organize vector, difference between adjacent elements of a vector

Hello, everything okay? How can I automate this sequence?

if A=[ a ,b, c ,d ,e, ..........]

outA=[ A(1)-2 , ((A(1)-2)+(A(2)-A(1))) , (((A(1)-2)+(A(2)-A(1))) +(A(3)-A(2))) , ((((A(1)-2)+(A(2)-A(1))) +(A(3)-A(2))) +(A(4)-A(3))), ..............]

for exemple

if A=[ 3 ,10, 55 ,100 ,888, ..........]

outA= [ 3-2, ((3-2)+(10-3)) , ((3-2)+(10-3) +(55-10)), ((3-2)+(10-3) +(55-10)+(100-55)), ((3-2)+(10-3) +(55-10)+(100-55)+(888-100) )]

outA=[ 1 , (1+7), (8+44 ) , (53+44), (98+788)]

outA=[ 1 8 53 98 886]

Answer 1

[Adam Danz]

 A=[ 3 ,10, 55 ,100 ,888];
 z = cumsum(diff([2,A]))

z = 1×5

1 8 53 98 886

[Fidele Adanvo]

you are doing

z=cunsum(A-2) ? i mean A(1)-2, A(2)-2,...........?

[Adam Danz]

It's doing what you described in your question:

outA=[ A(1)-2 ,  ((A(1)-2)+(A(2)-A(1)))   , (((A(1)-2)+(A(2)-A(1))) +(A(3)-A(2))) , ((((A(1)-2)+(A(2)-A(1))) +(A(3)-A(2))) +(A(4)-A(3))), ...

diff([2,A]) does this: [A(1)-2, A(2)-A(1), A(3)-A(2), A(4)-A(3), ..., A(n+1)-A(n)]

cumsum(__) adds them all together, cumulatively.

[Fidele Adanvo]

How would you please if it were like this?

diff([2,A]) does this: [min(1,(A(1)-2)), A(2)-A(1), A(3)-A(2), A(4)-A(3), ..., A(n+1)-A(n)]

[Adam Danz]

You want to subtract the first element of A by 2 which is why 2 is added to the beginning of [2,A].

If you want to apply the restriction min(1,...) to the first element of A,

 z = cumsum(diff([2,A]));
 z(1) = min(1,z(1))

Answer 2

[Ameer Hamza]

The rule you described in just A-2

A=[3 ,10, 55 ,100 ,888];
B = A-2;

Result

>> B
B =
     1     8    53    98   886

In the sum (A(1)-2) + (A(2)-A(1)) + (A(3)-A(2)) + ... + (A(n)-A(n-1)), after cancelling out the common term, you will only get A(n)-2.