I just to confirm my syntax of abs and mode

I have wrote it like this

x(mod(n, 2) == 0) = 5 * alpha .^ (abs(n(mod(n, 2) == 0))/2);

Is this one is right?

Answer 1

[Walter Roberson]

Not exactly.

syms alpha
n = [1 2 3 4 5]

n = 1×5

1 2 3 4 5

x(mod(n, 2) == 0) = 5 * alpha .^ (abs(n(mod(n, 2) == 0))/2)

x = 

length(n)

ans = 5

length(x)

ans = 4

[Zamer Chaudhary]

Can you please provide me the right syntax? How to write that equation

[Torsten]

Depends on what you expect as solution. Your sequence x(n) is defined for all integer values for n, also negative. In MATLAB, you cannot index arrays by negative values, i.e. x(-3), e.g., cannot be defined. What we can provide is - given an integer n - a function that returns the value x(n), not in array form, but as a single scalar value.

[Walter Roberson]

n negative or 0 is not a problem for Zamer's code, as there is no indexing by n

syms alpha
n = -3:5

n = 1×9

-3 -2 -1 0 1 2 3 4 5

x(mod(n, 2) == 0) = 5 * alpha .^ (abs(n(mod(n, 2) == 0))/2)

x = 

length(n)

ans = 9

length(x)

ans = 8

However notice the length of the output is not correct.

When you use logical indexing of a vector, the last item assigned to is the last location where the logical value is true. For example,

x([false, true, false]) = 7

would skip x(1), would assign to x(2), and would skip x(3).

When you assign to an element that is after the last defined element of an array, then the array is extended, and any locations between the previous last element and any now-assigned elements are filled with 0. So if x did not exist before, the "hole" at x(1) would be filled with 0, then x(2) would be explicitly assigned 7, then x(3) would only be assigned to (with 0) if there is at least one true index after x(3) . Since there is no such index in this case, x(3) would simply not be assigned to.

When you index with false then the element at the corresponding location is left alone if it already exists, and the array is not extended for positions corresponding to false that are beyond the last end of the array beyond the last true value.

So... if you want your x to come out the same size as n, then you will need to be sure to initialize x to be sure it is as large as n.

[Walter Roberson]

By the way (hint):

[false false true false] .* [1 2 3 4]

ans = 1×4

0 0 3 0

[Torsten]

n negative or 0 is not a problem for Zamer's code, as there is no indexing by n

I meant accessing the values of the sequence for n <= 0. The most natural way would be x(n) what does not work.