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8.1 The sum and cumsum Functions
8.2 The prod and cumprod Functions
8.3 The max and min Functions
8.4 The sort Function
8.5 The mean and median Functions
8.6 Moments and the Standard Deviation Function: std
8.7 The diff Function and an Approximate Derivative
There are a whole arsenal of function in MATLAB that will be useful
in your work. We will discuss and demonstrate the following functions
in this chapter.
|
Function |
Performs on vectors |
|
sum |
Sums the elements |
|
cumsum |
Gives a running sum of the elements |
|
prod |
Multiplies the elements together |
|
cumprod |
Gives a running product of the elements |
|
max |
Finds the maximum value |
|
min |
Finds the minimum value |
|
sort |
Sorts the elements |
|
mean |
Finds the mean value of the elements |
|
std |
Finds the standard deviation of elements |
|
diff |
Gives the differences between adjacent elements |
>> vr=[3 -3 -5.2 12.4];
>> vc=vr+i*[-2 3 -1 2.2];
>> mr=[3 14 -5
-3 15 -2
-5.2 1 10
12.4 16 11];
Now we will look at the way each function treats these arrays.
8.1 The sum and
cumsum Functions
Here is a test of the sum and cumsum functions.
>> disp(sum(vr))
7.2000
>> disp(sum(vc))
7.2000 + 2.2000i
>> disp(sum(mr))
7.2000 46.0000 14.0000
The sum function just sums the elements in both real and complex vectors. It sums each column in a matrix and gives a vector of these sums. If we want to sum over rows, we need to transpose the matrix first and transpose the result if we want a single column result.
>> disp(sum(mr))
12.0000
10.0000
5.8000
39.4000
Now that we have seen what the sum function does, it is not too hard to figure out what cumsum gives:
>> disp(cumsum(vr))
3.0000 0 -5.2000 7.2000
>> disp(cumsum(vc))
3.0000 - 2.0000i 0 + 1.0000i -5.2000 7.2000 + 2.2000i
>> disp(cumsum(mr))
3.0000 14.0000 -5.0000
0 29.0000 -7.0000
-5.2000 30.0000 3.0000
7.2000 46.0000 14.0000
The result is always an array of the same size as the array
operated on. The last element is always what sum gave. The
rest are partial sums over 1, then 2 then 3, etc. elements in the
vector or columns of a matrix. For example, operating on the real
vector: vr, we get:
3 as the sum of the first element 3+(-3)=0 as the sum of the first two elements 3+(-3)+(-5.2)=-5.2 as the sum of the first three elements 3+(-3)+(-5.2)+12.4=7.2 as the sum all four elements
These four results are then placed in a vector result.
8.2 The prod and
cumprod Functions
Next the prod and cumprod functions will be tested. The
prod function acts just like sum except the elements
are multiplied together:
>> disp(prod(vr))
580.3200
>> disp(prod(vc))
5.4444e+02 - 8.6268e+02i
>> disp(prod(mr))
1.0e+03 *
0.5803 3.3600 1.1000
The complex product may not be familiar, but the other results should not be surprising. You can check the two real cases with a calculator. You will find that the product over the columns of the matrix were displayed in short format mode and truncation shows only
0.5803 * 10^3 or 580.3
rather than 580.32.
The cumulative product function behaves in an analogous fashion to
the cumulativesum function:
>> disp(cumprod(vr))
3.0000 -9.0000 46.8000 580.3200
>> disp(cumprod(vc))
1.0e+02 *
0.0300 - 0.0200i -0.0300 + 0.1500i 0.3060 - 0.7500i 5.4444 - 8.6268i
>> disp(cumprod(mr))
1.0e+03 *
0.0030 0.0140 -0.0050
-0.0090 0.2100 0.0100
0.0468 0.2100 0.1000
0.5803 3.3600 1.1000
We will leave it to the exercises to figure out these results.
8.3 The max and min
Functions
The max and min functions find the largest and smallest
elements in a vector or each column of a matrix. This is done in the
algebraic sense and if the elements are complex, the magnitude of
complex numbers is used in the comparison. Here is what our example
arrays have for maxima:
>> disp(max(vr)) 12.4000 >> disp(max(vc)) 12.4000 + 2.2000i >> disp(max(mr)) 12.4000 16.0000 11.0000
The real arrays again offer no surprises, but the example chosen for the complex vector fails to tell us exactly what the criteria was since the last element had both the largest real part and largest magnitude. This can be easily remedied:
>> vc(3)=-5.2-12.4*i;
>> disp(max(vc))
-5.2000 -12.4000i
>> disp(vc)
3.0000 - 2.0000i -3.0000 + 3.0000i -5.2000 -12.4000i 12.4000 + 2.2000i
>> disp(abs(vc))
3.6056 4.2426 13.4462 12.5936
Thus we see that the criterion is based on the magnitude of the
numbers.
