katz basic activities of daily living test
welch overlapped segment averaging estimator

Some Muirhead Mean Operators for Intuitionistic Fuzzy Numbers and Their Applications to Group Decision Making

alpha 1, alpha 2..., alpha n=(1 n ! ∑ ϑ ∈ S n ∏ j=1 n alpha ϑ j p j) 1 ∑ j=1
alpha 1, alpha 2..., alpha n=∏ j=1 n alpha j 1
alpha 1, alpha 2..., alpha n=(1 n (n -1) ∑ i, j=1
Definition 6. Let alpha i=(u i, v i)(i=1,2
j=1 n u ϑ j p j, 1 - ∏ j=1 n (1 - v ϑ j p j),</disp formula>then<disp formula id="pone.0168767.e051">removedGRAPHIC ∑ ϑ ∈ S n ∏ j=1
pone.0168767.e052">removedGRAPHIC 1 n ! ∑ ϑ ∈ S n ∏ j=1
S n (1 - ∏ j=1 n u ϑ j p j) 1 n !, (∏ ϑ ∈ S n (1 - ∏ j=1 n (1 - v ϑ j p j) 1 n !),</disp formula>So, we have<disp formula id="pone.0168767.e053">removedGRAPHIC (1 n ! ∑ ϑ ∈ S n ∏ j=1 n alpha ϑ j p j) 1 ∑ j=1
disp formula>then<disp formula id="pone.0168767.e056">removedGRAPHIC 1 - ∏ j=1
following inequality<disp formula id="pone.0168767.e058">removedGRAPHIC u + v=(1 - (∏ ϑ ∈ S n (1 - ∏ j=1 n u ϑ j p j) 1 n !) 1 ∑ j=1
</disp formula>then<disp formula id="pone.0168767.e070">removedGRAPHIC 1 - ∏ j=1
pone.0168767.e071">removedGRAPHIC (∏ ϑ ∈ S n (1 - ∏ j=1 n u ϑ j p j) 1 n !<=(∏ ϑ ∈ S n (1 - ∏ j=1 n u ϑ j ' p j) 1
alpha 1, alpha 2..., alpha n=1 n ∑ i=1 n alpha i=(1 - ∏ j=1 n (1 - u i) 1 / n, ∏ j=1
alpha 1, alpha 2..., alpha n=(1 n ∑ i=1 n alpha i lambda) 1 / lambda=(1 - ∏ j=1 n (1 - u i lambda) 1 / n) 1
j 1 n (n -1) 1 / 2)</disp formula>- If removedGRAPHIC p=(1, 1, ⋯, 1 ︷ k, 0, 0, ⋯, 0
alpha 1, alpha 2..., alpha n=(∏ j=1 n alpha j) 1 / n=(∏ j=1
1 - (∏ j=1 n (1 - v j) 1 / n)</disp formula>- If removedGRAPHIC p=(1 n, 1 n, ⋯, 1
if<disp formula id="pone.0168767.e088">removedGRAPHIC ∑ j=1
j=1, 2, ⋯, n -1</disp formula><disp formula id="pone.0168767.e089">removedGRAPHIC ∑ j=1
Definition 7. Let alpha i=(u i, v i)(i=1,2
p j) 1 n ! 1 ∑ j=1 n p j,</disp formula><disp formula id="pone.0168767.e102">removedGRAPHIC u alpha +=(1 - (∏ ϑ ∈ S n (1 - ∏ j=1 n (1 - (1
Definition 8. Let alpha i=(u i, v i)(i=1,2
j=(1 - ∏ j=1 n (1 - u ϑ j p j, ∏ j=1 n v ϑ j p j),</disp formula>then<disp formula id="pone.0168767.e112">removedGRAPHIC ∏ ϑ ∈ S n ∑ j=1
j 1 n !=(∏ ϑ ∈ S n (1 - ∏ j=1 n (1 - u ϑ j p j) 1 n !, 1 - (∏ ϑ ∈ S n (1 - ∏ j=1 n v ϑ j p j) 1 n !),</disp formula>So,<disp formula id="pone.0168767.e114">removedGRAPHIC 1 ∑ j=1
Let<disp formula id="pone.0168767.e115">removedGRAPHIC u=1 - (1 - ∏ ϑ ∈ S n (1 - ∏ j=1 n (1 - u ϑ j p j) 1 n !) 1 ∑ j=1 n p j, v=(1 - (∏ ϑ ∈ S n (1 - ∏ j=1 n v ϑ j p j) 1 n !) 1 ∑ j=1
disp formula>then<disp formula id="pone.0168767.e117">removedGRAPHIC (1 - ∏ j=1 n (1 - u ϑ j p j) ∈ [0, 1 ], (1 - ∏ j=1 n (1 - u ϑ j p j) 1 n ! ∈ [0, 1]
following inequality<disp formula id="pone.0168767.e120">removedGRAPHIC u + v=1 - (1 - ∏ ϑ ∈ S n (1 - ∏ j=1 n (1 - u ϑ j p j) 1 n !) 1 ∑ j=1
1 - ∏ j=1 n (1 - v i) 1 / n)</disp formula>- If p=(lambda,0,⋯,0
j=1 n (1 - v i lambda) 1 / n) 1 / lambda)</disp formula>- If p=(1,1,0,0,⋯,0
j) 1 / C n k) 1 / k</disp formula>- If p=(1,1,⋯,1
p j) 1 n ! 1 ∑ j=1 n p j,</disp formula><disp formula id="pone.0168767.e147">removedGRAPHIC u alpha +=1 - (1 - (∏ ϑ ∈ S n (1 - ∏ j=1 n (1
is omega=(omega 1, omega 2,⋯, omega n) satisfying removedGRAPHIC omega j<0