To express (cos(x))^2 we have (using the sum/diff formulas):
(cos(x))^2 => 1/2*(1+cos(2*x))
But how can we express (cos(x))^3? Can this be done by using the same sum/diff formulas? I know the answer is:
1/4*(3*cos(x) + cos(3*x)) => But how to get here?
Use the complex form,
cos(x) = (z + 1/z) / 2
where z = exp(ix).
Then
cos³(x) = (z³ + 3z + 3/z + 1/z³) / 8 = (cos(3x) + 3 cos(x)) / 4
because
z³ = exp(i3x).
By the same method,
cos²(x) = (z² + 2 + 1/z²) / 4 = (cos(2x) + 1) / 2.
Related
So I have been given the following expression, but I cannot seem to solve it, can anyone do this and show the steps please?
Prove XY'Z + XYZ' + XYZ = XY + XZ
XY'Z + XYZ' + XYZ = XY + XZ
Notice X and Z are common factors between XY'Z and XYZ.
XZ(Y' + Y) + XYZ' =
Y' + Y is equal to 1 (if Y=0 then Y'=1 and so 0 + 1 = 1, that is 0 or 1 = 1. Similarly, if Y=1 then Y'=0 and so 1 + 0 = 1). Therefore, what you get is:
XZ·1 + XYZ' =
XZ·1 = XZ since A·1 = A (if A=0 then 0·1 is 0 and if A=1 then 1·1 = 1). Now the function is simplified to:
XZ + XYZ' =
Notice once again X is a common factor between XZ and XYZ'.
X(Z + YZ') =
Notice this time that Z + YZ' is a special case of the distributive law, which is A + A'B = A + B. This is because if we apply the general distributive law A + BC = (A + B)·(A + C) then we get A + A'B = (A + A')·(A + B) = 1·(A + B) = A + B. Following this reasoning we get to simplify the function even further:
X(Z + Y) =
All that's left is for us to use the distributive law and we finally arrive to the final result:
XY + XZ
Please note that nothing is written between variables, an AND operator (or "·" symbol) is assumed. It's just a way to save space.
I have recently had trouble with understanding the substitution method for solving reccurences. I watched few on-line lectures about the problem, but sadly it did not tell me much (in one of them I heard that it is based on guessing, which made me even more confused) and I am looking for some tips. My objective is to solve three different reccurence functions using substitution method, find their time complexity and their values for T(32).
Function 1 is defined as:
T(1) = 1
T(n) = T(n-1) + n for n > 1
I started off by listing first few executions:
T(2) = T(2-1)+2 = 1+2
T(3) = T(3-1)+3 = 1+2+3
T(4) = T(4-1)+4 = 1+2+3+4
T(5) = T(5-1)+5 = 1+2+3+4+5
...
T(n) = 1+2+...+(n-1)+n = n(n+1)/2
Then I proved by induction, that T(1) = 1 using the formula for sum of the first n natural numbers, and then that it is also true for n+1. It was pretty clear to me, but I am not sure whether this is substitution method. Also knowing the formula T(n) = n(n+1)/2 I easily calculated T(32) = 528 and counted the time complexity, which is O(n^2).
In examples (2) and (3) I only need solution for n=2^k when k is a natural number, but it would be nice if you recommended me any articles showing how to get these for all n as well (but I suppose it is way harder than that).
Function 2 is defined as:
T(1) = 0
T(n) = T(n/2) + 1 for even n > 1
T(n) = T((n+1)/2) + 1 for odd n > 1
As I was allowed to prove it only for n=2^k and based on my gained knowledge I tried to do it following way:
T(n) = T(n/2) + 1
= T(n/4) + 1 + 1 = T(n/4) + 2
= T(n/8) + 1 + 2 = T(n/8) + 3
= T(n/16) + 1 + 3 = T(n/16) + 4
= T(n/2^i) + i // where i <= k, according to tutorials
And this is the moment where I get stuck and I cannot proceed further. I suppose that my calculations are correct, but I am not sure how should I look for a formula, which would satisfy this function. After I get the right formula, calculating T(32) or time complexity will not be a problem.
Function 3 is defined as:
T(1) = 1
T(n) = 2T(n/2) + 1 for even n > 1
T(n) = T((n – 1)/2) + T((n+1)/2) + 1 for odd n > 1
My calculations:
T(n) = 2T(n/2) + 1
= 2(2T(n/4)+1) + 1 = 4T(n/4) + 3
= 4(2T(n/8)+1) + 3 = 8T(n/8) + 7
= iT(n/2^i) + 2^i - 1
And again it comes to the formula, which I am not sure how should be rewritten.
