Is this a logical Error in Matlab ? exp (pi*i*6) = 1.000000000000000 - 0.000000000000001i. WHY an i component ?, should just be 1

Is this a Logical Error in Matlab ?

exp (pi*i*6) = 1.000000000000000 - 0.000000000000001i.

WHY an i component ?, shouldnt the answer just be 1?

Please answer.

[Stephen23]

"Is this a Logical Error in Matlab ?"

Probably not.

"WHY an i component ?, shouldnt the answer just be 1?"

The answer may be exactly one if you are doing algebraic/symbolic mathematics:

exp(sym(pi)*i*6)

ans = 

1

But you did not tell your computer to do symbolic mathematics, you gave it a numeric task using numeric operations. All numeric operations have precision limited by their data types, take finite speed, have finite domains of validity. And when working with binary floating point numbers they easily accumulate floating point error.

The only "logical error" here is assuming that your computer has infinite precision when doing numeric calculations.

Answer 1

[John D'Errico]

Remember that pi in MATLAB is not truly pi, any more than pi is 3.14, or even 3.14159. pi is a transcendental number. There are infinitely many digits in pi.

format long
pi

ans = 
   3.141592653589793

And, yes, that looks vagely like pi. Many digits, but once we get past the first 16 digits or so, we need to recognize that pi there has only the first 16 digits correctly represented. We have an error in the least significant bits, as there must be in any finite representation of a transcendental number.

Double precision numbers are stored in a BINARY form. So they have 52 binary bits. In the case of pi, the binary form would look like

11.001001000011111101101010100010001000010110100011000

So there, the ones represent powers of 2. But the problem is, the true binary form for pi goes on forever, whereas in MATLAB, it stops at that point.

The result of all this is that pi as a double precision number is not truly pi. Close. But not truly pi, any more than pi==3.14. And if we try that simple aproximation for pi, we see the problem far more obviously.

exp(3.14*i*6)

ans = 
  0.999954342529211 - 0.009555776105247i

Of course you would not expect that result to be exact. But what you did was essentially the same thing.

If you want MATLAB to represent pi as the transcendental number pi, then you need to do something like this

exp(sym(pi)*i*6)

ans = 

1

where MATLAB now understands you want to use the actual number pi, instead of a floating point approximation to pi.

[Moses]

Thanks for taking the time to write this up. I appreciate it alot. I see that our tools are limited in the number of decimals they can represent for precision. For example, Matlab can represent up to 15 decimals (after the decimal point). Let me get your thoughts on "which tools do you know that can give you up to 50-100 decimals precision.

[Stephen23]

"I see that our tools are limited in the number of decimals they can represent for precision."

Actually your computer's (most likely integrated) FPU has limited precision. MATLAB, like most other numeric computing tools, uses the computer FPU because it is very efficient. Floating point operations in HW are much faster than doing any arithmetic in SW.

"which tools do you know that can give you up to 50-100 decimals precision."

Search this page for "SYM".

[Moses]

Thanks for your reply. I read about SYM, but it does not PRINT OUT the decimal (numeric) values, e.g., 50-100 decimals

[Torsten]

vpa(pi,150)

ans = 

3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940813

More about the choice of precision can be found here:

https://uk.mathworks.com/help/symbolic/vpa.html

[Moses]

That is what I needed. Thanks.

Answer 2

[Steven Lord]

πis a transcendental number. pi is the double precision approximation to that number. There are some functions in MATLAB that let you operate as though you were using πinstead of pi (like sinpi and cospi) but exp is not one of them.

So if you wanted to compute this without the roundoff error caused by using pi instead of π, you could use the identity .

format longg
y = [exp(6i*pi); ...
    cos(6*pi)+1i*sin(6*pi); ...
    cospi(6)+1i*sinpi(6)]

y = 
                          1 -  7.34788079488412e-16i
                          1 -  7.34788079488412e-16i
                          1 +                     0i

Answer 3

[Sulaymon Eshkabilov]

It is all about a display format:

format short
exp(pi*i*6) 

ans = 1.0000 - 0.0000i

% compare with this:
cos(6*pi)+1i*sin(6*pi)

ans = 1.0000 - 0.0000i

% Compare this:
format long
[exp(pi*1i*6);  cos(6*pi)+1i*sin(6*pi)]

ans = 
  1.000000000000000 - 0.000000000000001i
  1.000000000000000 - 0.000000000000001i