binomial theorem and MATLAB

Hello,

I habe 3 variables:

x= 0.988:0.0001:1.012;
y= (x-1).^7
y2=x.^7-7*x.^6+21*x.^5-35*x.^4+35*x.^3-21*x.^2+7*x-1;

y2 is the therm (x-1)^2 calculated with the binomial theorem.

So, to my understanding the solution for every x should be the same in y and y2:

But this is my Result:

Would be awesome if someone could explain this odd result to me :)

Answer 1

[Steven Lord]

Cleve Moler, the original author of MATLAB, wrote am article about this (with this specific example) in 1996.

https://www.mathworks.com/company/newsletters/articles/floating-points-ieee-standard-unifies-arithmetic-model.html

[MaJoBo]

Thank you very much!

Answer 2

[John D'Errico]

Actually, what you wrote for y2 is NOT (x-1)^2. It is (x-1)^7. 2 and 7 are not the same thing. Well, they look alike in some fonts and the way some people write the numbers. But 2 and 7 are in fact different numbers.

syms x
expand((x-1)^7)
ans =
x^7 - 7*x^6 + 21*x^5 - 35*x^4 + 35*x^3 - 21*x^2 + 7*x - 1

The plot you got is entirely reasonable for what you did, computing the result (x-1)^7 where x is in the interval [1-0.012,1+0.012].

So then x-1 lies in the interval [-0.012,0.012]. And seince:

0.012^7
ans =
   3.5832e-14

that plot should be scaled entirely in the range of +/- 5e-14. The fuzziness of the curve is due to floating point trash on the computation, sometimes called massive subtractive cancellation.

[MaJoBo]

I'm sorry. I mean (x-1)^7 of course.. Otherwise this whole thing would not make sense :)

[John D'Errico]

The plot you generated is quite correct. The difference between the two curves is simply due to massive subtractive cancellation. That does not happen for

(x-1)^7

But it does so when you try to compute things in the expanded form:

x^7 - 7*x^6 + 21*x^5 - 35*x^4 + 35*x^3 - 21*x^2 + 7*x - 1

It is MUCH more accurate to compute things using the form (x-1)^7, because there you subtract off 1 FIRST.

While the two forms are symbolically identical, in floating point arithmetic they are not. Look at the individual terms, for x=1.012.

format long g
x = 1.012;
(x-1)^7
ans =
      3.58318080000002e-14

x^7
ans =
          1.08708521100641

-7*x^6
ans =
         -7.51936410775185

21*x^5
ans =
          22.2906050625055

-35*x^4
ans =
           -36.71048264576

35*x^3
ans =
               36.27518048

-21*x^2
ans =
                -21.507024

7*x
ans =
                     7.084

-1
ans =
    -1

Remember that each of those terms has a tiny amount of noise in the least significant bits. But the least significant bits of each of those numbers vary by quite a bit.

eps(35*x^3)
ans =
        7.105427357601e-15

eps(-1)
ans =
      2.22044604925031e-16

So, when you add up all of those terms, you get noise that will be on the order of 1.e-14. Again, this is the concept of massive subtractive cancellation.