Basically, does substitution method for solving reccurences means finding and iterative formula?
After restudying the topic I found what I did wrong and not to leave my question unanswered, I will try to do it.
The first function is calculated well, induction proof is also correct - nothing to add here.
When it comes to the second function where I got stuck, I did not
pay attention that I was actually using a substitution n=2^k. This
is how it should look:
T(n) = T(n/2) + 1
= T(n/4) + 1 + 1 = T(n/4) + 2
= T(n/8) + 1 + 2 = T(n/8) + 3
= T(n/16) + 1 + 3 = T(n/16) + 4
= T(n/2^k) + k
= T(1) + k
= k
Induction proof that T(2^k) = k works:
Base case: k=1, then T(2^1) = T(2) = 1. (it cannot be k=0, as 2^0 is not bigger than 1)
Inductive step: assume T(2^k) = k, we want to show T(2^(k+1)) = k+1. As 2^k=n, then 2^(k+1) = 2*2^k = 2n.
T(2n) = T(n) + 1
= T(n/2) + 1 + 1
= T(n/4) + 2 + 1
= T(n/8) + 3 + 1
= T(1) + k + 1
= k + 1.
Time complexity: O(log n)
T(32) = T(2^5) = 5
In the third function I missed that every time the function called
itself, the value has doubled.
T(n) = 2T(n/2) + 1
= 2(2T(n/4)+1) + 1 = 4T(n/4) + 3
= 4(2T(n/8)+1) + 3 = 8T(n/8) + 7
= 8(2T(n/16)+1) + 7 = 16T(n/16) + 15
= 16(2T(n/32)+1) + 15 = 32T(n/32) + 31
= 2^k*T(n/2^i) + 2^k - 1
= 2^k*T(1) + 2^k - 1
= 2^k + 2^k - 1
= 2^(k+1) - 1
Induction proof that T(2^k) = 2^(k+1)-1 works:
Base case: k=1, then T(2^1) = T(2) = 3. Original function T(2) = 2T(1)+1 = 2+1 = 3, so base case is true.
Inductive step: assume T(2^k) = 2^(k+1)-1, we want to show T(2^(k+1)) = 2^(k+2)-1. Similarly, as in the second function, 2^k=n, so 2^(k+1) = 2*2^k = 2n.
T(2n) = 2T(n) + 1
= 2(2T(n/2)+1) + 1 = 4T(n/2) + 3
= 4(2T(n/4)+1) + 1 = 8T(n/4) + 7
= 8(2T(n/8)+1) + 1 = 16T(n/8) + 15
= 2^(k+1) + 2^(k+1) - 1
= 2*2^(k+1) - 1
= 2^(k+2) - 1.
We could also take a look at first few elements for T(n), which are 1, 3, 5, 7, 9, etc., so T(n) = 2n-1
Time complexity: O(n)
T(32) = T(2^5) = 2^(5+1) - 1 = 64 - 1 = 63
When solving a regular maximum weighted matching problem in GLPK, one would provide a DIMACS file and call glp_asnprob_lp with GLP_ASN_MMP. However, this will set all the constrains to <= 1, for example
Subject To
r_1: + x(1,9) + x(1,10) + x(1,12) <= 1
r_2: + x(2,10) + x(2,12) + x(2,13) <= 1
r_3: + x(3,11) + x(3,13) <= 1
r_4: + x(4,9) + x(4,12) + x(4,14) <= 1
I want each node to have a higher capacity (say 10), hence the constrains would be:
Subject To
r_1: + x(1,9) + x(1,10) + x(1,12) <= 10
r_2: + x(2,10) + x(2,12) + x(2,13) <= 10
r_3: + x(3,11) + x(3,13) <= 10
r_4: + x(4,9) + x(4,12) + x(4,14) <= 10
How do I go about it? I really don't want to have to build my own matrix.
