Major confusion for loops - precision not acquired

I have a loop that sums the values of function called pdfomegat

    omegat= -1.9999:0.01:+1.9999;
    sum1=0;

    for j=1:length(omegat)

       func(j)=pdfomegat(j)*0.01;
       sum1=sum1+func(j);
    end

        c1=sum1;

My second loop is slightly different loop and is as follows

    sum2=0;
    syms t 
    for j=1:length(omegat)

       func2(j)=t*pdfomegat(j)*0.01;
       sum2=sum2+func(j);
    end

        c2=sum2;

I know for fact that c1 is 1 .. However notice that in the second loop I change nothing but multiply by symbolic character t. Howver I get that

c2=0.99999464659732599597363744692302*t.

Why isn't c2=t?

Is there anyway I can fix it, I can't use the round function because it symbolic.. Any thoughts?

Answer 1

[A Jenkins]

Play around with vpa() and digits() and make sure to read the help files on Precision Control.

[DM]

Is there any way I can extract the number from the symbolic expression and round it off? My problem requires high accuracy.

[A Jenkins]

>> syms t
>> c2=0.99999464659732599597363744692302*t;
>> cf=vpa(c2,8)

cf = 

0.99999465*t

>> cf=vpa(c2,4)

cf =

1.0*t

Answer 2

[John D'Errico]

No. You cannot make a double precision number more precise than it is. There are simple many numbers that are not exactly representable using a double. For example, 0.1 is not so, since doubles are stored in binary form.

Anyway, my very good bet is that c1 is NOT 1, but that it was displayed in the command window as 1, using format short. So I think what you know for a fact is actually completely false. As well, it is entirely possible that you were using a single precision format to store pdfomegat, since c1 was so inaccurately NOT equal to 1.

However, you can do many things to improve your work. For example, in this line, rather than multiplying by 0.01, divide by 100, an integer. You do this because 0.01 is not stored exactly as a double, but 100 is.

As for the code itself, write it more efficiently. Learn to use vectors, and simple tools like sum.

Your first loop does nothing more than multiply the vector pdfomegat by 0.01, and then sum it.

omegat= -1.9999:0.01:+1.9999;

func1 = pdfomegat/100;
c1 = sum(func1);

You can also totally replace that second loop simply as

syms t 
func2 = t*func1;
c2 = t*c1;

As for making the sums themselves more accurate, you could convert the vector pdfomegat into a symbolic form (or my own HPF) but since they were originally stored as doubles, you simply won't gain much at all there, unless you create those numbers in a form that allows you to store them more accurately in the first place.

[DM]

My code is little more complicated than what I provided, for reason of simplicity I provided it that way. In fact my c2 is a product of two functions which are in turn functions of omegat for example

    syms t 
    func2 = t.*func1.*g;
    c2 = sum(func2);

I am guessing since both func1 and g and vectors this is where the round off error takes place?

Also when I type c1 in the command window I get

 c1= 1.0000

Any suggestions.

[John D'Errico]

But you are guessing completely WRONG.

That you get 1.0000 on the command line proves my point completely! It was displayed under format short rules.

help format

See what

format long g

would do instead. Then you would learn that your number was not in fact exactly 1, but slightly off. When the symbolic toolbox gets that number, it displays the number more accurately.

As for the code I showed, I would still contend that you need to learn to use vectors, that you could do what you are doing more efficiently and more cleanly. I cannot know that positively, but given the code you did show, it seems a very probable conclusion.

[John D'Errico]

Any rounding that you do AFTER the fact merely hides the numerical issues. It solves no problem. Instead, you need to show more clearly what you are doing, and do that computation more accurately.

Answer 3

[Star Strider]

It’s not straightforward, but you can round the coefficient:

syms t 
c2=0.99999464659732599597363744692302*t;        % Original Expression
cf = double(vpa(coeffs(c2)))                    % Get Coefficient
cf = fix(cf * 1E+8)/1E+8                        % Round To 8 Digits

[DM]

Thanks, but what if the function was not polynomial, my function is of the folllowing form

   func2= 1/(t^-3+1).*func1*g;

I am sorry I really tried to keep everything very simple. Coeffs doesnt work on this expression.

[Star Strider]

I would suggest that ‘keeping everything simple’ is not appropriate. We Answer the Questions posted, so if you do not state your actual problem, you will likely not get an Answer that will solve it.

I have no idea what ‘func1’ and ‘g’ are, so I can go no further. You might use the numden function to separate the numerator and denominator, then use coeffs or whatever function is appropriate. Consider also matlabFunction or similar functions if you want to use your function in other applications.