MATLAB: Sigma notation – difference of individual vector values, for loop or symsum

for loopsigma notationsymsum

Hey everyone,

I am trying to solve a sum in sigma notation with the difference in individual vector values, squared. I'm even having difficult explaining it, so here is the sum:

Where x(i) and y(i) are the two column vectors being used, and M is the size of each vector.

Thanks!

Best Answer

I think you really need to read the basic tutorials for MATLAB, because this is about as basic as you can get.

If you subtract two vectors, x and y, the result will be the difference you are trying to create, right? Thus ...

x - y

Now, can you square those elements? Of course. So you might do this ...

(x-y).^2

That squares each element of the difference. Note that I carefully used a .^ operator there. That is very important. If you don't understand why, then see my first comment. READ THE MANUAL!

How do you sum a vector? Hint:

help sum

So try this ...

sum((x-y).^2)

Finally, you need to divide by the number of elements. So to wrap it all up, you might do this...

M = numel(x);
MSE = sum((x-y).^2)/M;

If you already know the length of those vectors as M, you need not do the first line there.

But really, you need to do some basic reading. You will gain in the end, by not having to ask basic questions for everything you do. There are also some great books out there to teach you MATLAB. And when you see a problem that is too complex for you to handle, break it into small pieces. Solve each part, looking towards the solution that you need, as I did above.

Related Solutions

MATLAB: Create a matrix from a curve

Ok. So you have a CURVE, not a polynomial. A polynomial is something with a very specific definition in mathematics. So when you use that word, it implies something that you do not appear to have. In turn, that confuses people trying to answer your question.

plot(Curve_x,Curve_y)

Next, it appears that you want to create a matrix that will act as a 2-d lookup table. So, for ANY (x,y) pair you wish to use a simple index into the table to indicate is a point lies above(right) or below(left) that curve.

The fact is, this is something simply and trivially done using nothing more than a call to interp1. You never even need to create that lookup table.

For example, consider the (x,y) pair (x0,y0) = (2000,57).

x0 = 2000;
interp1([Curve_x,4095],[Curve_y,-1],x0)
ans =
          44.4959999999999

So, if y0 is greater than 44.4959999999999 then the (x0,y0) pair is to the right/above the curve.

We can now write the entire operation as simply as this:

curvetest = @(x0,y0,Curve_x,Curve_y) interp1([Curve_x,4095],[Curve_y,-1],x0) < y0;
curvetest(2000,57,Curve_x,Curve_y)
ans =
  logical
   1
curvetest(2000,23,Curve_x,Curve_y)
ans =
  logical
   0

Yes, I know this returns 0 or 1. You SAID 1. Not 2, even though that matrix contains 0 or 2. Trivially fixed of course. Just multiply the result by 2, if you really wanted 2 instead of 1.

In fact, curvetest is actually nicely vectorized.

curvetest(2000,30:50,Curve_x,Curve_y)
ans =
  1×21 logical array
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   1   1   1   1   1
curvetest(2000:100:3000,30,Curve_x,Curve_y)
ans =
  1×11 logical array
   0   0   0   0   1   1   1   1   1   1   1

You can test multiple points at once, as above, or like this:

curvetest([1000, 2000, 3500],[15 25 35],Curve_x,Curve_y)
ans =
  1×3 logical array
   0   0   1

So, do you really need any more than that? As you can see, I made it a function of the vectors Curve_x and Curve_y. So, if they change, then the function will see that change. Also, I made it go out to x=4095, which seems logical, although you can set that value to anything you want.

Do you REALLY need a matrix result? Yes, it is easy to do. But that also requires you to scale your inputs, and then to use interpolation on the matrix, so a call to interp2 anyway. While all of that can be made to be fast enough, the call to interp1 will still probably be as fast, unless you code it all very carefully.

So, IF you truly need a matrix result, and the function I showed you how to write is not sufficient, then I'll show you how to do it. But I'll observe that often someone thinks they know exactly what they want, when something simpler is sufficient, and in fact, better.

MATLAB: Sigma notation – two variables. ‘for’ loop or symsum

This runs:

N = length(x);
for k = 0:N/4
    Axx(k+1) = 0;
    for i = 1:N-k
        Axx(k+1) = Axx(k+1) + x(i)*x(i+k);
    end
end

I’ll let you determine if it gives the answers you expect.

Best Answer

Related Solutions

MATLAB: Create a matrix from a curve

MATLAB: Sigma notation – two variables. ‘for’ loop or symsum

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