Is there in sage, any instruction to solve a linear system equations
module p(x) (polynomial over finite field), where the system coefficients are polynomials over finite field in any indeterminate?. I know that for integers exists something like, example
sage: I6 = IntegerModRing(6)
sage: M = random_matrix(I6, 4, 4)
sage: v = random_vector(I6, 4)
sage: M \ v
(4, 0, 2, 1)
Here my code
F.<a> = GF(2^4)
PR = PolynomialRing(F,'X')
X = PR.gen()
a11 = (a^2)*(X^3)+(a^11)*(X^2)+1
a12 = (a)*(X^4)+(a^13)*(X^3)+X+1
a13 = X^2+(a^13)*(X^3)+a*(X^2)+1
a21 = X^3
a22 = X+a
a23 = X^2+X^3+a*X
a31 = (a^12)*X+a*(X^2)
a32 = (a^8)*(X^2)+X^2+X^3
a33 = a*X + (a^2)*(X^3)
M = matrix([[a11,a12,a13],[a21,a22,a23],[a31,a32,a33]])
v = vector([(a^6)*(X^14)+X^13+X,a*(X^2)+(X^3)*(a^11)+X^2+X+a^12,(a^8)*(X^7)+a*(X^2)+(a^12)* (X^13)+X^3+X^2+X+1])
p = (a^2 + a)*X^3 + (a + 1)*X^2 + (a^2 + 1)*X + 1 # is than 6 in the firs code
I'm trying
matrix(PolynomialModRing(p),M)\vector(PolynomialModRing(p),v)
but PolynomialModRing not exist ...
EDIT
another person talk me that I will make
R.<Xbar> = PR.quotient(PR.ideal(p))
# change your formulas to Xbar instead of X
A \ b
# ==> (a^3 + a, a^2, (a^3 + a^2)*Xbar^2 + (a + 1)*Xbar + a^3 + a)
this work fine but Now I'm trying to apply the Chinese Theorem Remainder after the code, then .... I defined
q = X^18 + a*X^15 + a*X^12 + X^11 + (a + 1)*X^2 + a
r = a^3*X^3 + (a^3 + a^2 + a)*X^2 + (a^2 + 1)*X + a^3 + a^2 + a
#p,q and r are relatively prime
and I'm trying ...
crt([(A\b)[0],(A\b)[1],(A\b)[2]],[p,q,r])
but I get
File "element.pyx", line 344, in sage.structure.element.Element.getattr (sage/structure/element.c:3871)
File "misc.pyx", line 251, in sage.structure.misc.getattr_from_other_class (sage/structure/misc.c:1606)
AttributeError: 'PolynomialQuotientRing_field_with_category.element_class' object has no attribute 'quo_rem'
I'm thinking that problem is the change Xbar to X
Here my complete example to integers
from numpy import arange, eye, linalg
#2x-3y+2z=21
#x+4y-z=1
#-x+2y+z=17
A = matrix([[2,-3,2],[1,4,-1],[-1,2,1]])
b=vector([21,1,17])
p=[17,11,13]
d=det(A)
dlist=[0,0,0]
ylist=matrix(IntegerModRing(p[i]),[[2,-3,2],[1,4,-1], [-1,2,1]])\vector(IntegerModRing(p[i]),[21,1,17])
p1=[int(ylist[0]),int(ylist[1]),int(ylist[2])]
CRT(p1,p)
Maybe... this is what you want? Continuing your example:
G = F.extension(p) # This is what you want for "PolynomialModRing(p)
matrix(G,M)\vector(G,v)
which outputs
(a^3 + a, a^2, (a^3 + a^2)*X^2 + (a + 1)*X + a^3 + a)
In your question you ask "where the system coefficients are polynomials over finite field in any indeterminate" so what I'm doing above is NOT what you have actually asked, which would be a weird question to ask given your example. So, I'm going to just try to read your mind... :-)
I need to write a cumulative summation function in R but I've been hitting a brick wall. The function has the following structure:
a*x1
a*x2 + a^2*x1
a*x3 + a^2*x2 + a^3*x1
a*x4 + a^2*x3 + a^3*x2 + a^4*x1
And so on. cumsum doesn't seem to work for this type of function. Is there any way this could be implemented in R?
Since your recursion is
u[n+1] = a * ( x[n+1] + u[n] )
i.e.,
u[n+1]/a = x[n+1] + a * u[n]/a,
you can use filter:
x <- 1:5
a <- 2
a*filter(1:5, a, method="recursive")
# Compare with the expected values
a*x[1]
a*x[2] + a^2*x[1]
a*x[3] + a^2*x[2] + a^3*x[1]
a*x[4] + a^2*x[3] + a^3*x[2] + a^4*x[1]