Note: This page has been translated by MathWorks. Please click here

To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

This example shows how to perform arithmetic and linear algebra with single precision data. It also shows how the results are computed appropriately in single-precision or double-precision, depending on the input.

Let's first create some data, which is double precision by default.

Ad = [1 2 0; 2 5 -1; 4 10 -1]

```
Ad =
1 2 0
2 5 -1
4 10 -1
```

We can convert data to single precision with the `single`

function.

`A = single(Ad); % or A = cast(Ad,'single');`

We can also create single precision zeros and ones with their respective functions.

n = 1000; Z = zeros(n,1,'single'); O = ones(n,1,'single');

Let's look at the variables in the workspace.

whos A Ad O Z n

Name Size Bytes Class Attributes A 3x3 36 single Ad 3x3 72 double O 1000x1 4000 single Z 1000x1 4000 single n 1x1 8 double

We can see that some of the variables are of type `single`

and that the variable `A`

(the single precision version of `Ad`

) takes half the number of bytes of memory to store because singles require just four bytes (32-bits), whereas doubles require 8 bytes (64-bits).

We can perform standard arithmetic and linear algebra on singles.

`B = A' % Matrix Transpose`

`B = `*3×3 single matrix*
1 2 4
2 5 10
0 -1 -1

`whos B`

Name Size Bytes Class Attributes B 3x3 36 single

We see the result of this operation, `B`

, is a single.

`C = A * B % Matrix multiplication`

`C = `*3×3 single matrix*
5 12 24
12 30 59
24 59 117

`C = A .* B % Elementwise arithmetic`

`C = `*3×3 single matrix*
1 4 0
4 25 -10
0 -10 1

`X = inv(A) % Matrix inverse`

`X = `*3×3 single matrix*
5 2 -2
-2 -1 1
0 -2 1

`I = inv(A) * A % Confirm result is identity matrix`

`I = `*3×3 single matrix*
1 0 0
0 1 0
0 0 1

`I = A \ A % Better way to do matrix division than inv`

`I = `*3×3 single matrix*
1 0 0
0 1 0
0 0 1

`E = eig(A) % Eigenvalues`

`E = `*3×1 single column vector*
3.7321
0.2679
1.0000

`F = fft(A(:,1)) % FFT`

`F = `*3×1 single column vector*
7.0000 + 0.0000i
-2.0000 + 1.7321i
-2.0000 - 1.7321i

`S = svd(A) % Singular value decomposition`

`S = `*3×1 single column vector*
12.3171
0.5149
0.1577

`P = round(poly(A)) % The characteristic polynomial of a matrix`

`P = `*1×4 single row vector*
1 -5 5 -1

`R = roots(P) % Roots of a polynomial`

`R = `*3×1 single column vector*
3.7321
1.0000
0.2679

`Q = conv(P,P) % Convolve two vectors`

`Q = `*1×7 single row vector*
1 -10 35 -52 35 -10 1

R = conv(P,Q)

`R = `*1×10 single row vector*
1 -15 90 -278 480 -480 278 -90 15 -1

`stem(R); % Plot the result`

Now let's look at a function to compute enough terms in the Fibonacci sequence so the ratio is less than the correct machine epsilon (`eps`

) for datatype single or double.

% How many terms needed to get single precision results? fibodemo('single')

ans = 19

% How many terms needed to get double precision results? fibodemo('double')

ans = 41

% Now let's look at the working code. type fibodemo

function nterms = fibodemo(dtype) %FIBODEMO Used by SINGLEMATH demo. % Calculate number of terms in Fibonacci sequence. % Copyright 1984-2014 The MathWorks, Inc. fcurrent = ones(dtype); fnext = fcurrent; goldenMean = (ones(dtype)+sqrt(5))/2; tol = eps(goldenMean); nterms = 2; while abs(fnext/fcurrent - goldenMean) >= tol nterms = nterms + 1; temp = fnext; fnext = fnext + fcurrent; fcurrent = temp; end

% Notice that we initialize several of our variables, |fcurrent|, % |fnext|, and |goldenMean|, with values that are dependent on the % input datatype, and the tolerance |tol| depends on that type as % well. Single precision requires that we calculate fewer terms than % the equivalent double precision calculation.

Was this topic helpful?