Thread Subject: how come matlab drew this?

Hi I saw a exercise in a book and asked me to plot these
graphs togheter as abs(y1-y2):

y1(x) =sqrt(x^2 + 1)- 1

and

y2(x) =x^2/sqrt(x^2 + 1) + 1

and I did the following:

delta=10^(-3);
x=linspace(10^(-4),10^(-3),100);
y1 = sqrt(x.^2+1)-1;
y2 = x.^2./(sqrt(x.^2+1)+1);
z = abs(y1-y2)
plot(x,z)
ylabel('value');
xlabel('x');
title('compare');

I get a graph but I don?t really understand why is that, I
mean if one solves abs(y1-y2) by hand everything becomes
zero, can anyone explain to me why matlab still graphs
something and is what I did wrong? I?m a beginner so please
understand that my reasoning is not that advanced

thanks

"Marcelo Tames" <jmarcelo.tb@mathworks.com> wrote in message <fps18k
$dh3$1@fred.mathworks.com>...
> Hi I saw a exercise in a book and asked me to plot these
> graphs togheter as abs(y1-y2):
>
> y1(x) =sqrt(x^2 + 1)- 1
>
> and
>
> y2(x) =x^2/sqrt(x^2 + 1) + 1
>
> and I did the following:
>
> delta=10^(-3);
> x=linspace(10^(-4),10^(-3),100);
> y1 = sqrt(x.^2+1)-1;
> y2 = x.^2./(sqrt(x.^2+1)+1);
> z = abs(y1-y2)
> plot(x,z)
> ylabel('value');
> xlabel('x');
> title('compare');
>
> I get a graph but I don?t really understand why is that, I
> mean if one solves abs(y1-y2) by hand everything becomes
> zero, can anyone explain to me why matlab still graphs
> something and is what I did wrong? I?m a beginner so please
> understand that my reasoning is not that advanced
>
> thanks
--------
  I assume from your remarks that you are aware that the two expressions in
question are mathematically identically equal to one another, so you are
apparently asking why you didn't get a graph showing a horizontal line of
height zero. If that is your question, the answer is round off error in
computing the two quantities. These involve differing mathematical
operations and will produce slightly different results because of this round off
error. Note that the values you are getting in the graph are of size
somewhere in the neighborhood of 10^(-16) which is a very small value
indeed. If you were to accompany this graph with another one that assumes
values in the neighborhood of 1, the first graph would be squeezed down to
the horizontal line you are expecting.

  If you are asking why the two expressions are identical, try multiplying the
numerator and denominator of y2 by (sqrt(x^2+1)-1) and simplify the result.
What you arrive at is precisely y1.

  By the way, in the first expression for y2, you appear to have left out a pair
of parentheses, though that is corrected in the code.

Roger Stafford

"Roger Stafford"
<ellieandrogerxyzzy@mindspring.com.invalid> wrote in
message <fpsjlc$q6q$1@fred.mathworks.com>...
> "Marcelo Tames" <jmarcelo.tb@mathworks.com> wrote in
message <fps18k
> $dh3$1@fred.mathworks.com>...
> > Hi I saw a exercise in a book and asked me to plot
these
> > graphs togheter as abs(y1-y2):
> >
> > y1(x) =sqrt(x^2 + 1)- 1
> >
> > and
> >
> > y2(x) =x^2/sqrt(x^2 + 1) + 1
> >
> > and I did the following:
> >
> > delta=10^(-3);
> > x=linspace(10^(-4),10^(-3),100);
> > y1 = sqrt(x.^2+1)-1;
> > y2 = x.^2./(sqrt(x.^2+1)+1);
> > z = abs(y1-y2)
> > plot(x,z)
> > ylabel('value');
> > xlabel('x');
> > title('compare');
> >
> > I get a graph but I don?t really understand why is
that, I
> > mean if one solves abs(y1-y2) by hand everything
becomes
> > zero, can anyone explain to me why matlab still graphs
> > something and is what I did wrong? I?m a beginner so
please
> > understand that my reasoning is not that advanced
> >
> > thanks
> --------
> I assume from your remarks that you are aware that the
two expressions in
> question are mathematically identically equal to one
another, so you are
> apparently asking why you didn't get a graph showing a
horizontal line of
> height zero. If that is your question, the answer is
round off error in
> computing the two quantities. These involve differing
mathematical
> operations and will produce slightly different results
because of this round off
> error. Note that the values you are getting in the graph
are of size
> somewhere in the neighborhood of 10^(-16) which is a very
small value
> indeed. If you were to accompany this graph with another
one that assumes
> values in the neighborhood of 1, the first graph would be
squeezed down to
> the horizontal line you are expecting.
>
> If you are asking why the two expressions are
identical, try multiplying the
> numerator and denominator of y2 by (sqrt(x^2+1)-1) and
simplify the result.
> What you arrive at is precisely y1.
>
> By the way, in the first expression for y2, you appear
to have left out a pair
> of parentheses, though that is corrected in the code.
>
> Roger Stafford
>

thanks Roger, I was suspecting that about the error but
wasn?t sure. I?ll discuss these with my classmates. thanks
again

best regards
Marcelo