Unexpected MATLAB Answer Explained: Solving Complex Valued Equations | Homework

[sara_87]

Homework Statement

I put the following code into MATLAB:

d=(0.2+0.1)-0.2

y=(asin((d)/(0.1)))

The answer gives:

d=0.100000000000000

and i know that d/0.1 is equal to 1. so, i expect the answer of y to be
y=1.570796326794897 (which is pi/2)
but, MATLAB gives:

y= 1.570796326794897 - 0.000000021073424i

Homework Equations

The Attempt at a Solution

if i put:
d=0.1
then it would give:
y= 1.570796326794897

But, i want to know why the answer is complex valued when it uses 0.1000000000000000 ?

Thank you in advance.

 

[gb7nash]

But, i want to know why the answer is complex valued when it uses 0.1000000000000000 ?

Thank you in advance.

The only reason I can think of is rounding error. The computer thinks .1000000000000 > .1, so when computing d/.1, you obtain a value greater than 1. Taking the arcsin of a value greater than 1, as you know, does not exist for real numbers. This is why there's a small complex term which contributes to the answer.

 

[uart]

Hi Sara87. Floating point numbers on a computer generally can't be represented exactly, but this is usually hidden from the user by limiting the number of significant digits to which results are displayed. What's unusual in this case is that you are operating precisely on the boundary of the real domain (and the extended complex domain) of the asin() function.

BTW. Are you familiar with complex numbers?

> format long

> d=0.3
d =  0.300000000000000
> ((d - 0.2)/0.1)*1e16
ans = 9999999999999998

> d = 0.2+0.1
d = 0.300000000000000
> ((d - 0.2)/0.1)*1e16
ans = 10000000000000002

 

[sara_87]

Thank you both.
Yes, i am familiar with the complex numbers.
So, how can i change my code so that is gives pi/2 instead of pi/2+ some complex number ?

 

[uart]

There are several possible ways.

The simplest, if you're certain your result should be real, is: y = real(asin(...))

OR you could test if the argument is in [-1-delta ... +1+delta], then throw an error if it isn't and clip it back to [-1...1] if it is.

 

[sara_87]

Thank you. I will use y=real(asin(...))