(¬B¬C + ¬BC + BC) <- see truth table below for the simplification of this
¬A¬C + A(¬B + C)
¬A¬C + A¬B + AC
truth table:
B C
0 0 = 1 + 0 + 0 = 1
0 1 = 0 + 1 + 0 = 1
1 0 = 0 + 0 + 0 = 0
1 1 = 0 + 0 + 1 = 1
which is ¬B + C