Now the min function is seen to give results based on the same
procedures used in the max function except we now look for the
smallest elements or those with the smallest magnitudes.
>> disp(min(vr)) -5.2000 >> disp(min(vc)) 3.0000 - 2.0000i >> disp(min(mr)) -5.2000 1.0000 -5.0000
There were no questions about what min would find, were
there?
8.4 The sort Function
Here is a function that goes along with the last pair. It gives the
complete order for all the elements in a vector or in each column of
a matrix. Again the magnitudes of complex numbers are used in the
comparison. The ordering is done from smallest to largest.
>> disp(sort(vr))
-5.2000 -3.0000 3.0000 12.4000
>> disp(sort(vc))
3.0000 - 2.0000i -3.0000 + 3.0000i 12.4000 + 2.2000i -5.2000 -12.4000i
>> disp(sort(mr))
-5.2000 1.0000 -5.0000
-3.0000 14.0000 -2.0000
3.0000 15.0000 10.0000
12.4000 16.0000 11.0000
If we want to sort the matrix over rows, we need to do a little transposing:
>> disp(sort(mr')') -5.0000 3.0000 14.0000 -3.0000 -2.0000 15.0000 -5.2000 1.0000 10.0000 11.0000 12.4000 16.0000
If we want to know where the elements in the sorted form stood in the original array we can use sort in the form:
>> m=[2 5 8
-1 7 5
0 2 10];
>> [ms,mi]=sort(m)
ms =
-1 2 5
0 5 8
2 7 10
mi =
2 3 2
3 1 1
1 2 3
8.5 The mean and median Functions
Now let's look at the difference between the mean and median of our arrays. Here are the mean values:
>> disp(mean(vr))
1.8000
>> disp(mean(vc))
1.8000 - 2.3000i
>> disp(mean(mr))
1.8000 11.5000 3.5000
In each case MATLAB simply adds the elements and divides by the
number of elements. We could have found the same result for the
vectors by:
>> disp(sum(vr)/length(vr))
1.8000 >> disp(sum(vc)/length(vc)) 1.8000 - 2.3000i
The matrix mean could also have been found as long as we use the length to get its number of rows:
>> disp(sum(mr)/length(mr))
1.8000 11.5000 3.5000
Now look at the median values applied to vectors with three elements:
>> disp(median([-1 5 6]))
5
>> disp(median([-1 -1 6]))
-1
It simply orders the elements and reports the middle one. In our example vectors, we have an even number of elements. Traditionally, we would expect to order them and find the mean of the two in the middle.
>> disp(median(vr))
0
>> disp(median(vc))
4.7000 + 2.6000i
Looking back at the sorted vectors, we see that the mean of the
middle elements is what is returned by the median function.
8.6 Moments and the Standard Deviation Function:
std
Let us now look at a single vector: x of n elements:
x1, x2, .... xn
The mean will be designated as: xmean. The kth moment about the mean
is defined as:
i=n µk =[sum(xi -xmean )k ]/n i=1
The standard deviation is then defined as:
s = µ2
The std function gives this value. Let's check to see that the
definitions given above are consistent with what we expect:
>> x=[.1 1 1.2 .5 .8 .9];
>> xbar=mean(x)
xbar =
0.7500
>> xmxb=x-xbar
xmxb =
-0.6500 0.2500 0.4500 -0.2500 0.0500 0.1500
>> n=length(x)
n =
6
>> mu2=sum(xmxb.^2)/n
mu2 =
0.1292
>> sqrt(mu2)
ans =
0.3594
>> std(x)
ans =
0.3594
It checks!
8.7 The diff Function and an
Approximate Derivative
We will demonstrate the diff function only for vectors. It can
be used on matrices and gives the same thing for each column of the
matrix that is does for a vector. For a vector with elements:
x1, x2, .... xn
diff gives a vector with elements:
x2 - x1, x3 - x2, .... xn - xn-1
If we apply diff to the vector:
>> x
x =
0.1000 1.0000 1.2000 0.5000 0.8000 0.9000
we get:
>> disp(diff(x))
0.9000 0.2000 -0.7000 0.3000 0.1000
As an application of diff look at calculating the derivative of the function sin(t) using the numerical approximation:
df/dt at tn ~ (f(tn+1) - f(tn))/(tn+1 - tn)
Here is a session used to do just that:
>> t=0:pi/32:pi; >> sint=sin(t); >> dsint=diff(sint); >> dt=diff(t); >> dfdt=dsint./dt;
The length of each of the vectors: sint, dsint, dfdt is:
>> disp(length(dfdt))
32
Thus we will make our comparison over the same number of values of t:
>> t=t(1:32);
>> plot(t,[dfdt;cos(t)])
>> title('Numerical Approximation to dsin(t)/dt')
>> xlabel('t')
>> text(1.5,.5,'___ Appx. with dt = pi/32')
>> text(1.5,.4,'--- cos(t)')
Here is what we got:
Figure 8.1 Numerical Approximation to a
Derivative
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