Raise a scalar to a descending series of powers in an array

I'm trying to compute a series of powers of a scalar.

This seems to work but the result is not what I expected:

0.8^1:3

ans =

0.8000 1.8000 2.8000

I expected:

ans =

0.8000 0.6400 0.5120

(It looks like it is returning 0.8:3)

This therfore doesn't work either:

0.8^3:-1:1

ans =

1×0 empty double row vector

Neither does this

0.8^(3:-1:1)

Error using ^ (line 51)

Incorrect dimensions for raising a matrix to a power. Check that the matrix is square and the power is a scalar. To perform elementwise matrix powers, use '.^'.

Neither of these work

0.8^[3 2 1]
0.8^[1 2 3]

Error using ^ (line 51)

Incorrect dimensions for raising a matrix to a power. Check that the matrix is square and the power is a scalar. To perform elementwise matrix powers, use '.^'.

So, how do I compute the scalar to the power of a series of descending values?

Answer 1

[Bill Tubbs]

Here is the correct answer I think. Element-wise operation:

0.8.^(3:-1:1)

Answer 2

[Steven Lord]

Your 0.8^1:3 example doesn't do what you think it does.

>> x1 = 0.8^1:3
>> x2 = [0.8^1 0.8^2 0.8^3]

x1 =
                       0.8                       1.8                       2.8

x2 =
                       0.8                      0.64                     0.512

The power operations in MATLAB are at operator precedence level 2 and 3, while the colon operator is at level 7. So your expression (which I used to define x1 above) is the equivalent of:

(0.8^1):3

What you want is to wrap the colon expression in parentheses. But if you try making that change alone you receive an error (line breaks added to avoid scroll bars.)

>> x3 = 0.8^(1:3)

Error using  ^  (line 51)
Incorrect dimensions for raising a matrix to a power. 
Check that the matrix is square and the power is a scalar.
To perform elementwise matrix powers, use '.^'.

So since you want to perform elementwise matrix powers, use the .^ operator instead of the ^ operator.

>> x4 = 0.8.^(1:3)

x4 will match x2.

Answer 3

[Bill Tubbs]

I found this solution:

power(0.8,3:-1:1)

ans =

0.5120 0.6400 0.8000

But still surprised if it can't be done with the ^ operator.